Punkin' Chunkin' with UltimaCalc - The Ballistics of Pumpkins

Punkin' Chunkin' with UltimaCalc
UltimaCalc If you haven't read the first part of this article, it would be worth doing so now.
Punkin' Chunkin' - The Ballistics of Pumpkins

When we neglected the effect of air friction, we found that we needed to throw a pumpkin at a speed of at least 244mph (or 394km/h) to achieve a range of 4000 feet (1219m) at the optimum angle of elevation of 45 degrees.

At such speeds, we surely cannot ignore air friction. Let's see if we can estimate it, and then go on to making a more realistic prediction of how fast we really have to throw the pumpkin. We will also check whether we need to adjust the angle of elevation to obtain the best range.

The effect of air friction is expressed by the drag equation:

Drag = 1/2 * density of medium * speed2 * reference area * drag coefficient

The drag equation is an approximation. (We will totally ignore drag caused by movement in a viscous medium, for example someone swimming in water.)

An object dropped from a great enough height will accelerate towards the ground until the drag on it equals the force of gravity. It then falls at a constant speed, the terminal velocity. It can be shown that the terminal velocity is

Vt = (2 * mass * g / (density * reference area * drag coefficient))1/2

Let's calculate the terminal velocity of a pumpkin. What's the mass of a pumpkin? I don't know, but I'll guess that it is about 10kg. The standard acceleration of gravity is about 9.81m/s2, and the density of air is 1.225kg/m^3 for an International Standard Atmosphere.

The reference area is related to the cross-sectional area of the pumpkin. I'll assume that the two are the same. Guessing that a pumpkin has a diameter of 0.3m, the reference area is then pi * diameter2/4, or 0.07m2. I guess that a pumpkin this size has a mass of 10kg.

Now for the drag coefficient. This is the hardest of all the terms to estimate. A typical car has a drag coefficient of about 0.25 to 0.45. A bullet might have a drag coefficient of 0.295. A Learjet 24 aircraft has a very low drag coefficient of only 0.022.

A smooth sphere has a drag coefficient of 0.1, but a rough sphere has a drag coefficient of 0.4. And a pumpkin? In the absence of actual data, your guess is as good as mine. My guess is 0.3.

Using these figures, the estimated terminal velocity of a pumpkin is 107m/s or 385km/hour. Looking back at the original estimate of the required muzzle velocity, we see that the pumpkin leaves the cannon's barrel at a speed greater than its terminal velocity in air. Clearly our first estimate will turn out to be much too low.

As previously, let's split the pumpkin's velocity into two components, namely a horizontal component Vx and a vertical component Vy. The names have been changed to suggest that we are using a coordinate system in which x measures how far the pumpkin is horizontally from the cannon's muzzle, and y measures the height of the pumpkin above the ground.

In the horizontal direction, the pumpkin's velocity, Vx decreases due to air friction. We can consider this a negative acceleration. The drag equation says that this 'acceleration' is proportional to the velocity. We can write this acceleration as Ax = -k.Vx2 where

k = 1/2 * density of medium * reference area * drag coefficient / mass.

In the vertical direction, things are not quite so simple, as the drag always opposes the movement of the pumpkin, whereas gravity always pulls it downwards. When the pumpkin is moving upwards, the drag is in the same direction as the force of gravity, and when it is falling, the drag becomes an upwards force counteracting the effect of gravity. This can be represented by the equation:

Ay = -k.Vy.abs(Vy) - g where is abs(Vy) is the absolute value of Vy, i.e. ignoring its sign. The value of k, being independent of the velocity, is the same in both cases.

We now know enough to be able to calculate the pumpkin's trajectory, the path it takes during its flight through the air. We know where it starts: at this point we can say that x is zero and y is the height of the muzzle above the ground. The initial vertical and horizontal components of the velocity are, as before, Vx = V0.cos(a) and Vy = V0.sin(a), where V0 is the muzzle velocity and a is the angle through which the barrel is raised above the horizontal.

Along with knowing the accelerations in each direction, we can now plot the trajectory.

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