Backwards math to optimize energy output

[Rohmpm]

akuric

Thanks, I got it all immediately and understand that calculations are exacts but results are approximation to help one decide if a change within reason MIGHT be worthwhile.

 

[Motorhead]

He who can not debate publicly or make any effort to substantiate any information shared ... is simply grandstanding and listening to them self wanting to sound intelligent.



DELIVERY is Key ( See TACT ) .... Offering support and a path to understanding what your selling is TEACHING. Don't or Can't do that your PREACHING.

 

[Guest]

There is so much more into it than SIMPLE math of calculating force generated by pressure times area and acceleration by barrel lenght. Also nothing limits port size to caliber size. Ppl have built bulletshooters with larger ports than caliber for years because you actually gain alot more power from it even if barrel port becomes choke point you get alot higher pressure shock. Other factors than those said above are projectile hardness, barrel friction and friction and surface quality shape and lenght of porting that air need move through to also dwell time bullet sizing etc play huge role its far from just calculating something and assuming something. 

 

[Guest]

1) You use diameter of port rather than area because you're not calculating the force being applied over the ports area, thats not how this calculation works, its approximating the max flow for your bore. So if you're ported to 75% of your bores diameter, you're essentially making 75% of your bores potential power output in fpe for your current configuration (pressure/barrel length/plenum volume)...

Dont' want to clutter up the thread so will quote each point individually

If I have a bore which is .25 cal. The cross sectional area of the bore is PI*r2. Simple enough. So (0.125^2)*3.141592 is 0.049 sq. in. That is the cross section of the bore and the area upon which the pressure acts. The diameter of that bore is 0.25. If I had a transfer port which was also 0.25 inches the ratio of the area of the port to the area of the bore would be 1 to 1. That part is true. Again I hope, simple enough. Now then, if I had a transfer port which was .20 the ratio of the diameter of the port to the diameter of the bore would indeed be 0.8; however, the ratio of the area of the port to the cross sectional area of the bore would be (0.049/((0.1^2)*PI)) ... Remember we calculated the sectional area of the bore above so we don't have to do it again. The new ratio of the area of the port to the area of the bore is 0.64 to 1. In other words that ratio decreases much, much faster than the the ratio of the diameters. For example if the diameter of the port were half the diameter of the sectional area of the bore the ratio of the diameters is 1:2. By your reasoning a port which is half the diameter of the sectional area of the bore would develop half the power (would move half the air necessary to develop that power) you could get out of the bore. A calculation of the ratio of the area of said port (0.125) and the bore (0.25) using the method above works out to (0.049)/((0.0625^2)*PI) which ratio is: 1:4 for all intents and purposes. I won't bother you with the self-evident fact this is because the formula for area of a circle squares the radius. This means that a port which is half the diameter of the bore will always have an area which is one quarter of the area of the bore.

Sorry for the long winded explanation of my thinking. Wanted to keep it as basic as I could and make sure I hit all the salient points. Some things are just not obvious to me and it helps me to work them out long hand. Check the math if you wish, there may be errors. The one thing I know by looking at the formula for calculating the area of a circle is that the area of a circle of radius 1 is four times as big as the area of a circle of radius one half. Basic geometry.

Now as to the exact effect of calculating energy transfer through a port into a vessel via a compressed fluid. I am not a chemist so I will simply speculate using the facts and concepts that I got in my chemistry classes, many years ago. We know that fluids under pressure will equalize across a barrier that has a port in it. We know that there is turbulance around the edges of the port as the fluid literally bumps into itself and the port. We know that turbulance extends into the surrounding fluid some distance and that eventually it gives way to laminar flow. This intuitively suggests that the relationship between the maximum volume of fluid a larger aperture will pass and the maximum volume a smaller aperture will pass is not linear. I assume you understand what I mean by "linear"? Not only that but that ratio becomes less linear as the size of the aperture is reduced. At some point practically no fluid will flow because of the turbulance at the port.

All that said, if you want to talk to someone who really, fully understands this stuff, you should cultivate an acquantence with Bob Sterne. If you are looking for a healthy discussion on fluid dynamics you need to talk to him, not me. I will say this though, I am fairly confident that I would be able to follow almost all of what he told me or showed me on the first pass.

Onward.

 

[Guest]

2) 5cc = constant (volume used per shot), 30 and 50 are the variables of plenum volume, 35 and 55 are the volumes all added together to calculate your pressure. This is how volume and pressure works...its pressure 2 = volume 1 * pressure 1 / volume 2. Its boyles law...a gas law...

Ok, I grok Boyles Law.

So you are suggesting that a set point of 2000 psi with a plenum of 30cc will have one pressure and the same set point will have a different pressure in a larger plenum? Right.

