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Z4K
06-26-2004, 08:30 AM
It's time for MULTIVARIABLE CALCULUS!! Hooray!

http://www.photodump.com/direct/Z4k/z_calc001_United.jpg

I need some help with a maths problem I have - I can't do it and don't have Matlab or Maple.

If any of you are mathsy types I'd really appreciate a hand with this, or even just a solution (including a graphical one).

There are a couple of ways of stating the problem. In vector notation (where a' is the derivative of a, a'' is the 2nd derivative of a, and a is the magnitude of vector a) it can be written:

(x'', y'') = L * sqrt [ x'^2 + y'^2 ] * (-y', x') + D * sqrt [ x'^2 + y'^2 ] * (-x', -y') + ( 0, -g )

where L, D and g are constants; x and y are postion vectors; x' and y' are the rate of change of x and y with time (ie. velocities in horizontal and vertical planes); similarly for x'' & y'' (acceleration).

Or in words:

I have a particle subject to the following accelerations:
- A constant acceleration vertically down;
- An acceleration proportional to the square of the velocity (= L*v^2) perpendicular to the path of the particle upwards (ie. there are two perpendiculars, it's on the upward one);
- An acceleration proportional to the square of the velocity (= D*v^2) parallel to the path of the particle and ******ing its motion (ie. slowing it down).

I want to be able to find two of the following; x, y, or velocity, from the other one. For instance "at a y displacement of 3 metres, the particle will have an x displacement of 2 metres and a velocity of 6 m/sec."

http://www.photodump.com/direct/Z4k/z_calc003_Train.jpg

Try as I might (having forgotten all my university calculus, and probably never having done any that could solve this anyway...) I can't solve it. If you have a look at it, you'll see that the acceleration in either direction is a function of the velocity in not only that direction, but the other as well ( a(x) = F(x, y) ). I can't separate it, can't make it an "exact differential equation" and hence don't think I can use integration of partial differentials to solve it.

I've got to the following random collection of expressions:

(B: angle path makes to horizontal;
i, j: x & y unit vectors;
v: velocity vector;
v: magnitude of velocity;
v{x}: horizontal velocity vector;
v{y}: magnitude of vertical velocity;
a: acceleration;
a{t}: magnitude of tangential acceleration;
a{n}: magnitude of normal acceleration;
L, D, g, m: constants)

IN NO PARTICULAR ORDER:

dv/dt = g sin(B) - Dv^2

dB/dt = (g cos(B) - Lv^2)/v = a{n}/v

d2x/dt2 = a{x} = (v^2/m)(L sin(B) - D cos(B))

d2y/dt2 = a{y} = (v^2/m)(L cos(B) + D sin(B)) - g

v = (dx/dt)*i + (dy/dt)*j

a = (sqrt ((dx/dt)^2 + (dy/dt)^2))*(( L(dy/dt) - D(dx/dt) )*i + ( L(dx/dt) + D(dy/dt) )*j ) - g*j

If you read this far, thank you.

- - - - - - - - - - - - - - -
Let's play Global Termonuclear War

[This message was edited by Tully__ on Sun June 27 2004 at 12:30 AM.]

Z4K
06-26-2004, 08:30 AM
It's time for MULTIVARIABLE CALCULUS!! Hooray!

http://www.photodump.com/direct/Z4k/z_calc001_United.jpg

I need some help with a maths problem I have - I can't do it and don't have Matlab or Maple.

If any of you are mathsy types I'd really appreciate a hand with this, or even just a solution (including a graphical one).

There are a couple of ways of stating the problem. In vector notation (where a' is the derivative of a, a'' is the 2nd derivative of a, and a is the magnitude of vector a) it can be written:

(x'', y'') = L * sqrt [ x'^2 + y'^2 ] * (-y', x') + D * sqrt [ x'^2 + y'^2 ] * (-x', -y') + ( 0, -g )

where L, D and g are constants; x and y are postion vectors; x' and y' are the rate of change of x and y with time (ie. velocities in horizontal and vertical planes); similarly for x'' & y'' (acceleration).

Or in words:

I have a particle subject to the following accelerations:
- A constant acceleration vertically down;
- An acceleration proportional to the square of the velocity (= L*v^2) perpendicular to the path of the particle upwards (ie. there are two perpendiculars, it's on the upward one);
- An acceleration proportional to the square of the velocity (= D*v^2) parallel to the path of the particle and ******ing its motion (ie. slowing it down).

