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JtD
05-27-2010, 09:33 AM
First question:

How can EAS change even if the thrust/drag ratio constantly remains at 1 (meaning they are equal all the time)? What happened in the graph below?

http://mitglied.lycos.de/jaytdee/LOL/Simple1.JPG

AndyJWest
05-27-2010, 10:08 AM
What happened? The pilot moved the throttle, thereby causing a temporary suspension of the laws of physics. http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

JtD
05-27-2010, 10:22 AM
By opening the throttle he'd just produce extra thrust. To remain at thrust=drag, there'd have to be extra drag. Where would the extra drag come from?

And laws of physics = TRUE for all of that. http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

AndyJWest
05-27-2010, 10:50 AM
Hum, yes. He can pull back on the stick to compensate for the extra thrust, but how does he keep T/D constant and increase his EAS? A barrel roll would do it, I think, though keeping the balance exactly right would be tricky.

Actually, you don't even need to do a roll. You can keep T/D constant in a turn while moving the throttle, by adjusting the rate of turn while you do it.

Can anyone tell me what T/D is for a glider?

BoB_GEN_Tazman
05-27-2010, 10:50 AM
The energy had to come from somewhere, most likely a loss of altitude.

JtD
05-27-2010, 12:34 PM
Andy, if you were increasing throttle and entering a turn, you could keep T/D constant. But you wouldn't reach a higher EAS. You could be getting a different IAS, though, as the position error might change depending on the angle of attack. T/D of a glider could be 0, but then it's a matter of definition. See below.

BoB_GEN_Tazman, you're absolutely right. That's the easy solution, you could simply dive, gain speed by doing so, adjusting throttle to keep up with the extra drag, and go up again, reducing throttle to adjust for the decreasing drag.

However, that's not what I am eventually looking for. Also, I'd like to include all forces resulting from climb and dive into thrust and drag, so that for instance you can determine a thrust = drag speed in a power off dive, thrust just coming from the angle of descent. Basically T/D = 1 means no net force on the plane in this case.

Now getting back to the original question - this situation can also occur in level flight.

AndyJWest
05-27-2010, 12:38 PM
Does the pilot raise the undercarriage and then lower it again?

JtD
05-27-2010, 12:42 PM
No, that would again change drag, which would have to be compensated by thrust, which would again keep the EAS constant.

You'll have to think about T/D=1 meaning no net force.

M_Gunz
05-27-2010, 01:03 PM
Change in height is neither thrust nor drag, a few meters dip and return to alt could make that as shown.
In fact it's hard if not impossible to stay at exactly the same height IRL.

TinyTim
05-27-2010, 01:08 PM
T/D = 1 with plane being in an ideally level flight at all times means that there is no net force on the aircraft - the aircraft won't accelerate or decelerate, there will be no change of absolute speed or TAS.

EAS however can change - a plane enters an area of higher outside pressure for eample, since EAS=TAS*sqrt(actual air density/standard air density).

JtD
05-27-2010, 02:15 PM
Well, you name it quiz and TinyTim is going to provide the right answer. I guess Dance will show up soon and tell me which plane I was assuming. http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

Yes, in the above example no acceleration / deceleration happens, the TAS stays constant, the plane is flying at the same speed, but the density of the air is changing, thus giving a different EAS conversion factor.
A high pressure zone as mentioned is one option.
Personally, I was thinking more of a local temperature difference. If you, on a hot summer day, at low altitude crossed over a cool lake you could get the effect illustrated above. You'd be flying from hot into cold (dense) into hot air, giving you low - high - low equivalent air speeds.
Also possible would be a different composition of air, for instance flying from humid into dry and back into humid air. Since H20 has a lesser molar weight than N2 or O2, the airs density will be reduced as H20 gets into it.

Getting back to BoB_GEN_Tazman's original idea of dive and climb, this is also true - even if you account for the forces resulting from dive and climb and have them cancel out with engine thrust and airframe drag. Because, as you dive down, you're entering denser air, and while your plane will proceed at the same speed, the EAS will increase because of the higher density down low.

Eventually, there are several ways to change EAS while having thrust equal drag, or, in the bigger picture, the sum of all forces along the longitudinal axis of the plane equal zero. Because the density of the air can change for many reasons.

I'll be thinking up a new tricky question, unless, the the style of the WW2 quiz, TT has a good one ready.

Edit: The below chart illustrates my above descriptions. Please note that while T/D remains constant at 1, drag actually changes and thrust has to be adjusted to maintain that ratio.

http://mitglied.lycos.de/jaytdee/LOL/asimple1.JPG

Dance
05-27-2010, 02:26 PM
Just don't say my name three times JtD http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

Good question, air pocket had crossed my mind, but wasn't sure, Tim is better qualified on this kind of stuff than me http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif.

Kettenhunde
05-27-2010, 02:48 PM
To remain at thrust=drag,

The condition was not just thrust = drag

It is at a constant thrust AND thrust equals drag.

The airplane is at Vmax such as the P47 in the conversation and the thrust remains constant.

Big difference....

Your changing EAS was wrong because you recalculated the density ratio for each altitude. The density ratio remains the same for the sea level conditions.

By definition EAS equals TAS at sea level.

If your coefficient of drag is .05 at 2000lbs of thrust at sea level, it will be .05 at 2000lbs of thrust at the TAS of any altitude examined.

the density of the air is changing, thus giving a different EAS conversion factor.

Once again, EAS is calculated with the density ratio at sea level for the conditions. That ratio does not change with altitude.

The answer to your question of how can it change is the simple fact your games models a dynamic atmosphere.

If you are recording that in your game using the same ratio, then you have proven your game models a dynamic atmosphere.

That means your EAS calculation is now wrong for those conditions, too. In order to get the correct EAS you must recalculate with the new sea level density ratio under the new non-standard conditions.

You can also use this property of EAS to determine the density ratio of the new atmospheric conditions if you know what you are doing.

M_Gunz
05-27-2010, 04:42 PM
Funny how the original definition keeps growing.

TinyTim
05-27-2010, 05:48 PM
No question ready, since my knowledge about flight physics does not extend much further than pure basics, so here's a simple one from structural integrity until JtD comes up with the next one.

On pics below we can see two very similar aircraft - one is a replica of the other, they both flew (with crew).

A gigantic Blohm & Voss BV 238, the biggest aircraft (heavier than air) to fly during WW2:

http://www.flyingboats.ca/FlyingBoats-old/german-BlohmUndVoss/German1944BlohmUndVoss-BV238-g.jpg

and here is its exactly 4 times smaller scale model, that was built first for testing purposes, the FGP 227:

http://www.shrani.si/f/1l/1k/3uOW5Nfj/q.jpg

Let's assume the idealized theoretical case of these two planes being perfectly identical apart from different size. They are both subjected to a radial acceleration with ever increasing G-force (tighter and tighter turn at constant speed) until their wings snap due to over-g. Let's also assume over-g is the only possible reason for these planes to loose structural integrity. Question: Which of the aircraft will snap its wings first (i.e. at lower G) and why?

