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Txflier

Differential equation Needed!!

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Hey all, I'm doing a project for Dynamic systems of modeling. I'm trying to plot the distance a jumper will be away from the plane after exit at a given time using matlab or similar program.

Assumptions: Plane Velocity 100 mph
Altitude 13.5 K
gravity 32.2 (m/s^2)

I've come up with a few equations.
Y(t) = -.5 *g * t^2 +Vo*t + Yo

Y(t)=altitude wrt to time
g-gravity
t-time
Vo initial velocity
Yo-initial altitude

But this is just assuming the objects initial velocity is 0 which isn't true.. So delving a little deeper. using drag coefficient

m * dv/dt = .5 * rho *Cd * A * V^2 -m*g

rho- density of air at 13.5k
Cd- drag coefficient
A - surface area
V- Velocity
m-mass
g-gravity.

but my goal is not to find terminal velocity or velocity wrt to time. I want to find distance away from plane after x seconds.

If you got any ideas I'd appreciate it. I'm trying to get this project done so I can study for finals on Mon. :o:)
I'm not the rope totin charlie Bronson wanna be that's getting us fucking lost.

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Distance horizontally, vertically, or both? On a cloudy day last year at the All-Kansas boogie, we did a pumpkin drop competition. Afterwards (ya i was too late) I came up with a diff eq to figure out the horizontal release point, given a set altitude and speed, so that the forward throw would end up on the target. I'll try to dig it up when I get home.

Either way, I think I went the route of finding t first (which you say will be given) then using the diff eq to find distance traveled during t seconds based on velocity decreasing because of drag. If that helps at all. It was awhile ago [:/]

"Are you coming to the party?
Oh I'm coming, but I won't be there!"
Flying Hellfish #828
Dudist #52

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Since humans can move around in the air, you might want to change it to plot a 'thing' instead of a jumper?

And, gravity is not 32.2 m/s^2 :P
Either 9.81 m/s^2 or 32.2 feet/s^2

Or is this the 420 pound male project? B|



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oops I was changing my units around and missed that.

I'm not the rope totin charlie Bronson wanna be that's getting us fucking lost.

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Since humans can move around in the air, you might want to change it to plot a 'thing' instead of a jumper?

And, gravity is not 32.2 m/s^2 :P
Either 9.81 m/s^2 or 32.2 feet/s^2

Or is this the 420 pound male project? B|



Quote


oops I was changing my units around and missed that.



LMAO - why don't you calculate what terminal velocity would likely be if the figures you originally stated were right.

:D:D

not going there - but my guess would be somewhere around a 1500MPH:D:D
I'm not usually into the whole 3-way thing, but you got me a little excited with that. - Skymama
BTR #1 / OTB^5 Official #2 / Hellfish #408 / VSCR #108/Tortuga/Orfun

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I think you need to add a little complexity to capture at least some of the dynamics of this problem. Since I really should get back to work, I'll just give you my initial impressions.

First, set this up as a vector problem. It doesn't necessarily need to be 3-dimension. You can assume no lateral motion and limit motion to the xz-plane (horizontal and vertical motion). This will allow you to quantify the velocity as a vector as well. Initial horizontal velocity is 100 mph, initial vertical velocity is 0 mph. The direction of the drag vector is directly related to velocity. It's in the opposite direction.

Second, start from the beginning to derive your equations of motion. Begin with the jumper's position in the xz space. You can set the (0,0) point as the exit point. Next take the derivative to get the velocity equation, and take the derivative again to get the acceleration equation. Depending on how you define everything, as in if you have a local coordinate system for your jumper, you may need to use the Transport Theorem to to make the derivatives work right. Use F=ma to get your equation of motion.

Third, once you get these derived, there are several equations you need to input into Matlab. In the xz coordinate system, you would need the dx/dt, d(dx/dt)/dt, dz/dt, and d(dz/dt)/dt with all the forces and whatnot included. Put these in an m-file, set up a driver m-file, and use ode45 to solve the system of equations. It will output x, dx/dt, z, and dz/dt. This will tell you your final position in xz space, and you can do a quick calculation to determine where the plane is if it stayed at its original speed and direction. From there you can get the final distance they are from each other.

Hope this helps. It seems like a fun problem. If I had more time (i.e., not at work), I'd derive it and give you some more concrete pointers.

**************
For once you have tasted flight you will walk the earth with your eyes turned skywards, for there you have been and there you will long to return.
~Leonardo da Vinci~

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Is the jumper head down or on his/her belly?



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exit on belly. Not really going to need to factor that in though. I'm just going to take an average drag coefficient.

I'm not the rope totin charlie Bronson wanna be that's getting us fucking lost.

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I came up with a diff eq to figure out the horizontal release point, given a set altitude and speed, so that the forward throw would end up on the target. I'll try to dig it up when I get home.



