Laplace Transforms
 

Can anyone help me by telling me how to get the Laplace transform of (dy/dx)^2 ? I'm trying to solve a take off equation but don't really know how to do it unless it's power of one. It's been ages since i've had to use these gastly things.

Cheers,

Ginger ;)

If I'm not mistaken the rule is:

L(f^(n))=(s^n)*L(f)-(s^(n-1))*f(0)-(s^(n-2))*f'(0)-- ... -- (f^(n-1)(0))

Laplace transform of the derivative of any order n

Barry

Edit: Syntax error

Yes, but the question is to find the laplace transform of (dy/dx)^2, not (d^2y/Dx^2)....

I thoroughly recommend Etkins 'Dynamics of Atmospheric Flight' - excellent for Laplace transforms (amongst other things)

Apologies,helps to read the problem properly.

Had a look in my maths textbooks, by Kreyzig and James, but could not find anything in either. My final year calculus formulas sheet had nothing similar either.

Do you have matlab/simulink would be easy enough to solve using it.

If you do find a way to solve it, please post your method I would like to see how its done.

No Barry, I suspect you might have interpreted it correctly. I just got a bit confused. Probably still confused!

Thanks for the replies chaps, i'm glad to see it's not straight forward and I havne't just been a dunce! I think i've found a way around the L transform bit of it, but it's still proving difficult. Basically i'm trying to find the take off roll for a propellor powered UAV, but with only friction, lift and drag coefficients and mass and all the other constants. So I end up with an equation with acceleration as an unknown, velocity as an unknown, and some constants. Some manipulation made me think I had a solution, but it turned out to include complex numbers :yuk: .

I have tried MatLab with a simple ordinary differential equation, but the chart of velocity against time it returned went up to 90m/s, which can't happen because the prop exhaust velocity is only 32m/s!

When I find a way of writing the formulas down so that I can post them, I will.

Cheers,

Ginger ;)

P.S. India Mike, you're correct, I was looking for the square of the first differential, not the formula for the second differential. Wow! Am I really talking about this at this time on a Saturday morning? Goodnight! :zzz:

I've got the same book but it doesn't seem to help in this case, despite being a very good book. It does actually have a modelling example for the take off run, but it's a bit too simplistic for what I want.

Simplifying Differential Equations
 

I don't know whether this helps, but the square of the first differential reminds me of v-squared, or (ds/dt)**2. In which case it can help if we keep it as v-squared, v**2, and write the second differential, d2s/dt**2 (which is dv/dt), as v.dv/ds.

With the v-squareds that crop up in many aero problems, this often rescues the situation by giving you a linear first-order differential equation in v and s rather than a non-linear one in v and t (or even a second order non-linear one in s and t).

Any help?

That is a help D120, thanks. I had already got there through the manipulation I mentioned above, which is how I got rid of the need for Laplace. I still ended up with a complex number in the mix though, which was most unwelcome.

Methinks that using Raymer I have found the solution:

Ground roll = (1/2*g*Ka)ln((Kt + KaVf^2)/(Kt + KaVi^2))

Where:

Kt = (T/W) - u

Ka = [rho/2(W/S)]*(uCL - CDo - KCL^2)

The formula I was using originally was from Raymer as well, but was looking for acceleration rather than the ground roll distance. I think this might be a case of 'keep it simple, stupid' :} .

Thanks for the help - it's nice to have such a wealth of expertise to tap into.

Cheers,

Ginger ;)