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BillSwagger
09-20-2010, 12:56 PM
I wondered what the difference between mushing in a turn and stalling was. From a bit of reading it seems mushing retains some sort of lift, while a stall is a complete loss of lift. Some aircraft designs have a wider mushing envelope while others have a very thin mushing envelope. Its probably best described as the onset of a stall. Usually, but not always, higher wing loading correlates with more mushing in harder turns, and generally this is why there is a loss of speed in turns for these aircraft. I would then assume that a lighter wing loaded plane is more efficient in similar turns, but the onset of stall is more sudden. In a general sense, it probably has more to do with the wing area than the loading, but in general terms, lower wing area has a tendency to reflect higher loading.

I wondered how this would effect initial turn on some aircraft. Some planes have the ability to cut inside a tight turn, although they are not capable of matching the entire radius.

K_Freddie
09-20-2010, 02:53 PM
Mushing is a good way to loose speed and force an overshoot.
The unwary pilot might be looking for a change in prop speed to indicate that his target is slowing down, and miss the mushing thing completely.
http://forums.ubi.com/groupee_common/emoticons/icon_cool.gif

M_Gunz
09-20-2010, 04:30 PM
Unless you get very deep into stall you don't lose lift. In fact you have a good deal of lift **for the speed you are moving**.

Problem #1 is that what you do get in stall is Loads Of Drag which makes it impossible to maintain your speed. Lose speed, lose lift.

Problem #2 is that once the outer wings stall your ailerons lose effect from lack of airflow over the top.

The warning signs that you are in stall, in order; buffets, loss of control, speed bleed, plane wants to spin.

LOL! Here's what Senior Member billswagger on a Warbirds forum has to say about mushing:

mushing: A plane usually mushes in a dive pull out or sudden changes in elevator deflection. I think it may stall the wing as well, but doesn't put the plane in a spin. If you can imagine putting a plane in a dive and pulling out to get your nose horizontal yet see the aircraft continue to drop another 200-300 meters. Some planes have better mushing characteristics than others. A plane that really flattens out while mushing can black out the pilot or even put the plane into a spin.

BillSwagger
09-20-2010, 04:55 PM
hey, you must be a fan of my online babblings.
http://forums.ubi.com/images/smilies/16x16_smiley-happy.gif

I still don't have a good understanding of mush, but i know what it is when i see it.

Its possible for a plane to mush through a turn with out slowing down, or at least it only slows to where the power/thrust can pull it through the turn despite the extra drag created by the mush.

It still sounds like the onset of stall to me.
I make that comparison because some planes don't mush at all and instead shutter, drop a wing, or spin.

Bill

M_Gunz
09-20-2010, 05:24 PM
I *knew* I had heard the term 'mushing' before I ever read Kit Carson's writing on the 190's! Long before.

Mushing is when you pull up hard to stall hard and the nose just 'mushes' down instead. When the wings have more AOA at the root and inner wing and less at the outer wing -just to tame the stall characteristics- then you can have the inner wings stalled making the nose drop before things get bad.

Example from here (http://www.airbum.com/articles/ArticlePerfectTrainer.html), also note the use of mush to describe control loss:


Most instructors would also like to see an airplane with a definite stall rather than simply mushing forward with the control yoke nailed to the student's chest. Most instructors would like to have the airplane approach the stall, give a characteristic buffet, let the controls turn to mush, and then break, if not sharply, at least noticeably, so the student actually sees the stall develop.

For a more technical explanation, there is one by an actual expert (check his creds) here in See How It Flies. (http://www.av8n.com/how/htm/energy.html)


What is less obvious to non-pilots is that at low airspeeds there is another regime with very high drag. This is called the mushing regime, and is labelled in the figure. The logic here is that it is more efficient to visit a lot of air and yank it down gently than to visit a small amount of air and yank it down violently. In this regime the airplane must fly at a high angle of attack in order to support its weight. This creates strong wingtip vortices that in turn produce huge amounts of induced drag, as discussed in section 3.12.3. Therefore if you are in the mushing regime, flying more slowly causes more descent rate, as can be seen in figure 1.13.

..

The dividing line between the mushing regime and the front side of the power curve is the highest point on the power curve. At this point, the airplane can fly with the minimal amount of dissipation; this is the “low-rent district”. The airspeed where this occurs is called the best-rate-of-climb airspeed and denoted VY.

..

The mushing regime and the stalled regime are collectively referred to as the back side of the power curve.


So by noted FAA Safety Adviser, Flight Instructor, etc, and Caltech graduate who loves to fly, you get the mush before the actual stall at least in most planes. I can tell you that there are some General Aviation planes that you can't stall (as in stall-proof) but can still mush right into the ground (not crash-proof).

If you want to see the diagram and full explanation, that's what the link is for. Use Edit to find the word "mush".

Romanator21
09-20-2010, 07:07 PM
Mushiness has to do with the degree of control effectiveness, ie "like a wet sponge" http://forums.ubi.com/groupee_common/emoticons/icon_cool.gif

It's not quite right to say that flying at slow speed is "mushing", although slow speed does make the controls "mushy".

M_Gunz
09-20-2010, 08:00 PM
IMO how many G's you are pulling matters as to the definition of slow speed in this case and many others. OTOH you could call it load and it would be the same way like what would be stall speed with a full load of bombs or cargo would be well above stall without.

The site has the context but I didn't cut and paste the whole page nor link images from it for viewing since what the hey I figured that one link to the page should be enough for inquiring minds. He refers to Vy and shows a diagram to illustrate an example for General Aviation Pilots, not for Warbird Sim Gamers. It's like reading Robert Shaw's Fighter Combat and finding uses for prop sims, the book was written for Modern Jet Combat Pilots so it is up to the reader to make the right interpretations.

Denker puts Mush and Stall together as back side of the power curve, aka the region of reverse control. That tells a lot more than naming the effects.

BillSwagger
09-20-2010, 11:01 PM
Originally posted by Romanator21:
Mushiness has to do with the degree of control effectiveness, ie "like a wet sponge" http://forums.ubi.com/groupee_common/emoticons/icon_cool.gif

It's not quite right to say that flying at slow speed is "mushing", although slow speed does make the controls "mushy".


I don't think mushiness of controls is the same thing as "mush" in this context.
It sort of answers the benefits of a lighter wing loaded plane.
Heavier wing loads are usually the result of smaller wings as to allow for more speed, and in some cases stronger support.
I notice that higher wing loads also correlate with better roll velocities, while lighter wing loads correlate with better turn performance.

I read an article a short time ago about racers they made during the 30s, and how faster designs had shorter wings because of lower drag, but that in turns, the planes with longer wings would catch up and sometimes surpass its counterpart because it had less mush in the turn. These were also the days where the engine output was closer to 500hp, so obviously a war machine with three times as much power might counter the effects of mush a bit better.
I often perceived wing loading in the context of turning and radius, but knowing how mush effects performance and drag gives more indication in how wing area and weight effect speed in turns, not only the stall speed or load limit.
Obviously if the plane is turning with in its intended envelope then its less likely to experience mush in the turns, but it was brought to my attention when talking about planes like the 190, P-40 and P-47s. Im sure there's other you can think of.

Bill

M_Gunz
09-21-2010, 02:46 AM
At 4 G's your stall speed is twice as high and you mush before you stall. At 6 G's stall is almost 2.5x faster.
High wing load, stall might be well over 300 kph and mush be more.

What Denker wrote, a high drag flight regime. It doesn't just put your path well under your nose, it slows you down. This is about high AOA.

Xiolablu3
09-22-2010, 03:31 AM
I always thought this was mushing, I am probably wrong though..

http://www.youtube.com/watch?v=JSXRPH9PVjM

http://www.youtube.com/watch?v...sx2k&feature=related (http://www.youtube.com/watch?v=4wlnnZZsx2k&feature=related)


http://www.youtube.com/watch?v=cHktfxe2T6Y

Isnt mushing what happens at the bottom of the loop and causes these crashes? (obviously lack of height too)

WTE_Galway
09-22-2010, 05:03 AM
Originally posted by Xiolablu3:
I always thought this was mushing, I am probably wrong though..

http://www.youtube.com/watch?v=JSXRPH9PVjM

http://www.youtube.com/watch?v...sx2k&feature=related (http://www.youtube.com/watch?v=4wlnnZZsx2k&feature=related)


http://www.youtube.com/watch?v=cHktfxe2T6Y

Isnt mushing what happens at the bottom of the loop and causes these crashes? (obviously lack of height too)



Those are all classic examples of accelerated maneuver stalls, one of the main causes of CFIT.

There is nothing mushing about that spit going in, it has well and truly stalled. You can clearly see its flight path does not change with the increased angle of attack. There is no lift component.

Mushing occurs at LOW airspeed, whilst those are very HIGH speed accelerated stalls.

When pulling out of a high speed dive its important to ease back on the stick and never EVER panic and pull it back suddenly as the ground looms at you. Too much stick to quick will see the aircraft changing its nose attitude but continuing on exactly the same flight path, often with the results seen in these videos.


Meanwhile this chart might help answer the original question http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

http://www.av8n.com/how/img48/power-curve-regimes.png

from here ... http://www.av8n.com/how/htm/energy.html



Note: You can buy a used airplane for about the same price as a new sports car.

Riddle: What’s the main difference between the sports car and the airplane?

Answer: If you speed up the sports car to about 75 miles per hour and pull back on the steering wheel, nothing very interesting happens.


That av8n web site is well worth a look through by the way http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

M_Gunz
09-22-2010, 05:50 AM
That's the chart. It shows a GA plane gliding. Same plane 4x as heavy (which is what 4 G's amounts to) would have to glide twice as fast to fall at the same rate for any point on that curve. Accelerated stall (or mush) happens at higher speed than 1 G stall so what is low speed for an unloaded plane is not low speed during a turn, loop or pullout.

Myself, I regard mushing as the very beginning of stall or at least a definite warning sign (look for the wingtip contrails in IL2). Looks like I was wrong, it's not quite stalled yet, just keep the stick back a bit longer.

Of course since that doesn't involve any mention of sex or violence then where's the gratification?

p-11.cAce
09-22-2010, 08:04 AM
Another factor that many pilots do not appreciate is how the lift vector adds or subtracts to their airspeed. With my crappy illustration you can see why being on the back side of the power curve can really dig you into a hole - figuratively and literally.

http://i25.photobucket.com/albums/c99/acmeaviator/horizontallift1.jpg

This illustrates one of aviation greatest misconceptions - why AoA affects speed.Many pilots do not understand that wings don't just work in the vertical - all that power of lift that holds us up can be used to speed us up or slow us down - and in almost every angle of flight it's doing one or the other. When you put the nose down your airspeed increases NOT because you are going down but BECAUSE the horizontal component of lift is pulling you forward. When you climb you slow NOT because you are climbing but BECAUSE the horizontal component of lift is pulling you back.

In mushing flight - or any regime on the backside of the power curve - the AoA is such that the vertical component of lift (what is holding you up) is now less than the horizontal component of lift (what is holding you back). Thrust can overcome this to a point - old timers called this "flying the motor instead of the wing". But because lift increases with airspeed, and because we have more lift pointed aft than up, when we increase power we DECREASE airspeed. Keep that up in anything with less than a 1.1+ power to weight ratio and eventually you dig a hole. The only way to get off the motor and back on the wing is to reduce AoA and get some of that lift back in the vertical.

Bremspropeller
09-22-2010, 09:58 AM
Sorry ace, but what you're writing there's just another simplicistic pilot-story one hears a lot around the airfields.
No offense, but there's a lot of that talking, and much of that is just plain wrong.

The lift-vector is an imaginary friend and helps keeping track of the equilibrium of forces.
By definition, you'd have to pull about an AoA of 45° for the lift-vector to overcome it's vertical component by it's horizontal component.
Drag rises on high AOA due to increased CL and thus increased induced drag and incresed parasitic drag (= more frontal area).
That's what causes the Thrust'rqd gap - it's not the "lift pulling backward".
"Lift" is the pressure-distribution around the wing, and on increased AoA, it's "largest" close to the leading-edge, till it crumbles at alpha'stall.
Now, while the net-effect is the same as in the oversimplified modell while pulling back, what actually happens is something different.

What actually helps you getting faster quicker is not the added thrust by the fwd-pointing lift-vector (there's no "thrust" coming out of the wing) upon pushing forward, but the decreased induced (CL close to 0) and parasitic drag (= less frontal area).
Also, on pushing forward, net-lift gets close to 0 or even negative, resulting in a negation of lift and thus, more drag again.
In your picture, both 109s are stalled, the one on the left has a negative stall (lift-vector pointing down) and the right one a positive. http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif


In short form (varying AoA only):

CL close to 0 => lots of excessive thrust
CL anywhere else => more/ less CL = more drag

You can actually run out of thrust in both ways - upside down or rightside up http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif


Xialo, the Eurofighter and the Flanker weren't "mushing", but they were pitted against their AoA-limiter.
Selecting afterburner got the Tiffie out; the Flanker was already dead, when starting that split-s.
The Flanker has a much wider range of T/W-ratios on internal fuel and too much fuel can ruin your airshow pretty much - just as seen in Lviv.

M_Gunz
09-22-2010, 10:03 AM
Originally posted by p-11.cAce:
Another factor that many pilots do not appreciate is how the lift vector adds or subtracts to their airspeed. With my crappy illustration you can see why being on the back side of the power curve can really dig you into a hole - figuratively and literally.

http://i25.photobucket.com/albums/c99/acmeaviator/horizontallift1.jpg

This illustrates one of aviation greatest misconceptions - why AoA affects speed.Many pilots do not understand that wings don't just work in the vertical - all that power of lift that holds us up can be used to speed us up or slow us down - and in almost every angle of flight it's doing one or the other. When you put the nose down your airspeed increases NOT because you are going down but BECAUSE the horizontal component of lift is pulling you forward. When you climb you slow NOT because you are climbing but BECAUSE the horizontal component of lift is pulling you back.

In mushing flight - or any regime on the backside of the power curve - the AoA is such that the vertical component of lift (what is holding you up) is now less than the horizontal component of lift (what is holding you back). Thrust can overcome this to a point - old timers called this "flying the motor instead of the wing". But because lift increases with airspeed, and because we have more lift pointed aft than up, when we increase power we DECREASE airspeed. Keep that up in anything with less than a 1.1+ power to weight ratio and eventually you dig a hole. The only way to get off the motor and back on the wing is to reduce AoA and get some of that lift back in the vertical.

The one on the right also shows why steady climb is about overcoming drag and not by excess lift. Sustained climb is also 'flying the motor'.

But... at low AOA your lift is less so your blue arrow should be shorter, much shorter at decent to high speed.

Also if stall is about 15 degrees, when ever is the 'horizontal component' (along your path?) going to be more than the 'vertical' (straight up from your path)? Well, okay, lift vector isn't straight up to begin with is it? But sine(15 deg/stall AOA) isn't so great either. So please, numbers?

Gaston444
09-22-2010, 10:44 AM
Mushing IS at high speed too...

Quote: "However, the FW-190 is never able to come out of a dive below 300 or 250 meters (930 ft or 795 ft). Coming out of a dive, made from 1,500 meters (4,650 ft) and at an angle of 40 to 45 degrees, the FW-190 falls an extra 200 meters (620 ft)."

And you can also put the rest the notion that is is merely an "accelerated stall", as after such a dive pull-out mushing, a disoriented, blind and unresponsive FW-190A pilot is often the result...

http://www.lonesentry.com/arti...an-combat-fw190.html (http://www.lonesentry.com/articles/ttt/russian-combat-fw190.html)

Gaston

p-11.cAce
09-22-2010, 11:33 AM
Sorry ace, but what you're writing there's just another simplicistic pilot-story one hears a lot around the airfields.

I guess because I learned from an old simplistic pilothttp://forums.ubi.com/groupee_common/emoticons/icon_smile.gif

http://ep.yimg.com/ca/I/stylespilotshop_2122_139887683

Leighton Collins is reflected in this book: practical, personable, realistic, and authoritative. Leighton first soloed in 1929 in an open-cockpit biplane. He founded Air Facts magazine, which he edited and produced almost single-handedly for 35 years, a magazine widely recognized as the pioneering force for improvement of the aviation safety record. While he contributed a chapter to Stick and Rudder, Langewiesche’s classic flight manual, Takeoffs and Landings was the first book in Mr. Collins’ long writing career.

Kettenhunde
09-22-2010, 01:12 PM
Mushing occurs at LOW airspeed, whilst those are very HIGH speed accelerated stalls.

Exactly and they have nothing to do with mushing.

Mushing is nothing more than flight in the region of reverse command above stall speed.

A WWII pilot would not fly in combat with a fighter aircraft in this region normally. It is below Vx and your best turn performance speeds for maneuvering. A pilot has screwed up and not paid attention to his airspeed. The airplanes maneuvering performance is greatly degraded in this realm of flight when compared to his best performance speeds.

The most common reason an aircraft is in this region is the take off and landing phase of flight.

Kettenhunde
09-22-2010, 01:19 PM
What actually helps you getting faster quicker is not the added thrust by the fwd-pointing lift-vector (there's no "thrust" coming out of the wing) upon pushing forward, but the decreased induced (CL close to 0) and parasitic drag (= less frontal area).


Correct.

Don't listen to that other stuff Ace, it is wrong. The amount of lift required changes with the vector and there is no component of lift that is thrusting forward or pulling back.

BillSwagger
09-22-2010, 01:47 PM
If it only occurs at slower speeds, I then wonder why pilots called it "mushing" because the sensation felt in the cockpit, mushing the pilot into the seat?

Also, the racer article referred to mush in turns experienced by fast moving racing planes.
I'd assume in a race, they weren't slowing to land or glide.

saying mush only occurs at slow speed is akin to saying a stall only happens at slow speed.
It seems mush is flying at higher AoA prior to a complete loss of lift. Something a plane is capable of at any speed.

Maybe mushing is another way of describing an accelerated stall, prior to dropping a wing or entering a spin.
There are as many opinions as there are pilots on the matter. One calls it mush, the other calls it stall, the other calls it buffeting. who knows....

"back side of the power curve"
This is the regime of flight where the engine is not keeping the plane in the air?


ADD:

If we take out the turning component, and look a plane with two weight configurations. In the heavier configuration it will have the tendency to sink, so the pilot has to increase the AoA to have the equivalent line of flight as with the lighter configuration. This increased AoA is not a stall, right? however it does create more drag and in order for the plane to match the speed of its lighter configuration it would need more power.
If we relate that concept to mushing in a turn, it seems that more AoA would be required to complete the same flight path as the lighter configuration in a turn. Higher AoA implies more drag, and more power needed to match the lighter configuration in speed through the turn.



Bill

Bremspropeller
09-22-2010, 02:24 PM
ADD:

Exactly.

p-11.cAce
09-22-2010, 02:45 PM
Don't listen to that other stuff Ace, it is wrong. The amount of lift required changes with the vector and there is no component of lift that is thrusting forward or pulling back

Really? My "Pilot's Handbook of Aeronautical Knowledge" published by the FAA states on page 4-5:

...downwash over the top of the airfoil at the tip has the same effect as bending the lift vector rearward; therefore the lift is slightly aft of perpendicular to the the relative wind,creating a rearward lift component. This is induced drag."

Everyone knows that the horizontal component of lift is what turns a plane, why do they not realize that the same component works in all horizontal directions - not just left and right?

M_Gunz
09-22-2010, 03:15 PM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Mushing occurs at LOW airspeed, whilst those are very HIGH speed accelerated stalls.

Exactly and they have nothing to do with mushing.

Mushing is nothing more than flight in the region of reverse command above stall speed. </div></BLOCKQUOTE>

In a 4+ G pullout, what do you call low (air)speed?

Kettenhunde
09-22-2010, 04:24 PM
,creating a rearward lift component


Induced drag is not lift. It is drag. While it is drag due to lift, it is not lift. Reducing drag due to lift is not adding thrust either. It is freeing up more of the finite thrust production of the engine to devote to aircraft performance.

Thinking of it in terms of "adding thrust" is conceptually incorrect. It is like "additional lift" or "energy conservation".

Lastly, none of the Pilots Handbooks of Aeronautical Knowledge I have use such terminology. If the latest edition uses such terms then it is the source of such confusion.

Would you like me to scan a few for you?

What you are trying to intellectually connect in aerodynamics is about as correct as saying lift is what propels a power producer and opposes drag. Technically, thrust is nothing more than the lift of the small wings mounted on the front of the engine and moving in a circular fashion.

Things would get confusing very quickly and as such, terms were developed to eliminate or reduce it. The lift those wings produce is called thrust in a power producer and not "lift" for this reason. In blade theory for example, this lift of our small wings moving in a circle at the front of the engine is calculated and then becomes thrust to propel the aircraft.

You can nitpick and go off on your own tangent but that does not mean it will be correct.


In a 4+ G pullout, what do you call low (air)speed?

At what speed??

I can pull 4 G at many speeds. However all of them above 1G stall speed will produce the same thing...an aggravated stall until my 4G stall speed is reached or exceeded.

Now I can slow down to the mushing realm and try to pull 4G's....

Again, I will just produce an aggravated stall. Do not confuse the two.

p-11.cAce
09-22-2010, 06:24 PM
I appreciate the offer - think I'll stick with the current FAA info (http://www.faa.gov/library/manuals/aviation/pilot_handbook/media/PHAK%20-%20Chapter%2004.pdf) http://forums.ubi.com/images/smilies/25.gif


Technically, thrust is nothing more than the lift of the small wings mounted on the front of the engine and moving in a circular fashion.

And lift from bigger circular moving blades is what picks up a helicopter, and by altering the AoA of those blades so that their lift vector is redirected the pilot moves where he wants to go.

