Can a Single Acting Spring Loaded Cylinder Launch a 1kg Body 5cm Upward?

[MadaraUchiha]

Hey guys
im planing on buying this single acting spring loaded cylinder with specs

BORE, 1 1/2"
STROKE, 5"
PORT, 1/4" NPT
ROD DIAMETER, 1/2"
ROD THREADS, MALE 1/2"-20, THREADS ARE 1 3/4" LONG

If i supply it with air at 100 psi
what force would it generate ?
and how far can it throw a body weighning 1kg, 5 cm away from rod end straight up into air ?

Thanks

 

[CWatters]

what force would it generate ?

The bore is known so you can work out the area of the piston (in square inches). The pressure (in pounds per square inch) acting on the piston is known so you can easily work out the force (in pounds) acting on the piston.

how far can it throw a body weighning 1kg

Using Newtons laws (eg F=ma) you can work out the acceleration the force would produce.

You also know the stroke so you can work out the distance over which the force is applied to the object.

Then (assuming the acceleration is constant) apply the SUVAT equations to work out the final velocity (eg the launch velocity).

http://en.wikipedia.org/wiki/Equations_of_motion#SUVAT_equations

Then knowing the launch velocity you can apply the SUVAT equations again to work out how high it will go.

The answer won't be exact because you don't know how much friction there is in the piston.

 

[MadaraUchiha]

You also know the stroke so you can work out the distance over which the force is applied to the object.

Then (assuming the acceleration is constant) apply the SUVAT equations to work out the final velocity (eg the launch velocity).

http://en.wikipedia.org/wiki/Equations_of_motion#SUVAT_equations

Then knowing the launch velocity you can apply the SUVAT equations again to work out how high it will go.

The answer won't be exact because you don't know how much friction there is in the piston.

Thanks,
I did but the answers just didnt seem right maybe i mixed up imperial and metric units (i want the answer in metric)...
Tell me if I am doing this right.

Bore= 1.5 inch = 0.0381 m
Area= pi* (o.0381/2)^2 = 3.14*0.00036 = 0.0011 m^2

Force=PressurexArea
Pressure= 100 psi=689476 pascals
So F= 689476*0.0011= ~780 Newton

Now F=ma
=>780=1*a
a=780 m/s/s

using v^2-u^2=2gh
here h= 5inch stroke = 0.127
since load is 5 cm away, h=0.127-0.05
h=0.077 m
at end of stroke
h=0.077m, g=9.8m/s/s, u=0 and v= to be calculated

So now do i use g=9.8 or the one calculated using f=ma i.e. 780 m/ss ?

 

[CWatters]

Bore= 1.5 inch = 0.0381 m
Area= pi* (o.0381/2)^2 = 3.14*0.00036 = 0.0011 m^2

Force=PressurexArea
Pressure= 100 psi=689476 pascals
So F= 689476*0.0011= ~780 Newton

Now F=ma
=>780=1*a
a=780 m/s/s

Ok so far.

using v^2-u^2=2gh
here h= 5inch stroke = 0.127
since load is 5 cm away, h=0.127-0.05
h=0.077 m
at end of stroke
h=0.077m, g=9.8m/s/s, u=0 and v= to be calculated

The acceleration during the piston stroke is "a" not "g" so the equation should be..

V²= U² + 2as

where
V= Velocity at end of stroke
U= Velocity at start of stroke = 0
a = 780m/s/s
s = 0.077m

so the launch velocity V is

V²= + 2*780*0.077
= 120m/s

That's rather fast so it might not be possible to ignore the effects of friction in the bore of the piston and air resistance of the projectile. Anyway...

So now you have the launch velocity you need to work out the height it will go to. You use the same equation modified because now the acceleration is that of gravity..

V²= U² + 2gh

V=0 (it stops at the top)
U=120m/s
g=-9.8m/s/s
h = height you are trying to find

So

h = 120²/(2*9.8)
=734m

This is an upper limit. In addition to friction losses I can see another problem that might reduce this.. Will the valve and pipework that turns on the gas be able to deliver and maintain 100psi during the stroke? If it let's gas into the cylinder relatively slowly the piston will start moving but won't accelerate as fast. I suggest you work out the time it should take the piston to travel the 5" and make sure the valve can open fully in say 1/10th of that time.

Sorry I had to edit this post a bit.