It was pretty close but good luck finding the equation. I think you did it on a napkin and I used it to wipe ketchup off my lip while we were watching the toss out by the gas tank and eating burgers.[:/]
It's called the Hillbilly Hop N Pop dude.
If you're gonna be stupid, you better be tough.
That's fucked up. Watermelons do not grow on trees! ~Skymama

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I think you need to add a little complexity to capture at least some of the dynamics of this problem. Since I really should get back to work, I'll just give you my initial impressions.

First, set this up as a vector problem. It doesn't necessarily need to be 3-dimension. You can assume no lateral motion and limit motion to the xz-plane (horizontal and vertical motion). This will allow you to quantify the velocity as a vector as well. Initial horizontal velocity is 100 mph, initial vertical velocity is 0 mph. The direction of the drag vector is directly related to velocity. It's in the opposite direction.

Second, start from the beginning to derive your equations of motion. Begin with the jumper's position in the xz space. You can set the (0,0) point as the exit point. Next take the derivative to get the velocity equation, and take the derivative again to get the acceleration equation. Depending on how you define everything, as in if you have a local coordinate system for your jumper, you may need to use the Transport Theorem to to make the derivatives work right. Use F=ma to get your equation of motion.

Third, once you get these derived, there are several equations you need to input into Matlab. In the xz coordinate system, you would need the dx/dt, d(dx/dt)/dt, dz/dt, and d(dz/dt)/dt with all the forces and whatnot included. Put these in an m-file, set up a driver m-file, and use ode45 to solve the system of equations. It will output x, dx/dt, z, and dz/dt. This will tell you your final position in xz space, and you can do a quick calculation to determine where the plane is if it stayed at its original speed and direction. From there you can get the final distance they are from each other.

Hope this helps. It seems like a fun problem. If I had more time (i.e., not at work), I'd derive it and give you some more concrete pointers.



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This is exactly what I'm looking to do.:)
main
n=2;
x0=zeros(n,1);
Vx(0)=85;
Vz(0)=9.81;

Tinit=0;
Tfinal=55;
Tspan=[Tinit,Tfinal];
[t,x]=ode45(@Johns9_25_equations,Tspan,V0);
plot(t,V(:,1),t,V(:,2))

grid
xlabel('time(s)')
ylabel('Response')


M function file
function Dx =Johns9_25_equations(t,x)
Dx=zeros(2,1);

rho=0.82; %(kg/m^3)
Cd=0.668402;
A= 0.9968; %(m^2)
Vxp= 44.7; %(m/s)
g= 9.81; %(m/s^2)

Vx(1)= Cd*rho*A*Vxp^2/(M*2);
Vz(2)= g;

I'm not the rope totin charlie Bronson wanna be that's getting us fucking lost.

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That one was rough, I did another one that next week that looked nice and pretty and showed it to martin. We decided it was pretty damn close to what we saw. Time to start looking for next year!
"Are you coming to the party?
Oh I'm coming, but I won't be there!"
Flying Hellfish #828
Dudist #52

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You are getting close. You do have to set it up as second order. You absolutely must have the acceleration equations. That is how your forces are going to get factored in. As you have it now, you have forces in your velocity equations. Not where they are supposed to be. Also, your initial conditions are a mix of a velocity (85) and an acceleration (9.81), and it looks like these should both be velocities.

The general framework of your m-files looks ok. I'll try to work through the equations real quick. I'm computer-less at the moment at home, so I'll see what I can come up with in the next few minutes before leaving work.

**************
For once you have tasted flight you will walk the earth with your eyes turned skywards, for there you have been and there you will long to return.
~Leonardo da Vinci~

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Hey all, I'm doing a project for Dynamic systems of modeling. I'm trying to plot the distance a jumper will be away from the plane after exit at a given time using matlab or similar program.

Assumptions: Plane Velocity 100 mph
Altitude 13.5 K
gravity 32.2 (m/s^2)

I've come up with a few equations.
Y(t) = -.5 *g * t^2 +Vo*t + Yo

Y(t)=altitude wrt to time
g-gravity
t-time
Vo initial velocity
Yo-initial altitude

But this is just assuming the objects initial velocity is 0 which isn't true.. So delving a little deeper. using drag coefficient

m * dv/dt = .5 * rho *Cd * A * V^2 -m*g

rho- density of air at 13.5k
Cd- drag coefficient
A - surface area
V- Velocity
m-mass
g-gravity.

but my goal is not to find terminal velocity or velocity wrt to time. I want to find distance away from plane after x seconds.

If you got any ideas I'd appreciate it. I'm trying to get this project done so I can study for finals on Mon. :o:)



It's actually quite tricky problem because the v^2 in the drag term results in the x and y distances being coupled instead of independent.

You can check your results against mine:

http://mypages.iit.edu/~kallend/skydive
...

