# Puck Launcher Calculations

## Main Question or Discussion Point

Hi, for a project at college i am building an ice hockey puck launcher for the development and training of ice hockey goaltenders. I am planning on using two wheels rotating in opposite directions to launch the puck. The pucks will be fed in by a pneumatic cylinder that is powered by a pressurized vessel. The wheels themselves will be driven by a starter motor from a car and six 12V batteries. I havent decided whether to use belt or gears to drive the wheels but i will get to that later in the project.

Now, im basing my system somewhat on the tennis ball launchers you see at tennis courts. They use two hard spinning wheels that compress the tennis ball to create more friction between the wheels and the ball. With my design I won't be able to do this because the hockey puck is made of a very hard rubber compound. This means i am using pneumatic tyres so that rather than the puck compress, the tyres will. However, i need to include some calculations in my project but im not sure what calculations there are for this sort of thing. Obviously i can work out things like the torque and rpm of the wheels but really i need some sort of calculations that translate this torque/rpm etc into the speed that the puck will be fired at. Anyone got an ideas?

Any help would be much appreciated. Thanks.

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welcome to PF!

This is an interesting problem. And not an easy calculation I don't think. But there are most likely ways that this can be simplified.

Let me try and isolate the system into halves. What would you have if you only had one wheel on one side and a wall on the other? There would be a torque that is applied to the edge of the puck causing it to rotate. Now, what you have on the other side is another wheel causing it to rotate in the opposite direction correct? So now there is pent up energy because two wheels are attempting to spin this puck, but neither of them can. This rotational energy is now being converted into translational energy.

How much do you know about the torque applied to the puck. and how much do you know about the laws of torque more specifically the directional relationship between force radius and torque

This is the force diagram that I have. The net vector of those two force vectors is always straight out if your puck launcher works correctly. At the beginning both of those force vectors are in opposite directions making no force. and as the puck gets shot out, those vectors come closer together. So you would really have to figure out the period of time that those wheels are in contact with the puck.

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What I would do, is turn RPM into torque. then torque to force. You need to do an integral for this problem. would you know how to start one?

Thanks for the welcome.

The problem is I dont know about the torque applied to the puck. I only have a basic knowledge when it comes to torque and speed calculations. All i know is the torque of the motor and the rpm at this point. I havent gone into detail on any gearing or anything yet as I just want to get some basic calculations as if both wheels are been driven directly off of the motor shaft. I can then work in the loss through gearing or reduction/increase in speed.

You are in some trouble with the speed calculations then. You have way too many components to take into account. The net torque applied to the puck is 0. each wheel applies a certain torque, and they cancel out and shoot the puck off. But if you want a very accurate calculation you will even need to take into account the difference in radius of the wheel when the wheel is cushioned in for friction. You also have changes in the magnitudes of force as it goes. counter-intuitively the force will increase as the puck reaches the end of the trajectory. There are two reasons this will happen. Those black arrows that i drew, they will start to point more upwards. which means most of the force will be in the upwards direction as opposed to the other direction in which they get canceled out. Also, as the force in the other direction(the ones getting canceled out) gets smaller, the wheels will start to expand back to their initial position. This expansion will propel the puck outwards again.

This is a calculation that is experimentally and theoretically in depth, and I believe you are not supposed to be able to do this. What I would do if i were you, is calculate a rough estimate of the speed. Then what you should do in an analysis section is include all of the calculations that you are unable to do, and explain what kind of affect they are going to have and why you weren't able to do them.

So you know the revolutions per minute. you also know the circumference of the wheel. with this you can figure out the speed that the outside of the wheel is rotating. Once you have that, you know the speed of the other wheel is exactly the same, so you don't even have to add them together, you just have to assume that in that time, the puck will get to the same speed as the perimeter of each wheel. This should give a fairly accurate result.

But I still strongly suggest thinking through the problem even more, not from a calculations point of view though. If you can explain why you're calculations are not right it should be sufficient.

Yeh i think you are right. That's probably the best way to go about it and i know the examiners will love that sort of thing.

Thanks for all your help and fingers crossed my project will all go to plan.

good good im glad i could help. I realize that some of those concepts about integrals may be hard to apply to a real life kinda thing. Im a physics student so im kind of interested to actually doing something in real life instead of just theorizing about things.