For an extra credit estimate the ratio of the G's at which planes are going to shed their wings.

na85
05-27-2010, 08:09 PM
If they're perfectly identical but for the scale, shouldn't one crack up at 1/4 the load factor of the other?

BillSwagger
05-27-2010, 08:37 PM
I don't know enough about the scale model.

My first guess was that the larger plane would fail sooner. My thinking is that ants can lift three times their own weight, but if they were as large as cars, then three times their own weight would crush them. Cars are also much stronger than ants.

Bill

Treetop64
05-27-2010, 09:36 PM
Originally posted by BoB_GEN_Tazman:
The energy had to come from somewhere, most likely a loss of altitude.

+1.

I'd go further and say that the throttle increased as he descended, then ******ed as he leveled off again. Drag increases with speed, and during descent the thrust would have to be increased to maintain the ratio with drag. Gravity does not count as thrust since they're both two of the four forces acting on an aircraft in flight (thrust, drag, lift, and gravity).

M_Gunz
05-27-2010, 09:57 PM
Originally posted by TinyTim:
Let's assume the idealized theoretical case of these two planes being perfectly identical apart from different size. They are both subjected to a radial acceleration with ever increasing G-force (tighter and tighter turn at constant speed) until their wings snap due to over-g. Let's also assume over-g is the only possible reason for these planes to loose structural integrity. Question: Which of the aircraft will snap its wings first (i.e. at lower G) and why?

Big one breaks first for the same reason you don't have 20 ft tall people. Ratio of cross-section area of structural
members changes by square of scale while volume and mass change by cube of scale. 4x as big has 16x cross section
area with 64x the mass. The smaller one might take 4x the strain.

However in flying a perfect 1/4 scale you might consider the effect of different Reynolds number on the wing surfaces
and remember the Mech Eng who could not after dinner explain how a bee can fly.

Kettenhunde
05-27-2010, 10:13 PM
Funny how the original definition keeps growing.

The definition has not changed at all.

WTF are you talking about? You are ignored.

-----------------------------

As for the scale model, it just depends on what they built it for as to what the load factor is going to be.

If the intent was a study of the airframe loads, then it is scaled to the same loads as the full sized one. Honestly I have never heard of that being done.

Scale models are more useful for determining q-limits as well as stability and control characteristics. They can be scaled to determine other aerodynamic features by adjusting the Reynolds number.

WTE_Galway
05-27-2010, 10:20 PM
Originally posted by BillSwagger:

Same G factor at same angle of bank.

But G factor is only a ratio. Same G factor does not mean same load.

Kettenhunde
05-27-2010, 10:31 PM
Same G factor does not mean same load.

Exactly! I have never seen a scale model used for load factor studies.

Generally only the wing is even tested. You calculate the load and literally pile sandbags on the wing, forklift with a scale, or a similar constructed apparatus.

http://www.sonexaircraft.com/a...v/xenosarchive3.html (http://www.sonexaircraft.com/aircraft/xenosdev/xenosarchive3.html)

If it does not break or deform too much, it passes the load test.

Boeing for example has a test rig with counterweights and hydraulic actuators:

Sometimes they will test the fuselage or other parts of the airframe as well but that is not as common as just testing the wing.

AndyJWest
05-27-2010, 10:44 PM
Yup, agree with Kettenhunde here. Use basic physics to design the wing (with a simple monoplane it is effectively a beam, plus some additional torsional loading), and then apply practical tests to ensure you've got it right. There are other things (flutter/resonance issues etc) that may be relevant too, but build it, then bend it until it breaks, has got to be a starting point. This way, you don't just test theoretical loadings, but practical factors like real material characteristics, manufacturing tolerances, and all the other things that make abstract theory less than reliable. You need then to allow a safety factor for all the variations you expect, plus a few you don't.

The Wright brothers worked like this. Sometimes they got it a little wrong, but they were near enough most of the time to learn from their mistakes. They'd figured out the craft of aircraft design. http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

julian265
05-27-2010, 10:47 PM
Originally posted by M_Gunz:
Big one breaks first for the same reason you don't have 20 ft tall people. Ratio of cross-section area of structural
members changes by square of scale while volume and mass change by cube of scale. 4x as big has 16x cross section
area with 64x the mass. The smaller one might take 4x the strain.

However in flying a perfect 1/4 scale you might consider the effect of different Reynolds number on the wing surfaces
and remember the Mech Eng who could not after dinner explain how a bee can fly.

In addition, beam stiffness/strength increases with the fourth power of the height of the beam.

I also remember using dimensionless variables for scale model comparisons at uni, but certainly haven't used it since, and hence I've forgotten it!

JtD
05-27-2010, 11:02 PM
TinyTim, that's a good question, can be addressed with any amount of detail. I'd agree with M_Gunz in the result, but would go into more detail.

I'd be reducing the problem to the strength of the main wing spar in a bending situation, which is probably the best approximation. In the big plane, it is 4 times as high and 4 times as wide. This gives it only 16 times the cross section, but it gives it 256 times the moment of inertia. The moment of inertia would be the indicative of the resistance to bending. So the big spar is 256 times as stiff as the small one. The peak stress in the spar would be equal to the ratio of the moment of inertia divided by the height of it, so we're back at 64. The big spar has 64 times the strength of the small one, thus can take 64 times the load of the small spar.

If the difference in load was just the difference in weight, the big and the small plane could take the same g load.

But the load ratio is higher than just the weight ratio, in a bending situation it's force * arm. The force is 64 times as high, and the arm 4 times, so we'll have 256 times the load of the smaller plane for the bigger one.

So the big plane has 64 times the strength, but needs to deal with 256 times the load of his small brother, so the small one can take 256/64=4 times the load.

While imo it is not just down to cross section and weight (in a shear situation it is), I agree with the 1:4 ratio M_Gunz arrived at. Maybe he just took a shortcut from experience. http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif

JtD
05-28-2010, 12:01 AM
Please note that I've added a chart to the answer (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/7991054468?r=4981025468#4981025468) of my first question, might help illustrating the point.

Hoping that TinyTims question has been answered as intended by him, I'll be providing another bit for entertainment:

Two identical airplanes, going at the same speed, are doing a level turn with the same angle of bank - but one turns faster. Why?

irR4tiOn4L
05-28-2010, 12:04 AM
Originally posted by TinyTim:
No question ready, since my knowledge about flight physics does not extend much further than pure basics, so here's a simple one from structural integrity until JtD comes up with the next one.

On pics below we can see two very similar aircraft - one is a replica of the other, they both flew (with crew).

