In my previous post on launching a pumpkin (punkin chunkin) I essentially just looked at what happens to the pumpkin after it is launched. How fast would you have to shoot it to go 1 mile? The answer seems to be around 1000 mph and they are currently shooting them around 600 mph.

The question for this post, how fast can you launch a pumpkin so that it does not smash into smithereens? First, why would it smash at all? Here is a diagram of a pumpkin being launched while still in the tube.

The pumpkin launcher works by releasing compressed air inside the tube. This means that the force from the pressurized side is greater than the force from the atmospheric pressure. The net resulting force makes the pumpkin accelerate. Well, why would it crush? Simply because the force that causes the thing to accelerate is not applied to all parts of the pumpkin. This is different that a gravitational force on the pumpkin, which pulls on all parts of the pumpkin. So, since the pumpkin is accelerating from one side, the “compression force” will be the same as the net force. The net force will be mass*acceleration. Since the mass does not change, I can just look at the maximum acceleration a pumpkin can have without having structural integrity problems (breaking up)

### Maximum acceleration of a pumpkin

There is an example of breaking a pumpkin because it accelerated. This is what would happen if you drop a pumpkin on the ground. SMASH.

Making some assumptions about dropping and smashing a pumpkin, I can get an estimate for the maximum acceleration. Suppose I drop a pumpkin from a height *h* and it stops in a distance *b* (but doesn’t smash).

I just realized a flaw in my plan. When the pumpkin hits the ground, it the area of contact with the ground is smaller than the area of contact when the pumpkin is being shot from the launcher. Oh well, maybe I will add an adjustment factor later. What if I dropped the pumpkin in water? Maybe that would be a better estimate – clearly I could drop a pumpkin from even higher if it landed in water (and I didn’t want it to break). Actually, the stopping distance would also be large. Let me just go ahead with the pumpkin on hard ground.

For this case, I am dealing with a change in motion over distances. This strongly suggests that I use the work-energy principle.

My work will be from the starting position at the point it was dropped to the end where it stopped moving after hitting the ground. If I pretend like the pumpkin is a point object, I can say that there is work done by two forces – the gravitational force from the Earth and the normal force from the ground. The gravitational force does positive work since the force and the displacement are in the same direction. The ground does negative work since the force and displacement are in opposite directions.

The only trick here is the displacement. The force from the gravitational interaction acts over the whole distance, but the floor only exerts a force over the distance *b*. (in case you can’t tell, I am using F_{N} to represent the force the floor exerts on the pumpkin in the vertical direction) Now, solving for F_{N}:

I really want the acceleration of the pumpkin. This is the net force over the mass. The net force will be F_{N}-mg:

Just a check – does this have units of acceleration? Well, *g* is 9.8 N/kg which is equivalent to m/s^{2}. Also hg/b also has units of m/s^{2} – so that is good. What if I stopped the pumpkin over the whole drop. This would means that as it falls, it is also being slowed down. Well, if it started from rest and ended at rest at took the whole distance to do that, it would move at a constant speed (of zero, I know that looks funny). If *b* = *h*, then a_{y} = 0. So that is good also.

What are the values for *b* and *h*? I don’t know. I guess someone could take a whole bunch of pumpkins and drop them to get some values. What if you dropped them on a tray of jello? That way the force would be applied to the whole side of the pumpkin? Anyway, if I had to guess, I would say *h* = 2 meters and *b* = 0.2 meters. This would give a max acceleration of 88 m/s^{2}.

I have another way of testing the acceleration. A top quality launcher has a barrel of about 30 meters and shoots a pumpkin at around 600 mph (let me just call that about 250 m/s). What is the acceleration for this case? Assuming a constant force and acceleration then (and in 1-dimension):

But I don’t know the time. However, I can assume that the following is true for the average velocity:

From this, I can get an expression for the change in time in terms of the change in position.

Now putting this into the expression for acceleration, I get:

I know what you are thinking “hey – that’s just one of the kinematic equations”. And you are correct. However, I like to derive it this way to show there is no magic involved and nothing up my sleeves. But anyway, I have an expression for acceleration in terms of distance and change in speed. The initial speed is 0 m/s, and using the numbers above I get an acceleration of around 1000 m/s^{2}. Quite a bit different from my first attempt. Let me pretend like this is the correct value. Also, let me assume that if you accelerate it more than this 1000 m/s^{2} value, it will break.

How long would a launcher have to be to get to the 2000 mph mark without breaking the pumpkin? I can just use the same expression I had for the acceleration, but solve for the change in position (delta s).

Starting from rest and speeding up to 2000 mph (about 900 m/s) with an acceleration of 1000 m/s^{2} would require a barrel length of around 400 meters. Not very practical, is it?

From my previous calculations, the magic mile range for a pumpkin would take a launch speed of about 800 mph (360 m/s). This would require a launcher that was over 60 meters long (over twice as long as the current devices). They are already long as it is – so I would be surprised if someone breaks the mile range mark.