# The Ballistic Coefficient of Pumpkins

1. Oct 21, 2009

### Grozny

A trebuchet is a type of catapult that converts the potential energy of a counterweight into the kinetic energy of a projectile. The simplest version operates like a see-saw, with the counterweight suspended from a hanger attached to the short arm and the projectile held in a sling attached to the throwing arm. When the short arm is raised and then released, the throwing arm rotates faster because it is longer and the sling rotates even faster as it whips around the end of the throwing arm.

At their inception during the Middle Ages, trebuchets were used to hurl boulders at or over castle walls in an attempt, usually successful, to batter them down. Modern trebuchet designers, lacking castles to besiege and fair maidens to rescue, must content themselves with hurling cooking pumpkins for distance. Competition, however, is fierce, and the winner of the contest can expect any fair maidens present at the pumpkin festival to hurl themselves at him. Thus, it behooves us to put as much study into the ballistics of pumpkins as old-time mathematicians put into the study of cannonballs.

Visit The Ballistic Coefficient of Pumpkins to read the rest of my paper.

Ron Toms, the owner of the Catapult Message Board, dares me to go talk to an engineering or physics professor regarding that nonsense I've been peddling about 45° being the optimal launch angle.

Very well. I will take that dare. Are there any engineering or physics professors on this forum who would like to comment on the accuracy of my paper?

2. Oct 22, 2009

### Bacat

The short answer is that he is right. You are correct that 45 degrees is best if there is no air resistance (which can be assumed for many projectiles), but for something as un-aerodynamic as a pumpkin, you cannot ignore drag effects and then the maximum range angle depends on what the drag coefficient, k, is.

From my Classical Dynamics book (Thornton and Marion 5th Ed, pg. 68), we have the formula for maximum range without drag:

$$r_{max}=\frac{v^2}{g}Sin(2\theta)$$ where v is the initial velocity of the projectile.

If you take the derivative of this you can easily see that the maximum range occurs when $$\theta=\frac{\pi}{4}$$ or 45 degrees exactly.

Now, using a perturbation method we can derive the effects of air resistance to a good approximation by expanding and keeping the highest order terms. If we do this we arrive at:

$$R_{max}=r*(1-\frac{4kV}{3g})$$ where r is the function for range without air resistance, k is the "retarding force constant" due to air resistance, V is the vertical component of the initial velocity, ie $$V=v*Sin(\theta)$$, and g is gravity.

Now, if we use Mathematica, we can find the angles that give maximum values for a given value of k. We keep V constant to find these values. You can also optimize the function by hand, but it was faster to just plug in a series of value and see what happens. So basically, I defined functions $$r[\theta]$$ and $$V[\theta]$$, then I ran:

NMaximize[{r*(1-(4*k*V)/(3*g), 0 <= Theta <= Pi/2, k = ?}, {Theta,k}]

where I have used the following values for k: 0.0, 0.01, 0.02, 0.03, 0.04, 0.05

This gives me an output that looks like this: {586.576, {k -> 0.05, Theta -> 0.573102}} (for k = 0.05). The first number is the maximum range obtained for v = 100 m/s. Then it gives the optimum value for Theta, which in this case is 0.573 radians.

Here is the table summarized:

k (Theta)
============================
0.00 (.785) (45 degrees exactly)
0.01 (.758)
0.02 (.752)
0.03 (0.680)
0.05 (0.573)
============================

Thus we see that the optimum angle for firing the trebuchet for a projectile with a lot of air resistance is lower than 45 degrees. The exact value depends on how much air resistance there is. I have no idea what the k-value for a pumpkin is, but I can tell you that it will be greater than zero.

Cheers.

3. Oct 23, 2009

### Grozny

Source: The Mathematics of Projectiles in Sports

Your equation flies in the face of Leonard Euler's assumption that drag is proportional everywhere to the square of the speed.

4. Oct 23, 2009

### Bacat

I make the same assumption as Euler.

If you still don't believe me, here is an article from USC:
http://physics.ucsc.edu/~peter/115/range.nb.pdf [Broken]

Near the top of the penultimate page, the author writes: "Note that the range increases rapidly as q increases and reaches a maximum at a value less than pi/4 (which is the value without air resistance)."

He solves this by integrating the equations of motion. I did it numerically by making a perturbation approximation. We both arrived at the same answer.

Why do you think my equation "flies in the face of...Euler's assumption"?

Last edited by a moderator: May 4, 2017
5. Oct 24, 2009

### Grozny

My analysis is based almost entirely on de Mestre’s book, The Mathematics of Projectiles in Sports. It is out of print, though much of it is available online here. The first four chapters are titled:

1) Motion Under Gravity Alone
2) Motion in a Linear Resisting Medium
3) Motion in a Non-Linear Resisting Medium
4) The Basic Equations and their Numerical Solutions

Chapter One is just the parabola that everybody learns about in Physics 101. In the introduction to Chapter Two (p. 24), we read:

Six pages later (p. 30) we arrive at this conclusion:

This just happens to be exactly the result that your Mathematica guy obtained.

I am not familiar with Mathematica, so I cannot follow this guy’s reasoning, but I believe that you have been misled. He starts out saying that drag is proportional to the square of velocity, but then he solves the linear and uncoupled equations for the simple case where drag is proportional to velocity.

This should be obvious since, even without an intimate knowledge of Mathematica, it is clear that this guy is not solving a system of coupled non-linear differential equations. That would be vastly more complicated than what I saw in his PDF file.

My paper is based on Chapter Three of de Mestre’s book, Motion in a Non-Linear Resisting Medium, and my Java Applet is based on Chapter Four of de Mestre’s book, The Basic Equations and their Numerical Solutions.

The professional version, which I have not yet written, will be based on Chapter Six of de Mestre’s book, Corrections Due to Other Effects, particularly wind corrections. Also, it will take hills into consideration.

6. Oct 25, 2009

### willem2

If there is air resistance the optimum angle is less than (pi/4).

The table of Bacat is of little value however, because k and V are so high, that the approximation isn't valid anymore. With a k of 0.01, a 100m/s pumpkin will decelerate with 100m/s^2. The approximation will only work if this value is small compared to g.

Numerically integrating $a_x = -k v_x v$, $a_y = -k v_y v - g$
for various initial angles, gives a distance of only 170m at a maximum angle of about 0.18$\pi$. The maximum distance without friction is 1020m.

I have $$k = \frac {F_{fric}} {v^2 m} = \frac {C_w A \rho_{air}} {2 m}$$

Grozny's pumpkins seem to have a k of about 0.002 and v = 54.3 m/s. (the maximum angle depends on v)

angle/pi range

0.2 205.9
0.22 209.3
0.23 210.0
0.24 210.0
0.25 209.3
0.3 195.5

the maximum is at 0.235 pi (42.3 degrees) and a range of 210.1 m
the difference in range with 45 degrees of 0.8m is probably unmeasurable with a trebuchet.
I think wind speed and pumpkin wobble will give much larger errors.
You could measure the difference between 0.2pi and 0.3pi to get an idea wich one is
closer to the maximum.
The experiment might be better done even more friction and less mass, for example with a soccer ball.

7. Oct 25, 2009

### Bacat

The original question for the physics community was whether an angle less than 45 degrees is really optimal for a projectile subject to air resistance. The answer is that this is true.

We can hash the numbers about pumpkins and talk about techniques for proving it, but the theory supports the reality that an angle less than 45 degrees is optimal.

Cheers.