In real life I mostly fly gliders and motor gliders. So far this month I have logged 10.5 hours on an ASK-21 at CCSC and the HK-36 at Sporty's (i69). I can promise you that it's the forward vector of lift that scoots me along after I pop the release in the ASK or shut down the engine on the HK-36.

It is beyond question that altering the lift vector horizontally with the ailerons is what turns an aircraft:

At a given airspeed, the rate at which an aircraft turns depends upon the magnitude of the horizontal component of lift. It is found that the horizontal component of lift is proportional to the angle of bank—that is, it increases or decreases respectively as the angle of bank increases or decreases.

That is straight out of the current handbook of aeronautical knowledge on page 4-19. You cannot possibly believe that the horizontal component of lift generated by rolling the wing just disappears when you pitch the wing! The same vector of lift that pulls you around a turn accelerates you when you pitch fwd and decelerates you when you pitch up - with or without the thrust from the spinning wings on the front.

M_Gunz
09-22-2010, 06:24 PM
I think that at 4 Gs you would mush at twice the speed as you would at 1 G. Or does the power curve no longer apply?

M_Gunz
09-22-2010, 06:46 PM
Originally posted by p-11.cAce:
It is beyond question that altering the lift vector horizontally with the ailerons is what turns an aircraft:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">At a given airspeed, the rate at which an aircraft turns depends upon the magnitude of the horizontal component of lift. It is found that the horizontal component of lift is proportional to the angle of bank—that is, it increases or decreases respectively as the angle of bank increases or decreases.

That is straight out of the current handbook of aeronautical knowledge on page 4-19. You cannot possibly believe that the horizontal component of lift generated by rolling the wing just disappears when you pitch the wing! The same vector of lift that pulls you around a turn accelerates you when you pitch fwd and decelerates you when you pitch up - with or without the thrust from the spinning wings on the front. </div></BLOCKQUOTE>

Rolling the lift vector to the side, just how much forward component does that make?

You pitch the nose down to lessen your AOA but still keep it zero or higher, you get less lift and less drag. The lift vector decreases but how does it gain a forward component along the path of flight? The drop in drag and possible loss of altitude.. you don't think that that accounts for your speed increase?

You pitch the nose below zero AOA and you get negative lift which points backward along your path.

p-11.cAce
09-22-2010, 07:01 PM
http://www.grc.nasa.gov/WWW/K-12/airplane/Images/glidvec.gif

This is right off the Nasa Glen Research Center website. That L is forward.

Kettenhunde
09-22-2010, 08:26 PM
I'll stick with the current FAA info

Which is the same thing I am saying....

The FAA is not confused.

You are confusing the definition of the induced drag with a creation of some mythical thrusting force from lift.

Kettenhunde
09-22-2010, 08:34 PM
That L is forward.

Work the math problems the Nasa Glen Research Center provides and examine the results.

WTE_Galway
09-22-2010, 08:36 PM
Originally posted by p-11.cAce:
http://www.grc.nasa.gov/WWW/K-12/airplane/Images/glidvec.gif

This is right off the Nasa Glen Research Center website. That L is forward.

aye tru enuf, relative to the GROUND ...

but still close enough to 90 degrees to the aircraft FLIGHT PATH http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

... and one would assume that CAS calculated relative to the flight path vector not relative to the ground. There is no lift component in that diagram in the direction of flight, in essence all the lift is doing is altering the angle of the flight path slightly.

Basically for your argument to work you need the flight path horizontal and the aircraft have a negative angle of attack.

M_Gunz
09-22-2010, 08:45 PM
Originally posted by p-11.cAce:
http://www.grc.nasa.gov/WWW/K-12/airplane/Images/glidvec.gif

This is right off the Nasa Glen Research Center website. That L is forward.

is it forward along the glide path? Because from what I am looking at, it is perpendicular to the glide path which has zero forward along the glide path component.

Horizontal: L * sin(a) - D * cos(a) = 0 <==== also from the same graphic.

Regardless, the flight path is at angle a to the horizontal.

PS: oooohhhh! Galway beat me to post!

WTE_Galway
09-22-2010, 08:48 PM
Originally posted by M_Gunz:


is it forward along the glide path? Because from what I am looking at, it is perpendicular to the glide path which has zero forward along the glide path component.

Horizontal: L * sin(a) - D * cos(a) = 0 <==== also from the same graphic.

Regardless, the flight path is at angle a to the horizontal.

PS: oooohhhh! Galway beat me to post!


Basically the only way I can see this working is with a vector thrust aircraft that allows sustained flight with a negative angle of attack.

http://www.usmilitary.com/wp-content/uploads/2008/11/vtol-us-marine-corps-osprey-careers.jpg

p-11.cAce
09-22-2010, 09:02 PM
You are confusing the definition of the induced drag with a creation of some mythical thrusting force from lift.

The semantics don't change the fact - shut off the motor or don't have one at all and there is no thrust just lift, drag, and weight. So what is creating forward movement? Weight alone will accelerate an object vertically but not horizontally. Drag is a resisting force by definition, so it's not moving us forward either. All we are left with is lift - which does create motion when it's vector includes a horizontal component.

WTE_Galway
09-22-2010, 09:39 PM
Originally posted by p-11.cAce:


The semantics don't change the fact - shut off the motor or don't have one at all and there is no thrust just lift, drag, and weight. So what is creating forward movement? Weight alone will accelerate an object vertically but not horizontally. Drag is a resisting force by definition, so it's not moving us forward either. All we are left with is lift - which does create motion when it's vector includes a horizontal component.

I may be confused but my understanding was weight can be broken down into a component that is opposed by lift and a component in the direction of glide path that causes an acceleration in the direction of travel.

Kettenhunde
09-22-2010, 09:57 PM
All we are left with is lift - which does create motion when it's vector includes a horizontal component.

Negative ghostrider....

Perhaps you should re-examine those formulas to resolve the forces on the vector The Glenn Research Center provided you.


With no engines, gliders move through the sky in much the same way as eagles or vultures--by balancing the forces of gravity (downward force), lift (upward force), drag (******ing force) and thrust (forward momentum).




Thrust is the force that propels the glider forward, working in direct opposition to drag. Because it has no engine, it needs help to acquire thrust initially. Otherwise, it will not go very far. Thrust is generated when the glider is launched into the air. The most common practice for launching gliders is called "tug," which means that an aircraft with an engine tows the glider for some distance. When the glider has acquired enough momentum, the connection is severed and the glider continues to fly on its own.



http://www.ehow.com/how-does_4965227_gliders-fly.html

http://books.google.com/books?...er%20forward&f=false (http://books.google.com/books?id=y1CCxCYbXEMC&pg=SA3-PA8&lpg=SA3-PA8&dq=What+propels+a+glider+forward&source=bl&ots=6X85zYtkyQ&sig=Nu0cvJ1cBUmHyIpoxcOf-DuUch4&hl=en&ei=nM2aTJf2GIqasAO73oz1BA&sa=X&oi=book_result&ct=result&resnum=6&sqi=2&ved=0CCkQ6AEwBQ#v=onepage&q=What%20propels%20a%20glider%20forward&f=false)


Once aloft, gravity (the weight of the hang glider and pilot) pulls the glider back toward Earth and propels the glider forward, continually causing air to flow over the wing.

http://adventure.howstuffworks.com/hang-gliding1.htm

Kettenhunde
09-22-2010, 09:58 PM
I may be confused but my understanding was weight can be broken down into a component that is opposed by lift and a component in the direction of glide path that causes an acceleration in the direction of travel.


That would be correct.

The weight is what converts the PE gained in the tow to KE expended to reach the ground.

p-11.cAce
09-22-2010, 10:12 PM
The weight is what converts the PE gained in the tow to KE expended to reach the ground.

EXACTLY. And what is the mechanism that converts that PE to forward KE? It is the forward component of lift!

Kettenhunde
09-22-2010, 10:15 PM
And what is the mechanism that converts that PE to forward KE? It is the forward component of lift!

No. There is no "extra lift" force propelling the aircraft forward. Lift only meets the amount of force required.

The force of gravity acting on weight is what converts PE to KE. This is why gliders carry water ballast.

p-11.cAce
09-22-2010, 10:25 PM
Negative Ghostrider....

Take a physics course, please.

The force of gravity acting on weight is what converts PE to KE.

Lift opposes weight, remember?

The force of gravity acting on weight does convert PE the KE - if its a rock the KE is straight down. But in an aircraft with power at idle or no motor at all that KE is forward. Why? The horizontal component of lift. Period.

WTE_Galway
09-22-2010, 10:51 PM
Forward (as in horizontal to the ground) is totally irrelevant in this frame of reference. For all intents and purposes the ground may as well not exist.

What matters is the vector in the direction of the glide path.

M_Gunz
09-22-2010, 10:53 PM
I see what you mean now, 'ghostrider'. And the NASA site does show that. It is very like why a ball rolls down hill.

Gaston444
09-23-2010, 12:28 AM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Mushing occurs at LOW airspeed, whilst those are very HIGH speed accelerated stalls.

Exactly and they have nothing to do with mushing.

Mushing is nothing more than flight in the region of reverse command above stall speed.

A WWII pilot would not fly in combat with a fighter aircraft in this region normally. It is below Vx and your best turn performance speeds for maneuvering. A pilot has screwed up and not paid attention to his airspeed. The airplanes maneuvering performance is greatly degraded in this realm of flight when compared to his best performance speeds.

The most common reason an aircraft is in this region is the take off and landing phase of flight. </div></BLOCKQUOTE>

Don't believe the above for a minute...

When Kurt Tank quotes, in a high speed handling test of the FW-190A, that at 400 MPH he could pull 7Gs, he also says the force required PER G was something UNDER 2 pounds per G... That's 14 pounds or 7 KG for a 7G pull-out at 400 MPH! Unbelievably low forces...

Now considering that the FW-190A was well understood as a low-speed fighter that enemy fighters had to keep as high a speed as possible against, that the Soviets described it as such "The FW-190A inevitably offers turning combat at a minimum speed", that 8th Air Force pilots described as: "if it didn't start to pull out by 8000 feet, it would then inevitably "pankake" itself belly first, trying and trying and trying in visible successive attempts to change its trajectory by sinking its tail..."

In THAT context, does the the Kurt Tank superlative 7G at 14 pounds quote make ANY sense?

It does if you realize that the 7Gs SO EASILY attained are so LIGHT because they only involve sinking the tail, loading up the airframe with perpendicular DECELERATION, hence the "tendency to black-out the pilot" WITHOUT A DECENT TRAJECTORY CHANGE...

In effect the high-speed tail-down sinking in this case is like an accelerated stall SLANTED TOWARDS THE INSIDE OF THE CURVE INITIALLY, then inevitably veering out in a semi-straight line (an "elongated curve") as it subjects the nose-up attitude aircraft to crushing deceleration throughout....

E.Brown: "Care must be taken on dive pull-outs not to kill speed by "sinking"...

The best clue of this is this quote: "The FW-190A tended to black-out the pilot", this in the context of describing its INFERIOR high-speed handling and "far inferior angle of pull-out"

http://img105.imageshack.us/img105/3950/pag20pl.jpg

If the FW-190A'S high speed (250 MPH +) handling is SO MUCH INFERIOR to a P-47D, WHY IS IT THE FW-190A THAT IS DESCRIBED AS "TENDED TO BLACK OUT THE PILOT"????

Because in this case a high speed, mushing, elongated curve DOES NOT MEAN LOW Gs...

An elongated curve SHOULD mean a low G load on the pilot, if it was an accelerated stall.

But it is NOT a true stall: The abruptly LARGE surface area presented by sudden tail-sinking creates a crushing "vertical" deceleration on the pilot, while only curving the trajectory "in an elongated loop"...

Hence, "Care must be taken not to kill speed by "sinking"

Hence a "tendency to black out the pilot" despite INFERIOR high speed behaviour and trajectory change...

You gotta love this account which shows how utterly uncompetitive the FW-190A is at high speed handling (except in roll of course), never competing in turns as it desperately dives switches and dives until:

"Finally he pulled streamers at the bottom of an elongated loop and seemed to have straightened out a little coming up" (meaning the pull-out angle was pretty flat compared to the nose-up attitude going in...)

Then: "I think he must have blacked-out right then and I scored many hits"

http://www.spitfireperformance...0-murrell-2dec44.jpg (http://www.spitfireperformance.com/mustang/combat-reports/20-murrell-2dec44.jpg)

Blacked out from "an elongated curve", doesn't that sound like a "tendency to black out the pilot" to you?

But don't worry: It's all a coincidence: There is no such thing as high speed "mushing", the math proves it... These guys saying there is such a thing as a "tendency to black out the pilot" don't know what they are talking about at all...

Non-tech rubes, you know...http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif

Gaston

Kettenhunde
09-23-2010, 05:07 AM
It is very like why a ball rolls down hill.

Exactly.


It is not the centripetal force provided by the ground surface that propels the ball forward....

M_Gunz
09-23-2010, 06:28 AM
No, it's the fact that the CoG of the ball is to one side of the point of contact between the ball and the surface it is on.

And that is very like a forward-tilted lift vector with the lift and glide path acting as the hill.

P-11.cAce (what is 'ghostrider' about?) has it right but about lift components and downhill gliding but the whole AOA thing he started with kind of gets in the way. Lowering AOA.. sure it can get you going downwards but doesn't in all cases. The downwards glide (with enough speed can be upwards) is about gravity which changing AOA is not always about.

WARNING! Put your drink down.

Once again, different people arguing about different things but http://forums.ubi.com/images/smilies/heart.gif *together* http://forums.ubi.com/images/smilies/heart.gif awwwwwwww!

p-11.cAce
09-23-2010, 06:50 AM
Once again, different people arguing about different things but Heart *together* Heart awwwwwwww!

Aww M_Gunz I http://forums.ubi.com/images/smilies/heart.gif you http://forums.ubi.com/images/smilies/16x16_smiley-happy.gif

Top Gun was great but I prefer "Acme" to Ghostrider http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

M_Gunz
09-23-2010, 08:13 AM
Enough to clear up or ie together the whole thing about gliding downwards, change in AOA and lift vector?

Seriously some parts of what you've posted don't stand as posted but would in unstated conditions.

Bremspropeller
09-23-2010, 08:19 AM
But in an aircraft with power at idle or no motor at all that KE is forward. Why? The horizontal component of lift. Period.

BS

It's a mind-game to make people get a mathematical hang of it.
But it's not "there" and it certainly won't "pull" the aircraft.

The speed of a glider comes exclusively from PE=>KE.

p-11.cAce
09-23-2010, 09:02 AM
It's a mind-game to make people get a mathematical hang of it.
But it's not "there" and it certainly won't "pull" the aircraft.

The speed of a glider comes exclusively from PE=>KE.

This keeps getting posted as if its the whole answer. All movement is due to the conversion of PE to KE the question is by what mechanism? I'm told to take a physics class (which I did long ago lol) and iir energy conversion requires a transducer. The mind game is acting as if this conversion just happens without a mechanism to achieve it - that is NOT possible.

PE of gasoline converted to KE by a engine and propeller.
PE of calories converted into KE by muscles.
PE of electricity converted into KE by a motor.

In an aircraft without a source of thrust what EXACTLY is the mechanism by which PE of weight is converted to KE?
Perhaps its so simple as to be lost to obviousness. The forward component of lift is transferred by the wings through the spar to the aircraft. This is the same force that pulls a sailboat to windward:
http://www.cncphotoalbum.com/other/racetrim/chp3fig456.gif

It's the same force that spins an autogyros rotor:

Any wind passing over an airfoil will create both lift and drag. The lift will be perpendicular to the airflow, and the drag will be parallel to the airflow. This is true for all airfoils, not just for the rotor in an autogyro. When the lift and drag vectors are added together, they create a Resultant Force. In autorotation, this resultant force is in front of the Axis of Rotation, so in addition to providing lift, it also pulls the rotor forward.

Bremspropeller
09-23-2010, 10:02 AM
In an aircraft without a source of thrust what EXACTLY is the mechanism by which PE of weight is converted to KE?

Gravity and heat.

p-11.cAce
09-23-2010, 10:28 AM
Gravity and heat.

A force and and energy. Where is your transducer? What is the mechanism that converts that gravitational force into kinetic energy?

It's like saying a car goes down the highway because of heat - which is a true statement but avoids the whole point of how the PE in the gasoline is converted to KE through the transducer of the engine and drive train.

Bremspropeller
09-23-2010, 12:02 PM
How does a stone change PE into KE?

K_Freddie
09-23-2010, 12:43 PM
Originally posted by Gaston444:
Don't believe the above for a minute...

When Kurt Tank quotes, in a high speed handling test of the FW-190A, that at 400 MPH he could pull 7Gs, he also says the force required PER G was something UNDER 2 pounds per G... That's 14 pounds or 7 KG for a 7G pull-out at 400 MPH! Unbelievably low forces...

This I attribute to the better design of the tailplane..
It's much longer lengthwise and a lot shorter widthwise, compared to other aircraft of the day.
Less stick force to get the same elevator angle. This is what, I would imagine, caught many inexperienced FW pilots and resulted in blackouts, and or mushing, after prolonged applied elevator.

This type of elevator action makes the FW good for 'jinking', giving it greater momentary turning ability.
http://forums.ubi.com/groupee_common/emoticons/icon_cool.gif

Bremspropeller
09-23-2010, 12:53 PM
When Kurt Tank quotes, in a high speed handling test of the FW-190A, that at 400 MPH he could pull 7Gs, he also says the force required PER G was something UNDER 2 pounds per G... That's 14 pounds or 7 KG for a 7G pull-out at 400 MPH! Unbelievably low forces...

I could get away with lower forces, provided I trimmed accordingly http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

p-11.cAce
09-23-2010, 01:03 PM
A falling stone and a gliding aircraft both posses KE from gravity pulling them towards the Earth. However, unlike a falling stone which has only parasitic drag to limit it's vertical velocity, a gliding aircraft converts a portion of gravities PE into forward motion. This conversion of PE from purely vertical into the horizontal to resist the force of gravity is "work" and must be accomplished by some transducer. The transducer is the wing and the forward velocity is due to the forward vector of lift the wing creates.

I have provided many examples to show how leading edge suction and the horizontal component of lift move airfoils of not only fixed wing aircraft, but gyrocoptor rotors, auto-rotating helicoptors, sailboats, and wind turbines. I have provided links to current FAA documents. I guess we've said all that can be said http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

M_Gunz
09-23-2010, 01:15 PM
Originally posted by Bremspropeller:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">But in an aircraft with power at idle or no motor at all that KE is forward. Why? The horizontal component of lift. Period.

BS

It's a mind-game to make people get a mathematical hang of it.
But it's not "there" and it certainly won't "pull" the aircraft.

The speed of a glider comes exclusively from PE=>KE. </div></BLOCKQUOTE>

If you bank the wings then the sideways component of lift pulls the AC around in a turn. Yet if you tilt the lift forward or back?

I know I've seen diagrams about how in a steady climb the titled back component of lift makes more drag. Perhaps you have some techno to make that yet more clear and explain how in a steady speed dive the tilted forward component of lift does not counter drag -as shown by NASA-?

The 'hard part' for me was separating his earlier post about lowering AOA from the later about glide slope. But I can't say that one has nothing to do with the other, only that as stated the first bit doesn't stand.

p-11.cAce
09-23-2010, 01:38 PM
http://i25.photobucket.com/albums/c99/acmeaviator/liftdistribtuion.jpg

This diagram shows that even in level flight relative to the airflow a wing generates a forward lift vector due to leading edge suction. Because the lift vector is perpendicular to the relative wind, a gliding aircraft has it's horizontal component of lift tilted even further forward as the air is striking the aircraft at an angle from below. In fact, due to the angle of incidence designed into most aircraft, you can have a negative deck angle and still have the wing at a slightly positive or flat AoA. This forward component of lift IS countering the drag, and is exactly why an aircraft has a steady speed in the glide which is dictated by the angle of attack. All lift is not lost when you go to a negative angle of attack, you just convert more of it into a horizontal component which increases your speed until the lift and drag are back in balance. You descend in a dive for the same reason you descend in a turn - you are trading off vertical lift for horizontal acceleration.

M_Gunz
09-23-2010, 01:45 PM
Oh well.....

Kettenhunde
09-23-2010, 01:46 PM
the tilted forward component of lift does not counter drag

There is no "tilted" vector of lift that is not cancelled by drag.

Notice the Glenn Facility equations equal zero.....

A 200lbs aircraft tilted 20 degrees has ~68lbs of thrust accelerating at 11 ft/s^2.

Where does this thrust come from....Gravity!


M_Gunz, what provides the force to keep you from flying off into space as you stand on the ground?

Why is it difficult to stand on a 60 degree slope of hill? You have no lift force propelling you down the hill. If not for the force of friction provided by your tennis shoes, you would fall down the hill. Just like a glider in fact.


Why is it impossible to stand on a 90 degree slope?

Where is the force providing the tilted vector for a ball rolling down hill coming from??

http://img29.imageshack.us/img29/175/u2l1d2.jpg (http://img29.imageshack.us/i/u2l1d2.jpg/)

Bremspropeller
09-23-2010, 01:53 PM
This diagram shows that even in level flight relative to the airflow a wing generates a forward lift vector due to leading edge suction

..if one blends out the red vectors on that diagram...

Kettenhunde
09-23-2010, 02:08 PM
This diagram shows that even in level flight relative to the airflow a wing generates a forward lift vector due to leading edge suction.


No that diagram shows the distribution of pressure.

p-11.cAce
09-23-2010, 02:12 PM
http://i25.photobucket.com/albums/c99/acmeaviator/gliderdynamics1.jpg

This is straight from the FAA Glider Flying Handbook - FAA-H-8083-13 page 3-2.
Here is the link to the FAA website for the pdf (http://www.faa.gov/library/manuals/aircraft/glider_handbook/media/faa-h-8083-13.pdf)


Where is the force providing the tilted vector for a ball rolling down hill coming from??