Anyways, if you need any help just post back on here and I'll probably see it. or give me a personal message.

sophiecentaur
Gold Member
Because the puck is not rotating, there is no net torque on it. As suggested earlier, I reckon the exit speed of the puck would be about the same as the peripheral speed of the wheels - except for the extra effect due to the squeezing of the wheels together. This would result in the same effect as squeezing of an orange pip between the fingers and shooting it forwards. For this, the wheels would. presumably, need to be lubricated a bit and the wheels pressed together fairly hard (as with the thumb and forefinger with the orange pip).

Ok so im back to this. Ive been working on some calculations today and i know how to work out the torque im getting and the surface speed of the wheels. One thing i need to know is torque required and i noticed a tennis ball launcher on this forum where someone needed the same thing. The guy who replied said this:

The principle is simple enough. The wheels supply a force to the ball over a small distance (the product of which is the energy imparted). The kinetic energy of the tennis ball is equal to the work done on the ball (Force x distance).

The force on the ball is a little complicated because it is a combination of the torque of the motors and the compression of the ball (which causes the ball to push off against the motor wheels). Ultimately, though, the energy comes from the motors.

$$\frac{1}{2}mv^2 = 2\tau\Delta\theta$$

where $\tau$ is the torque provided by each motor and $\theta$ is the angle of rotation of the motor while the motor wheel is in contact with the ball.

Lets' suppose that $\Delta\theta$ is 90 degrees or $\pi/2$ radians.

$$\frac{1}{2}mv^2 = 2\tau\pi/2$$

$$\tau = \frac{1}{2\pi}mv^2$$

$$v = \sqrt{\frac{2\pi\tau}{m}}$$

The mass of a tennis ball is .057 kg and if you want a speed of 30 m/sec, you would need a motor that produces a torque of:

$$\tau = \frac{1}{6.28}*.057 * 900 = 8.2 Nm$$

which is about 6 foot pounds (pounds-force) of torque.

AM

So all that makes sense and i can use that for my project, however... i was wondering two things about it. Did he use 2 X r X Delta Theta because he had two motors? So would i use just simply r X Delta Theta?

Also, he took the angle of rotation as 90 degrees or pi/2 radians... how do i know what my angle of rotation is?

sophiecentaur
Gold Member
I ask again, where does any net torque come from? IS the puck rotating after it leaves the launcher? Presumably you get the wheels to rotate then bring them together against the puck when you want to launch it.
Is this not a simple problem of momentum transfer? It would seem to be necessary just for the (already) rotating wheels to have significantly larger moment of inertia than the mass of the ball (times the radius squared)? That will maximise the amount of momentum transferred. The highest possible speed of the puck will be the peripheral speed of the wheels. A comparatively low power motor can provide, effectively, a high impulsive power, by storing energy in the rotating wheels.

This is similar to any bat and ball situation, where you use as massive a bat as is convenient to handle (and manouvre) and that gives you the maximum ball speed, after impact.

Yes, but remember the wheels have a force acting upon them which is going to try and stop them spinning. So surely if the motor has 0.00001lbs-ft of torque, the wheels would just stop spinning as soon as the puck makes contact?

sophiecentaur
Gold Member
That's like saying that a freewheeling cyclist stops moving when he hits a fly!
There is just a transfer of momentum and what happens depends upon the relative masses involved. That's why I suggested that the wheels should be relatively massive. This is elementary stuff.
And, btw, nothing, in Mechanics "tries" to do anything; it just follows Sir Isaac's laws.

there will be no net torque. If you want to calculate this using torque you will have to use integrals and so on. I suggest you do not do that. As sophiecentaur is saying, the puck should not spin, and the puck should also exit the launcher close to the speed of the wheels at the radius. When i say at a certain radius, i mean to tell you that the radius of the wheels will get a little smaller when the puck is pressed in against the both of them. you want to calculate the instantaneous velocity at that point and hope that its pretty close.

sophiecentaur
Gold Member
Any other treatment would be just 'standing up in a hammock', as they say.

lol does that mean very difficult?

sophiecentaur