Arrow Flight

  • Thread starter tomsiv
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I'm working on my engineering capstone design and part of our design requires a code we have written to calculate the appropriate zenith angle for a nerf gun to fire a projectile so that the dart hits a certain target.

We are feeding the code a target coordinate (x,y,z). One section of our code tracks the azimuth angle simply using the x and y coordinates(arctan(y/x)), then the zenith angle will be based on the projected range (r = [itex]\sqrt{x^{2}+ y^{2}}[/itex]) and the z value. The basic equations we are using are

Δr = vtotal*cos(θ)*Δt + .5*d*cos(θ)*Δt[itex]^{2}[/itex]

Δz = vtotal*sin(θ)*Δt + .5*(d*sin(θ) + g)*Δt[itex]^{2}[/itex]

where
Δr = change in position along the projected range
Δz = change in position along the z axis (elevation)
vtotal = total velocity (at muzzle)
Δt = time after firing
d = drag = -k*v2 = (we are going to assume this as a constant even though it depends on velocity)

Now since we are feeding our Matlab/simulink code a desired (x,y,z) value, we will have a desired Δr and Δz (or if you want to think in 2 dimensions a desired X and Y). So what we are trying to do is have our program solve these two equations simultaneously so that we can have a necessary θ value (zenith angle) to hit our target. Two equations, two unknowns (θ and Δt). I'm just running into problems when trying to set these up as a matrix to solve.

I feel like we may be over complicating this, because I've done similar calculations for rail gun targeting, but we had range, etc. as the unknowns and we were selecting theta values. Am I leaving something out or missing an obvious matrix solving method?

Thank you for any help you can offer, sorry if my wording is somewhat confusing.
 

Answers and Replies

  • #2
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What do you mean by "set these up as a matrix to solve"? Your equations are non-linear.
 
  • #3
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Yea, sorry, I realized that I had said that a little after posting my problem. I'm just trying to find the best way to solve for theta, using a method that can coded relatively easily (as far as computational time is concerned).
 
  • #4
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Multiple the second equation with ## \cos \theta ## and the first with ## \sin \theta ##; subtract the first from the second. You end up with ## \Delta z \cos \theta - \Delta r \sin \theta = 0.5 g \cos \theta \Delta t^2 ##. Divide this by ## \cos \theta ## and you obtain $$

\tan \theta = \frac {\Delta z - 0.5 g \Delta t^2} {\Delta r }

$$ bserve that $$ \cos \theta = \frac 1 { \sqrt{1 + \tan^2 \theta} }$$ so the first equation becomes $$

\sqrt {\Delta r ^2 + \left(\Delta z - 0.5 g \Delta t^2\right)^2}
= v \Delta t + 0.5 d \Delta t^2 $$ which is equivalent to an algebraic equation of the fourth degree.
 

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