I understand where the force is coming from. What you do not understand is that force must be converted by some means into forward motion.

Lighting a gallon of gasoline or running it through a motor both convert it to heat. Only the motor uses that heat and converts it into work. Dropping a rock or an airplane both converts PE to KE - only the airplane converts that KE to work through the mechanism of lift.

Kettenhunde
09-23-2010, 02:38 PM
What you do not understand is that force must be converted by some means into forward motion.


Again, were is the "tilted centripetal force driving forward" on your ball rolling down hill?

M_Gunz
09-23-2010, 03:15 PM
Originally posted by Kettenhunde:
[QUOTE]M_Gunz, what provides the force to keep you from flying off into space as you stand on the ground?

Ahhhhh? Ummmmm? Duh, gee, idn't dat ummmm, gravity huhhhhhhh?


Why is it difficult to stand on a 60 degree slope of hill?

Because I have to lean hard uphill and muh feets/ankles really feel it!


You have no lift force propelling you down the hill. If not for the force of friction provided by your tennis shoes, you would fall down the hill.

On ice or other slick surface, I would slide. That's what would happen without friction to keep me in place. Slide, not fall.


Just like a glider in fact.


Why is it impossible to stand on a 90 degree slope?

Where is the force providing the tilted vector for a ball rolling down hill coming from??

The CoG of the ball not being directly over the point of contact on the slope causes the ball to roll down the slope. If you draw a line from the center of the ball to the point of contact you would not only get a perpendicular to the slope but you would also be drawing a vector which you could split into components of what the slope holds up and the sideways force applied to the ball.

The slope supports the ball, but not directly under the CoG. The angle between straight down and the CoG to the contact point (OMG, same as the slope!) is what matters at the ball. It rolls because of unsupported weight (gravity acting on mass) which can be calculated at the center of the ball relative to the contact point off to the side of that center.

Now take a few moments to see how analogously lift works as the slope (glide path) and the lift vector being perpendicular to the glide path is analogous to the angle between the center of the ball and the contact point of the slope.

The ball rolls. On a slippery surface a block would slide and whoop-de-doo the glider glides.

I don't agree with how P-11.cAce explains whatever he is trying to explain (why the 'oh well' reply earlier, he got less clear about what I think I see him referring to not more but is that his problem?) but I do agree with some of it and learned about how/why balls roll downhill in more detail than "it falls down" back around 1970-71.

Kettenhunde
09-23-2010, 03:31 PM
Slide, not fall.


What is the difference between your slide over the surface of the earth and glider's flight path???

Nothing. The earth's surface meets the centripetal force required to keep you from flying off into space at all times and all angle short of 90 degrees

To sum it up...Two properties of lift are being confused here.

First lift always develops perpendicular to the path of flight. That is why the diagram shows the lift vector as tilted. Everyone sees that as proof of "tilted vector" driving the airplane.

Secondly, the lift force required meets the force required on the vertical and horizontal axis. It always balances like the centripetal force provided by the earths surface. There is no excess to drive anything.

However since we don't understand the second property of lift we are confusing that first property with the second.

Kettenhunde
09-23-2010, 03:49 PM
You have it right for the most part. We were double posting.

Just examine this part:


Now take a few moments to see how analogously lift works as the slope (glide path) and the lift vector being perpendicular to the glide path is analogous to the angle between the center of the ball and the contact point of the slope.

And keep in mind our second property of lift:


the lift force required meets the force required on the vertical and horizontal axis. It always balances like the centripetal force provided by the earths surface.

You will then understand it is not lift but just like our ball, gravity, that causes the glider to move down the flight path because of unsupported weight.


It rolls because of unsupported weight (gravity acting on mass) which can be calculated at the center of the ball relative to the contact point off to the side of that center.

M_Gunz
09-23-2010, 04:24 PM
And that NASA diagram and equations show?

Kettenhunde
09-23-2010, 05:44 PM
And that NASA diagram and equations show?

Exactly what I am telling you....



Crumpp says:

Secondly, the lift force required meets the force required on the vertical and horizontal axis. It always balances....

M_Gunz
09-23-2010, 10:04 PM
Then what is the force required on the horizontal axis? From the NASA diagram I gather it is equal to the drag in a steady speed dive, but what I think that p-11.cAce is saying is that there -is- a horizontal component which he is not balancing against the same drag.

IMO the lift vector should include induced drag since you can't have lift without induced drag, but that's just me.

Kettenhunde
09-24-2010, 05:47 AM
what I think that p-11.cAce is saying is that there -is- a horizontal component which he is not balancing against the same drag.

What is the sum of the forces on the vector?

....ZERO.

Kettenhunde
09-24-2010, 06:06 AM
Decomposing the mg vector gives a total of three force vectors at work in this diagram: the y-component of the gravitational force and the normal force, which cancel out; and the x-component of the gravitational force, which pulls the box down the slope. Note that the steeper the slope, the greater the force pulling the box down the slope.

http://www.sparknotes.com/test...apter8section3.rhtml (http://www.sparknotes.com/testprep/books/sat2/physics/chapter8section3.rhtml)

Gravity is what propels a glider forward, not some cockamamie extra lift force.

p-11.cAce
09-24-2010, 06:16 AM
I apologize for not being clear in what I am trying to say http://forums.ubi.com/images/smilies/blush.gif

What I was trying to say and illustrate in a very simple manner is that there is a horizontal component of lift that affects any airfoil.

There IS such a thing as "Leading Edge Suction" - it is what spins any auto-rotating rotary wing be it on an autogyro, auto-rotating helicopter, or wind turbine. It is what pulls along a sailboat when it is sailing upwind. It is what converts the force of gravity into forward motion in a gliding aircraft.

The sums on the NASA diagram come to zero in steady-state flight - let's say an aircraft gliding along at 70 knots. But by lowering the nose and directing more of the lift vector forward than vertically the aircraft accelerates to a new steady state speed.

The FAA states that "induced drag IS due to a REARWARD lift component". It only stands to reason that if there is a rearward lift component that slows an aircraft there is a forward lift component that accelerates it.

We cannot talk about falling objects that have only parasitic drag as equal to an aircraft affected also by the created resistance forces of lift. There are different energy conversions occurring in each.

A bonfire and a internal combustion engine both convert the PE of gasoline to heat, but that conversion in an engine creates KE and thermal energy. A gliding aircraft and a rock both convert the PE of gravity to KE and thermal energy, however the KE of the aircraft is ALSO converted to a resisting force that creates horizontal motion. There MUST be a mechanism for this creation of a resisting force. That mechanism is the airfoil, its resisting force is lift, and because the lift has a forward component the result is forward motion of the aircraft.

M_Gunz
09-24-2010, 06:27 AM
Originally posted by Kettenhunde:
extra lift force.

It's the words he used that I haven't agreed with, but I think I still "got" what he was getting at.

The forces may sum to zero which he didn't seem to be saying but there is a sideways component which is clearly there and just as clearly balanced by drag in a steady speed dive.

p-11.cAce
09-24-2010, 06:40 AM
The forces may sum to zero which he didn't seem to be saying but there is a sideways component which is clearly there and just as clearly balanced by drag in a steady speed dive.

And I am also saying that the "sideways component" can be manipulated by the pilot to determine what that steady state velocity is - within the limitations of the ability of the wing to convert the PE of gravity into horizontal KE. That is where the whole AoA thing comes into play.

Really what this comes back to is the ages old "stick or throttle?" question with the throttle part removed.

Kettenhunde
09-24-2010, 06:44 AM
The FAA states that "induced drag IS due to a REARWARD lift component".


Which is correct. It is drag due to lift.


he KE of the aircraft is ALSO converted to a resisting force that creates horizontal motion. There MUST be a mechanism for this creation of a resisting force.

There is a mechanism for the creation of that resisting force. The WING in the vertical component of lift. Notice in the Glenn Facility formulas WEIGHT is included in the vertical. The forces of lift and drag act against weight.

Digest that fact and then read the high school physics page I posted paying attention to the summation of the horizontal vector of the force of gravity on a slope.


x-component of the gravitational force, which pulls the box down the slope. Note that the steeper the slope, the greater the force pulling the box down the slope.

http://www.sparknotes.com/test...apter8section3.rhtml (http://www.sparknotes.com/testprep/books/sat2/physics/chapter8section3.rhtml)

The lift force acts just like the ground in your ball example. It resist's the force of gravity in the vertical axis.

Notice weight is NOT included in the Glenn Facility formula for force summation on the horizontal vector.

That is because lift and drag do not balance it in the horizontal axis.

Understand?

M_Gunz
09-24-2010, 06:47 AM
Originally posted by p-11.cAce:
It is what pulls along a sailboat when it is sailing upwind.

And not because the centerboard and rudder keep the boat on course against the push of the wind while the sail redirects the wind to some degree back along your path? Because without the centerboard or when sailing too closely into the wind you don't go forward at all.

If you stick an airfoil up instead of the sail, could you sail directly into the wind using this forward pull? You know that induced drag is more than that forward pull and both are due to circulation?

AndyJWest
09-24-2010, 07:20 AM
Am I missing something here?

Assuming still air, a glider in a steady descent has three forces we all seem to agree act on it:

(1) Gravity, pulling it downwards.

(2) Lift, balancing gravity and keeping the rate of descent constant,

(3) Drag, in two forms:

- (a) Induced drag, an inevitable consequence of using an aerofoil to produce lift.
- (b) Parasitic drag.

Now since the glider moves forward at a constant rate, rather than slowing down, there must be another force balancing out the drag components. The only rational explanation I can see is that the wing is producing more 'lift' than is necessary to balance the weight of the aircraft, with the surplus balancing out the drag. It can only do this if it produces a horizontal force.

p-11.cAce
09-24-2010, 07:22 AM
There is a mechanism for the creation of that resisting force. The WING in the vertical component of lift. Notice in the Glenn Facility formulas WEIGHT is included in the vertical. The forces of lift and drag act against weight.

THANK YOU. The only thing that you are not acknowledging is that the end result of the creation of that resisting force is forward motion. The same is true for the sailboat - it's resisting a sideways push instead of gravity but the mechanism and end result is the EXACTLY THE SAME - forward motion.

p-11.cAce
09-24-2010, 07:24 AM
Now since the glider moves forward at a constant rate, rather than slowing down, there must be another force balancing out the drag components. The only rational explanation I can see is that the wing is producing more 'lift' than is necessary to balance the weight of the aircraft, with the surplus balancing out the drag. It can only do this if it produces a horizontal force.

yes, yes, and yes. http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

Kettenhunde
09-24-2010, 07:30 AM
Notice weight is NOT included in the Glenn Facility formula for force summation on the horizontal vector.

You guys have fun discussing it.

Bottom line is gravity and conservation of momentum provide the thrust force to move a glider on the horizontal axis and not some miscued concept of excess lift due to the first property of lift in that it always develops perpendicular to the flow.

The second property of lift negates any notion of excess lift and that can be seen in the NASA Glenn Research tutorial posted on page two of this thread.

Good luck in your search.


Am I missing something here?

Yes.

AndyJWest
09-24-2010, 07:45 AM
Kettenhunde, the last time I checked, gravity operated vertically, not horizontally.

Do you agree that in our glider example there is a force ('drag') acting on it which has a horizontal component in the direction opposite to the forward flight of the aircraft? And if you do, can you explain why the glider won't decelerate without a balancing forward-acting force?

Kettenhunde
09-24-2010, 07:56 AM
Do you agree that in our glider example there is a force ('drag') acting on it which has a horizontal component in the direction opposite to the forward flight of the aircraft? And if you do, can you explain why the glider won't decelerate without a balancing forward-acting force?



Read the thread and your questions are answered.


Kettenhunde, the last time I checked, gravity operated vertically, not horizontally.


http://img832.imageshack.us/img832/8852/forceofgravityvectorson.jpg (http://img832.imageshack.us/i/forceofgravityvectorson.jpg/)


Save yourself some time and go back and read the last page I posted paying particular attention to the summation of the forces on the x-axis.

I am not really interested in repeating it or the time to spend correcting misconceptions on the internet.

Kettenhunde
09-24-2010, 08:20 AM
http://img213.imageshack.us/img213/9899/forcevectorsonaglider.jpg (http://img213.imageshack.us/i/forcevectorsonaglider.jpg/)

p-11.cAce
09-24-2010, 08:35 AM
Kette we are not talking about an object resting ON an inclined surface - we are talking about an object supported by an airfoil. They are NOT the same at all. The block sitting on surface is not generating its own resistance to gravity - that resistance is being proved by the surface on which it is resting.

JtD
09-24-2010, 09:10 AM
You guys know that stuff is relative, don't you?

Relative to Earth, lift has a forward component.
Relative to the plane, gravity has.

More than 99% of the world uses the plane as the reference object.

p-11.cAce
09-24-2010, 09:38 AM
Relative to Earth, lift has a forward component.
Relative to the plane, gravity has.

The question that we are dancing around is does an airfoil produce a forward component intrinsically to itself. That is why I keep bringing up autogyros, wind turbines, and sail boats. I think an autogyro rotor is probably the most obvious of these - the forward component of lift pulls the blade forward.

Kettenhunde
09-24-2010, 09:42 AM
You guys know that stuff is relative, don't you?

Relative to Earth, lift has a forward component.
Relative to the plane, gravity has.


Baloney. There is no forward component of lift relative to anything.

p-11.cAce
09-24-2010, 10:45 AM
Baloney. There is no forward component of lift relative to anything.

I'll agree to disagree.

JtD
09-24-2010, 10:59 AM
Originally posted by p-11.cAce:

The question that we are dancing around is does an airfoil produce a forward component intrinsically to itself.

An airfoil in the wind doesn't do ****. It just gets blown away.

M_Gunz
09-24-2010, 11:38 AM
If the lift is not straight up or down then it can be said to have a horizontal component. This is not extra. It is part of the lift and due to the lift not being totally vertical.

The wing is not being pulled along by a net 'forward lift' either regardless of the direction of pressure on -part- of the wing. If it was then you should be able to glide without loss of height given big enough light enough wings.

There is no mystery to lift tilted forward having a forward component just as there is no mystery to lift tilted to the side having a sideways component or lift tilted back having a backward component either.

Once again people argue different things and now people who do so unwilling to admit anything that might be 'used' to advance some idea not directly related to what they do not agree with, in advance. And I don't mean just one person.

M_Gunz
09-24-2010, 11:44 AM
Originally posted by Kettenhunde:
http://img213.imageshack.us/img213/9899/forcevectorsonaglider.jpg (http://img213.imageshack.us/i/forcevectorsonaglider.jpg/)

If you can split gravity into components then why not lift?

How does a plane turn? What provides the sideways force? It's not gravity.

M_Gunz
09-24-2010, 11:48 AM
Originally posted by JtD:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by p-11.cAce:

The question that we are dancing around is does an airfoil produce a forward component intrinsically to itself.

An airfoil in the wind doesn't do ****. It just gets blown away. </div></BLOCKQUOTE>

I've made flying wing models since 1968 that did fly even in noticeable breeze. Eventually they turn and fly in the direction of the breeze, having no pilot and only after the forward motion of the throw wore out.

All airfoil and... they flew.

p-11.cAce
09-24-2010, 12:11 PM
This is an excerpt of exactly what I am trying to say from the same discussion in a different forum (they were using chickens on a hill instead of rolling balls):


Sorry back to the hen house for you. Lift is *not* perpendicular to
the direction of motion. Lift is perpendicular to the airflow.

The chicken sliding down the hill is propelled forward by the component of the force pushing up on it from the sloped surface in response to gravity pushing it down onto that surface. Gravity operates vertically down through the center of mass and without that sloping surface the chicken would not move forward. There is a
component of this force vector pushing forward and the chicken moves forward down the slope.

This is exactly analogous to the lift vector pointed forward on the glider and the forward component of that vector providing the force to
overcome drag in forward flight. Gravity operates on the glider though it's center of mass, without that lift vector pointed forward the glider would not (continue to) move forward.

Gravity, without the thrust vector leaning forward, does not explain how a glider glides through the air, however Gravity is what provides the power/energy to do this (or store for that energy if you prefer).

The above discussion applies equally to chickens and African or European swallows.

idonno
09-24-2010, 12:20 PM
Originally posted by p-11.cAce:

The question that we are dancing around is does an airfoil produce a forward component intrinsically to itself. That is why I keep bringing up autogyros, wind turbines, and sail boats. I think an autogyro rotor is probably the most obvious of these - the forward component of lift pulls the blade forward.


There is no forward component to the lift generated by an airfoil. There is, in fact, quite the opposite. It's called Induced Drag.

http://en.wikipedia.org/wiki/Lift-induced_drag


An auto-gyro's rotor is not turned by a forward component of lift, but by the same forces that allow a helicopter to auto-rotate. The auto-gyro's rotor disc is pulled through the air at a slight rearward angle to provide airflow up through it. The rotor blades are angled down into the airflow. The result is that the lift being generated by the rotors (not the rotor disk as a whole) is angled slightly forward, causing the rotor to spin.

p-11.cAce
09-24-2010, 12:42 PM
An auto-gyro's rotor is not turned by a forward component of lift, but by the same forces that allow a helicopter to auto-rotate. The auto-gyro's rotor disc is pulled through the air at a slight rearward angle to provide airflow up through it. The rotor blades are angled down into the airflow. The result is that the lift being generated by the rotors (not the rotor disk as a whole) is angled slightly forward, causing the rotor to spin.

Sorry but the FAA disagrees with you (or whoever wrote the wiki - imagine that)

FAA-H-8083-21.pdf (http://www.faa.gov/library/manuals/aircraft/media/faa-h-8083-21.pdf)
Page 3-10
"The driving region, or autorotative region, normally lies between 25 to 70 percent of the blade radius. Part C of figure 3-22 shows the driving region of the blade, which produces the forces needed to turn the blades during autorotation. Total aerodynamic force in the driving region is inclined slightly forward of the axis of rotation, producing a continual acceleration force. This inclination supplies thrust, which tends to accelerate the rotation of the blade. Driving region size varies
with blade pitch setting, rate of descent, and rotor r.p.m."

http://i25.photobucket.com/albums/c99/acmeaviator/rotorlift.jpg

JtD
09-24-2010, 01:14 PM
Consider the "inflow" the "flight path" for that scenario, and you again have no forward component in the lift.

Like I said, relative. In this case, there's a difference between the airfoil section illustrated and the rotor disc. Same principle works with screws. You just rotate them and suddenly...they move longitudinally.

JtD
09-24-2010, 01:19 PM
Originally posted by M_Gunz:

I've made flying wing models since 1968 that did fly even in noticeable breeze. Eventually they turn and fly in the direction of the breeze, having no pilot and only after the forward motion of the throw wore out.

All airfoil and... they flew.

I'm willing to bet that there was gravity as well.

p-11.cAce
09-24-2010, 01:25 PM
Consider the "inflow" the "flight path" for that scenario, and you again have no forward component in the lift.

You are exactly rihgt, however are not talking about lift relative to the inflow. We are talking about lift relative to the wing, and ultimately relative to the aircraft. Way back on page 1 Kettenhunde asserted that:


there is no component of lift that is thrusting forward or pulling back.

I obviously disagree. What you are saying is exactly correct - lift is relative to the airflow, and in the case of autogyro blades (which really are just gliding wings with one fixed to the rotor hub) that lift is pulling them around.

Gaston444
09-24-2010, 01:26 PM
Originally posted by K_Freddie:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Gaston444:
Don't believe the above for a minute...

When Kurt Tank quotes, in a high speed handling test of the FW-190A, that at 400 MPH he could pull 7Gs, he also says the force required PER G was something UNDER 2 pounds per G... That's 14 pounds or 7 KG for a 7G pull-out at 400 MPH! Unbelievably low forces...

This I attribute to the better design of the tailplane..
It's much longer lengthwise and a lot shorter widthwise, compared to other aircraft of the day.
Less stick force to get the same elevator angle. This is what, I would imagine, caught many inexperienced FW pilots and resulted in blackouts, and or mushing, after prolonged applied elevator.

This type of elevator action makes the FW good for 'jinking', giving it greater momentary turning ability.
http://forums.ubi.com/groupee_common/emoticons/icon_cool.gif </div></BLOCKQUOTE>

-No, quite to the contrary I think the FW-190A's unusually narrow and small elevator plan view made it prone to tail stalling at high speed in the INITIAL portion of a turn...

Again, high speed initial turn and overall pull-out performance was very poor in the FW-190A, and to get the best you had to load it up very slowly at high speeds. Hence the common observation: "The FW-190A pilot seemed afraid to really reef it in in turns"

Or E. Brown's: "Care must be taken (in dive pull-outs) so as to not kill speed by sinking"

This poor high speed handling might very well have intimidated many inexperienced FW-190A pilots in low-speed turning as well, leading to the "afraid to reef it in" observation...

Poor FW-190A high speed handling also for some reason probably applied to any high altitude fighting, slow or fast. I doubt this was only due to lower engine performance... Aerodynamic effects are not inevitably linear with altitude...

This uncompetitive nature, even in unsustained high speed turns, is very visible here (note speeds were kept near 400 throughout):

"I turned inside them (FW-190s) without any trouble and the wingman snapped out of control. I stayed with the leader as he rolled, looped, snapped and tried to out-turn my P-51. HE INVARIABLY SNAPPED OUT OF CONTROL EVERY TIME HE REALLY TIGHTENED HIS TURNS UP... ...Estimated 400 MPH target"

http://www.spitfireperformance...0-murrell-2dec44.jpg (http://www.spitfireperformance.com/mustang/combat-reports/20-murrell-2dec44.jpg)

A slower G build-up through careful stick use might allow the FW-190A around 6 "real" Gs at high speed, hence "the effects of the Gs we were pulling must have thrown the shots behind him, although I held a 4 radii lead for an estimated 400 MPH target" but the loss in speed is slightly greater and gets worse the more Gs are pulled (or obviously the P-51 pilot would not consider his P-51 superior in high speed turns, and the German would not need to roll, snap and loop to "try" to out-turn him...), resulting in an inferior turn rate, until the tail drops abruptly and either the wing also drops ("snap"), or the turn gets much wider while the nose-up deceleration gets much worse...