Also trying to understand how the volume of air used per shot makes it's way into Boyles Law.

So lets take a moment to touch on Boyles Law. It really is very simple to understand.

It says if I have a given volume of a gas, say 1 cubic foot and I force it into 1/2 of a cubic foot the pressure will double. That is it in a nutshell. The rest is basic math. It is not nearly as hard as calculating the area of a circle. No magic numbers here.

Here you are trying to say if I have a plenum and I make it larger I will get more power. Then you trouble yourself to say, but only if I do something to the valve and/or port to use that power. That last part is exactly right. You could use a fifty five gallon barrel as your plenum and you power out would not change one bit because you valving system has not changed and therefore you are getting whatever it lets out of the plenum.

I guess I should add this. If you are talking about a regulated rifle and you are evacuating the pelnum when you shoot, then yes increasing the size of the pelnum will give you more power for obvious reasons. In that case you would be talking about a plenum which was about the size of the amount of air you are using per shot, I should think. Maybe I am wrong on that?

Does that make sense to you?

 

[Guest]

***Pressure increase/decrease***

Regulator pressure / Set point = 2000

Increasing to 2200

2200/2000 = 1.1

Energy output at 2k = 60 fpe

Energy output at 2.2k is therefore 60*1.1 = 66 fpe

To obtain this you will need your projectile to be roughly 1/2 grain per fpe, ie: 66/2 = 33gr. Additional hammer strike will be needed to overcome the additional pressure holding the valve closed. Increased air consumption to be expected.

Ok here. 2200/2000*60=66 yes. This is a linear equation and *might* give you a reasonable estimate of what you are going to see at the muzzle.

I suspect nearly everyone of us has done a similar calculation at some point. This sort of back of the envelope estimate is accurate when changes are incremental.

In this case you presuppose the valve is not restricting the flow of the air. Fair enough. I don't know where you are getting your pellet weight recommendation but I do see heuristics like this having some utility. Some of my graphs might even provide you data so that you could refine these heuristics. I think a Talon (or even a cheap Stormrider) would be a great gun for you because you like working the math and studying the guns and that's a really good thing.

What ways can this heuristic fail? When you get more than say ten percent or so away from the current values for you gun you will start to see these estimates diverge from reality because of the changes you suggest may be needed in the valve system. If your heuristic were a good model and you graphed the muzzle energy of a shot string you would get a nice straight slope with each shot dropping a bit more energy as you shot down the pressure in your tank. Typically we don't see that when we graph our guns, even unregulated guns. You will see some pretty interesting graphs and data if you look at the threads that detail my experiences with my Talondor. Sometimes no amount of additional air gives you any additional velocity, for example.

Onward.

 

[Guest]

***Barrel Length increase/decrease***

Old Barrel length 20"

New barrel length 22"

Energy output at 20" = 60 fpe

60 / 20 = 3

3 * 22 = 66

New energy output potential = 66 fpe

To obtain this you will need your projectile to be roughly 1/2 grain per fpe, ie: 66/2 = 33gr. Additionally you will be required to create more dwell/lift and use more air to make use of the additional barrel volume.

This one here is my favorite of the four because my Talondor taught me this lesson.

Suppose you are shooting a gun with a 10 inch barrel, you make that barrel 12 inches. You are getting 30 fpe out of the 10 inch barrel. You extrapolate that you will get 36 fpe with you new barrel. You shoot it on the chron and see only 33 fpe. What does that tell you? I'm pretty sure you know all this so I won't go into boring detail. It means you are not moving enough air to get the full benefit of your longer barrel. You touch upon that in your note. There are lots of ways to try to get more air to the back of that pellet while it is in that barrel.

Now suppose you set your dwell higher by increasing hammer spring tension and you don't see any additional increase? What does that tell you? Well the only thing it means is that you didn't change the amount of air that was getting to the pellet while it is in the barrel. So what do you do? You enlarge the port? Ok. Suppose you still are not getting your full 36 fpe. Time to stick a really heavy pellet on there and voila 38 fpe? WTF why? Barrel time. Pressure has more time to obey Boyle's Law with a heavier pellet. Turbulance in the valve and transfer port were eating up your additional power. Well, that would be my guess. To be honest I don't know much about PCPs or fluid dynamics. So I'll refer you to Bob again with the caveat that you spend more time listening to him than talking to him. You won't regret it.

So heuristics are nice, as far as they go. Mathematical modeling is nice as far as it goes. There are a whole lot of folks on this forum who have been doing this for many, many, years who might not be able to do that math but who can immediately figure out what needs doing and tell you about how much change is needed. They can do it in their heads without math.

So, maybe ... a little more of this ?