A gigantic Blohm & Voss BV 238, the biggest aircraft (heavier than air) to fly during WW2:

http://www.flyingboats.ca/FlyingBoats-old/german-BlohmUndVoss/German1944BlohmUndVoss-BV238-g.jpg

and here is its exactly 4 times smaller scale model, that was built first for testing purposes, the FGP 227:

http://www.shrani.si/f/1l/1k/3uOW5Nfj/q.jpg

Let's assume the idealized theoretical case of these two planes being perfectly identical apart from different size. They are both subjected to a radial acceleration with ever increasing G-force (tighter and tighter turn at constant speed) until their wings snap due to over-g. Let's also assume over-g is the only possible reason for these planes to loose structural integrity. Question: Which of the aircraft will snap its wings first (i.e. at lower G) and why?

For an extra credit estimate the ratio of the G's at which planes are going to shed their wings.

At a guess the bigger plane would break first - and guessing again, id say that as you scale up the growing number of atoms inside the material means the material loses strength as you scale up. Guessing once again, id say thats because quantised phenomena are getting less important the larger the object.

Im probably wrong too

EDIT: Hmm, just trying to get my head around the other explanations - are you guys saying that as you scale up, the volume/weight, and therefore stress, rises much faster than the cross section (ie, heightxwidthxdepth) and therefore the strength of the material?

Ok, no, i think i need someone to explain that.

irR4tiOn4L
05-28-2010, 12:14 AM
Originally posted by JtD:
Please note that I've added a chart to the answer (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/7991054468?r=4981025468#4981025468) of my first question, might help illustrating the point.

Hoping that TinyTims question has been answered as intended by him, I'll be providing another bit for entertainment:

Two identical airplanes, going at the same speed, are doing a level turn with the same angle of bank - but one turns faster. Why?

Lol, because one has downthrottled and is 'hanging on its prop' http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif http://forums.ubi.com/images/smilies/partyhat.gif
Be sure!

EDIT: If i had to take a gander, id say - maybe one of the aircraft has flaps enabled?

M_Gunz
05-28-2010, 12:15 AM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Funny how the original definition keeps growing.

The definition has not changed at all. </div></BLOCKQUOTE>

The original definition had nothing about how much power is applied, only that T=D.
In more than one post since conditions were added, last before I replied is that speed is Vmax.

M_Gunz
05-28-2010, 12:21 AM
Originally posted by JtD:
While imo it is not just down to cross section and weight (in a shear situation it is), I agree with the 1:4 ratio M_Gunz arrived at. Maybe he just took a shortcut from experience. http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif

Nahhh, I just didn't know any better.

Tully__
05-28-2010, 01:47 AM
Originally posted by JtD:
Now getting back to the original question - this situation can also occur in level flight.

Wind shear...

Edit: and next post after the one I quoted excludes this as an option.

TinyTim
05-28-2010, 02:12 AM
Yes guys, the simple mass vs cross section is what I had in mind with this question (hence the assumption about the over-g being the only reason possible for a breakup). Yup, the large one breaks 4 times sooner because of that, M_Gunz and JtD explained why.

Another, even simpler analogy would be: a cube with a certain mass is hanging below a thread with a square cross section with a certain thickness. When you enlarge the system 2 times, the mass of the cube will increase 2^3=8 times (since a cube with twice the lenght of the side can be assembled with 8 initial small cubes), while the carrying capacity of the thread only increases 2^2 times - you can assemble the new, thicker thread out of 4 initial ones. So only by enlarging system, we now effectively have two cubes hanging on each thread. In a nutshell, structural integrity rises with square, mass with cube.

I liked the remark about the ants and their ability to lift x time their own weight. Well, if an average human (who is in his natural size able to comfortly lift about half of its own weight) would be as small as an ant (let's say half a centimeter), it could lift 150 times own weight!

These facts about the bugs being able to lift 10 times own weight is nothing impressive at all when one considers this simple law (and yet it's trumpeted even on popular science media as an astounding fact). It's even shocking as to about their inability to lift more. If an ant would be as large as a human, its legs would crush in an instant.

TinyTim
05-28-2010, 02:44 AM
Originally posted by JtD:
Two identical airplanes, going at the same speed, are doing a level turn with the same angle of bank - but one turns faster. Why?

Is the turn coordinated (i.e. ball centered)?

Assuming it is, I cannot find other explanation than that the planes are flying in an areas with different gravity (one, for example flies over equator, the other one over one of the poles). This would make one turn 0,5% faster than the other, since that's difference in the values of the g. Now if we sent one up 16km high and keep the second one at the sea level, we could get another 0,5% of difference emerging from different g values at the two spots, for the 1% total. I strongly doubt however that's the answer JtD is after. http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

M_Gunz
05-28-2010, 04:47 AM
The one with the higher AOA (assuming both are able to sustain the turn) will have a higher thrust vector into the turn
and will turn faster. The hard part would be overcoming the higher drag especially given the same higher AOA means that
the forward thrust component will be less.

Izzat good?

Kettenhunde
05-28-2010, 06:26 AM
http://forums.ubi.com/images/smilies/partyhat.gif

The condition was at a constant thrust.

Readers the definition or conditions have not changed.

It has been the same conditions from the very beginning.

Crumpp says back in January:

It does not change with altitude. That is what makes it so useful. If we apply the correct compressibility correct it remains constant. That is why we work in ratios as well. It is simpler to figure and get the correct concept.

EAS is the most commonly used speed measurement in calculating an aircraft's performance envelope. It is easy to convert to TAS and greatly simplifies things. Even better, it is the correct way to do things.

EAS is just CAS corrected for compressibility error.

It works not matter what altitude or speed realm and does not change at a constant thrust.

http://forums.ubi.com/eve/foru...461051628#5461051628 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/2221055328?r=5461051628#5461051628)

JtD
05-28-2010, 07:54 AM
Originally posted by irR4tiOn4L:

Lol, because one has downthrottled and is 'hanging on its prop' http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif http://forums.ubi.com/images/smilies/partyhat.gif
Be sure!

http://forums.ubi.com/images/smilies/88.gif, but no. Also, if you're dropping flaps, you'll just be creating more lift. If you do that, and maintain angle of bank, you'll be gaining altitude and thus won't be doing a level turn anymore. So, this doesn't work.

TinyTim is again doing quite well:

Originally posted by TinyTim:

...I cannot find other explanation than that the planes are flying in an areas with different gravity (one, for example flies over equator, the other one over one of the poles). This would make one turn 0,5% faster than the other, since that's difference in the values of the g. Now if we sent one up 16km high and keep the second one at the sea level, we could get another 0,5% of difference emerging from different g values at the two spots, for the 1% total. I strongly doubt however that's the answer JtD is after.

This is brilliant, I honestly didn't think of that. Could be added that even locally differences in the region of 0.02 m/sē can often be found. The background of this is that the angle of bank is related to the ratio of radial acceleration and earth gravitation, so if you reduce the earth's gravitation while maintaining angle of bank, you'll also be reducing the radial acceleration and thus be reducing your turn rate.