While high speed pull-outs are often characterized by prolonged nose-up mushing, in turns, the sudden nose-up tail sinking creates more often an assymetrical wing load resulting in a sudden wing drop stall: "Snapping".

Pathetic FW-190A high speed handling performance is also well in evidence here:

http://img105.imageshack.us/img105/3950/pag20pl.jpg


As it is in the Soviet admonition: "Speeds must be kept as high as possible against the FW-190.", which you would think would give SOME people a clue... As does "The FW-190 inevitably offers turning combat at a minimum speed"...

http://www.lonesentry.com/arti...an-combat-fw190.html (http://www.lonesentry.com/articles/ttt/russian-combat-fw190.html)

Just what do you think they are talking about here? The FW-190 or the tooth fairy?


What I am still waiting for is for the high-flying "experts" around here to even TRY to explain to me what "FW-190A tended to black-out the pilot" means here:

http://img105.imageshack.us/img105/3950/pag20pl.jpg

It's all there in that little sentence folks...

Gaston

M_Gunz
09-24-2010, 01:46 PM
Originally posted by JtD:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by M_Gunz:

I've made flying wing models since 1968 that did fly even in noticeable breeze. Eventually they turn and fly in the direction of the breeze, having no pilot and only after the forward motion of the throw wore out.

All airfoil and... they flew.

I'm willing to bet that there was gravity as well. </div></BLOCKQUOTE>

Gravity and wind. I'm somehow pretty sure when you wrote about airfoils and wind that you did not mean wind without gravity so what difference does gravity make now?

JtD
09-24-2010, 02:10 PM
Sure I meant without gravity. Gravity is what makes gliders glide. Otherwise they'd just float, and go with the wind.

Kettenhunde
09-24-2010, 02:28 PM
How does a plane turn? What provides the sideways force?

The side force is provided by lift but in the case of a turn we are under acceleration changing lifts definition of "horizontal".

That acceleration vector is split between centripetal force vector and the vector of gravity.

That is why in a coordinated turn, if you drop your pen in the cockpit, it will fall straight to the floor just like 1G level flight.

M_Gunz
09-24-2010, 06:36 PM
I just won't bother any more.

p-11.cAce
09-24-2010, 07:03 PM
My wife printed this today and left it on my keyboard http://forums.ubi.com/images/smilies/51.gif

http://homepages.nyu.edu/~kmg357/pictures/xkcd2/xkcd22.png

K_Freddie
09-25-2010, 12:16 AM
They're aways moaning, when you find a 'hobby' that doesn't involve them http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif


-No, quite to the contrary I think the FW-190A's unusually narrow and small elevator plan view made it prone to tail stalling at high speed in the INITIAL portion of a turn...
Hmmm!..That is what I said,
.. and added if the elevator response is used properly this can be used to 'jink' the a/c around without that 'sinking feeling'
http://forums.ubi.com/groupee_common/emoticons/icon_cool.gif

Kettenhunde
09-25-2010, 07:47 AM
What you are saying is exactly correct - lift is relative to the airflow, and in the case of autogyro blades (which really are just gliding wings with one fixed to the rotor hub) that lift is pulling them around.


You know this is just not true.

First of all, rotor dynamics are extremely complicated.

This very much depends on the which portion of the blade and in what part of the disc your are examining.

You are correct in that the driving portion of the disc is using lift to pull the driven portion. You are wrong in thinking the vector of lift is tilted. The blade twist's to meet the relative wind and lift forms perpendicular to the relative wind meeting only the force required.

That is not the lift production of the disc or what supports the vehicle in the air.

I doubt we could have an informed discussion of the mechanics in this community.

Suffice it to say, your speculation is wrong and there is no "tilted vector of lift".

http://img840.imageshack.us/img840/9501/rotordynamics.jpg (http://img840.imageshack.us/i/rotordynamics.jpg/)

p-11.cAce
09-25-2010, 05:46 PM
Suffice it to say, your speculation is wrong and there is no "tilted vector of lift".

Whatever Kette - this clearly shows the lift vector as tilted forward relative to the blade. If a forward tilted lift vector can drive an autogyro blade (which it obviously does) it can also drive along a fixed airfoil in a glide. You can go off on a tangent, as you call it, or deny it out of hand without providing any source outside of your own speculation - but that does not mean you are correct. But since we are all obviously beneath your intelligence I'll let you take it up with the FAA http://forums.ubi.com/images/smilies/25.gif

http://i25.photobucket.com/albums/c99/acmeaviator/rotorlift.jpg

Kettenhunde
09-25-2010, 08:01 PM
this clearly shows the lift vector as tilted forward relative to the blade

No it does not show it tilted at all. Look at the relative wind and you will see the lift vector is perpendicular. That part of the rotor disc is just in very high angle of attack flight.

It is from a small portion of the disc termed the driving portion. It cannot be characterized as the sole representation of rotor blade dynamics.

The angle of attack changes from high to low as we move outboard of the disc.

http://img25.imageshack.us/img25/2349/highangleofattacko***yr.jpg (http://img25.imageshack.us/i/highangleofattacko***yr.jpg/)

http://img691.imageshack.us/img691/2349/highangleofattacko***yr.jpg (http://img691.imageshack.us/i/highangleofattacko***yr.jpg/)


At no point is lift tilted or not perpendicular to the relative wind. Lastly, a disc rotor has more in common with a propeller than a wing as it is a driven airfoil either directly attached or getting assistance from a power producing engine.

Because of that, a rotor disc does develop thrust and just like the blade of a propeller, that thrust is really lift in the horizontal axis. In that case, it is termed thrust.

The only lift production that is actual lift is in the vertical axis just like a fixed wing aircraft.

So the whole premise of tilted lift is not correct on any level.

M_Gunz
09-25-2010, 08:43 PM
Sigh...

so if you change AOA you can change the angle of lift relative to the wing which may or may not have some value in flight?

Kettenhunde
09-25-2010, 09:16 PM
Lift acts through the center of pressure of the object and is directed perpendicular to the flow direction.

http://www.grc.nasa.gov/WWW/K-12/airplane/lift1.html

M_Gunz
09-25-2010, 09:59 PM
If lift is always fixed perpendicular to airflow and changing AOA changes the angle between airflow and airfoil then certainly changing AOA must change the angle between lift and the airfoil itself.

Lift cannot be fixed to both when one changes in relation to the other. I hope the logic there is clear enough?

As for meaning, use, or consequences in actual piloting -- leave that out for now, it's not part of the statement, adding extraneous conditions or elements, whether stated or unstated, is not answering the above.

Col_SandersLite
09-26-2010, 02:41 AM
Having been lurking here for a long time I know the way kettenhunde comes off. Especially in the way he approaches things. In this case though, he's absolutely right, but just not explaining it very well.

Lets take this step by step. I'm going to try and explain this in laymans terms. It's 3 AM so the terminology may contradict the strictly technical terminology in places, but I'm going for understanding of the concept, not a formula.

The angle of lift is absolutely 100% perpendicular to the flight path.

There *is* a horizontal lift component, depending on the wing angle, but it is perfectly negated by the added drag. In practice it can be ignored.

So the question that keeps coming up is, how does the glider move forward then? kettenhunde maintains that it is the force of gravity, but he doesn't bother to clearly articulate this idea.


In order to demonstrate this effect, I propose a simple experiment that you can do in your own home, or even in your head (but it won't be as much fun).

Required materials:
1: Toy airplane with wheels on the bottom. If you lack a toy airplane, you can substitute a toy car with little wings glued on it, or just imagine the wings (hey a flying car!).
2: A piece of 2X4 of moderate length.
3: A raised surface (desk, bed, table, porch, whatever).

First place one end of the 2X4 on the edge of your raised surface, holding the other end in your hand so that you can move the end up or down to change its angle freely. You want the 2X4 to be level (perpendicular to gravity).

Place your toy plane on the 2X4. If the 2X4 is level, it should not move.

Next, tilt the 2X4 down at a 45 degree angle, and watch the toy plane move down the ramp right to your hand.

Note that the toy plane is generating a completely insignificant amount of lift (there's probably very small immeasurable amount there) but it manages to move forward despite this. What makes the plane move forward is other factors.

Doubtlessly some of you are thinking, but wait, a real plane doesn't sit on an inclined plane like that. In a real aircraft the wings acting against the air, in fact, creates the inclined plane (it's called an "air plane" for a reason you know). In this experiment, you are in effect substituting working wings with a 2X4.

The especially cool and astute among you didn't really even notice that, but instead noticed that I just gave you a reason to play with your toys http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif.

I hope that made things clear as mud as the saying goes. I'm going back to lurking now.



--
This space intentionally left blank.

Holtzauge
09-26-2010, 03:23 AM
Originally posted by p-11.cAce:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Consider the "inflow" the "flight path" for that scenario, and you again have no forward component in the lift.

You are exactly rihgt, however are not talking about lift relative to the inflow. We are talking about lift relative to the wing, and ultimately relative to the aircraft. Way back on page 1 Kettenhunde asserted that:


there is no component of lift that is thrusting forward or pulling back.

I obviously disagree. What you are saying is exactly correct - lift is relative to the airflow, and in the case of autogyro blades (which really are just gliding wings with one fixed to the rotor hub) that lift is pulling them around. </div></BLOCKQUOTE>

Leading edge suction is real. For example, designing a glider without taking this into account is a recipie for disaster.

As long as the flow over the wing is not separated there will be a forward force relative the chord line: The stagnation point will progressivly move from the front of the LE to below the LE as the AoA goes up. This forces the air forward and around the LE. This is turn pulls the wing forward relative the chord line. In a glider, with long wings this will create a substantial force that wants to pull the wings forward relative to the chord line which the structure has to bear somehow. This is in some designs done with a compression/tension bearing member behind the spar. As the AoA goes up, LE suction force will be overcome by the added drag due to flow separation and the force relative the chord line will once again start to pull the wings back.

Of course, at high speed, the stagnation point will again move towards the front of the LE and the load bearing member behind the spar will start to bear a compression load instead.

Notice that while a glider is a nice example to illustrate the point, LE suction and a forward load relative the chord line is present in many other cases just like p-11.cAce points out.

K_Freddie
09-26-2010, 04:33 AM
Originally posted by Col_SandersLite:
The angle of lift is absolutely 100% perpendicular to the flight path.
If the wing angle is parallel to the flight path (simple terms that is), otherwise this lift is a component of the total lift.
http://forums.ubi.com/groupee_common/emoticons/icon_cool.gif

Bremspropeller
09-26-2010, 04:45 AM
Notice that while a glider is a nice example to illustrate the point, LE suction and a forward load relative the chord line is present in many other cases just like p-11.cAce points out.

...only that:

a) this is a high CL phenomenon, not a low CL phenomenon as presented here

b) the force doesn't overcome overall-drag at any time, thus not creating any net-excess-power


There has been a french UL lost during flight-test due to the suction-effect.
However, it had been during a high-g pullout and they had the wings peel-off forward and upward.

The proclaimed suction is there and WILL pull your wings off if not taken into account during the design-phase, BUT it will not propell your plane forward at any time. It just doesn't overcome overall-drag at any CL.

Holtzauge
09-26-2010, 06:00 AM
Originally posted by Bremspropeller:
The proclaimed suction is there and WILL pull your wings off if not taken into account during the design-phase

Yes, good so we agree: There IS a forward component relative to the chord due to LE suction. I believe that was the point p-11.cAce was trying to get across.


Originally posted by Bremspropeller:
BUT it will not propell your plane forward at any time. I'm devastated: This means that my latest perpetum mobile design is not going to work http://forums.ubi.com/images/smilies/bigtears.gif

Bremspropeller
09-26-2010, 07:20 AM
I believe that was the point p-11.cAce was trying to get across.

No, he's trying to put accross that this unsignifigant component is gonna thrust your aircraft downhill.

Kettenhunde
09-26-2010, 08:07 AM
So the question that keeps coming up is, how does the glider move forward then? kettenhunde maintains that it is the force of gravity, but he doesn't bother to clearly articulate this idea.


Thank you for trying to clarify.


LE suction.

LE suction is a phenomenon that is significant only at very high angles of attack in large low aspect ratio wings.

As Bremspropeller already pointed out, like induced drag, it is a product of lift but in this case our vortex contributes to lift greatly. It does not create any kind of "tilted" left vector pulling the aircraft however. It just increases our wings CL significantly and helps to overcome the lack of flaps found on most low aspect ratio wings with large sweep angles.

It also pointed out it is very insignificant especially to this discussion.

In the pressure distribution graph Ace posted earlier as proof of his theory, he did not even use LE suction. He might have mistakenly thought he was using it.

In fact, LE suction is a totally different phenomenon from the diagram Ace posted.

He used what is called the stagnation point and attempted to get some force in the vertical axis from that.

http://en.wikipedia.org/wiki/Stagnation_point


In other words, it ``stagnates.'' The fluid along the dividing, or ``stagnation streamline'' slows down and eventually comes to rest without deflection at the stagnation point.

http://www.princeton.edu/~asmi...e_web/Bernoulli.html (http://www.princeton.edu/%7Easmits/Bicycle_web/Bernoulli.html)

The stagnation point is a great place to mount your LE devices such as Pitot tubes, slats, stall strips, or in your game, gun barrels. There is no movement and our flow velocity is zero. It is not a separation of flow so we do not have reversal or suction.

The stagnation pressure is found by summing the dynamic pressure and the static pressure. A pitot tube uses the principle of stagnation point pressure referenced with a static pressure from the static port to give you airspeed.

LE suction is just not relevant to gliders which have high aspect ratio wings nor to this conversation.

Additionally there is no leading edge suction on any airfoil in the region of normal command and in high aspect ratio wings I would be surprised to find LE suction in any realm of flight.

Of course mistaking LE suction for stagnation and then arguing the point it moves the airplane forward is just one of those things that makes you go...mmmmmmmm...UBIZOO circus....

Kettenhunde
09-26-2010, 08:28 AM
Here is the pressure distribution graph used to advance the theory of LE suction pulling our glider forward so readers do not have to go looking. It is on page three of the discussion.


Ace says:
This diagram shows that even in level flight relative to the airflow a wing generates a forward lift vector due to leading edge suction.

http://forums.ubi.com/eve/foru...981028388#1981028388 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/4721073388?r=1981028388#1981028388)


Crumpp says:
No that diagram shows the distribution of pressure.


http://forums.ubi.com/eve/foru...861048388#1861048388 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/4721073388?r=1861048388#1861048388)

Here is a primer on LE suction:

http://www.desktop.aero/applie...ous3d/highalpha.html (http://www.desktop.aero/appliedaero/viscous3d/highalpha.html)


At high angles of attack, several phenomena usually distinct from the cruise flow appear.


When the sweep is very large, separation tends to occur near the leading edge of the wing, but unlike in the low sweep situation, the separated region is not large and does not reduce the lift. Instead, the flow rolls up into a vortex that lies just above the wing surface.

http://www.desktop.aero/applie...ous3d/highalpha.html (http://www.desktop.aero/appliedaero/viscous3d/highalpha.html)

Now lets look at a forum expert on aerodynamics explanation for LE suction:


Leading edge suction is real. For example, designing a glider without taking this into account is a recipie for disaster.

As long as the flow over the wing is not separated there will be a forward force relative the chord line: The stagnation point will progressivly move from the front of the LE to below the LE as the AoA goes up. This forces the air forward and around the LE. This is turn pulls the wing forward relative the chord line. In a glider, with long wings this will create a substantial force that wants to pull the wings forward relative to the chord line which the structure has to bear somehow. This is in some designs done with a compression/tension bearing member behind the spar. As the AoA goes up, LE suction force will be overcome by the added drag due to flow separation and the force relative the chord line will once again start to pull the wings back.

Of course, at high speed, the stagnation point will again move towards the front of the LE and the load bearing member behind the spar will start to bear a compression load instead.

The fundamental error made by the local aerodynamics expert is confusing LE suction which is a flow separation event with the stagnation point that exist in any flow field.

The take away is there is no such thing as a tilted vector of lift when our motion is not accelerated.

Holtzauge
09-26-2010, 12:26 PM
Originally posted by Kettenhunde:
The fundamental error made by the local aerodynamics expert is confusing LE suction which is a flow separation event with the stagnation point that exist in any flow field.


No the fundamental error shows in you trying to paraphrase what I said above. The only question that remains is if you are doing this on purpose or if you simply don't understand what I post.


Originally posted by Kettenhunde:
The stagnation point is a great place to mount your LE devices such as Pitot tubes, slats, stall strips, or in your game, gun barrels. There is no movement and our flow velocity is zero.


You don't understand that the stagnation point moves with AoA do you?


Originally posted by Kettenhunde:
It is not a separation of flow so we do not have reversal or suction.


Again you don't get it or misquote on purpose. You do not need separation to get suction. Go back and read what I posted.


Originally posted by Kettenhunde:
A pitot tube uses the principle of stagnation point pressure referenced with a static pressure from the static port to give you airspeed.


How enlightening. Who ever would have guessed that?


Originally posted by Kettenhunde:
LE suction is just not relevant to gliders which have high aspect ratio wings


See, you have absolutely no idea what you are talking about. This statement proves it.


Originally posted by Kettenhunde:
Additionally there is no leading edge suction on any airfoil in the region of normal command and in high aspect ratio wings I would be surprised to find LE suction in any realm of flight.


Given the level of understanding you show here I'm not surprised you would be surprised.

Again your total lack of aerodynamic understanding shines through. Googling some texts, pasting some links and conconcting gobbledygock may impress a certain percentage of this forum but not someone who is actually familiar with aerodynamics.


Originally posted by Kettenhunde:
Of course mistaking LE suction for stagnation


How typically you: Change the context, take parts of someone's posts, put them togtether and make out like the person said something they did not.


Originally posted by Kettenhunde:
and then arguing the point it moves the airplane forward is just one of those things that makes you go...mmmmmmmm...UBIZOO circus....


No, it's one of those things that make you go...mmmmmmmm...classic Kettenhunde....

See comments above on not understanding other peoples posts/purposedly misquoting/gobbledygock.

Let's be honest: The only reason you post here is because in forums with a larger base of people with aerodynamic knowledge you are quickly exposed. Here you get some high quality opposition but in most cases you simply wear it down with quantity beating quality. It always seems to end the same way: You either get the last irrelevant post in or get the thread locked.

In many ways you reming me of Prostetnic Vogon (http://en.wikipedia.org/wiki/Vogon) Jeltz of the Galactic Hyperspace Planning Council: "Resistance is useless!"

AndyJWest
09-26-2010, 01:13 PM
http://i958.photobucket.com/albums/ae65/ajv00987k/GliderForces.jpg

Kettenhunde
09-26-2010, 03:17 PM
You don't understand that the stagnation point moves with AoA do you?

Ahh, Nobody ever said it did not move. That is one reason for a PEC.

However that has nothing to do with the fact LE suction has nothing to do with high aspect ratio wings even at high angles of attack. In fact, in high aspect ratio wings, the opposite occurs and lift is destroyed!!

It is a phenomenon that occurs in highly swept low aspect ratio wings at high angles of attack only.

The subject of LE suction is not even remotely relevant to anything in this thread. Your attempting to make some claim I did not understand the mechanics of your explanation and switch the focus on minutia most readers will not understand. Sorry, this fish won't bite.

The fundamental error is in even trying to use LE suction as having anything to do with a conversation on gliders. The correct response is "WTF does LE suction have to do....it is not applicable in any way."

Bremspropeller pointed this out to you, Holtzauge.

LE suction also has nothing to do with the fact there is never a tilted vector of lift outside of accelerated flight.

Even then lift force will only develop on the vector of apparent weight.

Basic stuff.

Kettenhunde
09-26-2010, 03:23 PM
AndyJWest,

Here is the answer to your diagram's questions from page 4 after I just spelled it out for folks:


What is the difference between your slide over the surface of the earth and glider's flight path???

Nothing. The earth's surface meets the centripetal force required to keep you from flying off into space at all times and all angle short of 90 degrees

To sum it up...Two properties of lift are being confused here.

First lift always develops perpendicular to the path of flight. That is why the diagram shows the lift vector as tilted. Everyone sees that as proof of "tilted vector" driving the airplane.

Secondly, the lift force required meets the force required on the vertical and horizontal axis. It always balances like the centripetal force provided by the earths surface. There is no excess to drive anything.

However since we don't understand the second property of lift we are confusing that first property with the second.



http://forums.ubi.com/eve/foru...521068388#9521068388 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/4721073388?r=9521068388#9521068388)

Here is where I pointed the participants in the right direction on page 3:


M_Gunz, what provides the force to keep you from flying off into space as you stand on the ground?

Why is it difficult to stand on a 60 degree slope of hill? You have no lift force propelling you down the hill. If not for the force of friction provided by your tennis shoes, you would fall down the hill. Just like a glider in fact.

http://forums.ubi.com/eve/foru...251048388#8251048388 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/4721073388?r=8251048388#8251048388)

Here is where Col_Sanders got it and pointed it out to the readers....


In a real aircraft the wings acting against the air, in fact, creates the inclined plane (it's called an "air plane" for a reason you know).

http://forums.ubi.com/eve/foru...001071488#1001071488 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/4721073388?r=1001071488#1001071488)

I will leave you to the host of aerodynamic experts on Ubizoo!

I am sure they will be glad to expound on the virtues of tilted lift and prove that the stagnation point has everything to do with LE suction on a high aspect ratio glider.