I want to be able to find two of the following; x, y, or velocity, from the other one. For instance "at a y displacement of 3 metres, the particle will have an x displacement of 2 metres and a velocity of 6 m/sec."

http://www.photodump.com/direct/Z4k/z_calc003_Train.jpg

Try as I might (having forgotten all my university calculus, and probably never having done any that could solve this anyway...) I can't solve it. If you have a look at it, you'll see that the acceleration in either direction is a function of the velocity in not only that direction, but the other as well ( a(x) = F(x, y) ). I can't separate it, can't make it an "exact differential equation" and hence don't think I can use integration of partial differentials to solve it.

I've got to the following random collection of expressions:

(B: angle path makes to horizontal;
i, j: x & y unit vectors;
v: velocity vector;
v: magnitude of velocity;
v{x}: horizontal velocity vector;
v{y}: magnitude of vertical velocity;
a: acceleration;
a{t}: magnitude of tangential acceleration;
a{n}: magnitude of normal acceleration;
L, D, g, m: constants)

IN NO PARTICULAR ORDER:

dv/dt = g sin(B) - Dv^2

dB/dt = (g cos(B) - Lv^2)/v = a{n}/v

d2x/dt2 = a{x} = (v^2/m)(L sin(B) - D cos(B))

d2y/dt2 = a{y} = (v^2/m)(L cos(B) + D sin(B)) - g

v = (dx/dt)*i + (dy/dt)*j

a = (sqrt ((dx/dt)^2 + (dy/dt)^2))*(( L(dy/dt) - D(dx/dt) )*i + ( L(dx/dt) + D(dy/dt) )*j ) - g*j

If you read this far, thank you.

- - - - - - - - - - - - - - -
Let's play Global Termonuclear War

[This message was edited by Tully__ on Sun June 27 2004 at 12:30 AM.]

FZG_Mined
06-26-2004, 08:33 AM
mmhhh....i have to put it on a paper to understand it well..it seems a normal newtonian calculus...i'll check if i can do that...

Davide

http://mined86.free.fr/banniere.jpg

Z4K
06-26-2004, 08:40 AM
Two things:

[1] How do I edit posts? I've either gone stupid and can't find the button, or there isn't one. If so, those blasted [ u ] tags will have to stay there.

[2] If anyone's wondering WHY I care about the above, it's for a glider I need to make to enter one of those Birdman competitions. The one where stupid people build a glider/wing/chicken suit and jump off a tower into the water and try and fly as far as they can.
I need to know the solution to the above to help work out which airfoil to use, because the velocity I reach in the initial dive must be high enough to start a proper, steady glide at the L/D ration glide slope.

- - - - - - - - - - - - - - -
Let's play Global Termonuclear War

Stormer777
06-26-2004, 09:04 AM
Oh my bad I must have stumbled into the revenge of the nerds forum by mistake...

J/K... Holy S

HART_dreyer
06-26-2004, 09:13 AM

Regards,
dreyer
the dreyer vs. Hartmann game! (http://www.dreyermachine.com/il2/)

Franzen
06-26-2004, 09:24 AM
Now you know why they invented video games. That's how they sorted the birdman wannabes. http://ubbxforums.ubi.com/images/smiley/16x16_smiley-wink.gif

Fritz Franzen

ploughman
06-26-2004, 09:27 AM
Hey, isn't the Union Flag upside down? You must be in distress. Luckily for me I flunked Algebra 2 - so I haven't a clue. http://ubbxforums.ubi.com/infopop/emoticons/icon_cool.gif

Aero_Shodanjo
06-26-2004, 11:22 AM

While i cant decypher anything about calculus, but ive got to say something about the posters.

They're rock http://ubbxforums.ubi.com/images/smiley/88.gifhttp://ubbxforums.ubi.com/images/smiley/88.gifhttp://ubbxforums.ubi.com/images/smiley/88.gif

trumper
06-26-2004, 11:22 AM
http://ubbxforums.ubi.com/infopop/emoticons/icon_smile.gifI've worked it out but i'm NOT telling http://ubbxforums.ubi.com/images/smiley/10.gif

ONLY JOKING http://ubbxforums.ubi.com/infopop/emoticons/icon_biggrin.gif http://ubbxforums.ubi.com/images/smiley/10.gif I don't even understand the question http://ubbxforums.ubi.com/images/smiley/11.gif

DD_NL
06-26-2004, 11:56 AM
Nice work on the war-propaganda posters http://ubbxforums.ubi.com/images/smiley/16x16_smiley-wink.gif

http://home.tiscali.nl/ddonline/IL-2/nighthawk.jpg

Dammerung
06-26-2004, 12:06 PM
Nice Pics =) I'm starting Precalc soon, can't help you now =(

Oh, there are no fighter pilots down in hell...
Oh, there are no fighter pilots down in hell...
The whole damn place is full of queers, navigators, and bombadiers...
Oh, there are no fighter pilots down in hell...

