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Physics/pumpkin chunking



I was going over a previous answer about pumpkin chunking and I seem to have a similar issue.  I have an 8lb [3.62kg] pumpkin to be shot from an 8" [0.2m] diameter barrel that is 8' [2.43m] long. My question is how much psi would it take to launch said pumpkin 25' [7.62m]?

Hello jon,

I recommend the barrel be at an angle of 45 degrees with horizontal. See
We can use the formula
d = V^2*sin(2*theta)/g
where d = 25 ft, theta = 45 degrees, and g = 32 ft/s^2. This tells us that we need a velocity of 28.28 ft/s coming out of the barrel.

To find the acceleration required to give your pumpkin that speed with a barrel length of 8 feet, we can use the kinematic fromula
Vf^2 = Vi^2 + 2*a*b
where Vf = 28.28 ft/s, Vi = 0, and b = 8 ft. This tells us that we need acceleration of 50 ft/s^2.

Newton's 2nd Law can tell us the net force required to provide this much acceleration.
F = m*a
where m is the mass of the pumpkin which is 8 lb/(32 ft/s^2) and a = 50 ft/s^2. This tells us that we need 12.5 lbs of net force. The pumpkin has 2 forces on it: its weight and the force due to the pressure you provide behind it. The force due to the pressure, I'll call it Fp, is the pressure*the pumpkin's cross sectional area. I'll assume the pumpkin's diameter is 7 inches, therefore the radius is 3.5 inches. We need that in feet because that's the length unit we're using. 3.5 inches = 0.29 feet. So the pumpkin's cross sectional area (pi*r^2) is 0.26 ft^2.

The net force is the vector sum of those in the direction of the axis of the barrel.
Fnet = 12.5 lbs = pressure*0.26 ft^2 - 8 lbs*sin45
Therefore Fp = pressure*0.26 ft^2 = 12.5 lbs + 8 lbs*sin45
pressure*0.26 ft^2 = Fp = 18.2 lbs
And therefore pressure must be 18.2 lbs / 0.26 ft^2 = 70 lbs/ft^2
In the more familiar unit psi, that's only 0.49 psi. That seemed a surprisingly low pressure, so I doubted myself and did some cross checking. The pumpkin's area in square inches is pi*(3.5in)^2=38.5 square inches. Multiply by pressure of 0.49 psi, and that's 18.8 lbs. I believe applying that much force over 8 feet will give it considerable speed. So I agree with what I did above.

Note that the pressure needs to be maintained as the pumpkin moves through the barrel.

I hope this helps,


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Steve Johnson


I would be delighted to help with questions up through the first year of college Physics. Particularly Electricity, Electronics and Newtonian Mechanics (motion, acceleration etc.). I decline questions on relativity and Atomic Physics. I also could discuss the Space Shuttle and space flight in general.


I have a BS in Physics and an MS in Electrical Engineering. I am retired now. My professional career was in Electrical Engineering with considerable time spent working with accelerometers, gyroscopes and flight dynamics (Physics related topics) while working on the Space Shuttle. I gave formal classroom lessons to technical co-workers periodically over a several year period.

BS Physics, North Dakota State University
MS Electrical Engineering, North Dakota State University

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