M_Gunz:
The planes are identical and going at the same speed, so they will be flying at the same AoA and have the same thrust axis. If they weren't identical, thrust vector would be important. It's also something that often gets overlooked.

What I'm after is:

Originally posted by TinyTim:
Is the turn coordinated (i.e. ball centered)?

No, it's not... http://forums.ubi.com/images/smilies/winky.gif

Kettenhunde
05-28-2010, 08:40 AM
This is brilliant, I honestly didn't think of that. Could be added that even locally differences in the region of 0.02 m/sē can often be found. The background of this is that the angle of bank is directly related to the ratio of radial acceleration and earth gravitation, so if you reduce the earth's gravitation while maintaining angle of bank, you'll also be reducing the radial acceleration and thus be reducing your turn rate.

The gravitational acceleration is a function of the radius the object is from the origin.

At a constant altitude, the radius of a hilltop beneath the airplane has absolutely no effect on the radius of the airplane from the origin.

http://teacher.pas.rochester.e...ure_g/Measure_g.html (http://teacher.pas.rochester.edu/PHY_LABS/Measure_g/Measure_g.html)

If the altitude is not constant the effect will occur but is extremely small unless we are discussing extremes.

No, it's not...

If it is uncoordinated turn, side force is what is causing the variation in speed but once again, our aircraft is no longer at equilibrium.

Thrust no longer equals drag and it will move to a new equilibrium point.

M_Gunz
05-28-2010, 08:44 AM
Is it impossible to hold a skidding or slipping turn steady?

TinyTim
05-28-2010, 09:00 AM
Originally posted by Kettenhunde:
The gravitational acceleration is a function of the radius the object is from the origin.

... assuming you are very far from the massive object (which in case of a plane flying above Earth surface isn't fulfilled), or that the massive object is perfectly spherically symmetrical (which also isn't entirely true).

Originally posted by Kettenhunde:At a constant altitude, the radius of a hilltop beneath the airplane has absolutely no effect on the radius of the airplane from the origin.

Well, it does, so the term absolutely should be switched for nearly in order for your sentence to be absolutely correct. There are anomalies in gravitational field even at a constant altitude. Yes, it's small, but it is measurable:

http://www.csr.utexas.edu/grace/gravity/ggm02/ggm02_03.jpg
Gravity anomalies as measured recently by GRACE (Gravity Recovery And Climate Experiment), a joint project of German Space Agency and NASA.

I do agree however that this is hairsplitting which has little to do with our debate.

Kettenhunde
05-28-2010, 09:10 AM
assuming you are very far from the massive object (which in case of a plane flying above Earth surface isn't fulfilled), or that the massive object is perfectly spherically symmetrical (which also isn't entirely true).

It is the point on the earth that has a gravitational difference due to the shape of the earth and its relief.

There is a difference in aircraft at altitude and that difference is included at higher levels of analysis.

It will increase towards the poles as well.

There are anomalies in gravitational field even at a constant altitude.

I was not aware of GRACE. That is really neat and perhaps moves us closer to figuring out the mystery of the graviton.

Yes there are holes in our physics models. At the level we are discussing they are accurate.

TinyTim
05-28-2010, 09:19 AM
Originally posted by Kettenhunde:
It is the point on the earth that has a gravitational difference due to the shape of the earth and its relief, not the air above it.

If the gravitational field has anomalies on the surface, it will retain them when moving away from it. They will slowly be diminishing with the altitude, but won't cease when you move 1cm above the surface of the Earth, due to gravitational field everywhere being a continuous function (and even continuously differentiable outside of the massive body). That's why the picture I posted can be used for anomalies at the ground or at 10 kilometers aloft.

Bremspropeller
05-28-2010, 09:51 AM
Also, if you're dropping flaps, you'll just be creating more lift. If you do that, and maintain angle of bank, you'll be gaining altitude and thus won't be doing a level turn anymore. So, this doesn't work.

He said he'd throttle down, thus decreasing q, which in turn needs a larger CL to maintain the same net-lift. The theory ain't wong at all.

That condition is not sustainable, however - he has to constantly make up thrust for drag or the contrary.

FatCat_99
05-28-2010, 09:59 AM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Crumpp says back in January:
EAS is the most commonly used speed measurement in calculating an aircraft's performance envelope. It is easy to convert to TAS and greatly simplifies things. Even better, it is the correct way to do things.

</div></BLOCKQUOTE>
Sorry,professionals have different opinion about EAS usefulness in aircraft performance analysis.
http://img265.imageshack.us/img265/7479/easornot.jpg

FC

M_Gunz
05-28-2010, 10:03 AM
Originally posted by TinyTim:
There are anomalies in gravitational field even at a constant altitude. Yes, it's small, but it is measurable:

http://www.csr.utexas.edu/grace/gravity/ggm02/ggm02_03.jpg
Gravity anomalies as measured recently by GRACE (Gravity Recovery And Climate Experiment), a joint project of German Space Agency and NASA.

I do agree however that this is hairsplitting which has little to do with our debate.

Aren't those differences mainly due to density differences in the crust, mantle and hydrosphere? Miles of water or
sedimentary rock being less dense than basalt or metal-bearing intrusions from below, edges of the plates, faults,
all that good stuff?
They were mapping such bumps by satellite in the 60's (mid to later 60's and since), and using data to find deposits
even then or so we were told in school back before the moon landing.

But the amount of difference is insignificant and the scale is almost regional, you won't find 'edges' in a plane.

tmp190
05-28-2010, 11:21 AM
Originally posted by JtD:
While imo it is not just down to cross section and weight (in a shear situation it is), I agree with the 1:4 ratio M_Gunz arrived at. Maybe he just took a shortcut from experience. http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif

Correct, although aircraft don't strictly obey the square-cube -law.

Kettenhunde
05-28-2010, 12:00 PM
Sorry,professionals have different opinion about EAS usefulness in aircraft performance analysis.

LMAO,

For specific performance he is absolutely right. I don't think that has been disputed.

However that does not mean that using TAS without the correct measured parameters is anymore accurate in a prediction.

Correctly done, the TAS will mirror any EAS prediction in performance trends.

You need "other professionals" that disagree including the entire FAA where EAS is used routinely to gauge aircraft performance.

Here is great primer for you on turn performance at altitude.