56th BMAC
06-26-2004, 12:16 PM
When did they start putting letters in math problems?? http://ubbxforums.ubi.com/images/smiley/59.gif http://ubbxforums.ubi.com/images/smiley/blink.gif

Dystopian
06-26-2004, 12:58 PM
You were right; I would'nt have read it.

Nice wimmen always drags a good post home, though. http://ubbxforums.ubi.com/images/smiley/16x16_smiley-tongue.gif

Steaky_361st
06-26-2004, 01:39 PM

Ask me in 2 years, Ill have taken calc by then...

Steaky

S 8
06-26-2004, 03:12 PM
IâÂ´m just a janitor.. http://ubbxforums.ubi.com/images/smiley/16x16_smiley-tongue.gif

_VR_ScorpionWorm
06-26-2004, 03:37 PM
I took PreCalculus(98-99) and Calculus(99-2000)...but alas after four years and two kids my brain isnt what it used to be.

Edit: Nice posters BTW, and you forgot the "H" in ThermoNuclear...oh, and where does this take place, I have always seen the 'RedBull' advertisement and always wanted to check it out?

"Soldiers, Sailors, and Airmen of the Allied Expeditionary force:
You are about to embark upon a Great Crusade toward which we have strived these many months. The eyes of the world are upon you. Good Luck! And let us all beseech the blessing of Almighty God upon this great and noble undertaking" - Gen. Dwight D. Eiseinhower-Supreme Allied Commander.

www.vultures-row.com (http://www.vultures-row.com)

DuxCorvan
06-26-2004, 04:03 PM
Ah, yes! I've heard this before.

http://ubbxforums.ubi.com/images/smiley/16x16_smiley-tongue.gif

06-26-2004, 04:15 PM

http://imageshack.us/files/copper%20sig%20with%20rank.jpg
310th FS & 380th BG website (http://www.310thVFS.com)

Z4K
06-27-2004, 12:46 AM
FZG_Mined:

Well?

Ploughman:

It's because it's flying to the left, rather than the right. But yes, I am in distress.

_VR_ScorpionWorm:

Thermonuclear's fixed - thanks.

As for where: all over the world. Apparently it started in a place somewhere in Britain (maybe) as a tourist attraction for an otherwise lame place and it (wait for it...) really took off (hahahaha). The one I'm gearing up for is in Sydney, Australia which will be run as a fund raising thing. I've seen it in Canberra a few years ago and I know that at least half a dozen other towns or cities in Australia have them.

I'd expect that wherever you are there'll be one somewhere in the country, sometime through the year (wow, how can he be so specific??!?!).

All who liked the piccies:

Thanks http://ubbxforums.ubi.com/infopop/emoticons/icon_smile.gif Here's another:

http://www.photodump.com/direct/Z4k/z_calc005_Dud.jpg

Anyone who can help otherwise:

This might assist (although it should say sinB = -vy/v):

http://www.photodump.com/direct/Z4k/z_calc000_Diag304.jpg

- - - - - - - - - - - - - - -
Let's play Global Thermonuclear War

Tully__
06-27-2004, 01:32 AM
<BLOCKQUOTE class="ip-ubbcode-quote"><font size="-1">quote:</font><HR>Originally posted by Z4K:
Two things:

[1] How do I edit posts? I've either gone stupid and can't find the button, or there isn't one. If so, those blasted [ u ] tags will have to stay there.
<HR></BLOCKQUOTE>

Gone now. The forum wont let you edit post after a time limit (60 minutes last time I looked, but the admins may have changed it).

