As the title suggests I am looking at making a hockey puck launcher so I can do some off ice technique training as a goal tender. I have no one to train with so I wanted to create a puck launcher similar to the tennis trainers they have that fire balls at you.

My idea was to run an electric motor that was attached to two spindles (via belt or chain drive) with spinning wheels. When the puck is inserted between them it will grab and take off. My problem is I'm unsure as to how to determine what size motor I would need as I don't know much about torque / force.

I want to be able to do the following:

  1. Fire a 100 gram (3.5 oz) inline puck.
  2. Speed should be able to be varied between 0 - 60 mph.
  3. Must be able to be plugged into 240v mains power (I live in Australia) via step down transformer or similar. (Note: I have qualified electricians who can do this for me).

I was looking at hobby RC motors as those cars likely weigh more than 100 grams and many of those can go extremely fast, they also have variable speed control to mimic a real car. My concern is though would they have enough power to grip the puck and throw it? Is there any math I can use to determine this at all?

If you need any more information or would like a design or something to help with the question to prevent it from being closed let me know.

  • \$\begingroup\$ Perhaps you should examine a pitching machine. \$\endgroup\$ Jan 5, 2014 at 23:55
  • \$\begingroup\$ Or clay pigeon launcher \$\endgroup\$
    – Matt Young
    Jan 5, 2014 at 23:57
  • \$\begingroup\$ I'm just looking at the cost of purchasing one of these outright as opposed to making my own as I assume they would be costly investments but I will have a look at both thankyou. \$\endgroup\$
    – Scott
    Jan 6, 2014 at 0:00
  • \$\begingroup\$ Do you need to have spin? If not, something much simpler, like a solenoid, might work. \$\endgroup\$ Jan 6, 2014 at 1:41
  • \$\begingroup\$ I don't really care about spin or anything I just want a projectile fired at me with speed, whatever the simplest and cheapest method to arrange that would be fine. Could you provide some examples of how a solenoid could work and I could do some further research? \$\endgroup\$
    – Scott
    Jan 6, 2014 at 3:30

3 Answers 3


You probably want a stored energy system to reduce required motor power.
Mean power can be low - if you lower the firing rate the mean power drops.

Easiest way to get consistent results is probably to have a rotating wheel with substantial mass that spins at the required circumferential velocity. If this is used to accelerate and "sling" the puck, and if Mwheel >> Mpuck then it will not slow much as the puck is accelerated to exit speed. Wheel is spun up by a motor of whatever power is desired. Larger motor = quicker recovery time. Note that if the wheel stores say 10 x the puck max energy you have a potentially lethal flywheel which would not be too hard to build safely, but care is required.

Wheel balancing machines spin the tires & wheels up with an electric motor. Some have quite a small Wattage motor and a slow run up time. Others use larger motors and faster runup. I've seen older ones with a 24 VDC motor and maybe a few 10's of Watts max power.

Spin up a wheel on such a machine to desired speed, place a sprung "floor" under the wheel, not touching but less than puck height separation, feed in a puck [from the BACK :-) ! ] and watch it vanish as it touches the tire. An old deadish wheel balancer may be available at minimal cost. Probably larger than what you had in mind :-).

Energy & Power:

  • Liberties: The following takes some liberties with power and energy profiles when accelerating a puck. If you accelerate at constant power then velocity change is not linear. If you accelerate at constant acceleration then power input is not linear. Time to accelerate depends on power input profile and is not the simplistic figure derived below. But, the following should give a good feel for power end energy levels involved.

Energy / Power / Acceleration / time in 'launcher' / ... :

Given puck with mass = 100g = 0.1 kg and
velocity_max = 60 mph = 27 m/s -> say 30 m/s then

Puck energy \$(E_K) = \frac{1}{2} \times m \times v^2 = 0.5 \times 0.1 \times 30^2 = 45 Joule = 45 Watt seconds. \$
(\$ E_K \$*=kinetic energy, m= mass of the object in kg, v= speed of object in m/s).*

Linear motor: If you accelerated it at constant power over a say 1 metre linear track them
Vmean = (30-0)/2 = 15 m/s so time to accelerate = 1/15s = 67 mS.
As puck energy = 45 W.s and you are delivering this in 1/15s the power during acceleration is ~= 45 Ws / (1/15s) = 675 Watts.
That's more Watts than you'd like to need in a linear launcher if you are driving it directly electrically.

"Crossbow": Instead you can wind up a "crossbow" type mechanism where a spring or torsion bar or whatever is wound up to store the desired amount of energy and then "tripped".
For a potential firing rate of 1/second you need 45 Watt of energy to storage transfer rate - say about double this to get electrical motor input or about 100W.
For a 5 second repeat rate you are down to 20W and 10W for 10 second cycle time.

"Putter:" A 10 Watt motor winding up a spring via a reduction box or screw thread or ...? is easy to build and play with. It would (probably) be easy to pull back a weight on a pivot (rigid shaft pendulum) and let it go so it swings and strikes the puck, golf putter style. Potential energy in a mass is mgh
or about 10 x kg x height.
You want 45 W.s max. 10 kg x 0.5 metre x 10 = 50 W.s
Height is the vertical height above rest level.
Velocities are now wrong and you are going to need to deal with impulse energy transfer, but it should be in the order of right.

Rocket: A potentially workable system would be to use compressed air and a pressure reservoir. A completely DIY system could use a "water rocket" type launcher as built by amateurs world wide. Liable to be somewhat noisy [tm] on release.

