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3DOF Reentry Trajectory Equations of motion 
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#1
Apr1013, 12:13 AM

P: 207

I'm working on a project for myself in regards to atmospheric reentry. I've come across some equations that describe the reentry trajectory. I decided to derive the equations using the diagram shown in the attached picture. Are these correct? The reason why I'm asking is that I'm getting small values for [itex]\phi[/itex] which is the angular displacement (range). I'm numerically integrating these equations. I've included plots from my program for altitude vs range and altitude vs velocity.
For this test I'm using the following initial values: α = 35° S = 12.97 m^{2} g = 9.815(R/(R + alt))^{2} m/s^{2} ρ = 1.752e^{alt/6700} kg/m^{3} m = 3855.54 kg θ = 0° V = 750 m/s alt = 120000 m [itex]\phi[/itex] = 0° time step size dt = 0.1 [itex]C_L = 2\sin^2\alpha \cos \alpha[/itex] [itex]C_D = 2\sin^3\alpha[/itex] [itex]L = 1/2C_d \rho V^2 S[/itex] [itex]D = 1/2C_d \rho V^2 S[/itex] [itex]\dot{V} = D/m  g\sin\theta[/itex] [itex]\dot{\phi} = V\cos\theta/(R + alt)[/itex] [itex]\dot{\theta} = (L/m  g\cos\theta)/V  \dot{\phi}[/itex] [itex]\dot{alt} = V\cos\theta[/itex] 


#2
Apr1013, 05:57 AM

P: 4

Hi roldy, good on you for working on your atmospheric reentry project just for fun ;)
It is difficult to assess whether or not your results are 'correct' without seeing how your code fits all the equations together. Just as an outside observer to your results, your graphs are saying that the spacecrafts descends with a constant gradient. So in other words, for any change in velocity in the normal direction (altitude), there is a correspondingly proportionate change in the tangential direction (range). This would mean that the resultant force vector (from adding lift, drag, gravity etc.) is always at a particular angle to the spacecraft (if using an nt coordinate system). Does this sounds like a realistic scenario? From your velocity graph it looks like your space craft can land safely after all! Have a good think through it and I'm sure you can solve it. :) 


#3
Apr1013, 02:38 PM

P: 207

The linear relationship of altitude to range leads me to believe that something is wrong with the range equation since the other plots of altitude vs velocity and altitude vs time look accurate. I've attached a zip file of the code I've developed thus far.
I've commented out some stuff intended to be used for actually finding the limits for the reentry corridor since I'm trying to figure out this range problem. Error while loop ends if the pilot penalty function (some function that adds up the total gforce loading on the pilot) is >= 1. I have an altitude penalty function setup for later if the capsule skips out which will be developed more later. ReentryCorr.m is the main program. I've been trying for a couple days now on this problem. Note: I believe that [itex]\phi[/itex] should be [itex]\phi = V\sin \theta /(R + alt)[/itex] and if I delete [itex]\phi[/itex] from the [itex]\dot{\theta}[/itex] equation I don't get a linear plot. However the order of magnitude for [itex]\phi[/itex] is 10^{4}. 


#4
Apr1113, 05:26 AM

P: 4

3DOF Reentry Trajectory Equations of motion
Thanks for posting your code.
Did you derive the system equations or are they from someone's text? Perhaps your trig ratios for Vdot and Theta dot should be: [itex]\dot{V} = D/m  g\cos\theta[/itex] [itex]\dot{\theta} = (L/m  g\sin\theta)/V  \dot{\phi}[/itex] ? I've briefly tested your code with the above change but am not convinced it gives any better results. Let me know how you go. It sounds like your model has some interesting things to come :) 


#5
Apr1113, 02:34 PM

P: 207

I derived them myself at first to fully understand how they came to be. I then checked in other sources. One of them is from the attached document on page 11. I've been trying to search for an example that has data with it so I can compare my plots and values.



#6
Apr1213, 04:56 PM

P: 207

I think I might of figure it out. I decided to move theta on the other side of the velocity vector and change the equations around to reflect this change. For the range I set up a ratio to find the angle. Vcosθ gives me the tangential distance. The ratio I set up is as follows.
[itex] \frac{2\pi(R + alt)}{360} = \frac{V\cos \theta}{\phi}[/itex] And I solve for [itex] \phi[/itex]. I now get a correct order of magnitude. However, the plot is still linearlike. I should note that the plot is not entirely linear; I went into photoshop and dropped a line from endpoint to endpoint and the plot sagged below the line. I've attached the new trajectory diagram and a plot of the range. 


#7
Apr1413, 02:51 PM

P: 207

Still not entirely sure about my correction since every document or website I come across has V*[itex]\cos \theta[/itex]/(R + alt).



#8
Apr814, 10:49 AM

P: 3

roldy can u please solve the 2 DOF equation of motion for reentry vehicle and numercally integrate it by euler integration or runge kutta 4 ODE



#9
Apr814, 02:45 PM

P: 207

Rd123, is this a question in regards to helping solve my problem or is this something that you are halving problems with?



#10
Apr914, 07:34 AM

P: 3

ya roldy i am little bit confused ,,,the first thing is that
1) derive the 2 DOF equations of motions for an reentry vehicle? 2)Numerically integrate the above equations of motions to obtain the trajectory of a reentry vehicle having following conditions:mass=3000kg flight path angle=2 deg velocity=7908 m/s phi=0 deg radius (r) = 100 km above the surface of earth surface area (s)=6 m^2 Cd=1.1 L/D= any choice so basically i have to numerically integrate that by euler or runge kutta 4 order ODE method ,,,, so i am stuck and don't know how to write that matlab code,,,please "roldy" help me out to solve it ..... Or if you could solve it .....it would be great helpful to me ,,,,,, 


#11
Apr914, 04:26 PM

P: 207

Upon rereading my original post and the equations, the equations presented can be used for the 2 DOF equations. In my case I just didn't use the 3rd dimension to describe translational movement.
I would approach coding this problem as follows: 1. Define reentry parameters 2. Define differential equation to be used in RungeKutta Method. This should replace the equation f = @(theta) 2.2067e12*(theta(1,1)^481e8). Do some research on this and try to see which differential equation you should use. My initial guess would be a differential equation in involving velocity. 3. You will need to change the FOR loop to a WHILE loop because you do not know the end time. You could do something like WHILE altitude > 0 ... end I attached some code on the RungeKutta Method for heat loss and the pdf describing the problem. 


#12
Apr914, 08:41 PM

P: 3

roldy ,,i hope it surely will help me ,,but in the mean time ,please can you post any link for the matlab coding of the trajectory equations,, needed to find various parameters....



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