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#1
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Re: pneumatic launch velocity
Aside from the actual speed that the piston moves (I hadn't thought about that until now), I don't think that it would be practical to launch the balls this way.
E=0.5mv^2 To get the ball to 12m/s, we need to supply it with 13J. This assumes no spin and minimal friction losses. W=Fd A 2" cylinder at 60psi will be able to give this 13J with a ~6" stroke. That's a huge volume of air. Your fire rate wouldn't be that great. Of course, if your goal isn't the full 12m/s, some pneumatic mechanism might work for you. |
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#2
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Re: pneumatic launch velocity
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the acceleration is 480M/S^2, and the time to get from 0 to 12m/S is 25mS |
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#3
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Re: pneumatic launch velocity
You are over constraining the problem. There is no need to have the piston directly provide the full range of motion required to accelerate the ball. Use a 2" diameter x 1" stroke piston. The piston is connected to a range-multipier lever with a 1:10 ratio. The piston will provide 188 pounds of force over 1" as input to the lever. The end of the lever will travel 10" with 18.8 pounds of force. As noted in Ken's post above, there is more than enough energy here to launch the ball.
-dave |
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#4
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Re: pneumatic launch velocity
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#5
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Re: pneumatic launch velocity
an 11" lever with the pivot point 1" in from one end will have a 10:1 multiplier
ten times the distance (speed) on one side vs the other, with 1/10th the force available. |
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#6
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Re: pneumatic launch velocity
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#7
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Re: pneumatic launch velocity
See attached.
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#8
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Re: pneumatic launch velocity
Ahhh! what if you were to take that 1" stroke piston 2" diameter and fire it directly to the ball. Would that be enough to fire it to the center goal??
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#9
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Re: pneumatic launch velocity
probably, if you have enough pressure in the cylinder
and you preload the cylinder with pressure, and use a mechanical release like I talked about on page one the only caveat, a 1" piston throw may not deform (dent) the ball enough to transfer all the energy quickly. You might need a bowel shaped plate to keep the ball from deforming as the cylinder punches it. a 1" piston throw into a 1:10 lever: definitely! |
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#10
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Re: pneumatic launch velocity
wow... Release latches, lever linkages etc.. Yall are making this way too complicated.
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#11
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Re: pneumatic launch velocity
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#12
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Re: pneumatic launch velocity
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Think about how you can maximize the airflow to the cylinders. Think about what sizes will move fastest. Think about all that wasted energy on the return stroke. |
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#13
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Re: pneumatic launch velocity
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Pnuematic shooters have some definate advantages: + The tank can be precharged before the match starts (a wheel must be spun up before it can launch a ball) + a pnuematic cylinder with a release pin is much simpler to build and mount than a motor/gearbox/wheel + much smaller + much lighter + the pressure control is simple: the regulator valve is very accurate + you can cycle pnuematics really quickly. I think we will definately see pnuematic launchers this year, in many configurations. |
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#14
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Re: pneumatic launch velocity
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Oh sheesh, I'll do the math v^2 = 2 * a * d (12 m/s) ^ 2 / (2 * 1in / 39.37in/m) = a a = 2834.6 m/s^2 so... 1/2 * a * t^2 = d t = sqrt ( (1/39.37) * 2 / 2834.6) t = 0.004 seconds. I'll say (from experience) I haven't seen pneumatics move 1 inch in 4 milliseconds. I think there's a lot of time required to have adequate airflow from the tanks to the piston, and like I alluded to before, I really think there are vicious damping issues that arise. Hence, it's my personal recommendation that you do not count on having a directly driven ball launcher using a pneumatic cylinder. However, I feel pretty confident that you could come up with something with some mechanical advantage ideas that was previously mentioned. Matt * an admittedly stolen disclaimer of Dr. Joe. Imitation is the greatest form of flattery. |
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#15
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Re: pneumatic launch velocity
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Ken's solution: F=ma => a=F/m 19lb = 84.5N a= 84.5/0.175 = 480m/s^2 V=at => t=V/a t = 12/480 = 0.025ms d=0.5at^2 d = 0.5*480*0.000625 = 0.15m = 59" You missed a decimal place in your conversion from meters to inches. The problem that I initially found was that 19lb (84N) over a distance of 6" (15cm) only supplies 1.26J. I was surprised I didn't notice that it was exactly 1/10 what it should have been. |
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