Brain teaser for any math wizards

jarmstrong said:

100 ? what does this number come from? shots taken?

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That is taking a pressure drop percentage as whole number (ie: 20%) and turning it into a decimal as to obtain the pressure drop. Say I want to know what 20% of a pressure drop from 1000 psi is. One approach is dividing 20 by 100 to get .2 then multiplying 1000 by that, to obtain 200, which is 20%.

I should of clarified that in the equation (that its a percent) so my apologies, I wrote two different equations to clarify.

-Matt

Here is another way to express the formula. Although I feel this almost gives away the answer.

What is x solving for?

V = Chamber Volume
P1 = Chamber Pressure
P2 = Plenum Pressure
P2D = Plenum Pressure Drop

x = V*(P1/(P2-P2*P2D))

Spartan said:

That calculates the dispersal rate for the flux capacitor. Well done.

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I'd give you a cookie, because man its definitely flux capacitor related! Bonus question clearly answered here.

Here is another way to express the formula. This makes it too easy..

AirGunShooter said:

Plenum Volume remaining after shot

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Definitely related to a new volume so you win! It's technically not *just* the plenum volume since that volume is generally static, but in this use case it does technically contribute to overall plenum volume, although marginally, so you're mostly right, and not wrong at all technically. A statistical anomaly if you will!

-Matt

drpietrzak said:

The math is less an issue than my understanding of what other than volume and pressure functionality are there.

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Its an equation to determine a new volume. Bernoulli's principle simplified.

v2 = (v1 * p1) / p2

One can use the same equation re-arranged for the pressure as well.

p2 = (v1 * p1) / v2


-Matt

shambozzie said:

Great stuff guys. That's what make the Nation great. "A seat at the table for everyone ".

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Very much so.

One could re-arrange the equation to determine approximately how far your pellet is down your barrel when your valve closes (and re-arrange again to find the pressure), since that volume is one that increases during the valves duration, you just need to know the relevant data to apply to the equation!

-Matt

Raider03 said:

The equation below don't use a Fpv correction, maybe they should....

v2 = (v1 * p1) / p2

One can use the same equation re-arranged for the pressure as well.

p2 = (v1 * p1) / v2

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What is FPV?

v2 = (v1 * p1) / p2

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This is Boyles law

"For a fixed mass of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional"

I mistook it for a derivative of Bernoulli's principle earlier, but its actually Boyles law.

Raider03 said:

Compressibility correction factor for when the ideal gas isn't a good approximation.

Gas Compressibility Factor

The gas compressibility factor or gas deviation factor is a measure of how close a real gas is to an ideal gas

www.engineersedge.com

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In the application that I am using this formula for, I am not sure its not necessary because its gas/pressure from one volume compressing gas/pressure in another volume via a piston. I'd imagine that correction factor would cancel itself out no? There are other factors to consider, such as the loss of thermal energy if the process is adiabatic.

-Matt

The majority of the 'work' done in this scenario is done in 50-100 micro seconds, very fast, so I keep the formula simple and sweet for myself, the compressive force continues to work for another 100-500~ micro seconds after the initial work is done so I treat the condition as ideal since over the duration of the event it becomes fairly 'ideal'.

-Matt