=================================================

http://members.optusnet.com.au/tully_78th/sig.jpg

IL2 Forums Moderator
Tully's X-45 profile (SST drivers) (http://members.optusnet.com.au/tully_78th/fb.zip)

Salut
Tully

icrash
06-27-2004, 05:45 PM
uuuummmm, I took cal awhile back and you triggered some interesting flashbacks from college life. Unfortunately I don't remember much. TMYTMB strikes again I guess. There were some crazies going on about some physics stuff here somewhere. You may check that out & pm one of them. (No offense crazies, all I need to know about physics is my plane goes down when I foul up http://ubbxforums.ubi.com/infopop/emoticons/icon_rolleyes.gif )

http://img38.photobucket.com/albums/v117/icrash/txraidersig.bmp

_VR_ScorpionWorm
06-27-2004, 07:16 PM
http://ubbxforums.ubi.com/images/smiley/cry.gif You got me thinking with this problem, after writing it down and seeing if I could help out I found out that Im going totally CRAZY http://ubbxforums.ubi.com/images/smiley/53.gif. I used to complete problems like this in a matter of hours, I cant believe I have forgotten so much. http://ubbxforums.ubi.com/images/smiley/16x16_smiley-indifferent.gif To be honest I actually had a DREAM of Calculus and math http://ubbxforums.ubi.com/images/smiley/blink.gif, was rather scary. Maybe I should start attending a Jr. college to try and relearn everything. http://ubbxforums.ubi.com/images/smiley/35.gif

"Soldiers, Sailors, and Airmen of the Allied Expeditionary force:
You are about to embark upon a Great Crusade toward which we have strived these many months. The eyes of the world are upon you. Good Luck! And let us all beseech the blessing of Almighty God upon this great and noble undertaking" - Gen. Dwight D. Eiseinhower-Supreme Allied Commander.

www.vultures-row.com (http://www.vultures-row.com)

T_O_A_D
06-27-2004, 07:26 PM
Well I am of no help but I sure enjoyed your Photoshop the Posters. Now if only your Maths was so much fun http://ubbxforums.ubi.com/images/smiley/35.gif

Have you checked your Private Topics recently? (http://forums.ubi.com/eve/forums?a=ugtpc&s=400102)
Commanding Officer of the 131st_VFW (http://www.geocities.com/vfw_131st/)
http://home.mchsi.com/~131st_vfw/T_O_A_D.jpg

Bwild
06-27-2004, 08:25 PM
You can always try a numerical approach. If you have the initial conditions you can obtain an approximation by using e.g. Euler or Runge-Kutta methods.

BBB_Hyperion
06-27-2004, 10:55 PM
I dont know what you are up too cause the formulars and drawings you did post were rather inaccurate about the intention. The vector diagram for example shows mixture of different units. You could use ax to define v and insert it in ay but if that was the intention behind it wasnt clear from your post.

I did somewhere a calculation for airdrag , lift for ballistics. Alone the simplyfied version has about 3 pages of reforming variables. So will not post this. Here is a indication how you can get 1 factor depending on other. Rest is
only Handywork. The Main Idea i try to follow is to get all components on x ,y ,z axis as function of speed vs time and range vs time. Then you can use this components to get the value x , y ,z .... depending on other components.

This is a for an unpowered throw at angles without airdrag but as simple example its useable.

http://www.butcherbirds.de/hypesstorage/calculation.jpg

High Ground is not only more agreeable and salubrious, but more convenient from a military point of view; low ground is not only damp and unhealthy, but also disadvantageous for fighting.

Sun Tzu : The Art of War

Regards,
Hyperion

[This message was edited by BBB_Hyperion on Sun June 27 2004 at 10:03 PM.]

HarryVoyager
06-27-2004, 10:59 PM
Ok, I have not had time to go over it and realy figure out how it works, but I would strongly advise hunting down an Electromagnitism text book, and reading up on how particals move in crossed E and B fields, because it sounds to me that that is very similare to what you are dealing with here, except with a drag term thrown in.

What I strongly suspect you are going to get in a net motion in a direction perpediculare to both the constant field, and the crossing field, but I could have the properties of the variable field gotten completely wrong.

What I am fairly certain of, is this should break down into a system of about three integro-differential equations, that are hopefully linear. If you have to get this solved before Monday evening, I would highly recommend heaing down to your local college book store and grabing copies of the Schaum's Outlines for Multi-variable/Vector calculus, Electromagnitism, Linear Intergro-Differential Equations, and possibly even Control Systems as most Control Systems texts involve methods for Linearising inconvenient non-linear systems.

I'll try to figure it out in more detail tomorrow after I get back from work.

