I’m making a direct motor catapult but have no idea what ratio to stick it on for long range flinging does anyone have a ballpark estimate? The arm is about 4” long
Team 118 in nothing but net did a 5:3 Torque gear ratio I believe that can shoot full-court into the nets. But, considering the fact that you’re sniping long range and only have two ball possession, I would suggest that you maybe try something like 5:1 High Speed to get a bit more torque so you can use less motors.
EDIT:// Wait, when you mean direct motor do you mean no slip gear?
No slip gear, no bands just the motor turning the catapult to fling it. Don’t know if it will work but giving it a try
So you are definitely still going to want some rubber bands. Just like on a lift the rubber bands just assist.
Ok we can do that just not pushing against the bands and then releasing like a standard catapult.
In terms of ratio I would start with 5:1 on a reasonable length lever arm. Change from there as appropriate. I dont have a great sense of how fast you need to go.
Actually, I’ve once built a catapult before that is directly powered by motors with no slip gear. I had very limited experience during that time, but I don’t believe it is possible to snipe that far with the type or motors VEX provides. I was able to do one that can do bankshots during that time, but never was able to snipe. Acceleration is just too slow.
There was a team in my region in NBN that found that it required 5 motors for a direct drive catapult to shoot full field and it requires rubber bands. Seems a bit infeasible to use 5 motors on a launcher this year. I have no idea what their ratio was.
Ok we’ll probably use slip gear then
Curious why that might be seen as unethical. It’s certainly within the rules of the game and the ethos of VEX.
He probably meant stupid not unethical
36:60 2m torque slipgear
Hello,
I saw your video. Very effective shooter. Great recommendation. Plans to make a two ball catapult?
-Sal
Maybe I’m making a statement about 5 motor launchers. Or maybe my brain stopped functioning while I was intending to write infeasible. ¯_(ツ)_/¯
Thanks for the clarification. Just thought I was missing something.
One motor has a stall current of 4.8 Amps, running at roughly 7.2 volts. Realistically, you won’t hit stall current. You’ll most likely hit 3.5- 4.0 Amps.
Approximate: 5 amps, 7 volts. That’s 35 watts. Or, 35 joules of energy. Please note that this means you need to have the motor accelerating the ball for 1 second, as 35 joules is 35 watts of power for 1 second.
Kinetic energy = 0.5 * mass * velocity^2.
So, if 100% of the energy is used to propel JUST the ball (without the catapults weight and lever), you would
have a velocity of: the square root of (35/(0.5*0.055)), which is 35 meters per second. If your catapult weighs twice as much as the ball, and assuming torque makes no difference to the work needed (it does), you would end up with 20 m/s.
If we assume the catapult weighs twice as much as the ball, and we are running at 3.5 amps, we end up with 17.5 m/s.
These calculations do not represent torque, energy loss to friction, etc. This ONLY represents what happens if we take 100% of the energy expended in 1 second by the motor, at 100% efficiency, and pore that energy into a purely mechanical, kinetic energy transfer. Realistically, we’re talking more along the lines of 4-5 m/s if you get lucky. At an angle of 45 degrees, 4-5 m/s will give you a flight time of 0.3 seconds, for a total distance of around 1 meter horizontally. (and about .8, .9 vertically.)