Math Spring powered catapult

[kjr]

Hi everybody,

I have a little catapult project, the physical thing is given to us and we need to build a mathematical model to predict the range of the projectile.

I'm attaching a sketch of the problem.


If i know the velocity and the angle that the projectile leaves the catapult, it's quite easy to find the distance it'll travel. The tricky bit is finding the velocity.

The way i see it, the easiest way is to use the energy method: use ElasticPotentialEnergy = RotationalEnergy to get angular velocity and then use it to find tangential velocity. Then do some experiments to determine the losses in energy to make the model more accurate.

I was wondering if someone could confirm that what i'm doing will give me the correct answer?

 

[C3PO]

Split the velocity into vertical and horizontal components. The vertical has an acceleration of 9.80665 m/s2. Calculate how long it takes the projectile to reach the ground, and multiply with the horizontal component.

 

[Quick_Nick]

Split the velocity into vertical and horizontal components. The vertical has an acceleration of 9.80665 m/s2. Calculate how long it takes the projectile to reach the ground, and multiply with the horizontal component.

I think he understands that much.

And with my limited knowledge, I agree with your tactic of using the elastic energy equation.

 

[Tommy]

Well, it won't be nearly as simple as "ElasticPotentialEnergy = RotationalEnergy", of course. You'll need to figure out the rotational inertia of the catapult arm (while it's "loaded") and consider the leverage of the arms, etc. Ideally, friction of the fulcrum bearing should also be taken into consideration, but that may not be required for this exercise.

Also, unless there's some kind of stop that limits arm travel, it may be difficult to determine the actual angle the projectile is ejected at. The shape of the cradle may come into play, if centrifugal force can cause the projectile to "slip" out of the cradle before it starts to deccellerate. I've made mid-evil style catapults, and it can be very tricky to get the payload to release according to plan.

If the catapult is designed and scaled like the diagram, the cradle should hold the payload until the cradle starts to slow down, so you'll need to figure out when the arm starts to slow. The force a spring exerts isn't uniform, and will decrease as the spring returns to it's "normal" position. In the diagram the "throwing" arm is longer than the "tension" arm, which would indicate more mass. The "throwing" arm will weigh more - and the effect of that weight disparity will be highest when the arm is horizontal - and decrease when the arm approaches vertical. At some point, the the spring's "pull" may no longer overcome the weight differential and the arm's rate of rotation will begin decreasing. The cradle will slow down faster that the payload, and once the payload passes the "lip" of the cradle it's trajectory will become ballistic - and ballistic trajectories are fairly simple to calculate, of course. For this exercise, you can probably assume that the "lip" effect is instant - and the payload will start the ballistic path as soon as the arm's rotation begins to slow.

If the catapult has a "hard stop" that limits the arm travel, you can most likely assume the payload will leave at an angle perpendicular to the arm when it hits the stop. You can measure the "pull" of the spring at it's starting position, and at it's "stop" position, and average the two to get a "mean" force. It won't be perfect (springs don't lose "pull" uniformly) but it should be close enough.

Worst case scenario is that the catapult doesn't have a cradle, but has a sling instead (like a trebuchet)!

 

[C3PO]

If i know the velocity and the angle that the projectile leaves the catapult....

There's no need to make it more difficult.

EDIT: Scratch that. I should read things more carefully. :blush:

I agree with Tommy. It all depends on if it has a hardstop or not.

 

[kjr]

Thanks for all your input, i didn't even think about the slowing down of the arm.

If the cradle wall is tall enough, the lever should maintain the trajectory, whilst it's not providing any more 'push' past the horizontal, as Tommy said.

So if i find out the difference in energy when the catapult is loaded and ready launch and when the lever is horizontal (when projectile leaves the cradle) that would give me the basis for the calculation, right?

 

[Tommy]

So if i find out the difference in energy when the catapult is loaded and ready launch and when the lever is horizontal (when projectile leaves the cradle) that would give me the basis for the calculation, right?

Assuming you are accounting for the effects of leverage and inertia, yes - assuming the projectile leaves the cradle when the arm is horizontal (which would indicate a "straight up" trajectory).

I can tell you how we did it with full size "historical reproduction" catapults - but we didn't require absolute accuracy. We just needed to get in the ballpark, from there normal artillery "bracketing" methods were used to fine tune the accuracy (which in an actual catapult intended as a weapon is just as important as maximum range). The nice thing about our methods is that they use fairly simple linear equations - and the results were pretty good.

Real catapults almost always have a way to limit the arm travel - if yours doesn't the hardest part will be determining the ejection angle.

There are few things we needed to determine to calculate distance. We determined the spring tension at the "firing" position and at the "release" position - and averaged them.

We determined mass for inertial purposes by using the following formula: ThrowingArmMass/3 + PayloadMass = InertialMass. This is a bit crude, but actually works quite well in practice. Dividing the arm mass by pi would probably be more correct, but using 3 worked better - it also helps account for friction losses at the fulcrum, etc.