Harry Voyager

diomedes33
06-28-2004, 01:15 AM
Here's what I got so far, first of all assume that v travels in the x direction and g acts an angle 0 from the y axis so it looks something like this

to make things easier vy is the velocity in the y direction and x is the velocity in the vx direction.
y
^
| D(vy)^2
|
|
L(vx)^2 |
-----------------------------+---------------------------&gt; x
| \
| 0\
| \
| \
| g

So from this diagram we get.
a = -L(vx)^2i + D(vy)^2j + g

From this we get the scalar equations

(ax) = -L(vx)^2 + gsin(0)
(ay) = D(vy)^2 - gcos(0)

From the definition of acceleration we know that a = dv/dt

so we get
d(vx)/dt = -L(vx)^2 + gsin(0)
d(vy)/dt = D(vy)^2 - gcos(0)

These are two independent non-linear ODEs. You can solve those assuming you know what theta will be and that it is constant. Solving this by hand is difficult to say the least. I would have to do some serious digging through my text books to figure out how to do this one. I'd probably just run it through matlab and see how it works out.

Now if my assumption is wrong and theta changes with time, then we have two lovely nonlinear PDEs &lt;shudder&gt; This you'll definatly need a computer to solve, trying to do this by hand is masicistic suicide.

If you want a specifc axis you can still use this way to solve the problem then use the dot product to project it onto the new axis.

If availability of matlab is a problem, heres a free clone.
octave (http://www.octave.org) it does almost everything that matlab does. Some better others not.

Here's also a nice symbolic solver that I found. I used it to check a lot of my work in my classes last term.
Its a little rough interface wise, but when I was solving the schrodinger equation the equations that it spit out where identical to the ones in the answer.

http://www.public.asu.edu/~guthriec/ubi_sig.jpg

[This message was edited by diomedes33 on Mon June 28 2004 at 01:34 AM.]

HarryVoyager
06-28-2004, 04:51 AM
If it's non-linear, we may not actually know how to solve it, but it should be possible to torque it into linearity under specific conditions, or to find a linear estimate over a certain range of solutions.

I know my Control Systems text has a method for linearizing equations, but I've got to leave for work in about 5 minutes. That's probably your best bet now, is to find a linear aproximation over a limited range.

Harry Voyager

Hope that helps.

Z4K
06-28-2004, 06:49 AM
_VR_ScorpionWorm:

Dreaming in Calculus is bad for you. I don't think you should try toooooo hard to solve it because as people are confirming - it's practically impossible without a calculator.

As for the pic, here's another (hooray! or 'stop encouraging him'):

http://www.photodump.com/direct/Z4k/z_calc006_Waste.jpg

T_O_A_D:

See 'stop encouraging him' above http://ubbxforums.ubi.com/infopop/emoticons/icon_wink.gif

Bwild:

I would love to try a numerical approach but I simply don't know where to start. Runge-Kutta sounds familiar but I haven't even touched on that, ever. I'd welcome any help...

BBB_Hyperion:

You're right, they were. I've crossed that one out now.

The following should give a better understanding of the actual problem I'm trying to solve (where the particle is the centre of mass of the pilot/wing assembly at any time somewhere on the -unknown- path shown in red):

http://www.photodump.com/direct/Z4k/z_calc000_Diag02.jpg

And I think that this one accurately reflects the problem mathematically:

http://www.photodump.com/direct/Z4k/z_calc000_Diag03.jpg

All the negative signs should be correct. If only ubb supported mathematical notation, eh?

That picture you posted wasn't much help for this problem. As far as I could see it simply modelled idealised ballistic motion, is that right? I'm trying to model the motion of an object subject to aerodynamic forces which act perpendicular to the motion on which they're dependent (ie. lift). As far as I can tell, there's no way of expressing the acceleration in x without referring to BOTH the velocity in x AND y, and similarly for a{y}.

I'd love to see the calculation for airdrag and lift you mentioned - that sounds like it suits the problem perfectly. I don't suppose you have a digital copy (scanned/pdf/.doc/whatever) you could email?

HarryVoyager:

At the risk of dismissing a perfectly valid representation, I don't think that an EM field is an applicable model. The Lift and Drag vectors change magnitude AND direction with time, while gravity stays constant (without gravity it'd be much easier in normal & tangential components but the constant g stuffs that up).

There's no immediate deadline for this, except "as soon as possible", so I can get the thing built so I can learn to fly it before the day (probably more important than an airworthy craft is an at least semi-competent pilot. That violin/violinist quote that's popular in people's sigs in here comes to mind). The Linearising of non-linear systems sounds good, I should look at that and Schaum's Outlines is now on my list.