Notice everything is in EAS to illustrate the effect of density:

http://www.nar-associates.com/...nalt_wide_screen.pdf (http://www.nar-associates.com/technical-flying/turning/turnalt_wide_screen.pdf)

JtD
05-28-2010, 12:02 PM
This seems to have gone off track a bit - current question being why one plane turns faster than the other in a level turn at the same speed and same angle of bank. Just waiting for someone to spell it out.

p.s. Regarding the off topic gravity thing, I have a gravitational map of the region around my town and I can definitely gain/lose a few grams by weighing myself at the proper spots. The military mapped the gravitation very accurately in order to achieve highest accuracy with their ballistic missiles. There are measurable differences within a few hundred meters distance.

tmp190
05-28-2010, 12:09 PM
Originally posted by JtD:
This seems to have gone off track a bit - current question being why one plane turns faster than the other in a level turn at the same speed and same angle of bank. Just waiting for someone to spell it out.

Did you not reveal it already ?
(Except gravity). Allowing no sideslip in turn is important. More slip->more drag->smaller turn rate.

Kettenhunde
05-28-2010, 12:11 PM
They will slowly be diminishing with the altitude, but won't cease when you move 1cm above the surface of the Earth, due to gravitational field everywhere being a continuous function (and even continuously differentiable outside of the massive body). That's why the picture I posted can be used for anomalies at the ground or at 10 kilometers aloft.

It does not change the relationship of radius nor does it effect an aircraft unless the altitude change is extreme.

Other than interesting and it is very neat.

It does not modify or change a thing about determining aircraft performance.

JtD
05-28-2010, 12:15 PM
Oh well, it's been kind of revealed. What I was looking for is rudder input, aka slip.

Rudder will also induce turning, independent from angle of bank. In order to maintain a level turn at the same speed and same angle of bank, a strong rudder input would also necessitate aileron input to compensate for the induced roll moment and more thrust to compensate for the extra drag - but you can eventually get two planes do different level turns at the same angle of bank and speed.

Kettenhunde
05-28-2010, 01:12 PM
Umm it has been spelled out...side force.

Don't get confused readers about the the effects of side force on a turn. Side force is a function of the fuselage area.

As a foundation for understanding turn performance, all airplanes at the same angle of bank and velocity will make the same turn.

http://www.nar-associates.com/...turn_wide_screen.pdf (http://www.nar-associates.com/technical-flying/turning/turn_wide_screen.pdf)

JtD
05-28-2010, 02:03 PM
The next one isn't really flight physics, but related:

MW50 or water injection as it was called by the Western Allies was used to increase the power output of high performance aero engines. The use of MW50/water injection allowed considerably higher manifold pressures / engine boosts to be used. It also had the effect of significantly reducing fuel consumption when enabled. Why?

na85
05-28-2010, 02:21 PM
Originally posted by JtD:
The next one isn't really flight physics, but related:

MW50 or water injection as it was called by the Western Allies was used to increase the power output of high performance aero engines. The use of MW50/water injection allowed considerably higher manifold pressures / engine boosts to be used. It also had the effect of significantly reducing fuel consumption when enabled. Why?

I don't know a lot about MW50 itself, but I would imagine that it functions similarly to most other ADI (anti-detonation injection) or "anti knock" systems, so I'll hazard a guess:

If methanol and water are injected into the cylinder just prior to combustion, the mixture will absorb heat from the charge in the cylinder. That would mean you wouldn't have to run the engine at full-rich to get the greatest amount of power from it.

Additionally, the water/methanol mix being injected will displace an equivalent amount of air/fuel mixture from the cylinder, thus reducing the amount of fuel per cycle.

Kettenhunde
05-28-2010, 02:35 PM
I thought the winner of quiz asked the next question.

I would give TinyTim the credit as he guessed the condition of flight from a chart which we found out represented any non-steady condition of flight.

Kind of like peering into a crystal ball....

Now we have two winners it looks like.

JtD
05-28-2010, 02:48 PM
Originally posted by na85:

If methanol and water are injected into the cylinder just prior to combustion, the mixture will absorb heat from the charge in the cylinder. That would mean you wouldn't have to run the engine at full-rich to get the greatest amount of power from it.

Additionally, the water/methanol mix being injected will displace an equivalent amount of air/fuel mixture from the cylinder, thus reducing the amount of fuel per cycle.

Both of this is true, but it is not the main cause. As a small addition to your statements, MW50 wasn't only injected into the cylinder (Germans did that), on the US planes it usually was injected into the supercharger.

The main reason is so painfully obvious that it is easy to miss.

na85
05-28-2010, 05:48 PM
Originally posted by JtD:

The main reason is so painfully obvious that it is easy to miss.

Thanks, that makes me feel so much better http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif

irR4tiOn4L
05-28-2010, 06:59 PM
Originally posted by Bremspropeller:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Also, if you're dropping flaps, you'll just be creating more lift. If you do that, and maintain angle of bank, you'll be gaining altitude and thus won't be doing a level turn anymore. So, this doesn't work.

He said he'd throttle down, thus decreasing q, which in turn needs a larger CL to maintain the same net-lift. The theory ain't wong at all.

That condition is not sustainable, however - he has to constantly make up thrust for drag or the contrary. </div></BLOCKQUOTE>

Well it was more of a joke on an earlier thread! http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

I actually interpreted the question as saying the aircraft was in a tighter, more 'wings vertical' turn than just a banking turn, because the reference was to AOA

If the question is saying the aircraft is maintaing altitude with, say, a 45 degree bank then i guess a different use of rudder, a bit of wind, flaps combined with more rudder - i think thered be a number of ways this could happen

M_Gunz
05-28-2010, 07:09 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 na85:

If methanol and water are injected into the cylinder just prior to combustion, the mixture will absorb heat from the charge in the cylinder. That would mean you wouldn't have to run the engine at full-rich to get the greatest amount of power from it.

Additionally, the water/methanol mix being injected will displace an equivalent amount of air/fuel mixture from the cylinder, thus reducing the amount of fuel per cycle.

Both of this is true, but it is not the main cause. As a small addition to your statements, MW50 wasn't only injected into the cylinder (Germans did that), on the US planes it usually was injected into the supercharger.

The main reason is so painfully obvious that it is easy to miss. </div></BLOCKQUOTE>

Burning the methanol as fuel?

Kettenhunde
05-28-2010, 07:11 PM
http://www.turbotuning.net/Artikel/naca-wr-e-264.pdf

JtD
05-28-2010, 10:27 PM
Originally posted by M_Gunz:

Burning the methanol as fuel?

Exactly. What initially had been added as an anti-freeze component is excellent fuel itself, and so was being used as a substitute for the actual fuel. Depending on the amount of WM50 and fuel injected, 20% of the burned substances could be the Methanol, which often was more than was needed to compensate for the higher power output. So fuel consumption was reduced. What's a little bit misleading is that the US referred to it as simply "water injection", while often 60% of the injected substance was Methanol.

M_Gunz
05-28-2010, 10:38 PM
Methanol does have less energy per volume than the gas but still not bad. Cooling the air charge though should allow
more efficient burn, more air (and fuel) and perhaps ability to run leaner without problems?

JtD
05-28-2010, 10:57 PM
Yes to all.

