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Unread 11-01-2006, 09:47
sciguy125 sciguy125 is offline
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Re: pneumatic launch velocity

Quote:
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.
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Unread 11-01-2006, 10:31
KenWittlief KenWittlief is offline
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Re: pneumatic launch velocity

Quote:
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)

Last edited by KenWittlief : 11-01-2006 at 10:50.
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Unread 11-01-2006, 11:32
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Re: pneumatic launch velocity

Quote:
Originally Posted by KenWittlief
ummmm... who missed a decimal?
Oops, sorry. Lesson learned: no physics between 10pm and 7am.
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Unread 11-01-2006, 20:26
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Re: pneumatic launch velocity

Quote:
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..
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