AndyJWest
09-26-2010, 04:09 PM
Thank you for not replying to the point raised by my diagram, Kettenhunde. http://forums.ubi.com/groupee_common/emoticons/icon_rolleyes.gif

Can I ask if you consider there to be rearward-acting aerodynamic forces on the glider (e.g. 'drag'), and if you do, do you agree that if the glider is not to change it's velocity over time, there must also be forward-acting aerodynamic forces?

Of course, it could be argued that the horizontally-acting forces cancel each other out, so there is no 'drag' at all, but this seems a little counter-intuitive.

Incidentally Kettenhunde, I'm sure you will find an object sliding down a hill exerts a horizontal force on the hill as it does it.

Kettenhunde
09-26-2010, 05:23 PM
Of course, it could be argued that the horizontally-acting forces cancel each other out, so there is no 'drag' at all, but this seems a little counter-intuitive.

Incidentally Kettenhunde, I'm sure you will find an object sliding down a hill exerts a horizontal force on the hill as it does it.

Sutdy the Glenn research center diagram and the high school physics pages posted on inclined planes.

It will answer all your questions. You can also refer to the various aerodynamic sites on glider flight I linked in my first postings.

Good Luck!

AndyJWest
09-26-2010, 05:35 PM
Come on, Kettenhunde, I'm asking questions with simple yes or no answers:

Do you consider there to be rearward-acting aerodynamic forces on the glider (e.g. 'drag')? Yes or no?

If your answer to the above is 'yes', do you agree that if the glider is not to change it's velocity over time, there must also be forward-acting aerodynamic forces? Yes or no?

Romanator21
09-26-2010, 05:46 PM
Can I ask if you consider there to be rearward-acting aerodynamic forces on the glider (e.g. 'drag'), and if you do, do you agree that if the glider is not to change it's velocity over time, there must also be forward-acting aerodynamic forces?

Of course, it could be argued that the horizontally-acting forces cancel each other out, so there is no 'drag' at all, but this seems a little counter-intuitive.

Forward acting force in a glider comes from gravity, in the same way that a ball rolling down a shallow ramp moves "forward". Drag on the plane (friction on the ball) matches this force if the velocity is constant. In still air (no updrafts) a glider will descend at a steady rate while moving forward. If the pilot were to raise the nose so that the plane would not descend (or if the ramp flattened out), drag would take over and start to act on the glider, decelerating it to the point that it would stall.

Just because the forces cancel does not mean you can say "there is no 'drag' at all."

JtD
09-26-2010, 05:51 PM
If you were to ask me I'd say yes, and the force is gravity.

Lift is perpendicular to the airflow, drag is perpendicular to lift in line with the airflow. Now you can turn this coordinate system any way you want, said so a couple of pages ago, and by choosing a suitable reference you can find "forward" lift.
It's like saying you fire a projectile forward, and someone tells you if you turn the gun upwards the projectile goes upwards, not forward. Well, fine. Doesn't mean the projectiles punches through the barrel.

AndyJWest
09-26-2010, 06:02 PM
Gravity acts vertically. How does this become a horizontal force, Romanator?

I still maintain that an object propelled down a ramp by gravity will exert a horizontal force on the ramp. This calls for a practical demonstration - I'll see if I can provide this tomorrow.

JtD, if you define lift as 'perpendicular to the airflow', then yes, there is a forward component. I've been saying this all along. I'd like to know for sure if anyone here suggests there isn't, hence the diagram.

Romanator21
09-26-2010, 06:08 PM
Gravity acts vertically. How does this become a horizontal force, Romanator?

http://forums.ubi.com/images/smilies/blink.gif

Uhh, well balls roll down hills, right? If we show the forces as vectors, gravity points down vertically. However, the ramp also exerts a force on the ball - otherwise the ball would fall through the ramp. This force is perpendicular to the surface of the ramp, and if I drew it, would be a slanted line. We can break the vector into its vertical and horizontal components, the latter of which is relevant to the forward motion of the ball or glider.

http://www.lowellphysics.org/beta/Textbook%20Resources/Chapter%2011.%20Rolling,%20Torque,%20and%20Angular %20Momentum/11-4/c11x11_4.xform_files/nw0577-n.gif

If we consider the above, for the ball to be moving without accelerating, F_g_cos(theta) is equal to F_N_, and F_s_ is equal to F_g_sin(theta).

Maybe I don't understand what you're getting at here.

p-11.cAce
09-26-2010, 07:07 PM
So this afternoon I'm chugging along in the ASK-21 trying to figure out how to come at this question in a straightforward way - so I did a little experiment.

I trimmed for 65 knots and let the speed stabilize. This puts me right in the Glen Center diagram - all forces balanced. I then pushed forward on the stick and re-trimmed at 100 knots. In the ASK this gives a pretty significant nose-down attitude by the way. Once again I'm back at the Glen Center diagram - just burning off a bit more altitude than before.

So then I thought - ok the force of gravity did not change, but I picked up 35 knots. Obviously this acceleration was provided by increasing the forward lift vector until it was balanced by the increasing drag. You can call this a decrease in induced drag (which it is) but since the FAA clearly points out that induced drag is the rearward acting component of lift, to imply that it' reduction is anything other than an increase in the forward component of lift is just playing at semantic games.

The I realized that while I was losing more height, the loss of height was not a linear increase in relation to the increase in airspeed as one would assume by Kette's assertion that all forces are always in balance. The reason the loss of height is not linear is due to the fact that overcoming parasitic drag at higher airspeeds requires converting an increasing amount of lift from vertical to horizontal, resulting in less lift available to resist the downward pull of gravity.

http://www.williamssoaring.com/images/fleet-images/ask21-polar-450x.jpg

So this sums it up:


Three principle forces act on aircraft when gliding:

* weight - gravity acts in the downwards direction
* lift - acts perpendicularly to the vector representing airspeed
* drag - acts parallel to the vector representing the airspeed

As the aircraft descends, the air moving over the wings generates lift. The lift force acts slightly forward of vertical because it is created at right angles to the airflow which comes from slightly below as the glider descend. This horizontal component of lift is enough to overcome drag and allows the glider to accelerate forward. Even though the weight causes the aircraft to descend, if the air is rising faster than the sink rate, there will be gain of altitude.

Kettenhunde
09-26-2010, 07:11 PM
So then I thought - ok the force of gravity did not change, but I picked up 35 knots.

The force of gravity did change at the Sine of the angle.


In the ASK this gives a pretty significant nose-down attitude by the way.

You took math in your education, right?

It is not linear because the aircraft accelerates to balance and is not in steady state flight until it balances.

During that time lift will meet the forces required in the acceleration vectors.

p-11.cAce
09-26-2010, 08:00 PM
The force of gravity did change

Really? Christ I did not realize that Starfleet built my ASK http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif

Now who needs the high school physics class?

Kettenhunde
09-26-2010, 08:09 PM
Really? Christ I did not realize that Starfleet built my ASK

You left out the important part in your quote.

.....at the sine of the angle.

Sin 10 = .174 * 32.2fpss = 5.6fpss

Sin 30 = .5 * 32.2 = 16.1 fpss

Or a 287% increase on the horizontal vector in the acceleration of gravity.

You see were you picked up the 35 knots now?

F=ma...right?

Romanator21
09-26-2010, 08:13 PM
Excuse me while I fix up a batch of popcorn...

Kettenhunde
09-26-2010, 08:16 PM
Excuse me while I fix up a batch of popcorn...


Make that two....

This is actually more amusing than I thought.

Col_SandersLite
09-26-2010, 08:18 PM
A better way to think of it (imo):

Gravity is not different between these to states. It is still 1g straight downwards.

The generated lift force generated by the wing did not change either (well, it increases with speed, but so does drag in the exact same proportion). Though the vector did.

What changed was drag.

When your nose is lower, your wing is generating the same amount of lift as it would be at any other glide angle for that speed. However, more of your lift is being counteracted by drag. Therefore you have less *usable* lift. When you have less usable lift to counteract the constant of gravity, the result is acceleration until you hit your new terminal velocity.

The answer is literally staring you in the face right here:


Originally posted by Kettenhunde:
http://img213.imageshack.us/img213/9899/forcevectorsonaglider.jpg (http://img213.imageshack.us/i/forcevectorsonaglider.jpg/)



Originally posted by Romanator21:
http://www.lowellphysics.org/beta/Textbook%20Resources/Chapter%2011.%20Rolling,%20Torque,%20and%20Angular %20Momentum/11-4/c11x11_4.xform_files/nw0577-n.gif

If we consider the above, for the ball to be moving without accelerating, F_g_cos(theta) is equal to F_N_, and F_s_ is equal to F_g_sin(theta).


I don't see how it can be any more clear. Did you try the experiment I suggested ace? It will really let you interact with the concept.


--
This space intentionally left blank.

p-11.cAce
09-26-2010, 08:22 PM
The generated lift force generated by the wing did not change either. Though the vector did.

That is exactly what I have been saying since page 1. http://forums.ubi.com/images/smilies/compsmash.gif


When your nose is lower, your wing is generating the same amount of lift as it would be at any other glide angle for that speed. However, more of your lift is being counteracted by drag. Therefore you have less *usable* lift. When you have less usable lift to counteract the constant of gravity, the result is acceleration until you hit your new terminal velocity.

You have less VERTICAL lift because you are increasing the HORIZONTAL component to overcome drag. Call it reducing induced drag by increasing the forward component. As you correctly pointed out the wing is generating the same force, you accelerate by changing where that force vector is pointed.

Oh and Kette unless the freaking mass of the earth changed its gravity is always the same.

M_Gunz
09-26-2010, 08:35 PM
Originally posted by AndyJWest:
Gravity acts vertically. How does this become a horizontal force, Romanator?

I still maintain that an object propelled down a ramp by gravity will exert a horizontal force on the ramp. This calls for a practical demonstration - I'll see if I can provide this tomorrow.

Of course there is by Newton's 2nd Law of Motion, at the point of contact the force of the object in normal to the ramp.
Funny, I posted about balls and blocks on inclined planes pages ago.


JtD, if you define lift as 'perpendicular to the airflow', then yes, there is a forward component. I've been saying this all along. I'd like to know for sure if anyone here suggests there isn't, hence the diagram.

The forward component is not forward with respect to the glide slope but is forward with respect to a gravity-aligned coordinate system which was also posted about by many pages ago.

Here is how moronic some of these arguments are:
Person A: "Where is the turn to the gas station?"
Person B: "You go down 3 blocks and turn left"
Person C: "But it was 2 blocks up and turn right on the way here so you must be wrong!"

M_Gunz
09-26-2010, 08:44 PM
Originally posted by p-11.cAce:
So then I thought - ok the force of gravity did not change, but I picked up 35 knots. Obviously this acceleration was provided by increasing the forward lift vector until it was balanced by the increasing drag.

You tilted the glide slope more downwards. How much speed increase came from increasing your vertical speed?

Forward lift vector.. forward of what? Bad choice of wording, too ambiguous.

M_Gunz
09-26-2010, 08:47 PM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">So then I thought - ok the force of gravity did not change, but I picked up 35 knots.

The force of gravity did change at the Sine of the angle. </div></BLOCKQUOTE>

More bad word choice. Force of gravity did not change. His alignment to gravity did.

M_Gunz
09-26-2010, 08:54 PM
Originally posted by p-11.cAce:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">The generated lift force generated by the wing did not change either. Though the vector did.

That is exactly what I have been saying since page 1. http://forums.ubi.com/images/smilies/compsmash.gif </div></BLOCKQUOTE>

The steeper you dive or climb at steady speed, the less lift required to keep that course. So yah the lift changes, it decreases.

I you dive straight down and make any lift at all, you will no longer dive straight down.. zero lift to dive straight down. The lift required varies with the cosine of the glide slope.

Kettenhunde
09-26-2010, 09:08 PM
Force of gravity did not change.

In the horizontal axis it certainly did change at the sine of the angle.

The choice of words was not poor.

Col_SandersLite
09-26-2010, 09:48 PM
Originally posted by p-11.cAce:
You have less VERTICAL lift because you are increasing the HORIZONTAL component to overcome drag.

No, the horizontal component is canceled by drag. End of story. Your acceleration is caused by moving down an inclined plane in same way gravity interacts with a slot car.



Originally posted by M_Gunz:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by p-11.cAce:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">The generated lift force generated by the wing did not change either. Though the vector did.

That is exactly what I have been saying since page 1. http://forums.ubi.com/images/smilies/compsmash.gif </div></BLOCKQUOTE>

The steeper you dive or climb at steady speed, the less lift required to keep that course. So yah the lift changes, it decreases.

I you dive straight down and make any lift at all, you will no longer dive straight down.. zero lift to dive straight down. The lift required varies with the cosine of the glide slope. </div></BLOCKQUOTE>

No, *usable* lift changes, because it is offset by drag. Lift itself does not change unless the speed does.


Frankly, this has been explained as clearly as humanly possible in multiple different ways. If these statements did not change your point of view, most likely the only thing that will is if Oleg Maddox himself magically descends from the heavens and smites you (close this book and never open again, dulang dulang). I dunno about ace, but gunz would probably still stubbornly refuse to see where he's gone wrong.

M_Gunz
09-26-2010, 11:30 PM
Originally posted by Col_SandersLite:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by p-11.cAce:
You have less VERTICAL lift because you are increasing the HORIZONTAL component to overcome drag.

No, the horizontal component is canceled by drag. End of story. Your acceleration is caused by moving down an inclined plane in same way gravity interacts with a slot car.



Originally posted by M_Gunz:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by p-11.cAce:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">The generated lift force generated by the wing did not change either. Though the vector did.

That is exactly what I have been saying since page 1. http://forums.ubi.com/images/smilies/compsmash.gif </div></BLOCKQUOTE>

The steeper you dive or climb at steady speed, the less lift required to keep that course. So yah the lift changes, it decreases.

I you dive straight down and make any lift at all, you will no longer dive straight down.. zero lift to dive straight down. The lift required varies with the cosine of the glide slope. </div></BLOCKQUOTE>

No, *usable* lift changes, because it is offset by drag. Lift itself does not change unless the speed does. </div></BLOCKQUOTE>

WRONG. Lift changes with speed and AOA. And to stay on path the plane only needs as much lift as to balance its weight times the cosine of the angle of its path.


Frankly, this has been explained as clearly as humanly possible in multiple different ways. If these statements did not change your point of view, most likely the only thing that will is if Oleg Maddox himself magically descends from the heavens and smites you (close this book and never open again, dulang dulang). I dunno about ace, but gunz would probably still stubbornly refuse to see where he's gone wrong.

Awwwww. The gratitous put-down. Too bad you're wrong Sparky. Here's some more for you to eat: Pitch Is Not Path. Learn to Read (example: impractical and impossible are two different words wth two different meanings). And a new one: Lift goes by Speed +and+ AOA.

And just because you quoted me and still can't figure it out, have a look again then get your crayons out.
I you dive straight down and make any lift at all, you will no longer dive straight down.

Think you can figure that out?

Have Fun With Yourself and quit playing Miss Representation with me. You're not that good.

M_Gunz
09-26-2010, 11:50 PM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Force of gravity did not change.

In the horizontal axis it certainly did change at the sine of the angle. </div></BLOCKQUOTE>

The gravity? On the horizontal axis? All by itself?

Even a block sliding on a hill, it is the hill that forces the sideways motion and gravity only supplying the energy.

Take the hill away, the block falls straight down. Either way, the gravity and the mass of the block are not changing, even if the slope of the hill changes. Replace the hill with lift and it's the same, same as pointed out pages ago.

You can't even admit that a banked plane has a sideways lift component which in the past you have pointed out to others. Why? Because then a glider on a downward slope might have a horizontal lift component. So -now- there is horizontal gravity!

How many aerodynamics links you want showing a banked plane with a sideways lift component? NASA good enough? How about Denker? Hmmmm?

Col_SandersLite
09-27-2010, 01:19 AM
Originally posted by M_Gunz:
I you dive straight down and make any lift at all, you will no longer dive straight down.. zero lift to dive straight down. The lift required varies with the cosine of the glide slope.

This is because, dun dun dun, if you're going straight down, your lift is 100% counteracted by drag in the horizontal plane!


Originally posted by M_Gunz:
Here's some more for you to eat: Pitch Is Not Path. Learn to Read

If you can quote where I said otherwise, you're not an illiterate idiot.

JtD
09-27-2010, 01:25 AM
You know what, this is an arrow to the left:

<-

Now you can all turn your screen any way you want, until you have it show up, left or down. And then fill 8 pages on an internet board about where the arrow points.

WTE_Galway
09-27-2010, 01:52 AM
Fun Toy :

http://www.oview.co.uk/cgi-bin...orces/fourforces.exe (http://www.oview.co.uk/cgi-bin/download.pl?url=fourforces/fourforces.exe)

http://www.oview.co.uk/fourforces/


BTW ... I have been ignoring this thread because I simply cannot understand where people are confused.

1. Sitting in a glider in a stable dive the lift is vertical to the glide path (give or take a degree or two).

2. This is countered by a component of the weight also vertical to the glide path but opposite to the lift.

3. In addition the "other" component of the weight (necessary for the total weight vector to point down) is along the glide path overcoming drag and moving the aircraft down the glide path.

Hence gravity causes the aircraft to move ... as commonsense would tell you.

JtD has the right idea http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif ... the frame of reference is the glide path not the ground.

M_Gunz
09-27-2010, 03:21 AM
Originally posted by Col_SandersLite:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by M_Gunz:
I you dive straight down and make any lift at all, you will no longer dive straight down.. zero lift to dive straight down. The lift required varies with the cosine of the glide slope.

This is because, dun dun dun, if you're going straight down, your lift is 100% counteracted by drag in the horizontal plane! </div></BLOCKQUOTE>

Try again. Drag is opposite the direction of flight.


<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by M_Gunz:
Here's some more for you to eat: Pitch Is Not Path. Learn to Read

If you can quote where I said otherwise, you're not an illiterate idiot. </div></BLOCKQUOTE>

I'm not illiterate and I'm not an idiot regardless. I also don't think much of your idea of logic, btw. Keep proving what you do about yourself.
As for learning to read, perhaps you can work on the difference between 'impractical' and 'impossible'. Then maybe apologize to Waldo.

M_Gunz
09-27-2010, 03:25 AM
Originally posted by JtD:
You know what, this is an arrow to the left:

<-

Now you can all turn your screen any way you want, until you have it show up, left or down. And then fill 8 pages on an internet board about where the arrow points.

LOL, it points southward on my screen. But I won't say that means it doesn't point left, which is how fubar this thread has gone! http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

p-11.cAce
09-27-2010, 06:10 AM
Hence gravity causes the aircraft to move ... as commonsense would tell you.

Yes- as M_Gunz has correctly stated many many times gravity supplies the energy but not the horizontal force:


Even a block sliding on a hill, it is the hill that forces the sideways motion and gravity only supplying the energy.

Take the hill away, the block falls straight down. Either way, the gravity and the mass of the block are not changing, even if the slope of the hill changes. Replace the hill with lift and it's the same, same as pointed out pages ago.

Or in another example from a few pages back:


The ball rolling down the hill is propelled forward by the component of the force pushing up on it from the sloped surface in response to gravity pushing it down onto that surface. Gravity operates vertically down through the center of mass and without that sloping surface the ball would not move forward. There is a component of this force vector pushing forward and the ball moves forward down the slope.

This is exactly analogous to the lift vector pointed forward on the glider and the forward component of that vector providing the force to
overcome drag in forward flight. Gravity operates on the glider though it's center of mass, without that lift vector pointed forward the glider would not (continue to) move forward.

Gravity, without the thrust vector leaning forward, does not explain how a glider glides through the air, however Gravity is what provides the power/energy to do this (or store for that energy if you prefer).

Kettenhunde
09-27-2010, 06:44 AM
without that sloping surface the ball would not move forward.

I highlighted the key part for you. It is the force of lift that provides the sloping surface for the glider.


I have been ignoring this thread because I simply cannot understand where people are confused.

Me either....

It is proof of the dumbing down of western society. Our education standards are in the toilet compared to other places in the world. Soon everything will be built by Indian Engineers, financed by Middle Eastern money, and manufactured in China.

The vector alignment and math are very straight forward. It is amazing to me that people cannot see that.


Once aloft, gravity (the weight of the hang glider and pilot) pulls the glider back toward Earth and propels the glider forward, continually causing air to flow over the wing.

http://adventure.howstuffworks.com/hang-gliding1.htm

p-11.cAce
09-27-2010, 06:48 AM
...and - TADA!

From:

there is no component of lift that is thrusting forward or pulling back.

to


It is the force of lift that provides the sloping surface for the glider.

in only 8 pages http://forums.ubi.com/images/smilies/53.gif

Kettenhunde
09-27-2010, 06:53 AM
Yes- as M_Gunz has correctly stated many many times gravity supplies the energy but not the horizontal force:


Gravity is a fundamental force in the universe. That is very basic and taught in any physics class.

http://hyperphysics.phy-astr.g...e/forces/funfor.html (http://hyperphysics.phy-astr.gsu.edu/hbase/forces/funfor.html)

Force = Mass * Acceleration

Acceleration due to gravity = 32.2 fpss

At a 30 degree dive angle with an object of 200lbs exerts a force in the horizontal axis of:

Force = 200lbs * 32.2fpss <Sin 30> = 3220 pdl

Kettenhunde
09-27-2010, 07:02 AM
in only 8 pages


You don't get it at all.

The properties of lift....


What is the difference between your slide over the surface of the earth and glider's flight path???

Nothing. The earth's surface meets the centripetal force required to keep you from flying off into space at all times and all angle short of 90 degrees

To sum it up...Two properties of lift are being confused here.

First lift always develops perpendicular to the path of flight. That is why the diagram shows the lift vector as tilted. Everyone sees that as proof of "tilted vector" driving the airplane.

Secondly, the lift force required meets the force required on the vertical and horizontal axis. It always balances like the centripetal force provided by the earths surface. There is no excess to drive anything.

