pneumatic launch velocity

[Keeler836]

I was wondering if anyone knew the velocity that a pneumatic cylinder would actuate at, I was thinking of using one on this years robot but i dont know if it would produce the necessary velocity to launch the balls as far as we would like.

[greencactus3]

Originally Posted by Keeler836

I was wondering if anyone knew the velocity that a pneumatic cylinder would actuate at, I was thinking of using one on this years robot but i dont know if it would produce the necessary velocity to launch the balls as far as we would like.

i dunno the numbers but no. i tried with a 3/4 bore cylinder with 4 FIRST tanks at 100psi and a tennis ball but it only launched it a couple feet at best.
but if you use an arm to extend the distance (And velocity) of the ball it will shoot a lot more. i had a phyz project that had size limits of 25cm high x 50cm long and i was able to shoot a tennis ball 10meters using a simple lever arm with 2 3/4 bore i think 6in stroke cylinders at initial 100psi with 4 FIRST tanks. with small sling arm.... sling arm would be cool but quite hard to reload in our robot's case tho

[KenWittlief]

dont say no too quickly

the problem with using a pneumatic cylinder to launch things is the small size of the tubing and valves. You cant get a fast enough inrush of air to move the cylinder piston as fast as you need to

but dont give up, there are ways around this

one would be to rig up a stop, or trigger/release pin about halfway down the cylinders length. To shoot something you would pressurize the cylinder against the trigger pin. That will build up a volume of air inside the cylinder at 60PSI

then you pull the trigger out of the way. There will be nothing to slow down its movement this way, except the ball that is being launched.

I must point out, these cylinders are not designed to be slammed into their own ends, so if you do something like this you must put a mechanical bumper or stop that limits the motion of the piston, so it cannot slam the end internally. If not, sooner or later the end will blow out of the cylinder, with a great deal of force: someone could be seriously injured.

there are other ways - brainstorm the concept: use the cylinders to load up some other form of energy storage/release mechanism.

[Rickertsen2]

Originally Posted by KenWittlief

I must point out, these cylinders are not designed to be slammed into their own ends, so if you do something like this you must put a mechanical bumper or stop that limits the motion of the piston, so it cannot slam the end internally. If not, sooner or later the end will blow out of the cylinder, with a great deal of force: someone could be seriously injured.

Of course, BIMBA does make cylinders with two types of internal bumpers so that they can tolerate this sort of use, but FIRST does not allow the. With some trickery we have gotten a ball to a cylinder to launch a ball at 8m/s, with plans to go higher in an entirely FIRST legal setup. It can be done.

[sciguy125]

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.

[KenWittlief]

I should add: you would have to leave the fitting out of the far end of the cylinder, or it will restrict the air as it flows out.

this means the cylinder can only be pressurized from one end, so you will need a weak spring to retract it for the next shot. maybe surgical tubing?

[KenWittlief]

Originally Posted by sciguy125

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.

using F=M*A and V=A*t my calculations indicate you only need 19 lbs of force over a distance of 6" to accelerate a 0.175 kg ball up to 12m/s

the acceleration is 480M/S^2, and the time to get from 0 to 12m/S is 25mS

[dlavery]

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

[sciguy125]

Originally Posted by KenWittlief

using F=M*A and V=A*t my calculations indicate you only need 19 lbs of force over a distance of 6" to accelerate a 0.175 kg ball up to 12m/s

the acceleration is 480M/S^2, and the time to get from 0 to 12m/S is 25mS

There's something wrong here. All the equations make sense. However, the solutions that use said equations don't. Actually...nevermind, in the middle of writing this post I found the descrepency.

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.

[KenWittlief]

Originally Posted by sciguy125

There's d = 0.5*480*0.000625 = 0.15m = 59"

You missed a decimal place in your conversion from meters to inches.

.

ummmm... who missed a decimal?

0.15 meters = 59 inches?

so 1 meter = 32 feet?!

0.15 meters = 5.9 inches (what I said the first time :^)

(32 is the difference between slugs and pounds: pounds = force, and slugs = mass)

[sciguy125]

Originally Posted by KenWittlief

ummmm... who missed a decimal?

Oops, sorry. Lesson learned: no physics between 10pm and 7am.

[Arkorobotics]

Originally Posted by sciguy125

Oops, sorry. Lesson learned: no physics between 10pm and 7am.

Not a good excuse for school .. Teachers don't buy that anymore..

[greencactus3]

Originally Posted by KenWittlief

I should add: you would have to leave the fitting out of the far end of the cylinder, or it will restrict the air as it flows out.

this means the cylinder can only be pressurized from one end, so you will need a weak spring to retract it for the next shot. maybe surgical tubing?

nah. have the next ball push it back.

[Arkorobotics]

Originally Posted by dlavery

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

What is a range-multipier lever with a 1:10 ratio???? what do you mean?

[KenWittlief]

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.