The only sure way to survive a canopy collision is not to have one.

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Alrighty, try this. It's assuming all sorts of things about the drag coefficient, but we'll just assume it's constant all the way through. I doubt your professor is going to care about such vaguaries. Anyway, here's the derivation:

p=xi+zk (p is your position vector, i and k are unit vectors)

dp/dt=dx/dti+dz/dtk (velocity vector)

d(dp/dt)dt=d(dx/dt)/dti+d(dz/dt)/dtk (acceleration vector)

Now let's look at Newton's 2nd Law, F=ma

a is your acceleration equation.

The forces you have are drag and weight. Drag has two components. One in the x-direction (Dx), one in the z-direction (Dz). Let's set the coordinate system at the point the jumper leaves the plane with the x-axis pointed out the nose of the aircraft and the z-axis pointed down. This means that the Dx is in the negative x-direction and Dz is in the negative z-direction. Consequently, weight (m*g) is in the positive z-direction.

So your forces are F=-Dxi+(mg-Dz)k.

Solving for a, you get a=F/m

So your two equations of motion are

d(dx/dt)/dt=-Dx/m
d(dz/dt)/dt=(mg-Dz)/m

To set this up for ode45 to solve in Matlab, you need these two equations and velocity equations. So let's use y to set up all of these equations. The output vectory will be [x, dx/dt, z, dz/dt]. So in the file called by ode45 set up ydot:

ydot(1)=y(2);
ydot(2)=-Dx/m; (where Dx is dependent on y(2), the x-velocity, as well)
ydot(3)=y(4);
ydot(4)=(mg-Dz)/m; (where Dz is dependent on y(4), the z-velocity)

With this setup, the first line of the file would be

function ydot=Johns9_25_equations(t,y)

Then set up your variable, add the equations, and let it run.

This should work pretty well. There maybe a couple typos that you can troubleshoot, but it gives you a place to start. You can make it much more complex by playing with the drag coefficient more, but that's a little overkill.

Have fun!

**************
For once you have tasted flight you will walk the earth with your eyes turned skywards, for there you have been and there you will long to return.
~Leonardo da Vinci~

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Oh, and your initial vector, y0, in your main file is [0 85 (convert to proper units) 0 0]

**************
For once you have tasted flight you will walk the earth with your eyes turned skywards, for there you have been and there you will long to return.
~Leonardo da Vinci~

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Hey all, I'm doing a project for Dynamic systems of modeling. I'm trying to plot the distance a jumper will be away from the plane after exit at a given time using matlab or similar program.

Assumptions: Plane Velocity 100 mph
Altitude 13.5 K
gravity 32.2 (m/s^2)

I've come up with a few equations.
Y(t) = -.5 *g * t^2 +Vo*t + Yo

Y(t)=altitude wrt to time
g-gravity
t-time
Vo initial velocity
Yo-initial altitude

But this is just assuming the objects initial velocity is 0 which isn't true.. So delving a little deeper. using drag coefficient

m * dv/dt = .5 * rho *Cd * A * V^2 -m*g

rho- density of air at 13.5k
Cd- drag coefficient
A - surface area
V- Velocity
m-mass
g-gravity.

but my goal is not to find terminal velocity or velocity wrt to time. I want to find distance away from plane after x seconds.

If you got any ideas I'd appreciate it. I'm trying to get this project done so I can study for finals on Mon. :o:)



It's actually quite tricky problem because the v^2 in the drag term results in the x and y distances being coupled instead of independent.

You can check your results against mine:

http://mypages.iit.edu/~kallend/skydive


Thanks I was looking at your website. I'll check it out again when I get this matlab rolling
I'm not the rope totin charlie Bronson wanna be that's getting us fucking lost.

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Oh, and your initial vector, y0, in your main file is [0 85 (convert to proper units) 0 0]



Thanks for the help I'll give it another go.. I was going to ignore the drag in the Z direction though. since the time I'm looking for is in the first couple of seconds. Though maybe I'll factor that in just to be able to run a whole simulation for actual jump time.
I'm not the rope totin charlie Bronson wanna be that's getting us fucking lost.

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You are getting close. You do have to set it up as second order. You absolutely must have the acceleration equations. That is how your forces are going to get factored in. As you have it now, you have forces in your velocity equations. Not where they are supposed to be. Also, your initial conditions are a mix of a velocity (85) and an acceleration (9.81), and it looks like these should both be velocities.

Quote


I think your right.. I think my initial condition in the Z direction would start at 0 and in the x direction should be the speed of the plane.

I'm not the rope totin charlie Bronson wanna be that's getting us fucking lost.

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You need to solve two second-order coupled differential equations like the ones in the attachment (solved using MathCad). The graph shows vertical position (in metres) relative to the aircraft, red trace is for a plane and the blue is for a balloon (no horizontal velocity at exit).

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