Following that link (and not reading a whole lot - have to get up for work at 5:30 tomorrow morning http://ubbxforums.ubi.com/infopop/emoticons/icon_frown.gif ) it seems promising, but I haven't yet tried to use it. My guess is I'll need to look in some textbooks to get the hang of it properly.

diomedes33:

Theta DOES change with time, which is why I'm having conniptions because of this problem.

Thanks for the links. I found Octave a few days ago, but can't work out (read: haven't put the effort into learning) how to express my problem in a format Octave can use - and solve.

I found MuPAD last night, after my previous post. Looks neat (my 56k could only handle the Light version) but despite figuring out the basics (and therefore being miles ahead of Octave) I can't get it to solve it - bear in mind I've put the grand total of 30 minutes into it.

<pre class="ip-ubbcode-code-pre">odeEquations := ode({x''(t) = (((x'(t))^2 + (y'(t))^2)^(1/2))*( -l*y'(t) - r*x'(t) ),
y''(t) = (((x'(t))^2 + (y'(t))^2)^(1/2))*( l*x'(t) - r*y'(t)) + g,
x''(0) = 0,
y''(0) = g,
y'(0) = 0,
x'(0) = 2,
y(0) = 7,
x(0) = 0}, {x(t), y(t)})

solve(odeEquations)</pre>

...was certainly optimistic and unsurprisingly just gave the same thing back out as output (incidentally, 'r' is used instead of 'd' because I was paranoid it was interpreting 'd' as differential).

ALL:

Thanks for you interest and help!

- - - - - - - - - - - - - - -
Let's play Global Thermonuclear War

BBB_Hyperion
06-28-2004, 09:29 AM
So far we have start values y0 , and v0 . But now comes the problem the cl and cd values are depending on aoa of the wing section also stallspeed and such needs to be considered.
As well c0 and K values.

The theoretical solution to the problem is a function that delivers a possible function within a function range. But that would be hard to get.

By limiting the starting angle you can integrate the given function and get a result function for 1 possible path.

From all the function chars
There can be x , y values that are visited 2 times so the function x vs y doesnt exist as bijectiv projection all the time. Also same for v vs x, v and x can exist more than 1 time.

So you need to build a function that doesnt have this effect and does solve the formular.

However you can avoid the problems above when you set cl as static and assume it is always same aoa at the wing and it just falls linear on glidepath.

Hope i didnt oversee something.

High Ground is not only more agreeable and salubrious, but more convenient from a military point of view; low ground is not only damp and unhealthy, but also disadvantageous for fighting.

Sun Tzu : The Art of War

Regards,
Hyperion

diomedes33
06-28-2004, 11:06 AM
<BLOCKQUOTE class="ip-ubbcode-quote"><font size="-1">quote:</font><HR>Originally posted by Z4K:
_diomedes33_:

Theta _DOES_ change with time, which is why I'm having conniptions because of this problem.
<HR></BLOCKQUOTE>

hehe now that you posted that picture everything seems clearer. Still not sure how to solve it, but at least I know what's going on know. I'll play around with it today and see if there's something I can come up with.

http://www.public.asu.edu/~guthriec/ubi_sig.jpg

diomedes33
06-28-2004, 12:28 PM
Here's an m-file that I got to work to find the velocities. I used the functions that you found above, except g should be negative.

vel.m
-----------------------
function a=vel(v,t)
#a{x} = a(1)
#a{y} = a(2)

#change these to fit your problem
g = 9.8;
D = 1;
L = 1;

a = zeros(2,1);
vmag = sqrt ( v(1)^2 + v(2)^2);

a(1) = vmag * (-D*v(1) - L*v(2));
a(2) = vmag * (-D*v(2) + L*v(1)) - g;

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

In octave or matlab, run
#t = linspace (a, b, number of steps)
t = linspace (0, 10, 200);

#v0 = [vx0; vy0]
v0 = [0; 0];

For octave
v = lsode ("vel", v0, t);

For matlab
v = ode45 ("vel", v0, t);

You can then use a numerical integrator to find the displacement. Matlab has trapz, but octave doesn't seem to have one.

