Need help with D.E to account for air drag of my projectile

  • Thread starter guipenguin
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  • #1

Main Question or Discussion Point

This past weekend, for fun, I made a pneumatic air cannon, using compressed air to shoot out a small projectile at very high velocity. I'd post my video, but I have not made 15 posts on this forum.

What I need to do is find as close as I can to the actual initial
velocity of a projectile, once it leaves the barrel. If I have the air
cannon pointed at 90 degrees to the ground, than I could shoot the
projectile vertical and time hang time, factoring in known
acceleration of gravity, to find an approx initial velocity. The only
problem is, with a projectile going this fast, with significant
surface area, there will be significant air drag to account for.

As air resistance effects the velocity of the projectile, and as
decreased velocity coincides with a decrease in drag, I believe I will
be needing to use differential equations.

Do you have any idea how I could setup a differential equation to
account for air resistance in my quest to find as close as I can to
the projectile's initial velocity? I have never done anything with air drag before. I just finished Calculus II, but I can put in the nessesary time and effort if someone can get me started here.

Thanks,
John
 

Answers and Replies

  • #2
HallsofIvy
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How are you modeling the air resistance? Two commonly used ways are "linear";F= -kv, where F is the friction force due to air resistance, v is the velocity vector, and k is a constant, or "quadratic"; F= -k |v|v, where |v| is the 'length' of the velocity vector or speed. This is "quadratic" because |F|= -k|v|2.

The first case is fairly simple: the x coordinate of velocity is given by mvx'= -kvx and the y coordinate of velocity is given by mvy'= -kvy- (1/2)mg. Unfortunately that is only accurate for very low speeds.

The quadratic case is harder: the x coordinate of velocity is given by

[itex]mv_x'= -k\sqrt{v_x^2+ v_y^2}v_x[/itex], [itex]mv_y'= -k\sqrt{v_x^2+ v_y^2}v_y- (1/2)mg[/itex].
 

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