# Cam launcher energy transfer

I’m trying to find how much energy would be transferred from a pinball launcher to a Nbn ball. I have calculated the potential energy, but I want to find a way to see how much of that energy disappears when the launcher hits the ball. Any ideas?

Measure the balls exit velocity - hard to do but will give you the most accurate answer

Alternately,
See how far the ball goes and use that to determine its initial velocity/kinetic energy - easier to do but less accurate because of drag

Why do you want to do this?

+1 to this. The deformation on the ball is hard to measure, so measuring the velocity as it just gets hit/leaves the launcher is your best bet.

Slow mo cameras on smartphones against a backdrop with measured lines might be good enough. Measure the ball across a piece of paper with some fine ruled lines to see how far the ball moves in the first tenth of a second or so.

Like you see on Mythbusters… (best example I could find easily on google) Alright, here is the update. I measured an idea of how far we shoot normally. By using basic geometry ideas, I figured that we shoot around 8 1/2 feet each time we shoot, and it takes about 1 1/2 seconds from the second the ball is struck to when it gets into the rim of the high goal. This is a velocity of 5.78 ft/s. putting this into the KE formula with 50 grams as the mass which is 835.31 J. The only thing is I measured the potential energy to be below 100 J. I used 14.408163592 lb/in as k and 6 as the spring stretch legnth… Where did I go wrong?

There are some simple ballistics equations that should allow you to determine the launch velocity vs. distance. See this article where is shows equations for distance travelled:

These ignore drag, but that’s probably okay for giving you a rough idea. Note that you must know something about the launch angle.

Off the top of my head, I think the kinetic energy of an NBN ball traveling fast enough to make it across the field is in the neighborhood of about 2 joules. To put that into perspective, a single 393 motor theoretically will output about 4 watts or 4 joules/second, ignoring friction, etc.

With an iPhone my kids tried to video their puncher launching a ball, but it happens so fast, it was just a green blur. I’d like to know how to set up a flash to do this with a normal digital camera just to get some cool photos.

The article looks great! skimmed through it and will read it throughly tomorrow, but if the energy required to shoot full court is only 2 joules, why am I getting insanely high readings through equations? The potential energy of an elastic isn’t affected by the angle, so the energy transfer shouldn’t either, but then again I’m supposed to start with more potential energy than kinetic.

Keep in mind that the velocity of the ball is one thing and the velocity of your puncher might be something very different from that. I think the puncher velocity and resulting ball velocity would be the same only if the puncher had the same mass as the ball, and that’s probably very unlikely.

When being stretched, the elastics are essentially storing energy that the motors are pumping into them. When released, the elastics are accelerating the entire mass of your puncher. This puncher mass has energy and momentum, and some of that energy and momentum will get taken by the ball when it gets punched. A lot of momentum will probably remain with the puncher after it hits the ball, and that left over puncher energy will end up being dumped into your puncher’s stop. That’s where lots of energy probably gets wasted. I’m guessing that the ideal puncher could have the relationship of its mass “tuned” with the elastics in such a way that less energy is wasted, but I haven’t done any kind of analysis of that and my kids couldn’t understand it even if I showed it to them, so I’m not bothering. I know their own design kicks the living heck out of their robot and their stops often get hammered to bits.

Have you actually measured your spring constant personally to see what it is over the full range of your elastics? Also, things might depend on how your elastics are pre-stretched (how much they are stretched before the action of the motors starts to stretch them toward the firing position.) It’s helpful to understand how your elastics actually behave so you will know if they are running into a non-linear region where your spring constant won’t be so easily calculated.

So I’m not sure how you got 835.31 J. If the velocity was 5.78 ft/s then that amounts to .078 J. Remember it’s kg*(m/s)^2. I’m guessing the units tripped you up (or I’m making a mistake - either one) As for the elastic potential energy, your rubber bands likely are not hookian unfortunately, and if they were then I’m getting 29.3 J with 6 inches of stretch.

Wow… I was kinda hoping my intro to physics class could carry my through this, but if the surgical tubing doesn’t follow hookian, then I have no idea what I’m doing. Also, I did get my units mixed up and you are correct Highwayman. So the only question now is, if the puncher is putting the force into the ball, then I need to be measuring the kinetic energy in the puncher, correct?
Thanks for all the help guys! I really needed this otherwise who knows what crazy results I would be ending up with.

That might work better than the potential energy of your elastics, yeah.

Don’t panic. One approach is to get a piece of graph paper and create a Force vs. Displacement curve for your elastic. Plot it with meters on the X scale and newtons on the Y scale. Work = Force x Distance, so the amount of work done on the elastic will be the area under that curve, and this work will be the amount of potential energy that is stored in the elastic (ignoring any inefficiencies). The units of work (= potential energy) will be in newton-meters.

If the Force vs. Displacement curve for your elastic were linear, then you get
Work = 1/2 Force x Distance
because the area under the curve would be a triangular area. (Area of triangle = 1/2 base x height).

But if the Force vs. Displacement curve is not linear, then you could count all the squares of your graph paper that are under that Force vs. Displacement curve and get a number for Work (= potential energy).

Note that a newton-meter is the same thing as a joule. And a watt is one joule per second.

If you had an equation for your non-linear Force vs. Displacement curve, then you could use integral calculus instead of counting all of your graph squares under the curve.

When thinking about this problem, be careful you keep very clear exactly what you’re thinking about. Don’t confuse force with momentum with kinetic energy. The puncher transfers kinetic energy, but it also transfers momentum, which is a different concept and uses a different calculation. Of course, forces are involved, but you probably will get the most insight out of this problem by considering the transfer of energy and/or momentum from puncher to ball.

One problem you might have with determining how much energy gets transferred to the ball is that you probably don’t know what your energy losses are when the ball gets compressed during a punch. Some of that compression energy will get returned as kinetic energy to the ball as the compression springs back. But some amount will be lost as heat. Perhaps one experiment you could easily do is drop the ball from, say, 10 feet or so and see how far up it bounces back. That might give you a crude sense of how much energy is lost/returned during rapid compression/decompression.

How would you recommend that he measures the force?

Also, what are we trying to accomplish, again? Finding the efficiency of the launcher, right?

Comparing momentum and energy could be an interesting exercise in this case.

Before we got some cheap spring scales, I had the kids use a plastic milk jug and fill it with various amounts of water as they measured the displacement. Water is about 1 gram per cubic centimeter (cc), so you can get some rough ideas from that.

100 cc of water = 100 grams = 0.1 kilogram, etc.

Be aware that the force can change over time and that the force can change depending on how many times the elastic is cycled. It can get somewhat complicated if you try to factor in all the things to worry about.