Here's a simple integrator that I wrote that uses the trapizoid rule.

trapint.m
-------------------------------------
function I = trapint(x,y,c)
#x is the independent variable
#y is the dependent variable
#c is the intial condition

I = zeros(length(x), 1);
I(1) = c;

for index = 2:length(I)

I(index) = (x(index) - x(index - 1)) * (y(index) + y(index -1) / 2 + c;

end
---------------------------------------------------------------------------------------------

run
x = trapint (t, v(:,1), x0);
y = trapint (t, v(:,2), y0);

I know you're looking for an anylitical solution, but you can find fairly accurate answer by interpolating the data generated above. If nothing else it will give you an idea if your solution that you eventually find is accurate.

http://www.public.asu.edu/~guthriec/ubi_sig.jpg

[This message was edited by diomedes33 on Mon June 28 2004 at 12:04 PM.]

JJaguar
06-28-2004, 03:12 PM
I think you're over-complicating things. First of all, it may not even be necessary to solve this equation. An estimate of desired glide speed is all you need. Airplane design is a very circular process. Basically you start by estimating various parameters and feeding your results back into your design until your numbers start to converge. You may find a book on aircraft design to be useful, one that takes a practical rather than a theoretical approach. If you can find one specifically for glider or sailplane design, then even better.

You can approximate an answer that's probably good enough using simple energy equations. Potential energy is represented by PE = mgh, and kinetic energy by KE = .5mv^2. Kinetic energy at the bottom of your descent will equal your potential energy at the top of the tower. Setting PE = KE and solving for v, you get v = sqrt(2gh).

Since we're neglecting drag, this will give you an upper bound, actual speed would be lower. If you want to take drag into account then you'll wind up with one messy equation with two unknowns that you would only be able to approximate using numerical methods. It would be reasonable however to neglect lift. Think of a ball rolling down a ramp. For a given height, the speed at the bottom of the ramp is the same regardless of the angle and shape of the ramp. Lift is just providing the shape of the ramp, so to speak.

Also, you wouldn't want to drop all the way to the surface of the water anyway. You want to design your glider to fly at some speed lower than this potential top speed so as to reserve some altitude to extend your glide. The lower the distance you are falling before establishing a glide, then the closer to reality this approximation becomes.

BTW, is this contest going for distance or time aloft? That difference is important because you will need to optimize different design parameters.

Z4K
06-30-2004, 07:00 AM
BBB_Hyperion:

x0, y0, x'0, y'0, x''0 and y''0 are all known.

CL and CD are determined independently. CL/CD is, for the airfoils I'm considering and practically the entire flight regime, essentially constant given a particular AoA - close enough to be a reasonable simplification anyway.

I am treating CL as essentially static and assuming that the pilot is good enough to maintain (within some margin) a constant AoA. However, that doesn't mean that the glide path is linear.

I can't assume that the glidepath is linear because this is accelerated flight.

diomedes33:

I think I want to marry you http://ubbxforums.ubi.com/infopop/emoticons/icon_wink.gif Thank you so much for those .m files - I'd never have stuck with Octave long enough to be able to write them. When this thing flies, your name will be on the list of indispensable contributors.
While you're right and I'd 'like' an analytical solution, my primary aim is to get a numerical answer which I can actually use. Which makes these perfect.

I think I'm having trouble with them though.

I changed a line in trapint.m from:

<pre class="ip-ubbcode-code-pre">...+ y(index -1) / 2 + c;</pre>

to

<pre class="ip-ubbcode-code-pre">...+ y(index -1)) / 2 + c;</pre> (inserted a parenthesis)

And set L = 0.12 and D = 0.01 as rough but representative values.

I then gave Octave the following:

<pre class="ip-ubbcode-code-pre">
&gt;&gt; t = linspace (0, 20, 1000);
&gt;&gt; v0 = [2; 0];
&gt;&gt; v = lsode ("vel", v0, t);
&gt;&gt; x = trapint (t, v(:,1), 0);
&gt;&gt; y = trapint (t, v(:,2), 7);
</pre>

(I experimented with a range of numbers of steps for t, and 1000 doesn't cause problems.)

Here's my output (with plot command at top left):

http://www.photodump.com/direct/Z4k/z_calc000_OctaveOut01.jpg

http://www.photodump.com/direct/Z4k/z_calc000_OctaveOut02.jpg

v against t
http://www.photodump.com/direct/Z4k/z_calc000_OctaveOut03.jpg

I'm sure the plot of y vs. x is incorrect - that linear portion isn't right; it shouldn't be rising that high above the starting point; and finally it reaches a stable, hovering altitude (magic!, if I can do that in the competition I might not go far enough but it'll be impressive).

