OAL question

[Panhead Bill]

All other factors being the same, how does different OAL affect pressure?  Although I'm speaking generally, I'm loading .45 ACP cast LRN.

I'm assuming shorter OAL means higher pressure, and conversely longer OAL means lower pressure. Also, relatively speaking, how much of a difference in length is within acceptable margins (not loading for extreme precision)?  For example, will .05" make a substantial difference in pressure?

Thanks,

Bill

[Devereaux]

Somewhere I believe there was an NRA paper published about the variation of pressure with bullet seating depth. It showed, IIRC, that pressure rises quite dramatically with shortening the OAL (or deeper seating of the bullet if all else is held the same), but I can't recollect just how much change there was with a particular change in depth. I would think that 0.05" is a lot of difference if you are near max load, and not so much if you are shooting a light load.

[Jim Fleming]

Bill...

A physicist explained it to me, but I'm not smart enough to repeat what he said. Nor do I have the space, time, or a proper keyboard.

So I'll try an analogy.

Think of a sealed brass walled box, exactly 12x12x12" square. 1728 cu.in. quite a bit of space in that box, eh? Let's put a compressed air fitting on the side of the box. Starting out at 0 psi. Attach an air hose, and raise the chamber pressure to perhaps 10 psi.

Now we remove the air fitting without losing any of the molecules of air. (It's an analogy, we're just pretending a few things.)

Now we shrink that sealed brass box by 1/2 of the original size. Maintaining the _ volume _ of air. If we keep the same volume, with no leakage at all, we increase the internal pressure exponentially!

I'm saying I can't repeat the numbers accurately. But it seems to me that, (I'm probably on numbers but I'm certain about the concept!)

I think that if we reduce the volume by by half, we increase the pressure by the square of the original pressure! In other words the original 10 psi increases to 100 psi!


Dev, you've had more physics classes than I have. Am I correct?

Sent from my Evil Black Droid that sneers at all iPhones!

[SteveZ]

The Ideal Gas Law PV=nrT is a linear relationship...no exponents involved. 

Solving this for P yields P =nrT/V.  As V (volume) goes to zero, the Pressure goes to infinity...but it does it linearly.....I think this is right...I'm a software engineer..we didn't deal much with pressure in school...except when I took 3 quarters of physics.

[Devereaux]

Steve Z is right. I think it's Boyle's Law, but the gist is that gas acts in response to 3 factors - Pressure, Volume, and Temperature. The others are constants relating to the particular gas being discussed, but they don't change the relationship. So if you assume the temperature is basically constant (same powder, same case, same bullet), then if you cut the volume in half, you would double the pressure. Note that while cartridges don't run at their max designed pressure, they DO run somewhere close. So cutting the volume will raise the pressure by the same RATIO. Ergo, if you cut 25% off the volume (so now you have 3/4 the volume), you increase the pressure by 1/3 (4/3 x pressure).

Decreasing OAL is basically cutting volume. You see where you can get into trouble pretty quickly if you aren't very careful.

[Teamklr2bar]

My Head Hurts! haha

[Devereaux]

Unfortunately you asked a basic math question.

You need to understand some of the basic issues involved. For any calibre that you pick, your increase in pressure will relate to how much you decrease the volume. Since volume is the area of the circle of the inside of the case X height of the space, and since the radius of the circle stays the same, your volume will vary with the height of the available cartridge space. So if you decrease the height 5%, you will decrease the volume 5%. If you then do the simple math, you increase pressure by 5.26%. If you take a cartridge like a .40, your operating pressure is about 40,000, so your pressure increase is approximately 2100. If OAL of a .40 cartridge is 1.135 and the case length is 0.850, and you assume 1/2 the bullet is IN the case, then you have approximately 0.710" of height of case for powder, etc. If you decrease the height by your suggested 0.05, you decrease volume by 7+%. This will increase your pressure by 7.6%. At the above stated pressure, you will get about 3050 increase in pressure. That gets you to 43500 cup - a sizable increase.

There are, of course, some assumptions in this example. Like we are not accounting for any case taper, the thickness of the base is not excluded from the volume calculation, and  you are running at the top of the scale for pressure, so you are shooting a hot load. But it is just an example.  So you can expect that for this example, the numbers approximated are actually somewhat low. But we are only trying to give you a sense of the amounts considered. Notice that a % decrease in volume gives you a larger % increase in pressure - because we are talking a smaller volume in the denominator. The ratio is linear, but not a direct replacement. Of course when you talk about a lower pressure cartridge you talk about a lower pressure increase. BUT usually low pressure rounds (eg. .45 ACP) have lower pressure tolerances.

?Did I muddy up the waters with all this.

[Jim Fleming]

As a matter of fact, Dev, you and Steve clarified and corrected me. Thank you. Seriously, thank both of you.

My concept was correct, but my specifics were wrong.

Sent from my Evil Black Droid that sneers at all iPhones!

[Panhead Bill]

My brain hurts!  Actually, strangely this all makes sense to my mathematically-challenged mind.

Thanks for the comments. My question was more theoretical than anything else, as I try to keep my loads well within the published data, with some built in margin for error, so that I'm not right on he edge of the safe levels, just in case there's some variance. But, I've got a chrono on order and I'm trying to learn some of the why's and how's of reloading to get a little more precise with my loads.

Bill

[SteveZ]

As a matter of fact, Dev, you and Steve clarified and corrected me. Thank you. Seriously, thank both of you.

I'm just glad you didn't ask a more challenging question like......

You're on the roof of a house...and you've got a roll of tar paper which is 100 feet long in a roll that measures 20" in diameter and weights 50lbs, the pitch of the roof is 25 degrees and the bottom of the roof is 35 feet off the ground.  If you let the roll of tar paper un roll while standing on the tail of it...how far from the house will it hit the ground and at what velocity?

Yes....I actually did have this question on one of my physics finals....and I got the answer right...but if I had to solve that problem again today... 

[Devereaux]

Hmm. Now THERE you have an interesting problem. Strikes me that it's basically a velocity/energy issue, with the pitch of the roof giving you the vertical rise to the potential energy at the top. Then you figure out the speed at the bottom and then the time to hit bottom from vertical accel, and the distance horizontally from speed (and if you are good, you even include the vectors since the roll is moving on a 25 degree slope). Easy - if you have a slide rule and time. Not so fun doing it in your head.

[Devereaux]

Actually knowing the old physics equations about velocity and accel can allow you to calculate your own range tables if you know the velocity of your round. I did that for my bow after chronoing the arrow, and the results were darn accurate concerning the drop at 20', 30', and 40'. Also if you are a hunter you can do some calculations about drop AND lead if you know the bullet speed and the distance you are going to possibly be shooting at. Deer run at 30 mph, which is 44 ft/sec. So once you calculate the time it takes your bullet to get to that distance, you will be able to figure out how far the deer ran in that time, so you know the lead. Simple (heh, heh!)   

[Jim Fleming]

 

Skull's bursting!