In particular with the US concept of injecting MW50 into the supercharger you're getting a lot of benefits out of it.

BillSwagger
05-28-2010, 11:57 PM
And your willing conclude that the methanol is the main reason for better fuel economy?

I would've thought that using a leaner mixture than what was previously required, (ie auto rich to prevent detonation) would be the biggest contributor for better fuel economy. I guess it depends if your goal is max output or a steady cruise.

I also read that the water can burn if the compression reached is high enough.
Not sure if that is the case inside a WW2 piston engine.

Bill

JtD
05-29-2010, 12:41 AM
Originally posted by BillSwagger:
And your willing conclude that the methanol is the main reason for better fuel economy?

Yes, I am. Because if you include the methanol into the equation, the mixture with MW50 isn't necessarily any leaner than without. This varied with the engineers choice. So the savings here weren't that huge.

M_Gunz
05-29-2010, 12:41 AM
Splitting and recombining water is done at a net loss. Yet there's people still selling just that idea.

Cooling the supercharged air is key to getting maximum power Bill. IMO nitrous oxide is better though,
it chills on release and provides more oxygen. People use it on land racing vehicles too, it's baaaad
shize on a crotch rocket!

na85
05-29-2010, 02:55 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 BillSwagger:
And your willing conclude that the methanol is the main reason for better fuel economy?

Yes, I am. Because if you include the methanol into the equation, the mixture with MW50 isn't necessarily any leaner than without. This varied with the engineers choice. So the savings here weren't that huge. </div></BLOCKQUOTE>

Do you have a source for that? The energy density of methanol is something like a quarter of the energy density of avgas.

WTE_Galway
05-29-2010, 03:57 AM
Originally posted by BillSwagger:

I also read that the water can burn if the compression reached is high enough.

Bill

With what to form what ? I would be very surprised if this turned out to be true.

If you mean ionising the water and then recombining it, that seems rather pointless.

tmp190
05-29-2010, 05:46 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 BillSwagger:
And your willing conclude that the methanol is the main reason for better fuel economy?

Yes, I am. Because if you include the methanol into the equation, the mixture with MW50 isn't necessarily any leaner than without. This varied with the engineers choice. So the savings here weren't that huge. </div></BLOCKQUOTE>

From F4U-1 engine chart:
War emergency, SL, 2250hp, 245 gal/hr
Military, 2000hp, 290 gal/hr

Interpolating MIL power to WEP, we get 290*2250/2000= 326.25 gal/hr
To gain the power difference from gasoline we need a fuel flow of 326.25-245 =81.25 gal/hr.
But as the energy content is 114,000 BTU/gal for gasoline and 56,800 BTU/gal for methanol, we need a methanol fuel flow of 114000/56800*81.25 = 163.07 gal/hr.

The size of the MW-50 tank is 10.3 gal, half of it methanol. The endurance is thus
5.15/163.07 hour, or less than 2 minutes. How long was it in reality ?

Edit: Found it: 8 and a half minutes.

TinyTim
05-29-2010, 06:08 AM
Originally posted by BillSwagger:
I also read that the water can burn if the compression reached is high enough.

Burn with oxygen? No go. Water molecule already is in an energy extremely poor state, that's why it's so inert - no matter the pressure.

I'd be eager to read the source.

Kettenhunde
05-29-2010, 07:36 AM
I would've thought that using a leaner mixture than what was previously required, (ie auto rich to prevent detonation) would be the biggest contributor for better fuel economy.

That is the largest contributor to improved fuel economy. That is what the NACA concluded in the report I posted as well.

If you look at the SFC graphs included, "fuel economy" is a very relative term.

It has better fuel economy than auto rich straight fuel could achieve because it allows the mixture to run leaner.

However it has worse fuel consumption than any lower power settings if the engine is at the same condition.

Edit: Found it: 8 and a half minutes.

Good stuff btw. I like your approach to the problem.

I looked at several F4U Flight Operating Charts and the engine settings are NOT the same.

The 290 gph for take off is at sea level in auto rich. Auto rich is designed at Take Off power to deliver more fuel than the engine can burn. It uses the extra fuel as coolant sort of like the German C-3 Einspritzung.

The 245 gph for WEP is in auto lean.

That falls inline with the NACA conclusions that leaning the mixture is primary reason for fuel economy.

tmp190
05-29-2010, 08:06 AM
Originally posted by JtD:
As a small addition to your statements, MW50 wasn't only injected into the cylinder (Germans did that), on the US planes it usually was injected into the supercharger.

Afaik the Germans injected the water into the supercharger intake, And Americans into the carburettor, that was between the the auxiliary and main blower (in R-2800)

M_Gunz
05-29-2010, 08:15 AM
Originally posted by WTE_Galway:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by BillSwagger:

I also read that the water can burn if the compression reached is high enough.

Bill

With what to form what ? I would be very surprised if this turned out to be true.

If you mean ionising the water and then recombining it, that seems rather pointless. </div></BLOCKQUOTE>

It's a net-loss process. It takes more electron-volts to split water than you can get back out in recombination.

There's a huge scam that's been running based on ... people who don't know better wanting to save real money,
willing to believe pseudo-science and 'testimonials' from those netting the money (pyramid style, you get paid for
every new dupe you bring in after you've paid of course) called water4gas. They even have their own 'debunk' sites
that say oh no, we tried to prove it's wrong but it's either A) real or B) mostly real.
The crux of the con is that the dupe wants to believe based on the end results.

Holtzauge
05-29-2010, 09:16 AM
I find it upsetting that this thread only seems to be concerned with increasing the performance of combustion engines and totally disregards the huge environmental impact associated with Dihydrogen Monoxide!

Actually, thousands of deaths each year can be traced back to the usage of DHMO!

Some of the dangers:

* Death due to accidental inhalation of DHMO, even in small quantities.
* Prolonged exposure to solid DHMO causes severe tissue damage.
* DHMO is a major component of acid rain.
* Gaseous DHMO can cause severe burns.
* Contributes to soil erosion.
* Found in biopsies of pre-cancerous tumors and lesions.

And here are some of it's uses:

* as an industrial solvent and coolant,
* in nuclear power plants,
* in biological and chemical weapons manufacture,
* in the development of genetically engineering crops and animals,

Read more and join the movement at:

Dihydrogen Monoxide Reserch Division (http://www.dhmo.org/)

Note that even AlQuada uses DHMO!

It's not to late! Join the movement to ban DHMO now!

TinyTim
05-29-2010, 09:31 AM
@Holtz

http://forums.ubi.com/images/smilies/88.gif http://forums.ubi.com/images/smilies/11.gif

tmp190
05-29-2010, 10:01 AM
Originally posted by TinyTim:
@Holtz

http://forums.ubi.com/images/smilies/88.gif http://forums.ubi.com/images/smilies/11.gif

That is one way. I personally wouldn't bother with trolls. This is a good thread.

