The tree catapult is a staple of cartoon physics, and I think I’ve even seen it acted out in one or two live-action movies. But according to the Mythbusters, it may have been actually used by medieval armies laying siege to castles. I actually thought preconstructed wheeled catapults would have been the norm, but still, you have to wonder: does it work?

The underlying concept behind a tree catapult is simple enough: trees are stiff. If you pull one down and let it spring back up, it stands to reason that you could potentially fling a dead body pretty far. But that very same stiffness that gives a tree its flinging power also makes it resist being pulled down in the first place. This is a pretty straightforward application of the principle of conservation of energy: whatever energy the tree is able to expend in flinging its cargo into the air has to come from the work you do while pulling it back in the first place.

I can make a rough calculation by treating the top of the tree as a spring, known in physics terms as a simple harmonic oscillator (SHO), which is subject to a restoring force of the form

The force you apply to pull the tree out of equilibrium (straight up) works against the restoring force, so the negative sign disappears,

The last expression on that line makes it easy to calculate, at least approximately, the energy that Tory, Jesse, and Grant put into their tree catapult. They applied a force of \(\unit{2400}{\pound}\) to pull it down by \(\unit{45}{\degree}\). Now, if the entire tree had been tilted at that angle, that would correspond to a distance of \((\unit{70}{\foot})(\unit{45}{\degree}) = \unit{17}{\meter}\) (that’s \(\unit{55}{\foot}\)), but since the tree bent gradually over its length, I’ll estimate that the actual distance it was pulled back was about half of that, \(\unit{8.4}{\meter}\). So \(W = \frac{1}{2}(\unit{8.4}{\meter})(\unit{2400}{\pound}) = \unit{45}{\kilo\joule}\). If all that energy had been transferred to Buster, who reportedly weighs about \(\unit{185}{\pound}\), we’d have

or

Plugging in the appropriate values, the equation says that Buster would have left the tree at \(\unit{33}{\frac{\meter}{\second}}\). The time it takes him to fall to the ground, assuming he was released horizontally, is given by the solution of

which is \(t = \sqrt{2h/g}\), and the horizontal distance he’d travel in that time is

If you plug in \(h = \unit{70}{\foot}\) and the previously calculated velocity, you’d expect to find Buster on the ground \(\unit{68}{\meter}\) (\(\unit{220}{\foot}\)) away! He would have easily cleared the \(\unit{40}{\foot}\) makeshift parapet the Mythbusters constructed. Obviously, that didn’t happen; Buster landed at the foot of the castle wall, only \(\unit{100}{\foot}\) away. Why?

Well, there’s one obvious thing I haven’t accounted for: the tree doesn’t instantaneously stop at the top of its swing and transfer all its energy to the dummy — it keeps moving, and that takes kinetic energy too. So instead of setting the work, \(\unit{45}{\kilo\joule}\), equal to \(\frac{1}{2}mv^2\), we should set it equal to the total energy of Buster and the tree,

In the second term I’ve used the expression for rotational kinetic energy because the tree is not an object moving with a single uniform velocity. It’s not exactly rotating either, but that’s going to be a closer approximation than assuming it’s moving straight.

Anyway, to calculate the moment of inertia of the tree, we need to know its mass. For that, I’m going to use the density of Douglas Fir, \(\rho = \unit{443}{\frac{\kilo\gram}{\meter^3}}\) (source), and estimate the diameter \(d\) at about \(\unit{1}{\foot}\). I can plug these into the formula for moment of inertia of a rod rotating around its end,

Using \(v = r\omega\), the equation involving the work done becomes

which can be solved for the velocity at the top of the tree, i.e. Buster’s release velocity

Plugging in the appropriate values gives \(v = \unit{17}{\frac{\meter}{\second}}\), and that corresponds to a horizontal distance traveled of \(\unit{35}{\meter}\), or about \(\unit{115}{\foot}\). Not bad, especially for a rough estimate!

For most purposes, a simple analysis like that using basic mechanics is good enough, but this story is not complete yet ;-) It should come as no surprise that there’s a whole area of physics devoted to the study of bending beams, just like this tree catapult, and in a future post I’m going to apply the heavy machinery of Timoshenko beam theory to our tree. I’ll see if it can do any better than the simple harmonic oscillator analysis. Stay tuned!