View Full Version : pneumatic launch velocity
Keeler836
10-01-2006, 22:15
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
10-01-2006, 22:57
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
11-01-2006, 00:29
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
11-01-2006, 00:41
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
11-01-2006, 00:43
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
11-01-2006, 00:45
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
11-01-2006, 01:06
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
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
11-01-2006, 09:47
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
11-01-2006, 10:31
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
11-01-2006, 11:32
ummmm... who missed a decimal?
Oops, sorry. Lesson learned: no physics between 10pm and 7am.
Arkorobotics
11-01-2006, 20:26
Oops, sorry. Lesson learned: no physics between 10pm and 7am.
Not a good excuse for school .. :( Teachers don't buy that anymore.. :(
greencactus3
11-01-2006, 21:51
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
16-01-2006, 16:58
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? :confused:
KenWittlief
16-01-2006, 17:17
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.
Arkorobotics
16-01-2006, 17:57
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.
Is this a mechanical lever or is some kind of pneumatic device, because I am seriously lost. :confused:
Matt Adams
16-01-2006, 18:32
See attached.
Arkorobotics
16-01-2006, 18:50
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??
KenWittlief
16-01-2006, 20:07
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!
Rickertsen2
16-01-2006, 20:09
wow... Release latches, lever linkages etc.. Yall are making this way too complicated.
wow... Release latches, lever linkages etc.. Yall are making this way too complicated.
Do you have some alternative to offer?
KenWittlief
16-01-2006, 20:21
wow... Release latches, lever linkages etc.. Yall are making this way too complicated.
as complicated as it needs to be.
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.
Rickertsen2
16-01-2006, 20:22
Do you have some alternative to offer?
Well yes but thats a secret ;)
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.
Matt Adams
16-01-2006, 23:23
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??
This is going to be honestly more of a hunch, so your mileage may vary*, but I'm going to say outright that I think there will be some velocity damping going on in these cylinders, and I'll venture to say that there just won't be a 12 m/s end velocity on these cylinders. Someone can feel free to do the math, but accelerating to 12 m/s in 1 inch...
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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