However since we don't understand the second property of lift we are confusing that first property with the second.

http://forums.ubi.com/eve/foru...521068388#9521068388 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/4721073388?r=9521068388#9521068388)

Go back and look at your Glenn Research Facility diagram you so proudly posted...

Vertical Axis = Lift <Cos a> + Drag <Sin a> - WEIGHT = ZERO

Notice drag and lift are added to the force offsetting the component of weight. Our weight would be equal to weight x Sine of the angle.

It is the force of lift and drag that provide our sloping surface in the vertical axis.

Horizontal Axis = Lift <Sin a> - Drag <Cos a> = ZERO

No weight is included in the horizontal axis. Conservation of momentum and the force of gravity pull the aircraft on that horizontal vector just like your ball rolling down hill.

Kettenhunde
09-27-2010, 07:20 AM
http://img835.imageshack.us/img835/3739/horizontallift.jpg (http://img835.imageshack.us/i/horizontallift.jpg/)

M_Gunz
09-27-2010, 10:56 AM
Originally posted by p-11.cAce:
http://www.grc.nasa.gov/WWW/K-12/airplane/Images/glidvec.gif



Go back and look at your Glenn Research Facility diagram you so proudly posted...

Vertical Axis = Lift <Cos a> + Drag <Sin a> - WEIGHT = ZERO

Notice drag and lift are added to the force offsetting the component of weight. Our weight would be equal to weight x Sine of the angle.

It is the force of lift and drag that provide our sloping surface in the vertical axis.

Horizontal Axis = Lift <Sin a> - Drag <Cos a> = ZERO

No weight is included in the horizontal axis. Conservation of momentum and the force of gravity pull the aircraft on that horizontal vector just like your ball rolling down hill.



Weight being gravity x mass only points in the direction of gravity, straight down. It's not just an idea, it's the Law!

They clearly show the tilt of the lift and drag as the only vectors with horizontal components. What drives them is not pushing the plane to the side.

My problem with the term "forward lift" is that it seems to infer to a net lift component forward along the flight path which that same diagram shows there is not. The term by itself not saying forward of what is ambiguous and easy to turn into something it is not, standard practice here and elsewhere. But that's no reason to deny that the lift vector tilted with respect to gravity does account as the mechanism for motion sideways to gravity.

K_Freddie
09-27-2010, 10:58 AM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Force of gravity did not change.
In the horizontal axis it certainly did change at the sine of the angle.
</div></BLOCKQUOTE>
That's Heavy !!

Col_SandersLite
09-27-2010, 11:29 AM
Originally posted by M_Gunz:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Col_SandersLite:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by M_Gunz:
I you dive straight down and make any lift at all, you will no longer dive straight down.. zero lift to dive straight down. The lift required varies with the cosine of the glide slope.

This is because, dun dun dun, if you're going straight down, your lift is 100% counteracted by drag in the horizontal plane! </div></BLOCKQUOTE>

Try again. Drag is opposite the direction of flight.
</div></BLOCKQUOTE>

Try again, there is more than one drag component.



Originally posted by M_Gunz:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content"><BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by M_Gunz:
Here's some more for you to eat: Pitch Is Not Path. Learn to Read
If you can quote where I said otherwise, you're not an illiterate idiot. </div></BLOCKQUOTE>
I'm not illiterate and I'm not an idiot regardless. I also don't think much of your idea of logic, btw. Keep proving what you do about yourself.
As for learning to read, perhaps you can work on the difference between 'impractical' and 'impossible'. </div></BLOCKQUOTE>
Fail.

Kettenhunde
09-27-2010, 03:55 PM
Weight being gravity x mass only points in the direction of gravity, straight down. It's not just an idea, it's the Law!


http://img337.imageshack.us/img337/8852/forceofgravityvectorson.jpg (http://img337.imageshack.us/i/forceofgravityvectorson.jpg/)


http://img265.imageshack.us/img265/9899/forcevectorsonaglider.jpg (http://img265.imageshack.us/i/forcevectorsonaglider.jpg/)

It is just simple vector resolution from high school physics, M_Gunz.

AndyJWest
09-27-2010, 04:05 PM
Yes Kettenhunde, you can do that with vector maths. So what?

In any case, this inclined plane analogy is just that, an analogy. And a bad one, since the air isn't a frictionless solid...

"Proof by analogy is fraud": Bjarne Stroustrup.

WTE_Galway
09-27-2010, 04:39 PM
Originally posted by M_Gunz:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by JtD:
You know what, this is an arrow to the left:

<-

Now you can all turn your screen any way you want, until you have it show up, left or down. And then fill 8 pages on an internet board about where the arrow points.

LOL, it points southward on my screen. But I won't say that means it doesn't point left, which is how fubar this thread has gone! http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif </div></BLOCKQUOTE>

Actually points North here.

Is that because I am in the Southern Hemisphere ?

M_Gunz
09-27-2010, 04:39 PM
Yup, it is the tilt of the force normal to the inclined plane that makes the sideways force happen.

Jr High physical science when I went to school, before the Reagans raped the school budget.

M_Gunz
09-27-2010, 04:43 PM
I like the one about horizontal drag when the plane has only vertical motion. Imagine having drag without motion! Now that's about as "I wanna win!" as you can get. Gaston's got some real competition at fantasy-land building there!

Kettenhunde
09-27-2010, 06:21 PM
Yes Kettenhunde, you can do that with vector maths. So what?

I am surprised at the number of folks who cannot....

Vertical Axis = Lift <Cos a> + Drag <Sin a> - WEIGHT = ZERO

Notice drag and lift are added to the force offsetting the component of weight. Our weight would be equal to weight x Sine of the angle.

It is the force of lift and drag that provide our sloping surface in the vertical axis.

Horizontal Axis = Lift <Sin a> - Drag <Cos a> = ZERO

No weight is included in the horizontal axis. Conservation of momentum and the force of gravity pull the aircraft on that horizontal vector just like your ball rolling down hill.

There is no "tilted vector of lift". Lift is perpendicular to the RW as with the first property.

Lastly there is no "excess lift" under any steady state condition.


this inclined plane analogy is just that, an analogy.

It does illustrate the point but it wasn't my analogy, it was M_Gunz. The force of lift in the vertical plane does act like an inclined surface.

The first three pages of conventional explanations simply were not working.

http://forums.ubi.com/images/smilies/partyhat.gif

Kettenhunde
09-27-2010, 06:32 PM
And a bad one, since the air isn't a frictionless solid...


BTW, It is just meant to be a less complicated way to illustrate the point.

The vector resolution of gravity is not just in the vertical plane and you don't have the friction vectors to deal with using the frictionless diagram.

It is like turning.

The vector of acceleration shifts from the vertical in 1G level flight and our lift force is resolved in two components.

However to the airplane in a coordinated turn, the apparent weight is still through the vertical axis.

http://img638.imageshack.us/img638/5600/vectorofaceleration.jpg (http://img638.imageshack.us/i/vectorofaceleration.jpg/)

http://img201.imageshack.us/img201/1414/jjgsturnill155.jpg (http://img201.imageshack.us/i/jjgsturnill155.jpg/)

M_Gunz
09-27-2010, 07:03 PM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content"> Yes Kettenhunde, you can do that with vector maths. So what?

I am surprised at the number of folks who cannot....

Vertical Axis = Lift <Cos a> + Drag <Sin a> - WEIGHT = ZERO

Notice drag and lift are added to the force offsetting the component of weight. Our weight would be equal to weight x Sine of the angle. </div></BLOCKQUOTE>

The Weight that NASA shows is gravity x mass, straight down. Lift and drag components are the -complete- force offsetting the weight.. what do you see them +added+ to? There is nothing else in the diagram and nothing else in the equations.


It is the force of lift and drag that provide our sloping surface in the vertical axis.

Horizontal Axis = Lift <Sin a> - Drag <Cos a> = ZERO

No weight is included in the horizontal axis. Conservation of momentum and the force of gravity pull the aircraft on that horizontal vector just like your ball rolling down hill.

There is no "tilted vector of lift". Lift is perpendicular to the RW as with the first property.

The lift vector is tilted with respect to gravity as the diagram clearly shows.


Lastly there is no "excess lift" under any steady state condition.

And here is the mystery, did what anyone post actually say there is? Because with all the loose term usage I sure can't say! That doesn't count use of the term 'usable lift' which was just total BS.


<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">this inclined plane analogy is just that, an analogy.

It does illustrate the point but it wasn't my analogy, it was M_Gunz. The force of lift in the vertical plane does act like an inclined surface.

The first three pages of conventional explanations simply were not working.

http://forums.ubi.com/images/smilies/partyhat.gif </div></BLOCKQUOTE>

Lift is perpendicular to the flight path. The flight path is at an angle to vertical. The lift vector is at an angle to vertical. That is the conventional explanation as NASA would have it.

How about, just for laughs, trying to understand what NASA is communicating there and not trying to turn it into something else or remap it to some other view?

Kettenhunde
09-27-2010, 07:54 PM
That is the conventional explanation as NASA would have it.

Nasa does not say that lift propels the aircraft anywhere....

And neither does the FAA.

The forces are balances at all times M_Gunz as lift only meets the force required. That is why the formula equals ZERO.

Where are you getting any excess lift to propel the Glider?

You are not and it is gravity that brings the aircraft to the ground.

That is why gliders carry water ballast!

If you don't believe me, then lets end all the games and refer to the basic definition of gliding flight...

http://img72.imageshack.us/img72/8660/glidingflight.jpg (http://img72.imageshack.us/i/glidingflight.jpg/)

http://books.google.com/books?...#v=onepage&q&f=false (http://books.google.com/books?id=V3SZXFWuCIgC&pg=SA3-PA16&lpg=SA3-PA16&dq=Engine+out+glide+gravity&source=bl&ots=HkRJKijIlz&sig=0bvmyLUTOqtN4u9-xq2Ox8ndxc4&hl=en&ei=dUmhTL76FZGisQPF5fVs&sa=X&oi=book_result&ct=result&resnum=4&sqi=2&ved=0CBwQ6AEwAw#v=onepage&q&f=false)

Kettenhunde
09-27-2010, 08:20 PM
There are some real clowns on these forums....

http://forums.ubi.com/images/smilies/partyhat.gif

M_Gunz
09-27-2010, 08:56 PM
If you take the lift out of it, the plane falls straight down. The only vectors in that NASA diagram or those NASA equations with a horizontal component are lift and drag.

At this point we are just arguing viewpoints and where to assign forces. Some books say one thing, others say something else.

It's the same with turns:

What would a Caltech physicist who is also an FAA official know about aircraft forces? (http://www.av8n.com/how/htm/4forces.html#sec-defer-forces)


4.3 Forces During a Turn

The most important non-aerobatic situation where you have to worry about the forces on the airplane is during a turn. In a steeply-banked turn, the lift vector is inclined quite a bit to the left or right of vertical. In order to support the weight of the airplane and pull the airplane around the turn, the lift must be significantly greater than the weight. This leads us to the notion of load factor, which is discussed in section 6.2.3.

The bottom line is that thrust is usually nearly equal (and opposite) to drag, and lift is usually nearly equal (and opposite) to weight times load factor.

In a turn, it is sometimes useful to express the total lift as a sum of two components.

* The vertical component of lift, as usual, is what opposes weight, so there is no net vertical force, so that the airplane does not accelerate upwards or downward.
* The horizontal component of lift is what provides a horizontal force that changes which way the airplane is going.1

In a steeply-banked turn, the horizontal component of lift is quite large. In the pilot’s frame of reference, that means the airplane is subject to very significant centrifugal forces. This important and interesting topic will be discussed in section 6.2.

Also:

Far and away the most powerful technique2 for changing the direction an airplane is going is to put it into a bank, so that the horizontal component of lift forces a change in flight-path, as mentioned in section 4.3 and especially section 6.2. This is not yaw; bank by itself will not change the direction you are pointing.

Also in more than a dozen places on that site. Yet you can find a view in a book that says different.

And here in figures 4.2, 4.3 and 4.4 the lift vector is shown at an angle to gravity. Do you suppose that makes it gravity? (http://www.av8n.com/how/htm/4forces.html)

What I am pointing out is that this view is correct and no amount of different viewpoint changes that.

Kettenhunde
09-27-2010, 09:10 PM
It's the same with turns:

I have no idea what your point is M_Gunz. All the stuff you posted on turns says exactly the same thing I posted.

It is acceleration in turning flight that provides the apparent weight that lift must meet the force required in the horizontal and vertical axes.

You do understand that turning flight MUST have an acceleration while gliding flight does not?

What is the point???

As for gliding flight, there is no tilted vector of lift driving the aircraft forward along along the horizontal axis.

That is a fact no matter what point of view or reference you want to take.

The forces of lift and drag are always in balance in a glider on the horizontal axis.

It is the force of gravity that propels the aircraft forward in the horizontal axis.

Kettenhunde
09-27-2010, 09:14 PM
In a steeply-banked turn, the lift vector is inclined quite a bit to the left or right of vertical.

Kind of like this diagram....

http://img651.imageshack.us/img651/1414/jjgsturnill155.jpg (http://img651.imageshack.us/i/jjgsturnill155.jpg/)

M_Gunz you have drank the kool aide my friend.

There are not multiple frames of reference in that Glenn Research diagram. There is only ONE frame of reference, the aircraft.

It uses what is called the body axis to resolve aerodynamic forces on the CG.

There are multiple vectors to resolve those forces acting on the CG but only ONE frame of reference.

The fact it was even brought up is amusing. I am surprised "the experts" did not know that.

Between that and Holtzauge's Leading Edge suction which is about as relevant to this discussion as dancing bears, it is no wonder you are confused.

p-11.cAce
09-27-2010, 09:40 PM
http://www.gifsoup.com/view/552541/bunny-eats-popcorn-o.gif

Kettenhunde
09-27-2010, 09:45 PM
Something to read while you eat....


Airplane Flying Handbook
By Federal Aviation Administration

http://img72.imageshack.us/img72/8660/glidingflight.jpg (http://img72.imageshack.us/i/glidingflight.jpg/)

http://books.google.com/books?...#v=onepage&q&f=false (http://books.google.com/books?id=V3SZXFWuCIgC&pg=SA3-PA16&lpg=SA3-PA16&dq=Engine+out+glide+gravity&source=bl&ots=HkRJKijIlz&sig=0bvmyLUTOqtN4u9-xq2Ox8ndxc4&hl=en&ei=dUmhTL76FZGisQPF5fVs&sa=X&oi=book_result&ct=result&resnum=4&sqi=2&ved=0CBwQ6AEwAw#v=onepage&q&f=false)

M_Gunz
09-28-2010, 03:15 AM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">It's the same with turns:

I have no idea what your point is M_Gunz. All the stuff you posted on turns says exactly the same thing I posted.

It is acceleration in turning flight that provides the apparent weight that lift must meet the force required in the horizontal and vertical axes.

You do understand that turning flight MUST have an acceleration while gliding flight does not?

What is the point??? </div></BLOCKQUOTE>

The acceleration is provided by the tilted lift vector. It is acceleration because it is not balanced out by an opposing sideways force.


As for gliding flight, there is no tilted vector of lift driving the aircraft forward along along the horizontal axis.

Only when the gliding flight is at a steady speed. The tilted lift vector is balanced by drag then. However gliders can and do accelerate by changing pitch and for a time the forces are unbalanced. I'm pretty sure it's possible to loop a glider and turn lift upside down though the AOA would likely be kept very small at the top.


That is a fact no matter what point of view or reference you want to take.

The forces of lift and drag are always in balance in a glider on the horizontal axis.

No they're not. They are only in balance in steady state flight.

If I put my hand on the wall and push, the wall pushes back equally hard and opposite my push. Unless I push harder than the wall can push back in which case the wall moves, or a piece of it does. I've been able to calculate such things, still remember most of it and had to know the whys as well. Hard to do tool design otherwise.


It is the force of gravity that propels the aircraft forward in the horizontal axis.

Yah well the steam in the piston drives the steam engine but it is the crank that turns the reciprocating motion into circular motion. And speaking of circular motion....

Has this thread mushed or stalled?

M_Gunz
09-28-2010, 03:18 AM
Originally posted by Kettenhunde:
M_Gunz you have drank the kool aide my friend.

There are not multiple frames of reference in that Glenn Research diagram.

Not in that diagram. There are in this thread.

M_Gunz
09-28-2010, 05:08 AM
So since lift and drag combine to counter weight and the horizontal component of drag cancels the horizontal component of lift the lift is then not what moves the glider forward. It moves forward because it moves down along the path.

Nice explanation.

The lift is still tilted wrt gravity. How about when the forces are unbalanced and the glider accelerates?

Kettenhunde
09-28-2010, 05:21 AM
http://forums.ubi.com/images/smilies/partyhat.gif

M_Gunz
09-28-2010, 05:32 AM
Of course since gravity is balanced by components of lift and drag it is not gravity that moves the glider either.

The glider moves because it has an initial motion that is not acted upon by any net external forces.

p-11.cAce
09-28-2010, 06:04 AM
As for gliding flight, there is no tilted vector of lift driving the aircraft forward along along the horizontal axis.

That is a fact no matter what point of view or reference you want to take.

A ball rolling down a slope has a horizontal component because of the slope. Take away the slope and it falls vertically.

A gliding plane has a horizontal component of motion because of lift. Take away the lift (stall the aircraft) and it falls vertically.

The horizontal motion of the ball is due to the horizontal push from the slope. The horizontal motion of the plane is due to the horizontal pull of lift.

No slope = no horizontal motion.

Kettenhunde
09-28-2010, 09:15 AM
A glider uses gravitational energy to fly. Gravity doesn’t just pull the glider down, it also pulls it against air resistance. Gravity
powers the glider. From the information we have, we can calculate the energy and power provided by gravity.

http://www.scienceguy.org/Arti...idersCurriculum.aspx (http://www.scienceguy.org/Articles/ModelGliderArticles/SimpleGlidersCurriculum.aspx)


Glide—a basic maneuver in which the aircraft loses altitude in a controlled descent with little or no engine power; forward motion is maintained by gravity pulling the aircraft along an inclined path, and the descent rate is controlled by the pilot balancing the forces of gravity and lift.

http://avstop.com/ac/weight_shift_control/6-14.html


Aeronautical Knowledge Handbook

The pilot's handbook of aeronautical knowledge introduces pilots to the broad spectrum of knowledge that will be needed as they progress in their pilot training.



GLIDES—A glide is a basic maneuver in which the airplane loses altitude in a controlled descent with little or no engine power; forward motion is maintained by gravity pulling the airplane along an inclined path and the descent rate is controlled by the pilot balancing the forces of gravity and lift.

http://ma3naido.blogspot.com/2...escending-turns.html (http://ma3naido.blogspot.com/2009/10/descent-and-descending-turns.html)


Airplane Flying Handbook


GLIDES [16]--A glide is a basic maneuver in which the airplane loses altitude in a controlled descent with little or no engine power; forward motion is maintained by gravity pulling the airplane along an inclined path and the descent rate is controlled by the pilot balancing the forces of gravity and lift.

http://en.wikiversity.org/wiki...sic_flight_maneuvers (http://en.wikiversity.org/wiki/Airplane_Flying_Handbook/Basic_flight_maneuvers)


A descent, or glide, is a basic maneuver in which the airplane is losing altitude in a controlled descent with little or no engine power; forward motion is maintained by gravity pulling the airplane along an inclined path, and the descent rate is controlled by the pilot balancing the forces of gravity and lift.

http://avstop.com/ac/flighttra...ndbook/descents.html (http://avstop.com/ac/flighttrainghandbook/descents.html)

And finally from your own source, the FAA Glider Handbook:


WEIGHT
Weight is the third force that acts on a glider in flight. Weight opposes lift and acts vertically through the center of gravity of the glider. Gravitational pull provides the force necessary to move a glider through
the air since a portion of the weight vector of a glider is directed forward.

http://img713.imageshack.us/img713/8765/faagliderflyinghandbook.jpg (http://img713.imageshack.us/i/faagliderflyinghandbook.jpg/)

Notice that lift DOES NOT provide a force to move the glider forward....


LIFT
Lift opposes the downward force of weight and is produced by the dynamic effects of the surrounding airstream acting on the wing. Lift acts perpendicular to the flight path through the wing’s center of lift.

http://www.memphis-soaring.org...ining/faaChap_03.pdf (http://www.memphis-soaring.org/Training/faaChap_03.pdf)

M_Gunz
09-28-2010, 09:33 AM
A glider at steady speed on a straight path needs no net force to keep moving as it is. An object in motion...

Kettenhunde
09-28-2010, 09:54 AM
An object in motion...


Unless acted upon by an outside force....

You can disagree with me, NASA, Science Departments, and the FAA all you want, it is not a reflection on them.

UBIZOO!!!

http://forums.ubi.com/images/smilies/partyhat.gif

From the FAA's Glider Flying Handbook...


Gravitational pull provides the force necessary to move a glider through the air since a portion of the weight vector of a glider is directed forward.

I have been telling you that since what? Page 2??

p-11.cAce
09-28-2010, 10:11 AM
Kette everything you post just affirms this:


The ball rolling down the hill is propelled forward by the component of the force pushing up on it from the sloped surface in response to gravity pushing it down onto that surface. Gravity operates vertically down through the center of mass and without that sloping surface the ball would not move forward. There is a component of this force vector pushing forward and the ball moves forward down the slope.

This is exactly analogous to the lift vector pointed forward on the glider and the forward component of that vector providing the force to
overcome drag in forward flight. Gravity operates on the glider though it's center of mass, without that lift vector pointed forward the glider would not (continue to) move forward.

Gravity, without the thrust vector leaning forward, does not explain how a glider glides through the air, however Gravity is what provides the power/energy to do this (or store for that energy if you prefer).