Treadmill motor / Direct drive rotary flinger: A potentially excellent excellent motor for a flinger is a treadmill motor. These are typically rate in the 0.5 to "several" HP range and 200 VDC permanent magnet units powered from 230 VAC rectified mains (and no doubt the 110 VAC based versions in the US) are common. Speed control by PWM or other voltage variation allowing wide range speed control. An arbitrarily large disk could be attached directly to suit the motors preferred RPM maximum. eg an 1800 RPM motor = 30 RPS will need a 1 metre circumference disk to deliver 30 m/S = a diameter of 318 mm or about 1 foot.

  • \$\begingroup\$ Thankyou very much extremely informative. I definitely like the idea of the stored energy device. The less moving parts the less someone or something could get tangled up in it. \$\endgroup\$
    – Scott
    Jan 6, 2014 at 12:05
  • \$\begingroup\$ @alexan_e - mistake -> typo given following line. Whatever :-). | Here in the antipodes, on the edge of the Empire, at the dawning of every day, The Queen's English, she is still spoke, and tire is spelled tyre, as it should be. And meter is spelt metre, when referring to the unit of length in the metric aka MKSA aka SI system, as it very most certainly should be. But I'll avoid the pedantic pettiness of reconverting spellings to their correct form and leave them in their quaint colonial variants. [[[ :-) ]]]. \$\endgroup\$
    – Russell McMahon
    Jan 6, 2014 at 18:51
  • \$\begingroup\$ @RussellMcMahon Oh, sorry. I wasn't aware of that. I hope a similar rule doesn't apply to tyres :-) \$\endgroup\$
    – alexan_e
    Jan 6, 2014 at 19:03
  • \$\begingroup\$ @alexan_e - Afraid so :-). Tires there, tyres here. But we are well used to that difference and it's not important. Meter for metre tends to be a mental saddle burr - it catches on the edge of one's consciousness annoyingly - BUT I am in fact well used to North American Colonials [tm] (as opposed to we Antipodean Colonials) using that spelling. | BUT your observation that 1/1 <> 1/2 was entirely correct in all systems :-). FWIW - Google thinks that these are tyres \$\endgroup\$
    – Russell McMahon
    Jan 6, 2014 at 22:26

You want to impart energy to a puck weighing 100 grammes. You want to accelerate it to 60mph or 96.5km per hour (22.8 metres per second). The energy received by the puck will be: -

\$\dfrac{m\cdot V^2}{2}\$

Where m is mass (0.1 kg) and V is velocity (22.8 m/sec). This is an energy of about 36 joules.

Given your description of the device you envisage, the puck will likely accelerate to that speed in about 0.1 seconds.

This means a power of 360 watts and can be either delivered by the motor or, if the motor is lower power, the momentum can be built up using flywheels.

I'd probably favour going for a 1 h.p. motor. I'd use a dc motor with a variable speed control either on an external filed winding or directly onto the armature if no field winding is available.

Hope this helps.


A linear motor could provide what you want with efficiency, but they are complex if you don't make your own.

linear motor comparative with rotary motor
Exemplifying a linear motor from a rotary motor

Anyway, for a lower cost belt system translating the rotational motion from the motor to linear, could provide what you want.

belt drive linear actuator
(source: pacific-bearing.com)

Belt drive linear actuator from http://www.pbclinear.com/Blog/How-to-Adjust-Belt-Tension-on-an-MT-Series-Actuator

You can made a more simple version of this. Use the actuator to move a "paddle" with the puck on low friction means.

The motor power is dependent on some variables, but you certain need a circuit to vary the current to the motor relative to the speed you want.

Another option is to use a rotary motor using the tangential force applied to the puck, and releasing at the correct time. Depending on speed and number of rotation before release you can vary speed. This would release the puck in some different directions in a regulated angle (from the launcher) depending on the controller precision. BUT: - You need a very good balance on the rotating part, if not, it's very dangerous. - You need a programmer and a micro-controller to release the puck at the correct time, as well to control the current to the motor.

So, I think the other options are more simple. A solenoid as suggested is an option, but they normally have low stroke, so it needs to have a high acceleration with this load. You could even try making you own. With a neodymium magnet and a coil (need go calculations), you can get pretty results. Anyway, if the cost is not a potential question, you can buy the correct stroke, force, solenoid, that shouldn't be cheap to impulse a 100g at 60mph, maybe at some degrees of inclination.

Other options are like the already projected ones, like this:

puck launcher http://i1.ytimg.com/vi/OKYWDKFZ6mY/hqdefault.jpg
From http://www.youtube.com/watch?v=OKYWDKFZ6mY

In this case you could use a universal motor's, but you need replace the brushes, probably they can hold more than a year without problems, or even more. Power tools and home mobile vacuum cleaners are examples of use of this motor, you can vary the speed with phase angle or "dimmers", although that inject some noise in the network that can be attenuated with filtering.

Another option is a induction motor. That's used in much places, there's near no part to wear with use, mostly the bearings, and a 2 pole one on a 60Hz system, would go to 3600 RPM (a bit less) without load. The problem goes controlling their speed, you would need an inverter that can be costly (some projects use phase-control put that's has a series of drawbacks).

(I read this is your idea, although I think you could put directly in the motor axle, or at least if using other mechanical transmission, you probably don't need to change ratios. I leave the calculations on motor power to other collaborators)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.