3DOF Reentry Trajectory Equations of motion

Click For Summary

Discussion Overview

The discussion revolves around the equations of motion for a three-degree-of-freedom (3DOF) atmospheric reentry trajectory. Participants are exploring the derivation and numerical integration of these equations, addressing issues related to the accuracy of the trajectory predictions and the relationships between various parameters such as altitude, range, and velocity.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant is deriving equations for atmospheric reentry and numerically integrating them, questioning the correctness of their results based on observed small values for angular displacement.
  • Another participant notes that the graphs indicate a constant gradient in altitude versus range, suggesting a potential issue with the range equation.
  • A participant expresses concern about the linear relationship observed in the altitude to range plot, indicating a possible error in the range equation.
  • There is a suggestion that the equations for velocity and angle of attack may need adjustment, though the effectiveness of these changes remains uncertain.
  • One participant mentions deriving the equations independently and then cross-referencing with external sources, seeking examples with data for comparison.
  • Another participant proposes a new approach to calculating the range based on the velocity vector and angle, reporting improved results but still observing a linear-like plot.
  • There is a request for assistance in deriving and numerically integrating the equations of motion for a reentry vehicle under specific conditions, indicating a need for MATLAB coding help.
  • One participant reflects on their initial post, clarifying that their equations can indeed be used for the 2DOF equations of motion, suggesting a structured approach to coding the problem.

Areas of Agreement / Disagreement

Participants express various concerns and suggestions regarding the equations and their integration, but no consensus is reached on the correctness of the equations or the best approach to resolve the issues presented. Multiple competing views and uncertainties remain throughout the discussion.

Contextual Notes

Participants highlight potential limitations in their equations and numerical methods, including the dependence on specific assumptions and the need for further validation against known data. Some equations are noted to be derived from personal understanding, while others reference external sources, indicating variability in foundational approaches.

Who May Find This Useful

This discussion may be useful for individuals interested in atmospheric reentry dynamics, numerical methods for solving differential equations, and those seeking to understand the complexities of modeling trajectories in aerospace applications.

roldy
Messages
206
Reaction score
2
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 m2
g = 9.815(R/(R + alt))2 m/s2
ρ = 1.752e-alt/6700 kg/m3
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]
 

Attachments

  • Untitled-1.jpg
    Untitled-1.jpg
    14.8 KB · Views: 829
  • alt vs range.jpg
    alt vs range.jpg
    13.2 KB · Views: 779
  • alt vs vel.jpg
    alt vs vel.jpg
    11.2 KB · Views: 536
Physics news on Phys.org
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 spacecraft s 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 n-t coordinate system). Does this sounds like a realistic scenario?

From your velocity graph it looks like your spacecraft can land safely after all!

Have a good think through it and I'm sure you can solve it. :)
 
Last edited:
Reentry program

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 g-force 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.
 

Attachments

Last edited:
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 :)
 
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.
 

Attachments

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 linear-like. 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.
 

Attachments

  • Untitled-1.jpg
    Untitled-1.jpg
    14.5 KB · Views: 662
  • range.jpg
    range.jpg
    11.7 KB · Views: 677
Still not entirely sure about my correction since every document or website I come across has V*[itex]\cos \theta[/itex]/(R + alt).
 
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
 
Rd123, is this a question in regards to helping solve my problem or is this something that you are halving problems with?
 
  • #10
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
Upon re-reading 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 Runge-Kutta Method. This should replace the equation f = @(theta) -2.2067e-12*(theta(1,1)^4-81e8). 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 Runge-Kutta Method for heat loss and the pdf describing the problem.
 

Attachments

  • #12
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...
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 76 ·
3
Replies
76
Views
7K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 1 ·
Replies
1
Views
3K