M_Gunz
05-29-2010, 10:09 AM
Originally posted by TinyTim:
@Holtz

http://forums.ubi.com/images/smilies/88.gif http://forums.ubi.com/images/smilies/11.gif

That used to be such a nice town too. But.. DuPont.. need I say more?

JtD
05-29-2010, 10:42 AM
Originally posted by tmp190:

From F4U-1 engine chart:
War emergency, SL, 2250hp, 245 gal/hr
Military, 2000hp, 290 gal/hr

Interpolating MIL power to WEP, we get 290*2250/2000= 326.25 gal/hr
To gain the power difference from gasoline we need a fuel flow of 326.25-245 =81.25 gal/hr.
But as the energy content is 114,000 BTU/gal for gasoline and 56,800 BTU/gal for methanol, we need a methanol fuel flow of 114000/56800*81.25 = 163.07 gal/hr.

The size of the MW-50 tank is 10.3 gal, half of it methanol. The endurance is thus
5.15/163.07 hour, or less than 2 minutes. How long was it in reality ?

Edit: Found it: 8 and a half minutes.

This is interesting, but according to the Corsair pilot handbook the fuel consumption at Military is just 230 gallons per hour. Nothing given for WEP, though. So nothing to extrapolate it to. What charger gears are your numbers for? P-47 numbers I have are similar to yours, though. But there I don't have MW50 consumption rates (for the given boost level).

Also, sometimes ethanol was being used and the ratio sometimes was up to 60%. Could make it last 3 minutes. http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

JtD
05-29-2010, 10:45 AM
Originally posted by tmp190:
Afaik the Germans injected the water into the supercharger intake, And Americans into the carburettor, that was between the the auxiliary and main blower (in R-2800)

The carburettor on the R-2800 injected into the main stage of the supercharger. That's where the water went, too. The R-2800 had a ring fuel distributor directly in front of the impeller of the charger.
I'm not sure about the Germans, you could be right.

tmp190
05-29-2010, 11:53 AM
Originally posted by JtD:
This is interesting, but according to the Corsair pilot handbook the fuel consumption at Military is just 230 gallons per hour. Nothing given for WEP, though. So nothing to extrapolate it to. What charger gears are your numbers for? P-47 numbers I have are similar to yours, though. But there I don't have MW50 consumption rates (for the given boost level).

Also, sometimes ethanol was being used and the ratio sometimes was up to 60%. Could make it last 3 minutes. http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

OK, I think I posted the picture, different thread.http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif

JtD
05-29-2010, 12:54 PM
Yes, you did. Thanks for doing so. I'd agree with your interpretation. Numbers from Corsair handbook, which I guess you know, too:

http://mitglied.lycos.de/jaytdee/f4ufuel.jpg

Max continuous is 10% less than what you've got, Military 20% less. Odd.

kimosabi79
05-29-2010, 01:05 PM
Originally posted by JtD:
The next one isn't really flight physics, but related:

MW50 or water injection as it was called by the Western Allies was used to increase the power output of high performance aero engines. The use of MW50/water injection allowed considerably higher manifold pressures / engine boosts to be used. It also had the effect of significantly reducing fuel consumption when enabled. Why?

When you add WEP, the first thing that happens is that you cool down the fuel/air mix. That results in more dense air entering the cylinders, thus adding more oxygen molecules per every ccm of air. Since an old combustion engine usually is set up with a certain timing at certain rpm's, you'd also need to add something that prevents detonation(or knocking) from that increased density. Hence the Methanol which has less energy therefore it is less likely to self-detonate from higher combustion temperatures from increased density and from the higher cylinder pressures that occurs when you add water, which transforms into steam inside the cylinders.

An engine is primarily based on burning oxygen, fuel is only a part of the prosess to make that happen.

So, what you get is higher cylinder pressures, which = more power which again = into less need of fuel. You're not really decreasing fuel consumption strictly by adding WEP but the fuel need decreases as the cylinder pressure increase provides more power which again usually results in a state where leaner fuel mix is the way to go to save fuel. You don't need the same amount of gasoline anymore since the Methanol CAN substitute some of it to prevent detonation from higher cylinder pressures and combustion temperatures. On engines which does not have adjustable air/fuel mix, you won't see much consumption decrease either.

Hope I was clear enough. http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif

Holtzauge
05-29-2010, 01:12 PM
"Nothing is more curious than the almost savage hostility that Humour excites in those who lack it."

George Saintsbury

http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

Kettenhunde
05-29-2010, 02:52 PM
The USN Corsair manual list's 59.5 inHG at 2700 rpm for take off for 245 gph in auto lean.

The British F4U-1 lists ~52 in HG for 230 gph.

They are not the same boost levels.....

Kettenhunde
05-29-2010, 02:53 PM
kimosabi79

It s funny that several questions have already been answered yet the "quiz" never seems to get passed along.

kimosabi79
05-29-2010, 03:56 PM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">kimosabi79

It s funny that several questions have already been answered yet the "quiz" never seems to get passed along. </div></BLOCKQUOTE>

I wouldn't say "never" Thou impatient one. I'll cook up something nice by the end of tomorrow.

*edit* Or maybe this is up to JtD? Looks like he's the one asking the questions around here. http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

Outlaw---
05-29-2010, 09:09 PM
The vaporization of the alcohol/water mixture reduces the charge temperature, thus preventing detonation at higher boost pressures. Without it (assuming nothing else changes) the engine will suffer at higher boost pressures due to detonation. That's really all there is to it.

--Outlaw.

M_Gunz
05-30-2010, 02:03 AM
Originally posted by Outlaw---:
The vaporization of the alcohol/water mixture reduces the charge temperature,

And greatly increases the density of the charge as well including the oxygen content.

kimosabi79
05-30-2010, 02:35 AM
Originally posted by Outlaw---:
The vaporization of the alcohol/water mixture reduces the charge temperature, thus preventing detonation at higher boost pressures. Without it (assuming nothing else changes) the engine will suffer at higher boost pressures due to detonation. That's really all there is to it.

--Outlaw.

That's not accurate. Although the WEP reduces air/fuel mix temperatures, the decreased charge temperature alone has nothing to do with knock resistance. If they would add water only, to cool down the mix and provide a higher density mix, you'd see crazy detonation. The methanol is what prevents detonation, not the charge temperatures. Lower temps and more oxygen molecules per ccm, without adding methanol, actually increase the risk of detonation.

BillSwagger
05-30-2010, 03:03 AM
Originally posted by WTE_Galway:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by BillSwagger:

I also read that the water can burn if the compression reached is high enough.

Bill

With what to form what ? I would be very surprised if this turned out to be true.

