Pneumatic Cylinder Force

Hi,

I have a pneumatic cylinder which i want to calculate the force of and i just can't logically think of how. I want to say that i need to use the pressure of the air entering the cylinder somehow..

So as an example consider the following image...




The image is a couple of walls where one of them has a pneumatic cylinder attached to it. So let's i want to calculate the force the piston will hit the other wall. How would it?
I can calculate the volumetric flow rate and therefore the velocity of the piston.

Using the image below for the abbreviations.

I now believe it's P1xA1 - P2xA2 = Force ?





But how do i then convert force to kinematic energy? I know the two units aren't compatible as they are but could i add to the mix to be able to do it.

Or could i calculate it straight off using....

KE = 0.5xMassxVelocity^2

Where the velocity is the velocity of the piston and the mass is the mass of the piston(estimated from volume of it) but what about the pressure of the pneumatic air?



ohhh......i'm so confused with this. lol.

this might help :

http://www.engineeringtoolbox.com/pneumati...rce-d_1273.html

Huh. I would have simply thought inlet pressure minus atmospheric divided by piston area - at least as far as how much work the cylinder can do (or how much weight to 'balance' it). Or you're looking to calculate the force - in foot pounds I guess - as if the rod were throwing a punch?

F = P A

Gives you the pressure on the rod attached to the cylinder.

Work = F s

(s is displacement).

That's it, except you don't need to subtract atmospheric pressure, because virtually all pressure guages are calibrated to take this into account, or all guages would read 15PSI with no pressure input. Therefore, the atmospheric pressure amounts cancel out.
It's simply F=PxA. F is the force, P is the pressure as measured, & A is the area of the piston.
Note that this is for static force, & that the pulling force of a double acting cylinder is always lower than it's pushing force, because of the cross sectional area of the rod..
Dynamic force, such as when the rod strikes a stop, is something else again, as it has to take into account the mass of the moving components & the stopping distance involved.
Interestingly, dynamic force is much lower for hydraulic cylinders, because they always move at the same speed & they do not store energy.

Calculation depends also on:
How fast does the pressure build up
Does the pressure remain constant as the stroke increases the volume

Is this a static or dynamic thing?

If you want to know the peak pressure of the piston (already in motion) hitting the wall (not in motion) at the moment of impact, you'll need to define a whole lot more parameters (mass, velocity, elasticity, surface evenness, etc.) before any guess can be given.

Tim

The developed force is F = P x A

F = force exerted in Newton
A = application surface pressure expressed in mē
P = pressure in Pascal

Don't forget .. it is only the base, it takes into account other performance etc ... but nothing worth the experience of a professional

Well i want to create a mini projectile launcher, something that will fire fruit/veg across the garden on to a target. Something interesting to do with food that's grown rather mouldy.... I just didn't want to mention it at first, just in case anyone got funny about it but now i've decided to come clean about it, as it will make things easier.

So i have the equations for mass flow rate and volume flow rate, velocity of the piston and i've calculated the piston chamber areas. I have the equations of motion for the fruit/veg once it's been hit by the piston. So now all i have to do is calculate the velocity of the veg to make sure it doesn't destroy my garden and to keep it within the boundaries of my garden as i wouldn't want it to leave my premises. If i can find the velocity of the fruit/veg i can then adjust the pressure accordingly to do this. From the velocity i can calculate the maximum height and distance.

So my linking equation is......
Velocity of the fruit/veg= sqrt(kinetic energy (joules) / 0.5 x mass)

So as you can see i'm attempting to get a value of kinetic energy of the pneumatic piston. I know i haven't taken in to account any energy losses but that can be my safety factor And i do realise i've simplified the force of the piston but i thought i would just use the maximum force (found at the very of the stroke) to make things easier. As i think calculating the force, millimetre by millimetre would be a little excessive for what i want it for.


So would you say that the kinetic energy is the force at the end of the stroke multiplied by the distance the piston pushes the fruit? So say i allow the piston to move the fruit by 10mm before it reaches the very end of the stroke and the fruit leaves the piston. I would then multiple the maximum force by 10mm?

Sounds more like fodder for a small catapult or trebuchet...

If you assume that 10mm is enough for the system to stabilise, the velocity of the piston will slow from it's calculated value due to the added weight. At the end of stroke the piston must be stopped and this is waste energy. The piston assy must be made as light as practicable to minimise this. It may also be better to accelerate the veg and piston together, especially with soft stuff. If you have variable air pressure, just use the empirical method,otherwise known as suck it and see.

lol!

Good points GPG! I didn't think about the added weight of the fruit. I'll take that in to consideration. Cheers.