We had a "hard" stop, so calculating the arc traveled (distance traveled by each side of the arm) and ejection angle was easy enough.

From this we could calculate the force applied to the payload using simple linear leverage equations, from there we calculated the acceleration (using the inertia calulated earlier), and that was used to determine release velocity. That and the angle were all we needed for a basic ballistic trajectory.

In practice, the catapult springs were attached to a winch (and had a tension gauge) so we could de-tension the arm for cocking and loading. Tension was only applied when ready to fire. The stop was adjustable to set the ejection angle (to change the range) and we had pre-calculated the velocity for ejection angles of 30 degrees, 40 degrees, and 50 degrees. We just interpolated for angles in between. We had uniform payloads (factory reject 18 pound bowling balls), so the only calculations we needed to make in the field were interpolating the velocity (if we were at an "in-between" angle) and the basic ballistic calculation. We used an artillery "slide rule" for that - fast, easy, and no worry about!

I don't know if these methods will be sufficient to please your math teacher, but most math teachers seem to like simple, "elegant", solutions that are reasonably accurate.

 

[kjr]

Thank you, Tommy. Well, i've incorporated a hard stop for the catapult. It's a bit crude (a pin inbetween the supports, constant release angle 40deg).
I've built the mathematical model using the energies as i've stated before. I've got a strange issue though, it's dead accurate when firing from horizontal position (within 20mm), but the accuracy deteriorates rapidly when the angle of travel is increased. I'm wondering if that's due to the spring releasing its energy unevenly?

Is there a way to account for that? Moment of inertia seems to be pretty spot on, since i can fire projectiles of different masses with precision (but only from horizontal). Apart from that, there aren't that many other variables that could have such a huge effect on the release velocity. It seems that energy in the spring just vanishes and that's not possible; it has to be going somewhere...

 

[Notebook]

Just a thought, have you checked that the friction is constant over the arms travel?

N.

 

[kjr]

I'm not sure about that. There isn't a great deal of it to be fair. I don't think it would cause such a big difference. The arm is pinned through two pylons on either side of it with an M6 bolt (the catapult is really small, total arm length - 330mm, height - 200mm, powered by two parallel springs with combined spring constant of 490N/m).

Also, the bolt is tightened only enough to prevent the device from rattling, there is very little resistance to its motion.

I'll look into it though, I have been wrong before... I'll lubricate the joint and see if it makes any difference.

 

[Notebook]

No, I doubt it too, probably your spring is non-linear, or something is causing it to behave like that. Don't know how you would check it though.

N.

 

[kjr]

That's what i'm thinking now. I determined the spring constant by running it through the tensile testing machine, the graph it produced is quite standard and fits Hooke's law. However, only small extension was used during the test, so that's not going to prove that that it behaves linearly throughout a much larger extension.

 

[Tommy]

In a tension spring, (as opposed to a compression spring), force is sort of bell curved. Force will be low when the spring is shortest, and when it gets long enough to begin "over stretching" the spring. A coil spring has a torsion effect, not just a "bend" effect.

Definitely lube the fulcrum, there is a very large load on that when firing - remember that the fulcrum is redirecting the entire force by 180 degrees. That's a lot of energy is a short time, so fulcrums are much more vulnerable to friction than you might think.

Assuming there's no serious wind, the only reason results will vary is if the arm isn't accelerating uniformly. It's OK that the power curve is uneven - as long as it's the same every time. Given that you are using the same spring, same starting position, and a stop, friction is the most likely variable to be causing the problem.

 

[Notebook]

I suppose you could build a second model, with a different bearing design.
Sleeve type brass/steel, or ball-bearing.
Then you could provide a total-loss lubrication system, maybe a gas-generator-turbo-pump powered system, LOX-KERO propellants, and...I'll get my coat.

N.

 

[kjr]

Redesign isn't an option at this point :/ It's a shame we were given these as-is. I would have loved to design my own catapult. Some issues i'm having with this one drives me and my group crazy (especially the release; or more specifically the lack of one)...

Since i'm massively running out of time, what i decided to do is go and spend a whole day testing, get lots of data and then put predicted and actual values on a graph and get some sort of correction curve. It's not pretty, but it should work

 

[Notebook]

If thats what you have to work with, no problem. You may find something from your graphs, once you have the data.

N.

 

[kjr]

Just thought i would report on the competition.

I didn't manage to get the consistent results from testing as i planned, so just decided to use what i have and it wasn't too bad. Our team was only about 5cm off to the side from the targets, with the distance pretty much perfect.

I think the reason behind the inaccuracy was that the spring was close to its maximum length, and since we did all testing with the heaviest projectile, so once the mass was reduced the accuracy improved.

In the end we came joint 2nd. Thank you all very much for your help.