Energy transfer/loss in catapult

[Twellmann]

TL;DR Summary: In a classic catapult the arm usually meets a cross brace that stops it and helps define the exit angle. This seems a bit wasteful.

I have been thinking about making a small catapult for my sons class.

In a classic catapult the arm usually meets a cross brace that stops it and helps define the exit angle. This seems a bit wasteful.

When looking at trebuchets, optimization can be done to maximize the energy transfer to the projectile, eg. matching counterweight to projectile, allowing the frame to move on wheels, and other things.

I haven't really found anything similar for a simple catapult.

Any ideas?

 

[berkeman]

Welcome to PF.

Since this is for schoolwork (even your son's schoolwork), it has been moved to the schoolwork forums section of PF.

What ideas have you been thinking of so far? You say you haven't found anything, but have you tried brainstorming some with your son? I can think of at least one mechanism that would help the efficiency a fair amount (initial guess is 10-20%)...

https://www.britannica.com/technology/catapult-military-weaponry

 

[Twellmann]

Hi Berkeman,
Thank you for your reply, but this is not a school project, even if it is for his class to enjoy.

I didn't say that I haven't found anything, but more that a lot of discussions focus on getting good performance.

Talks about efficiency seems to be more in the trebuchet domain, as it already has advantage over simple catapults.

A free swinging catapult with a low inertia arm is probably my first approach. The challenge is then to optimize for power input in relation to the mass of the projectile.

My goal is to have a 20g projectile launch around 15m. So I need input energy in the order of 1.4J. Assuming low efficiency, <50%, I need around 3J input.

I haven't started a sketch yet.

Br

 

[berkeman]

Thank you for your reply, but this is not a school project, even if it is for his class to enjoy.

No worries. At PF we treat all schoolwork-like questions the same, and we want them in the schoolwork forums with the "student" (no matter if for self-study or similar) to show lots of effort. We like to help folks "learn how to learn" here as much as possible.

BTW, my idea to increase the efficiency was to recover some of the energy that you mention is lost when the rapidly-moving arm hits the top stop. I agree that that energy is lost, so why not try to recover as much of that energy and use it to lower the amount of energy needed to input to the arm for the next shot... Can you think of a way to recover that energy at the end and then be able to use it to make it less effort to load the next shot?

Also, my definition of "efficiency" here may be different from what you want. I'm thinking "energy per shot for multiple shots in a row", and you may be thinking "get the most distance out of each shot with an initial investment of Potential Energy (PE)".

 

[haruspex]

Can you be a bit more specific about the design? What is the energy storage… something elastic I assume, either twisted or stretched.
If it can be modelled as a spring, ##\Delta F=k\Delta x##, then the stretch varies from ##x_0## to ##x_1## and the energy stored is ##\frac 12k(x_1^2-x_0^2)##.
You may think that this wastes ##\frac 12k(x_0^2)##, but as @berkeman notes, that energy is a constant, never used, never lost. Attempting to use it takes up available range of x, limiting ##x_1##. (In the same way, a bow is strung such that there is considerable tension even before the string is drawn back. Note that a bow cannot be modelled as a simple spring, though, because the angle between the string and arrow changes during release.)

So the only wasted energy is the KE left in the arm.
When I toss a ball of paper into a distant bin, I position my left hand so that my right wrist hits it. That flicks my hand forward, giving an extra kick to the paper. The crossbeam on the catapult could achieve the same.
Suppose the arm is length L with linear density ##\rho##, the crossbeam is at h from the top end, and we put a joint in the arm at that point so that it can flick forward freely. Payload has mass m.
If the angular velocity just before hitting the beam is ##\omega## then the angular momentum about the crossbeam of that upper portion of the beam and the payload then is ##mhL\omega+\rho h^3\omega/12+\rho h(h/2)(L-h/2)\omega=h\omega(mL+\rho hL/2-\rho h^3/6)##.
If the angular velocity just after hitting the crossbeam is ##\omega'## the angular momentum is now ##mh^2\omega'+\rho h^3\omega'/3##.
Since the impulse from the crossbeam has no moment about itself, these two angular momenta are equal.
The velocity of the payload changes from ##\omega L## to ##\omega' h##.
Sanity check: with ##h=L## we get ##\omega'=\omega##.

Finding h to maximise the velocity of the payload I leave as an exercise. I get a quadratic for h, and with the arm having a fifth the mass of the payload, e.g., h is approximately L/2.