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Hello, I was trying to derive the equation of velocity for a rocket subject to gravity and quadratic air resistance, but I was stuck and I realized it might be really hard to solve the differential equation so I just did linear velocity, which is bit unrealistic for a rocket. I used a linear differential equation to solve it and was wondering if anyone could check to see if my calculations are correct because I can't find anything on google.
Also to define the variables, -k=dm/dt, m=m0-kt, b=a constant for the drag force, f(v)=-bv, Vex is exhaust velocity. I'd like to make sure my calculations are correct for linear resistance, but does anyone know how to solve for quadratic drag? Thanks.
>>7928220
>equation of velocity for a rocket subject to gravity and quadratic air resistance
http://www.phys.ufl.edu/~thorn/homepage/cmuglectures.pdf
p16
>>7928234
>http://www.phys.ufl.edu/~thorn/homepage/cmuglectures.pdf
A rocket is different than a projectile because the mass in a rocket decreases as the fuel burns.
I am asking,
1. Are my current calculations for a rocket subject to linear drag and gravity correct?
2. How do you solve the differential equation of motion, mdv/dt=-dm/dtVex -cv^2 -mg?
Does this have to be done by hand? Rocket flight is meant to be solved numerically, I've never tried to predict a rocket's flight any other way because it's usually just not solvable without making a lot of assumptions (linear velocity included) that make it hopelessly inaccurate
Anyways I'm not really sure how to follow your work and I don't really want to try
You're over-complicating your solution procedure for the assumptions you're making.
Just write a program that solves for time, distance, velocity, acceleration, air density, thrust, aerodynamic drag, and mass on an incremental basis.
I wrote a similar program for ground vehicle kinematic simulations. Shouldn't take you more than an hour or so.
Check limit cases. That should work. In mechanics (and some other areas of physics, but mainly mechanics) you often have the advantage that you can use your intuition to check your calculations. This is obvious especially in the limit cases (for example, what happens when all the fuel has been burned? Does it make sense?).
I don't think you can solve analytically the differential equation if you assume quadratic air resistance.
>>7928265
That's a nonlinear differential equation, mang.
If you find a way to analytically solve it, lemme know.
>>7928435
Another issue, your first line should really be
[math]
m \frac{dv}{dt} = -\frac{dm}{dt}v_ex - mg - cv^2
[/math]
>>7928445
If you want to get something analytical out of this, solve the equation for a constant velocity.
That means
[math]
\frac{dv}{dt} = 0
[/math]
so
[math]
v_{ex}\frac{dm}{dt} = -mg - cv^2
[/math]
so we can see this is already significantly easier; the second term is a constant, as is v_{ex}.
>>7928461
Yeah, pretty much.
>>7928461
If you really want to solve it by hand try linearizing it using a small perturbation term and keeping only the first order terms. You know rockets go up, the interesting dynamics happen usually linearly anyway
>>7928220
Try an exponential approach and check it's derivative to find missing factors
But I can tell you right now that will be one hell of an ugly equation
>>7928484
Right because there's only one way to solve a non-analytic equation...
I found an error anyways. When finding the integrating factor I did definite integral on [t,0] instead and I came out with a different final result.
>>7928403
>>7928424
came here to say this, if you are actually trying to build a rocket only an idiot would try working it out by hand. Before computers they read data off of experimental charts. You simply cannot accurately describe rocket flight with any model simple enough to have an analytical solution.
>>7928220
Is this why rocket science is hard?
>>7928220
iight homie hows this?
>>7932890
Ok lil bruh lemme explain:
m(t)=mass at time
m'(t)=rate of mass lost/gained
anything else u need help wit?
You can easily account for more forces wit dis format, da only reason u need a diff eq is cuz you wanted to use a drag function, which depends on da velocity.
>>7932911
Also you wanna treat the v_ex as da speed of average of all mass you throw off, including unspent fuel / stages / etc if you wanna be real accurate otherwise it will give you an unrealistically high acceleration for your engine
(higher m'(t)/mass loss means da velocity gotta be lower for the same effect)