Gravity is what provides the power - but it is the forward vector of lift that is creating horizontal movement. In all those quotes you emphasize gravity and ignore the slope.

M_Gunz
09-28-2010, 10:14 AM
NASA diagram and equations show forces in the vertical add to zero and forces in the horizontal add to zero.
So how am I disagreeing?
The only thing that weight is doing is balancing lift and drag in steady state glide.

When you don't have steady state then you you can talk about what drives the acceleration.

Kettenhunde
09-28-2010, 10:22 AM
but it is the forward vector of lift that is creating horizontal movement.

There is no forward component of lift driving anything. In the horizontal axis, Lift is always balanced by drag and it is the component of gravity that provides the force to change them.

When you change the angle of descent in a glider, the force of gravity is what gives the glider the ability to change its motion.

Lift is not "driving" anything, it is meeting the force required to offset the force of gravity. Lift force will never exceed gravitational force, real or apparent, on the horizontal axis.

Refer to the second property of lift I already have explained to you several times.

It is not tilted either. Lift is always perpendicular to the relative wind.



ignore the slope.


No Ace I do not ignore the slope.

I have told you since page 2 that lift creates that slope by acting in the vertical axis. Lift provides the force in the vertical axis that offsets a component of weight.

You can see it plainly illustrated for you in the Glenn Research diagram YOU posted.

I honestly do not know how to explain this any clearer.

You have the correct information now. In fact it was a small misunderstanding of what both NASA and the FAA were illustrating that led you to the wrong conclusions. You would have seen it had you read further down the page in the Glider Flying Handbook pages you posted.

I think you are an intelligent guy who is trying to look at some complicated things that are difficult to explain over the internet.

p-11.cAce
09-28-2010, 10:36 AM
From the FAA's Glider Flying Handbook...


Gravitational pull provides the force necessary to move a glider through the air since a portion of the weight vector of a glider is directed forward.



I have been telling you that since what? Page 2?? There is not tilted lift force that drives the glider forward in steady state flight.

From the same FAA document:
http://i25.photobucket.com/albums/c99/acmeaviator/gliderdynamics1.jpg

This can go on forever - like M_Gunz has ably pointed out we are just circling around vectors and semantics. Though I am curious what exactly is directing that lift vector forward. Beyond that - what force is the pilot even able to change? I mean the mass of the glider is not changing, gravity certainly is not changing from 1G...something is a variable here that the pilot is able to control. What is it Kette?

Kettenhunde
09-28-2010, 10:37 AM
directing that lift vector forward

There is no forward component of lift driving anything. That is a fundamental misunderstanding of the basic properties of lift.

In the horizontal axis, Lift is always balanced by drag and it is the component of gravity that provides the force to change them.

When you change the angle of descent in a glider, the force of gravity is what gives the glider the ability to change its motion.

Lift is not "driving" anything, it is meeting the force required to offset the force of gravity. The wing is just an airfoil with tips and cannot drive anything. It requires energy to be dumped into it in order to convert it to lift.

Lift force will never exceed gravitational force in a descending glider, real or apparent, on the horizontal axis.

Refer to the second property of lift I already have explained to you several times.

It is not tilted either. Lift is always perpendicular to the relative wind.

Last time I post this as I assume we double posted.

The glider moves forward not because of the horizontal component of gravity just as the Glider Flying Handbook states....


Gravitational pull provides the force necessary to move a glider through the air since a portion of the weight vector of a glider is directed forward.

http://www.memphis-soaring.org...ining/faaChap_03.pdf (http://www.memphis-soaring.org/Training/faaChap_03.pdf)

p-11.cAce
09-28-2010, 10:47 AM
Lift is always perpendicular to the relative wind.

Kette you do know that in a glide the relative wind is not perpendicular to the pull of gravity...right?

JuHa-
09-28-2010, 11:06 AM
My €0.01:

This is a sweeping simplification of the matter, but still.

The forward motion for the glider is provided by its' aerodynamic shape. When it is pulled (by gravity) downwards, its' shape turns the (vertical) airflow very efficiently to the horizontal plane. This has the effect of "pushing" the glider forwards, as the mass flow of the air is directed backwards.

If one drops a ball, it'll fall in a straight line to the ground.
If one drops a paper plane. it'll fall down in an interestingly curved path.

The big difference is the aerodynamic shape. Think the glider as a very efficient wedge, that tries to go sideways when pulled downwards through a medium (air/water/something with proper viscosity and mass).

p-11.cAce
09-28-2010, 11:23 AM
JuHa you are exactly correct.

K_Freddie
09-28-2010, 12:20 PM
You old grannies still at this... http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif

M_Gunz
09-28-2010, 12:42 PM
F = M A

In steady state flight where is the acceleration?

Set A = 0 then F = 0 as well.

V = V0 + AT

In steady state flight V = V0 since A = 0.

When forces are balanced what is pulling anything that is not countered? Does it matter the pull that is countered? Does it affect the motion? All the forces are pulling but none is pulling anything anywhere until it is not countered 100%.

You have to include the start condition of a steady state glide of being at speed and direction. That speed and direction are maintained as long as no net force acts upon it. The glider in a steady state glide moves by inertia alone. The glider does not need to be pulled anywhere because it is already in motion. No acceleration, no net force, steady state. If there was more than inertia involved, the glider would accelerate. So why posit extra anything in a steady state glide?

Holtzauge
09-28-2010, 12:58 PM
While arguing with a Vogon is as an exercise in futility, maybe the following will help:

Imagine a parachutist. He is falling vertically. There is a lifting force L opposing the fall. I think we can agree that L, while not as large as m*g, is directed in the opposite direction and there is no movement in the horizontal plane.

The parachutist has a sudden attack of inspiration and pulls the strings on one side.

Now what will happen? Will m*g suddenly get a trigonometric component or will the vector L be a tiny winy shifted to one side? Will we suddenly have a glider?

Mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm..........

To forestall the obvious objection that a parachute is not a plane perhaps we can add a paraglider to the discussion? Last time I looked they had quite reasonable glide ratios.

Also consider how the canopy get's off the ground: You stand in a slope facing the wind with the canopy on the ground behind you. Give it a tug and it lifts off and moves FORWARD above you. Certainly seems to be a FORWARD component to the lift there.

No wait! It's must be a trigonometric component of gravity. Here are some more pictures and links I Googled!

Romanator21
09-28-2010, 01:22 PM
Give it a tug and it lifts off and moves FORWARD above you. Certainly seems to be a FORWARD component to the lift there.

You do realize a parachute is attached to you by not-stretchy strings. If it rises it HAS to move forward - it can't just rise straight up.

Kettenhunde
09-28-2010, 01:27 PM
http://img716.imageshack.us/img716/5452/gravitypowersaglider.jpg (http://img716.imageshack.us/i/gravitypowersaglider.jpg/)

From the FAA, confirmed by NASA and everybody else in aviation or aeronautical sciences....


Gravitational pull provides the force necessary to move a glider through the air since a portion of the weight vector of a glider is directed forward.

http://www.memphis-soaring.org...ining/faaChap_03.pdf (http://www.memphis-soaring.org/Training/faaChap_03.pdf)

They must be wrong and Ubizoo denizens correct, right?

It has to be excess tilted lift driving the airplane forward combined with Leading Edge Suction in a plutonium frame of reference.

I mean that is why we have to tie down airplanes at the airport and I have to really be careful getting my airplanes out of the hanger. At any moment the wing could generate excess lift and plane float away.

http://forums.ubi.com/images/smilies/partyhat.gif

Kettenhunde
09-28-2010, 01:33 PM
It's must be a trigonometric component of gravity.


There is no such thing, right Mr Wizard? I would get my money back from the institution that you claim gave you an aeronautical sciences degree.

Wait, maybe it is Leading Edge Suction that pulls the glider forward and tilts the vector of lift???

http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif

p-11.cAce
09-28-2010, 01:53 PM
You know Kette - you've got a gift for obfuscation I'll give you that. So when I push the stick forward on my ASK is it increasing gravity that accelerates me to a higher airspeed?

Holtzauge
09-28-2010, 02:11 PM
Originally posted by Romanator21:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Give it a tug and it lifts off and moves FORWARD above you. Certainly seems to be a FORWARD component to the lift there.

You do realize a parachute is attached to you by not-stretchy strings. If it rises it HAS to move forward - it can't just rise straight up. </div></BLOCKQUOTE>

Sorry if I was a bit unclear: It will lift off the ground behind you and move forward radially as dictated by the lines to a position above you at which time you take a few steps down the slope and then you're airborne. http://forums.ubi.com/images/smilies/16x16_smiley-happy.gif

I been thinking about starting paragliding myself but so far I've been put off by the horrendous accident figures....

Kettenhunde
09-28-2010, 02:14 PM
Just like leading edge suction is opening the slats on my airplane....

http://www.youtube.com/watch?v=-vbqgfjyW2Q

Holtzauge
09-28-2010, 02:14 PM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">It's must be a trigonometric component of gravity.


There is no such thing, right Mr Wizard? I would get my money back from the institution that you claim gave you an aeronautical sciences degree.

Wait, maybe it is Leading Edge Suction that pulls the glider forward and tilts the vector of lift???

http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif </div></BLOCKQUOTE>

Looks like we can add lacking in understanding of irony to the long list of your shortcomings http://forums.ubi.com/images/smilies/88.gif

Kettenhunde
09-28-2010, 02:20 PM
So when I push the stick forward on my ASK is it increasing gravity that accelerates me to a higher airspeed?



What does the FAA, NASA, and USN say....

"Gravitational pull provides the force necessary to move a glider through the air since a portion of the weight vector of a glider is directed forward."

JuHa-
09-28-2010, 02:22 PM
Originally posted by M_Gunz:
In steady state flight where is the acceleration?
That speed and direction are maintained as long as no net force acts upon it. The glider in a steady state glide moves by inertia alone. The glider does not need to be pulled anywhere because it is already in motion.

Friction. It resist all movement, even glider's flow through the air. Thus inertia alone will not keep the glider going on forever.
Also known as drag, and it's also known as negative acceleration.

Kettenhunde
09-28-2010, 02:49 PM
Ace,

All baloney aside, it is really pretty simple.

Only three things really act on your glider, Lift, Drag, and gravity.

If you don't believe gravity is what powers your glider, wheel your glider out to the runway, hop in it, and under its own power, take off.

JtD
09-28-2010, 03:11 PM
Standing on the runway, there's gravity but no lift.

Kettenhunde
09-28-2010, 03:20 PM
Standing on the runway, there's gravity but no lift.

Exactly!!! We have a winnah!

...explain why is there no lift, Kemo sabe.

Just to be clear:

We have the force of gravity present but lift and drag forces are not present sitting still on the runway, right?

What is balancing the force of gravity on the runway??

Maybe the light will come on and certain individuals will start to understand the properties of lift.

http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

Despite some very clear explanations including blatantly spelling it out...NASA, the FAA, and every aeronautical institution I can think of is not getting the point across.

p-11.cAce
09-28-2010, 07:44 PM
Standing on the runway, there's gravity but no lift.

So I am not moving when there is gravity but no lift? Is that what you are saying?


What is balancing the force of gravity on the runway??

I'm gonna say the runway is balancing the force of gravity.

WTE_Galway
09-28-2010, 08:30 PM
this thread has got silly

Kettenhunde
09-28-2010, 08:41 PM
I'm gonna say the runway is balancing the force of gravity.

Correct. The runway is pushing back against gravity to provide the centripetal force required.

In level 1G flight we know lift equals weight, right?

Why?

Why is that no matter how fast our airplane goes, lift will only equal weight in 1G level flight?

In other words, when dynamic pressure reaches the point the wing can generate enough lift force to equal weight it stays only at that amount in 1G level flight no matter how much the dynamic pressure increases.

What property of lift force is that demonstrating?

Don't worry about the co-efficient or angles of attack, just explain the forces.

AndyJWest
09-28-2010, 09:30 PM
In level 1G flight we know lift equals weight, right?
True.

In any unaccelerated flight (climbing, descending, or level), lift must equal weight. Otherwise it would accelerate. http://forums.ubi.com/groupee_common/emoticons/icon_rolleyes.gif

Or is the 'lift' that actually works in two different directions (vertically, against gravity, or at right angles to the flight path), depending on how you want to confuse the issue?

Breaking aerodynamic forces into a 'lift' vector and an orthogonal 'drag' vector is a mathematical convenience. If a glider is in an unaccelerated descent, the sum of aerodynamic forces must balance the gravitational force (which acts vertically downwards). Like I said earlier, unless you want to argue that 'drag' equals zero, you have to explain where the counteracting horizontal force comes from - and without using imaginary inclined surfaces....

M_Gunz
09-28-2010, 10:41 PM
Close Andy. Please allow I take some exception?

In any unaccelerated flight (climbing, descending, or level), lift must equal weight. Otherwise it would accelerate. Roll Eyes

In unpowered descent the weight is offset by vertical components of both lift and drag.

An unaccelerated climb -must- involve thrust (unless we're talking about flying through rising air or LTA craft) and then you have thrust countering weight and vertical components of lift and drag.

If weight is balanced out, does it still affect the movement of the glider or plane? I can't see how.

JtD
09-29-2010, 01:12 AM
Originally posted by WTE_Galway:
this thread has got silly

On page two.

Kettenhunde
09-29-2010, 05:41 AM
Otherwise it would accelerate.

And there is the flaw in your thinking.

Lift reacts and meets the force required, it does not cause the acceleration.

Anymore so than your tires accelerate your car around a turn or can drive the vehicle. They are there to provide a friction force that meets the centripetal force required and convert momentum and energy to that friction force.

When my airplane sits on the runway, I have an engine converting the energy from combustion to thrust by whirling two wings in a circle. It can move the craft to provide the dynamic pressure required to create the lift I need to fly.

A glider only has the force of gravity until it is hauled into the air. It then converts that Potential Energy to fly.

PE = Weight x height

My engine has a throttle that allows me to regulate the thrust production.

The glider pilot uses the angle of the inclined plane which the wing converts some of that energy and provides the inclined plane. The pilot uses his controls to adjust the force of gravity on the horizontal vector to regulate "thrust".

Drag and lift are both functions of the weight and the angle the pilot chooses. You can see that in the force resolution diagram I posted.

http://img716.imageshack.us/img716/5452/gravitypowersaglider.jpg (http://img716.imageshack.us/i/gravitypowersaglider.jpg/)

There is no portion of lift that is driving anything in a glider. Lift force is always reacting to the force of momentum and gravity meeting the force required.

It is really that simple.

That is why the definition of gliding flight is:


GLIDES—A glide is a basic maneuver in which the airplane loses altitude in a controlled descent with little or no engine power; forward motion is maintained by gravity pulling the airplane along an inclined path and the descent rate is controlled by the pilot balancing the forces of gravity and lift.

http://ma3naido.blogspot.com/2...escending-turns.html (http://ma3naido.blogspot.com/2009/10/descent-and-descending-turns.html)

And why the FAA's Glider Flying Handbook says:


WEIGHT
Weight is the third force that acts on a glider in flight. Weight opposes lift and acts vertically through the center of gravity of the glider. Gravitational pull provides the force necessary to move a glider through the air since a portion of the weight vector of a glider is directed forward.

http://www.memphis-soaring.org...ining/faaChap_03.pdf (http://www.memphis-soaring.org/Training/faaChap_03.pdf)

It is not lift that moves the glider through the air anymore so than a tire can move itself down the road.

It is gravity that powers the glider in the horizontal axis.

p-11.cAce
09-29-2010, 06:34 AM
In level 1G flight we know lift equals weight, right?

A gliding aircraft is not in level 1G flight - it is descending.


The pilot uses his controls to adjust the force of gravity on the horizontal vector to regulate "thrust".

This is where you lose me Kette. The mass of the glider does not change, nor does gravity ever pull with more than 1G. Flying level or diving at the ground, gravity always pulls with 1G.


If you don't believe gravity is what powers your glider, wheel your glider out to the runway, hop in it, and under its own power, take off.

For the 5th time:


Gravity, without the thrust vector leaning forward, does not explain how a glider glides through the air, however Gravity is what provides the power/energy to do this (or store for that energy if you prefer).

But really what makes me question your argument the most (other than your need to include slights and insults) is this:


GLIDES—A glide is a basic maneuver in which the airplane loses altitude in a controlled descent with little or no engine power; forward motion is maintained by gravity pulling the airplane along an inclined path and the descent rate is controlled by the pilot balancing the forces of gravity and lift.

Of those two forces - gravity and lift
which one can be controlled by the pilot? Because whichever one is variable at the command of the pilot is determining the rate at which the PE of gravity is converted into horizontal and vertical KE. So what is the pilot in control of Kette - gravity or lift?

Kettenhunde
09-29-2010, 07:25 AM
This is where you lose me Kette. The mass of the glider does not change, nor does gravity ever pull with more than 1G. Flying level or diving at the ground, gravity always pulls with 1G.


Ok, good at least we have the issue cornered.

You do understand about sine, cosine, tangent, and vector resolution? Not a joke or dig, it is a concept that is essential to understanding what is happening.

Look over the force resolution diagram I posted.

In the horizontal axis, lift and drag are both functions of the aircraft weight and the angle the pilot chooses to descend.

The acceleration of gravity in the horizontal axis is also a function of the angle the pilot chooses to descend.

We know from Newtons second law of motion that FORCE equals MASS <independent of acceleration> x ACCELERATION <gravitational or apparent>.

A glider pilot uses his descent angle to change the FORCE of gravity on the horizontal axis. That FORCE change in turn establishes the force required that drag and lift move to meet.

There is no excess lift or drag force to drive anything in your glider. It all comes from the force of gravity and the other forces move to meet that force requirement.

Gravity is your engine and its properties combined with the second property of lift <Lift will meet the force required> are your throttle in a glider giving you control over the force of gravity in the horizontal axis.

Now you can go back to the airfield and when your pilot buddies start talking about lift propelling the glider.....you can set them straight.

Kettenhunde
09-29-2010, 07:43 AM
So what is the pilot in control of Kette - gravity or lift?


Are you in control of your tires on a car?

Do tires propel the vehicle or adjust the amount of power the car transfers to the road when you step on the gas?

No, tires just meet the forces required until they lose friction and slide.

Kettenhunde
09-29-2010, 08:10 AM
BTW,

This was all explained to you on page 1 by Bremspropeller. I told you to listen to him.

http://forums.ubi.com/eve/foru...051056388#2051056388 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/4721073388?r=2051056388#2051056388)

WTE_Galloway also very nicely explained it on page 2:


I may be confused but my understanding was weight can be broken down into a component that is opposed by lift and a component in the direction of glide path that causes an acceleration in the direction of travel.


Col_Sanders then tried to set you straight for several pages to no avail.

There are people that have an understanding and are trying to help.

Unfortunately they get drowned out by "experts" that cloud the issue and drag out a fairly simple concept for 11 pages.

That is not an insult. It is an observation of the dynamics of these forums.

Normally the discussions are on cartoon airplanes. It does not matter and most people who know what they are doing will just walk away after a while and let the clownfest percolate.

Real gliders can make you real dead if you don't understand what is going on.

AndyJWest
09-29-2010, 08:34 AM
M_Gunz, I think I made my point more clearly in the latter part of my last post: "If a glider is in an unaccelerated descent, the sum of aerodynamic forces must balance the gravitational force (which acts vertically downwards)". It is this unnecessary distinction between 'lift' and 'drag' that is clouding the issue. For unaccelerated gliding flight, the net aerodynamic forces must be equal to the gravitational force, and operating in the opposite direction - vertically upwards. This is nothing to do with the aerodynamic properties of aircraft, it is simple Newtonian physics.

Kettenhunde is just plain wrong, where he makes sense at all, but I can't be bothered to argue with him any more....

Kettenhunde
09-29-2010, 08:44 AM
For unaccelerated gliding flight, the net aerodynamic forces must be equal to the gravitational force, and operating in the opposite direction - vertically upwards.

MMMMMMM

The FAA, NASA, USN, and every aeronautical institution I know of are wrong too!

http://forums.ubi.com/images/smilies/partyhat.gif


Gravitational pull provides the force necessary to move a glider through the air since a portion of the weight vector of a glider is directed forward.

http://www.memphis-soaring.org...ining/faaChap_03.pdf (http://www.memphis-soaring.org/Training/faaChap_03.pdf)

p-11.cAce
09-29-2010, 09:15 AM
Kettenhunde is just plain wrong, where he makes sense at all, but I can't be bothered to argue with him any more....

+1

Bremspropeller
09-29-2010, 10:56 AM
A gliding aircraft is not in level 1G flight - it is descending.

So it's not 1g then? http://forums.ubi.com/images/smilies/53.gif

Can somebody already tilt the "gliding aircraft"-diagram, so everybody sees that lift has no component into the aircraft's direction of travel (= no thust there), but weight does (= thrust there)?

p-11.cAce
09-29-2010, 11:06 AM
So it's not 1g then?

No - it's not level. If you want http://forums.ubi.com/images/smilies/53.gif see Kettes assertion that the stick controls gravity.


The ball rolling down the hill is propelled forward by the component of the force pushing up on it from the sloped surface in response to gravity pushing it down onto that surface. Gravity operates vertically down through the center of mass and without that sloping surface the ball would not move forward. There is a component of this force vector pushing forward and the ball moves forward down the slope.

This is exactly analogous to the lift vector pointed forward on the glider and the forward component of that vector providing the force to overcome drag in forward flight. Gravity operates on the glider though it's center of mass, without that lift vector pointed forward the glider would not (continue to) move forward. (which is also why I cracked up about Kette pointing out that a glider on the ground -still experiencing a 1G pull - is not moving. No lift=no horizontal movment)

Gravity, without the thrust vector leaning forward, does not explain how a glider glides through the air, however Gravity is what provides the power/energy to do this (or store for that energy if you prefer).