I can't for the life of me figure out what's going on. I'm just not literate in Octave/Matlab, not even enough to really play around with your files to start trying to figure it out. If you've got time to either: find out what's going on; or tell me what I'm doing wrong it'd be fantastic.

Thank you again for your help - if I knew how that bowing smilie was done, I'd put it in http://ubbxforums.ubi.com/infopop/emoticons/icon_smile.gif

Jet Jaguar:

The contest tests distance. Time aloft has no bearing on the outcome.

I wish I were over-complicating it. My glide speed (once I've reached it - which will only happen AFTER I dive to accelerate) is easily calculated from CL (known, from tables), CL/CD (also known) and mass. See the following:

http://www.photodump.com/direct/Z4k/z_calc000_Diag04-JetJaguar.jpg

Incidentally, CL and CD are also affected by the proximity to the ground. I forgot to mention that in the above piccy, but it can pretty much be neglected in the accelerating portion of the flight anyway

I don't plan to drop all the way to the surface of the water (although I understand it looks like that in the diagrams). It's just I'm not having a problem with the gliding part, it's the first bit I can't do. In the glide phase (steady flight. ie. no acceleration), all I need is as high a CL:CD ratio as possible.

For the initial portion of the flight I CAN'T make any of those simplifying assumptions. For normal aircraft it can (usually) be done because in flight, the displacements associated with these effects are less than a fraction of a percent of the distance the aircraft travels that the accuracy I want is not required. Either that, or an approximation using a circular path or similar is close enough.
I can't do that, because this part of my flight is likely to approach 25% of the total displacement. As you can see below, if I want to choose a linear glide path, the possible range is so great as to make the exercise meaningless:

http://www.photodump.com/direct/Z4k/z_calc000_Diag05.jpg

I'm treating the flight as one made of two distinct parts. An accelerating phase, where I trade altitude for speed, and a glide phase where I maintain my velocity and descend only very slowly and at a constant rate.

Z4K
07-02-2004, 08:44 PM
diomedes33:

http://www.photodump.com/direct/Z4k/z_calc007_OerTheRamparts.jpg

Hooray. I found it.

I had to change the following line in trapint.m:

<pre class="ip-ubbcode-code-pre">I(index) = (x(index) - x(index - 1)) * (y(index) + y(index -1) / 2 + c;</pre>

to

<pre class="ip-ubbcode-code-pre">I(index) = ((x(index) - x(index - 1)) * (y(index) + y(index -1)) / 2) + I(index - 1);</pre>

Now, given the following input:

<pre class="ip-ubbcode-code-pre">

&gt;&gt; t = linspace (0, 20, 1000);
&gt;&gt; v0 = [3; 0];
&gt;&gt; v = lsode ("vel", v0, t);
&gt;&gt; x = trapint (t, v(:,1), 0);
&gt;&gt; y = trapint (t, v(:,2), 7);
&gt;&gt; sqv = ((v(:,1)).*(v(:,1)))+((v(:,2)).*(v(:,2)));
</pre>

I get:

plot (t, (sqrt(sqv)))
http://www.photodump.com/direct/Z4k/z_calc000_Diag08.jpg

and

plot (x, y)
http://www.photodump.com/direct/Z4k/z_calc000_Diag07.jpg

Which is much more reasonable. Of course, I hit the water about 1.5 seconds after jumping off the platform, which gets me 7 metres (less than 10% of the target distance), but now I've got a tool with which to design a decent craft.

Thank you!

- - - - - - - - - - - - - - -
Let's play Global Thermonuclear War

Z4K
07-10-2004, 01:35 AM
Just a bump to make sure diomedes33 sees my thanks.

http://www.photodump.com/direct/Z4k/z_calc009_DoItRight_MakeItBite.jpg

And the fun things I've been doing with his m files:
http://www.photodump.com/direct/Z4k/e344-e440.jpg

- - - - - - - - - - - - - - -
Let's play Global Thermonuclear War

Black Sheep
07-10-2004, 04:59 AM
I've forgotten pretty much everything I ever learned about calc at uni but those propaganda posters absolutely rock ! http://ubbxforums.ubi.com/images/smiley/11.gif

Hope to see some more sometime http://ubbxforums.ubi.com/images/smiley/16x16_smiley-wink.gif

Friendly_flyer
07-10-2004, 05:04 AM
Where do you find all those posters?