If you mean ionising the water and then recombining it, that seems rather pointless. </div></BLOCKQUOTE>

The discussion came up before when we were talking about WEP, and i remember reading that the water in the mixture can burn. It doesn't mean there is a gain from it. Its just a part of the reaction from water vapor sent through the combustion process. Obviously you aren't running the engines on water alone if you thought that's what i was implying. Nor does all of the water used in a MW50 system get burned.

First the heat and the compression separate the molecule and it becomes hydrogen and oxygen. The oxygen is used is the combustion process just like the regular intake air, while the hydrogen burns. In the burning process new chemicals are formed, it doesn't necessarily have to recombine back into water.

Bill

M_Gunz
05-30-2010, 04:23 AM
What do you think the hydrogen 'burns' with Bill? Answer is that 2 hydrogen 'burn' with 1 oxygen and do so more
strongly than anything else present given air, gas and methanol are what is present. The hydrogen and oxygen want
to combine very strongly which is why hydrogen and oxygen make more powerful rocket fuel than gas or methanol and
oxygen.

Beyond that, the kind of temperature it takes to split water IIRC is well beyond what you get in a running engine.
And no, the engine pressure doesn't make any magic happen there. Of course there will be -some- water molecules
splitting... but not enough to be more than a statistical chance with a -lot- of zeros between the decimal point
and the first non-zero number. 18 grams of water has over 6 with 23 zeros after it molecules in it so yeah just
by room temperature and probability -some- of those are at high enough energy to split all the time making it safe
to say "some water splits" but it's much less than a penny compared to the USA yearly budget, hope that gives you
some idea of the numbers involved. If the chance was noticeably higher there would be no water on Earth for a long
time now.

I wouldn't write so much about it but this is the third-plus time it's been brought up here as if it matters.

JtD
05-30-2010, 04:50 AM
Originally posted by kimosabi79:

...If they would add water only, to cool down the mix and provide a higher density mix, you'd see crazy detonation...

No, you wouldn't. Detonation is caused by high peak cylinder temperatures and pressures during the combustion of charge in the cylinder. Both is very effectively reduced by water injection. This has been tried, proven and reported by various engine companies and several independent laboratories at the time. Your opinion stands out.

-----

Could we close this now? It is meant to be a quiz, not a discussion. And while I my last question allowed for more correct answers then the one I was aiming for, all the necessary things have been said. To sum it up:
- compared to Military setting allows a considerably leaner mixture
- Alcohol is burned in the process as well
- efficiency gains

kimosabi79
05-30-2010, 08:39 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 kimosabi79:

...If they would add water only, to cool down the mix and provide a higher density mix, you'd see crazy detonation...

No, you wouldn't. Detonation is caused by high peak cylinder temperatures and pressures during the combustion of charge in the cylinder. Both is very effectively reduced by water injection. This has been tried, proven and reported by various engine companies and several independent laboratories at the time. Your opinion stands out.

-----

Could we close this now? It is meant to be a quiz, not a discussion. And while I my last question allowed for more correct answers then the one I was aiming for, all the necessary things have been said. To sum it up:
- compared to Military setting allows a considerably leaner mixture
- Alcohol is burned in the process as well
- efficiency gains </div></BLOCKQUOTE>

To put it very simple so that everyone understands:

Methanol is much better at chargecooling than water. Water alone would only be usable as a direct cylinder injection since water alone would freeze in your induction system. It's a team effort in more ways than one. Octane increase is another team effort. How do you come to the conclusion that cylinder pressures are reduced during WEP BTW? It's a bit contradictory since engaging WEP/MW does increase the cylinder wall friction and wear quite considerably, and the reason to why things like that happen is usually increased piston ring pressure and/or piston/cylinder materials expanding. You're right about the cooling but WEP was implemented solely to be able to run higher boosts. Higher boosts and WEP engaged still have it's limits. Detonation is one of them so using water alone in your aircraft engine would be useless, considering the reasons I described above.

I'm curious about those labs and engine manufaturer reports though. Got any links?

tmp190
05-30-2010, 09:55 AM
Originally posted by kimosabi79:
Methanol is much better at chargecooling than water.

If that is true, then why use water at all ?

kimosabi79
05-30-2010, 10:14 AM
Originally posted by tmp190:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by kimosabi79:
Methanol is much better at chargecooling than water.

If that is true, then why use water at all ? </div></BLOCKQUOTE>

Because water doesn't turn into a gas prior to the cylinders. It's inside the cylinders, where temperatures make the water expand, that water is of any real use. When the water turns to gas, or steam, that's when it instantly absorbs large amounts of heat and carries it out through the exhaust channels.

ggb123
05-30-2010, 01:43 PM
Originally posted by BillSwagger:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by WTE_Galway:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by BillSwagger:

I also read that the water can burn if the compression reached is high enough.

Bill

With what to form what ? I would be very surprised if this turned out to be true.

If you mean ionising the water and then recombining it, that seems rather pointless. </div></BLOCKQUOTE>

The discussion came up before when we were talking about WEP, and i remember reading that the water in the mixture can burn. It doesn't mean there is a gain from it. Its just a part of the reaction from water vapor sent through the combustion process. Obviously you aren't running the engines on water alone if you thought that's what i was implying. Nor does all of the water used in a MW50 system get burned.

First the heat and the compression separate the molecule and it becomes hydrogen and oxygen. The oxygen is used is the combustion process just like the regular intake air, while the hydrogen burns. In the burning process new chemicals are formed, it doesn't necessarily have to recombine back into water.

Bill </div></BLOCKQUOTE>

Well, in addition. while compression separate the molecule to becomes Hydrogen and Oxygen, such process "takes" energy(in this case more fuel) and time(so it affects ignition timing and allow for higher compression ratio to more efficient burn). Hydrogen and Oxygen then later react with fuel to provide more total reactants in the formular. therefore more hoursepower. I think MW50 does boost a bit high altitude performance, too. http://forums.ubi.com/images/smilies/25.gif

M_Gunz
05-30-2010, 03:48 PM
Tickets to Clueland....

If you actually split enough water to be noticed then the volume change from vapor to gas would raise the
engine pressure enormously. A tiny bit of water splits into a large amount of hydrogen and oxygen.

High pressure steam turbines:

The boilers in a typical power plant generate steam at around 2,400 pounds per square inch--about 163 times atmospheric pressure. Some "supercritical" steam plants operate at pressures over 4,000 PSI. Here the steam is heated far above the boiling point, generally to 1,000 degrees Fahrenheit or higher.

The water injected absorbs great amounts of heat when cooling the air charge as it goes through state change
from liquid to vapor. Check the latent heat of vaporization of alcohol vs water here (http://en.wikipedia.org/wiki/Latent_heat)
and see that water absorbs over twice the energy per mass (and is more dense, the weight per volume kind of dense)
which makes it ideal for the purpose.

The methanol is there to lower the freezing point greatly which at altitude in thin pipes is desired. Nuff sed?