Bremspropeller
09-29-2010, 11:10 AM
Ok, so you'll watch the g-meter next time you fly an ASK and you'll owe me a brew http://forums.ubi.com/images/smilies/25.gif

p-11.cAce
09-29-2010, 11:16 AM
Ok, so you'll watch the g-meter next time you fly an ASK and you'll owe me a brew

As I said - 1G but not level. Anytime you are in Cinci I'd be happy to fly with you - either in the ASK, Grob, or 2-33's at CCSC or the HK-36 at Sporty's (i69).

Bremspropeller
09-29-2010, 11:52 AM
I'll be happy to come by http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

But still, it doesn't matter of you're flying level or if you're in a stationary (= trimmed) glide.
1g is 1g.

p-11.cAce
09-29-2010, 12:23 PM
But still, it doesn't matter of you're flying level or if you're in a stationary (= trimmed) glide.
1g is 1g.

I know - I apologize for including that in the quote. It's not the 1G - hell it's not even gravity that Kette has got wrong. It's that the relative wind of a glider or any gliding aircraft is at an angle from the front. You can see that in every diagram, work that in every calculation.

My argument all along is that Gravity is supplying the force but it is the wing and the lift it produces that converts that force into forward motion. Remove lift by stalling the wing or landing and the horizontal motion stops - but the 1G pull is still there.

Replace the lift from the wing with a sloping runway and the horizontal motion returns, not because the 1G has changed, but because the slope imparts a horizontal force to the wheels of the aircraft and not just a vertical one.


Gravity, without the thrust vector leaning forward, does not explain how a glider glides through the air, however Gravity is what provides the power/energy to do this (or store for that energy if you prefer).

Holtzauge
09-29-2010, 12:57 PM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content"> For unaccelerated gliding flight, the net aerodynamic forces must be equal to the gravitational force, and operating in the opposite direction - vertically upwards.

MMMMMMM

The FAA, NASA, USN, and every aeronautical institution I know of are wrong too!

http://forums.ubi.com/images/smilies/partyhat.gif
</div></BLOCKQUOTE>

Let's see on who's head the party hat truly belongs:

This is the statement that according to Kettenhunde deserves a party hat:

"For unaccelerated gliding flight, the net aerodynamic forces must be equal to the gravitational force, and operating in the opposite direction - vertically upwards."

Let's formulate this in mathematical terms:

W=R Where R is the resultant of L and D

By Kettenhunde's own admission and upheld as the only truth: L=W*siny and D=W*cosy where y is glide angle. (See double underline in figure 8.1 posted by Kettenhunde above)

So W=R=sqrt(L^2+D^2)

So let's plug in Kettenhundes L=W*siny and D=W*cosy:

W=sqrt(W^2*cos^2y+W^2*sin^2y)

W=W*sqrt(cos^2y+sin^2y)

W=W

Thus we can conclude that the statement is true.

Q.E.D

Now where does the party hat belong?

http://forums.ubi.com/images/smilies/partyhat.gif

Edit: So what's wrong is not the FAA, NASA, USN, and every aeronautical institution. What's wrong is you lacking the ability to connect the dots and see things in any other perspective than texts you google and dogmatically parrot until the opposition just gives up and dies.

AndyJWest
09-29-2010, 01:29 PM
http://forums.ubi.com/images/smilies/25.gif http://forums.ubi.com/images/smilies/partyhat.gif

M_Gunz
09-29-2010, 01:43 PM
Originally posted by Kettenhunde:
A glider only has the force of gravity until it is hauled into the air.

Although you may find a web page that names gravity as a force it is not a force. Gravity is an acceleration. Weight is the force.

F = M A
Weight = Mass x Gravitational Constant

If gravity was a force then heavier objects would fall slower, wouldn't they?
A = F / M

Weight -is- the driving force for the glider (in still air at least). Take the weight away and the glider would not overcome drag even though it would need no lift.

Just did a simple check: in their defense, these sites are not teaching physics but rather using physical principles to try and explain gliding flight. All this means is that you can't take everything these guys write literally.

ma3naido.blogspot.com (http://ma3naido.blogspot.com/2009/10/descent-and-descending-turns.html)

forward motion is maintained by gravity pulling the airplane along an inclined path and the descent rate is controlled by the pilot balancing the forces of gravity and lift.

Memphis-Soaring.org (http://www.memphis-soaring.org/Training/faaChap_03.pdf)

The center of gravity of the glider is the point where the total force of gravity is considered to act.

M_Gunz
09-29-2010, 01:54 PM
Originally posted by AndyJWest:
M_Gunz, I think I made my point more clearly in the latter part of my last post:

I don't get that from re-reading the entire post yet again. If you meant different than the part I quoted then why didn't you change that part?

And WTH is the 'force of gravity'? Gravity is an acceleration. 9.81 m/s/s. The more mass it acts upon, the more force is generated.

AndyJWest
09-29-2010, 02:05 PM
Originally posted by M_Gunz:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by AndyJWest:
M_Gunz, I think I made my point more clearly in the latter part of my last post:

I don't get that from re-reading the entire post yet again. If you meant different than the part I quoted then why didn't you change that part?

And WTH is the 'force of gravity'? Gravity is an acceleration. 9.81 m/s/s. The more mass it acts upon, the more force is generated. </div></BLOCKQUOTE>

Ok, I could perhaps have been clearer. I was only trying to repeat what my earlier diagram illustrated anyway.

And no, gravity isn't an acceleration. It is a force that can cause an acceleration. Gravity is currently pushing me down onto my chair, but I'm not accelerating....


Gravitation, or gravity, a non-contact force, is one of the four fundamental interactions of nature (along with the strong force, electromagnetism and the weak force), in which objects with mass attract one another.
http://en.wikipedia.org/wiki/Gravity

M_Gunz
09-29-2010, 02:34 PM
Fine then, 'force of gravity' near Earth surface means Weight.

Kettenhunde
09-29-2010, 04:03 PM
What we commonly call weight is really just the gravitational force exerted on an object of a certain mass.

http://csep10.phys.utk.edu/ast...tory/newtongrav.html (http://csep10.phys.utk.edu/astr161/lect/history/newtongrav.html)

Kettenhunde
09-29-2010, 05:11 PM
W=sqrt(W^2*cos^2y+W^2*sin^2y)

W=W*sqrt(cos^2y+sin^2y)

Huh?

SQRT (W^2 + W^2) is not W

SQRT (2*W^2) = SQRT(2) * SQRT(W^2)

= SQRT (2) * W

= 1.4142 W

Of course that in not even getting into the other operations...

Where did you learn Algebra Holtzauge?


http://forums.ubi.com/images/smilies/partyhat.gif

WTE_Galway
09-29-2010, 05:22 PM
There is far to much magical thinking in this thread.

1. You have an aircraft in a stable unpowered descent. We do not care how it got there. The pilot has found a good angle of attack vs the glide path to maintain that descent at a constant speed and descent rate.

2. We do not CARE about the earths surface or if it even exists. Until we crash into it, its entirely hypothetical. So the concept of HORIZONTAL is incorrect ground bound thinking. What we do care about is the total weight vector when we combine the forces of the earth moon sun and any other random nearby masses on the object.

3. The aircraft by its very nature is effected by lift which is vertical to the flight path and tries to make it loop and drag (including any backward component of the lift) which tries to slow it down.

4. In an equilibrium descent the total weight vector MUST exactly cancel the lift+drag. Breaking this down we get a component of weight vertical to the flight path opposing lift and a component of weight forward along the flight path opposing drag.


There is NO magical component of lift FORWARD along the flight path. There cannot be in a stable equilibrium descent. It is impossible.

The fact that there IS a forward component of lift "horizontal" to the hypothetical earths surface is IRRELEVANT and ground based bad thinking. There is no such thing as "horizontal" in terms of this system. there is the glide path and the forces acting on the aircraft relative to it.



Note that just because you disagree with Kettenhunde's attitude and discussion techniques does not make him automatically wrong http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

I am going to ignore this thread again for a few days http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

M_Gunz
09-29-2010, 05:51 PM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">W=sqrt(W^2*cos^2y+W^2*sin^2y)

W=W*sqrt(cos^2y+sin^2y)

Huh?

SQRT (W^2 + W^2) is not W

SQRT (2*W^2) = SQRT(2) * SQRT(W^2)

= SQRT (2) * W

= 1.4142 W

Of course that in not even getting into the other operations...

Where did you learn Algebra Holtzauge?


http://forums.ubi.com/images/smilies/partyhat.gif </div></BLOCKQUOTE>

How did you get from:
sqrt(W^2*cos^2y+W^2*sin^2y)

to

SQRT (W^2 + W^2)

Kettenhunde
09-29-2010, 08:22 PM
How did you get from:

duh....Pythagorean Identities....

sqrt(W^2*cos^2y+W^2*sin^2y)

Sqrt [W^2(cos^2y + sin^2y)]

I was thinking:

cos^2 x = (1 + cos 2x)/2

sin^2 x = (1 - cos 2x)/2

Holtzauge is correct in that it equals Weight.

Which is exactly what it should be as he has combined all the vectors which should equal weight.

In this case we are looking for the components along the vector.

The FAA and every aeronautical institution is not wrong nor am I.


Gravitational pull provides the force necessary to move a glider through the air since a portion of the weight vector of a glider is directed forward.

http://www.memphis-soaring.org...ining/faaChap_03.pdf (http://www.memphis-soaring.org/Training/faaChap_03.pdf)


Note that just because you disagree with Kettenhunde's attitude and discussion techniques does not make him automatically wrong

Thank you and good post. You have nicely summed it up for a second time.


http://img815.imageshack.us/img815/2116/forcesinaglide.jpg (http://img815.imageshack.us/i/forcesinaglide.jpg/)

http://img535.imageshack.us/img535/6536/forcesinaglide2.jpg (http://img535.imageshack.us/i/forcesinaglide2.jpg/)

http://img823.imageshack.us/img823/7105/forcesinaglide3.jpg (http://img823.imageshack.us/i/forcesinaglide3.jpg/)

Romanator21
09-29-2010, 10:04 PM
There is far to much magical thinking in this thread.

1. You have an aircraft in a stable unpowered descent. We do not care how it got there. The pilot has found a good angle of attack vs the glide path to maintain that descent at a constant speed and descent rate.

2. We do not CARE about the earths surface or if it even exists. Until we crash into it, its entirely hypothetical. So the concept of HORIZONTAL is incorrect ground bound thinking. What we do care about is the total weight vector when we combine the forces of the earth moon sun and any other random nearby masses on the object.

3. The aircraft by its very nature is effected by lift which is vertical to the flight path and tries to make it loop and drag (including any backward component of the lift) which tries to slow it down.

4. In an equilibrium descent the total weight vector MUST exactly cancel the lift+drag. Breaking this down we get a component of weight vertical to the flight path opposing lift and a component of weight forward along the flight path opposing drag.


There is NO magical component of lift FORWARD along the flight path. There cannot be in a stable equilibrium descent. It is impossible.

The fact that there IS a forward component of lift "horizontal" to the hypothetical earths surface is IRRELEVANT and ground based bad thinking. There is no such thing as "horizontal" in terms of this system. there is the glide path and the forces acting on the aircraft relative to it.

+1

p-11.cAce
09-29-2010, 10:21 PM
The ball rolling down the hill is propelled forward by the component of the force pushing up on it from the sloped surface in response to gravity pushing it down onto that surface. Gravity operates vertically down through the center of mass and without that sloping surface the ball would not move forward. There is a component of this force vector pushing forward and the ball moves forward down the slope.

This is exactly analogous to the lift vector pointed forward on the glider and the forward component of that vector providing the force to overcome drag in forward flight. Gravity operates on the glider though it's center of mass, without that lift vector pointed forward the glider would not (continue to) move forward. (which is also why I cracked up about Kette pointing out that a glider on the ground -still experiencing a 1G pull - is not moving. No lift=no horizontal movment)

Gravity, without the thrust vector leaning forward, does not explain how a glider glides through the air, however Gravity is what provides the power/energy to do this (or store for that energy if you prefer).

WTE_Galway
09-29-2010, 11:17 PM
Originally posted by p-11.cAce:
The ball rolling down the hill is propelled forward by the component of the force pushing up on it from the sloped surface in response to gravity pushing it down onto that surface. Gravity operates vertically down through the center of mass and without that sloping surface the ball would not move forward. There is a component of this force vector pushing forward and the ball moves forward down the slope.

This is exactly analogous to the lift vector pointed forward on the glider and the forward component of that vector providing the force to overcome drag in forward flight. Gravity operates on the glider though it's center of mass, without that lift vector pointed forward the glider would not (continue to) move forward. (which is also why I cracked up about Kette pointing out that a glider on the ground -still experiencing a 1G pull - is not moving. No lift=no horizontal movment)

Gravity, without the thrust vector leaning forward, does not explain how a glider glides through the air, however Gravity is what provides the power/energy to do this (or store for that energy if you prefer).

This is seriously confused in two areas:

a) confusing the forces necessary to initiate a descent versus the forces maintaining an equilibrium descent.

b) confusing the vectors acting horizontal to the ground versus the vectors acting down the glide path.


But I give up, its totally not important unless you are trying to pass aeronautics at college. I will just wander off and feed the llamas and play some early Rolling Stones http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

M_Gunz
09-29-2010, 11:59 PM
My apologies for my, was it two?, posts a number of hours ago yesterday afternoon (it's almost 2AM here now). I had just gotten home from what for me was a long hard trip, was in a good bit or hurt and yet zooming along from the wayyyy too much caffeine I drank to make sure I would get home and couldn't just crash. Upshot was not thinking before posting and wow that was a massive mistake! Big mistake was to come here and post in that state. Sorry, I will try and not do that again. After rest and popping my back more into shape I woke with the thought "I wrote WHAT?". So two kinds of ouch for Max in under 12 hours, not one of my better days.

Again, sorry for that.

..

All the vectors are there simultaneously. Stopped in any moment one can look at them and predict motion. But only one force is driving the glider and we know what it is regardless of how you interpret the 'picture'.

Kettenhunde
09-30-2010, 10:59 AM
Guess we can lay this one to rest, finally.

It is gravity that powers our glider not Leading Edge suction, forward component of lift, excess lift, or tilted lift.

AndyJWest
09-30-2010, 11:22 AM
As I undertand it, the net aerodynamic forces are exactly counterbalancing the gravitational force, so the glider carries on moving at the velocity it has through inertia. If you define 'lift' as acting upwards at right angles to the flightpath, and 'drag' as acting backwards along the flightpath, a little vector maths (which we've been over several times) seems to show that the horizontal forward component of lift is equal to the backward horizontal component of drag.

Yes, ultimately gravity is necessary to maintain this equilibrium, and the fact that the glider cannot be 100% efficient (i.e. drag free) means that it loses potential energy as it descends - without drag the flight path could be horizontal. Note however that gravity is pulling the glider vertically downwards, the horizontal component of drag is pulling it backwards, and the vertical components of lift and drag are pulling it upwards. Since we're defining lift as perpendicular to the flightpath, it isn't vertical, and has a forward horizontal component. I thought the argument was whether this forward horizontal component of lift was equal to the opposing horizontal component of drag. Is it or isn't it?

p-11.cAce
09-30-2010, 12:04 PM
It is gravity that powers our glider not Leading Edge suction, forward component of lift, excess lift, or tilted lift.

The fact that gravity is supplying the potential energy has been acknowledged all along and reposted multiple times:

Gravity, without the thrust vector leaning forward, does not explain how a glider glides through the air, however Gravity is what provides the power/energy to do this (or store for that energy if you prefer).

The question has been through what mechanism is that PE converted to kinetic energy in a forward direction relative to the aircraft - i.e. airspeed.

Your explanation is:

The pilot uses his controls to adjust the force of gravity on the horizontal vector to regulate "thrust".

My explanation is:

The pilot uses his controls to adjust the vector of lift on the horizontal vector to regulate "thrust"

Potato / Potahto ....Peace.

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Holtzauge
09-30-2010, 12:17 PM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">W=sqrt(W^2*cos^2y+W^2*sin^2y)

W=W*sqrt(cos^2y+sin^2y)

Huh?

SQRT (W^2 + W^2) is not W

SQRT (2*W^2) = SQRT(2) * SQRT(W^2)

= SQRT (2) * W

= 1.4142 W

Of course that in not even getting into the other operations...

Where did you learn Algebra Holtzauge?


http://forums.ubi.com/images/smilies/partyhat.gif </div></BLOCKQUOTE>

You've already earned one party hat. Look's like this time you've earned the mother of all party hats.........

Since you are a bit slow on the uptake and obviously mathematically challenged I'll walk it through with you, only SLOWER this time:

W=R Where R is the resultant of L and D.

L=W*siny and D=W*cosy where y is glide angle.

W=R=sqrt(L^2+D^2)

The concept that R=sqrt(L^2+D^2) is known as Pythagoras theorem. You familiar with this or am I going to fast?

Plug in L=W*siny and D=W*cosy:

W=sqrt(W^2*cos^2y+W^2*sin^2y)

This seems to be the step where you get lost so I'll go really slow from here.

W=W*sqrt(cos^2y+sin^2y)

So here I extract the W^2 from the sqrt() and I am left with W. You with me so far?

Next, from basic trigonometry (which you obviously do not understand) we know that:

cos^2y+sin^2y=1

Now if you don't belive me, just google trigonometric functions and eventually you are bound to find a page which explains this. I suggest you copy the figure and underline the equation in red like you usually do.

So W=W*sqrt(cos^2y+sin^2y)

Which, based on the above, may also be written as:

W=W*sqrt(1)

Also, to convince yourself, google the square root of 1. You will find that it is 1.

So W=W right?

If your brow is creased by now I'm not surprised, since you think this comes out to 1.4142 W instead of W you will obviously need some time to digest this.

If you still don't get it don't worry: I suggest a night class in basic high school maths.

And don't forget to bring this to class:

http://forums.ubi.com/images/smilies/partyhat.gif

You have certainly earned it.

DrHerb
09-30-2010, 12:20 PM
Man, now I wanna go gliding again http://forums.ubi.com/images/smilies/16x16_smiley-sad.gif

M_Gunz
09-30-2010, 12:44 PM
Originally posted by AndyJWest:
As I undertand it, the net aerodynamic forces are exactly counterbalancing the gravitational force, so the glider carries on moving at the velocity it has through inertia.

Ya know, I think you might have something there. Something about bodies in motion and outside forces....


If you define 'lift' as acting upwards at right angles to the flightpath, and 'drag' as acting backwards along the flightpath, a little vector maths (which we've been over several times) seems to show that the horizontal forward component of lift is equal to the backward horizontal component of drag.

Yes, ultimately gravity is necessary to maintain this equilibrium, and the fact that the glider cannot be 100% efficient (i.e. drag free) means that it loses potential energy as it descends - without drag the flight path could be horizontal. Note however that gravity is pulling the glider vertically downwards, the horizontal component of drag is pulling it backwards, and the vertical components of lift and drag are pulling it upwards. Since we're defining lift as perpendicular to the flightpath, it isn't vertical, and has a forward horizontal component. I thought the argument was whether this forward horizontal component of lift was equal to the opposing horizontal component of drag. Is it or isn't it?

Funny thing but I have this idea that with the pilot controlling the path that any net force not perpendicular to the path would tend to accelerate the glider along that path regardless of how many vectors were along for the ride. Why should gravity at an angle to the flight path be wrong while a horizontal component of lift also at an angle to the flight path be right?

Truth I see is that -only- a net force vector should matter as to acceleration and when net forces are zero then none is making any difference except in playing its part of keeping the balance. In a steady-state glide, gravity isn't pulling the glider along and neither is some component of lift. What gravity, lift and drag are doing is keeping each other from changing the equilibrium and nothing more.

Kettenhunde
09-30-2010, 12:50 PM
Reality just does not seem to be a part of your world, Holtzauge.

http://forums.ubi.com/images/smilies/partyhat.gif

Perhaps you just missed the post below?



Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">How did you get from:

duh....Pythagorean Identities....

sqrt(W^2*cos^2y+W^2*sin^2y)

Sqrt [W^2(cos^2y + sin^2y)]

I was thinking:

cos^2 x = (1 + cos 2x)/2

sin^2 x = (1 - cos 2x)/2

Holtzauge is correct in that it equals Weight.

Which is exactly what it should be as he has combined all the vectors which should equal weight.

In this case we are looking for the components along the vector.

The FAA and every aeronautical institution is not wrong nor am I.


Gravitational pull provides the force necessary to move a glider through the air since a portion of the weight vector of a glider is directed forward.

http://www.memphis-soaring.org...ining/faaChap_03.pdf (http://www.memphis-soaring.org/Training/faaChap_03.pdf)


Note that just because you disagree with Kettenhunde's attitude and discussion techniques does not make him automatically wrong

Thank you and good post. You have nicely summed it up for a second time.


http://img815.imageshack.us/img815/2116/forcesinaglide.jpg (http://img815.imageshack.us/i/forcesinaglide.jpg/)

http://img535.imageshack.us/img535/6536/forcesinaglide2.jpg (http://img535.imageshack.us/i/forcesinaglide2.jpg/)

http://img823.imageshack.us/img823/7105/forcesinaglide3.jpg (http://img823.imageshack.us/i/forcesinaglide3.jpg/) </div></BLOCKQUOTE>

http://forums.ubi.com/eve/foru...471057488#7471057488 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/4721073388?r=7471057488#7471057488)

p-11.cAce
09-30-2010, 12:54 PM
Man, now I wanna go gliding again

The popularity is increasing again - finally! My club (CCSC) has over 200 members and we gained 7 more in August http://forums.ubi.com/images/smilies/25.gif A lot of guys realize that dropping big $$$ to putt around in a 40 year old Cessna for an hour or two is not a heck of a lot of fun. Personally I like flying the HK-36 motorglider out of Sporty's - it's not the best glider by any means, not the best powered aircraft - but it's fun having the option either way.