This is a very important part to building a trebuchet! To big and you weaken the swing arm, to small and you will have a damaged axle.

Since we are going to be swinging 600 lbs around we wanted to make sure that it wasn’t going to fail. We also wanted to make sure that AS MUCH of the energy is transfered to the pumpkin and not the axle. To do this you just need a really rigid axle, but how do you size it?

First thing is to find out how much energy you gain by droping the counterweight. So, we have a counterweight with mass M. We also have a beam with mass B. For simplicity I am going to say that the beam, though it’s size is going to be considerable and should be considered for a complete analysis, is negligible. I am also going to simplify the model even further by saying that the weight is a fixed weight on the swing arm itself. This reduces one degree of freedom and simplifies the problem immensely.

What we are actually looking at is now just a simple pendulum. This isn’t a true representation, but should get us in the right ball park.

Ok the force on the axle as it swing will be the centripetal force acting on it. We know that

F=ma (1)

but that is for a linear force. Since our weight is swinging we want angular, and that where the centripetal acceleration comes in.

a = v^2/r (2)

a = acceleration

v = velocity

r = radius

Substituting 2 into 1 yeilds

F=m*v^2/r (3)

We still have too many variables to use the equation directly. We don’t know the velocity, yet. We do know that velocity is dependent on height, and gravity. We know that

1/2 * mv^2 = mgh

Since the mass doesn’t change the mass cancels out.

1/2 * v^2 = g*h

solving for v^2 (normally you would want to solve for v not v^2, but it makes sense. Trust me!)

v^2 = 2*g*h (4)

Ok, we are getting somewhere now! Sub 4 into 3

F= m*(2*g*h)/r = 2mgh/r (5)

Now this is better, this is something we can work with. However, we can simplify this even further since h, the height is dependant on the angle of the trebuchet and the length from the axle to the counterweight.

From magic, or voodoo (whatever you wish) you get that

h = r*(1-cos(Θ)) (6)

So we can sub 6 into 5

F = 2*m*g*(r*(1-cos(Θ))/r = 2*m*g*(1-cos(Θ))

So

F= 2*m*g*(1-cos(Θ))

or

F = 2W(1-cos(Θ))

So, in my case, I am using a 600 lbs counterweight, and ideally we will be dropping it 135 degrees (measured from the resting position).

So

F = 2*600*(1-cos(135deg))=2048lbs

That’s a pretty significant force. If you were able to do a full rotation your axle would be experiencing 4 times the weight of the counterweight. That’s good to remember when designing your trebuchet!

Other Resources

Trebuchet Design

Cement Blocks

Counterweight

Sizing the Axle (Part 1)

Sizing the Axle (Part 2)

A-Frames

Car Throwing Trebuchet

Cool Trebuchet Pictures

The bucket