Tennis Ball Cannon Pressure & Exit Velocity

AI Thread Summary
The discussion centers on designing a tennis ball launcher powered by compressed air, with a focus on calculating the exit velocity of the ball based on chamber pressure and volume. A static launch angle of 45 degrees is proposed, but the relationship between pressure and exit velocity remains unclear. Participants suggest using the work-energy principle and basic physics equations to derive a simplified formula for exit velocity, while acknowledging the complexities introduced by factors like air resistance and ball spin. The conversation also touches on the potential use of a sabot to minimize spin effects. Overall, the need for a practical equation to guide the design process is emphasized, alongside the challenges of achieving accurate predictions in real-world conditions.
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I'm designing a tennis ball launcher using compressed air that will hopefully be able to hit a target at some distance away. We won't know the exact distance until the day of the competition, but I was curious if there was some relatively simple equation relating the pressure of the compressed air and the initial volume of it to the exit/muzzle velocity of the tennis ball as it exits the barrel.

Given the range, and I was thinking of just using a static launch angle of 45 deg., it would be fairly simple to calculate the required exit velocity but then relating that to what pressure I would need to fill the chamber up to is eluding me. Any help would be greatly appreciated. Thank you.
 
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Part of the issue will be any spin put on the ball as it's fired from the cannon. Most likely it will be top spin which will curve the ball downwards and decrease range, even with a higher than 45 degree launch angle. It might be possible to use a "sabot" which would would be a very light but strong and slick cannister or piston like device that holds the tennis ball to prevent spin, but gets left behind once the tennis ball is a few feet beyond the end of the cannon.

I'm not sure how consistent the range would be due to other variations.
 
Yes this is far from an idealized situation and I figured I would take these things along with drag and friction into account when I actually have a working model to test varying pressures/distances with but for the time being I'm just looking for an expression to analytically determine the exit/muzzle velocity. I need to put together a midproject report and I would like to have a distance vs pressure plot
. Thanks for the ideas though I will keep those in mind to improve my accuracy.
 
In the real world, actual test are done for artillery, with significant steps between angle and charge (pressure), and computers are used to fill in the "in-between" data via curve fitting between sample points.
 
Unfortunately I do not have a working model built yet, and won't for quite some time. Am I to believe there is not such equation in existence? I highly doubt that. I've seen a couple somewhat complicated differential equations that were used to arrive at the result I'm looking for but I can't say those are my strong suits. I was looking for something simpler even if it isn't 'prefect' or going to give me 100% accurate results in the real world. This I understand is impossible to achieve and is not what I was expecting.


Edit:
How terrible would it be to use something like

Work done by the expanding gas = 1/2*m*v^2 ?

Been a while since I took thermodynamics. Could I use something like -P(dV) for the work done in this case?
 
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I was trying to work out a similar thing a couple of years ago, but it was more to do with optimizing the tank volume/pressure for given muzzle velocity. But I got stuck as it involved finding the rate of change of pressure and stuff.

Simple way as you asked:

If you assume a constant pressure (large air tank, small barrel volume, good valve), then:

F = PA, where A is the area of the barrel will give you the force on the ball, excluding friction and air resistance of course.

so,
F = P * pi.r², where r is barrel radius.

a = (pi.r².P) / m, where m is tennis ball mass.

assume acceleration only for the length of the barrel

v² - u² = 2as, where s is barrel length.

but u = 0:

v = √(2.pi.r².P.s / m)

Might be interesting to plot v vs P vs barrel length as well?
 
Air resistance is going to be a significant factor if pressure along with acceleration and velocity of the ball in the cannon are high. Think of it this way, imagine a huge high pressure tank connnected to the cannon. Even without the tennis ball, there's still going to be a significant amount of pressure at the open end of the cannon. I'm not sure how to get the parameters you'd need for air, the cannon, ..., to be used to calculate the pressure loss in an open cannon (no tennis ball).

http://en.wikipedia.org/wiki/Darcy–Weisbach_equation

With the tennis ball, there will be an increase in pressure behind the ball, and a decrease in front of the ball, but the pressure in front of the ball will be above ambient, so I'm not sure of an easy way to calculate the net pressure differential on the ball.
 

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