Curious Questions

[Phil3]

As a curious type, I have been wondering about a few things that I don't see answers to in my limited Google searches.

1) Just how hot is the flame in the chamber when a round is fired? I expect the temperature rises with pressure, but if the chamber is say around 50,000 psi, then what might the temperature inside the chamber be at that moment?

2) I have long wondered what amount of actual pressure it takes to force a bullet from the grip of the case neck. Or how much pressure it takes to force the bullet into the engraving and continue to push it down the barrel with the friction of the rifling trying to hold it back. A .224 diameter bullet has an area of .03940814 square inches. If 50,000 psi exists behind it, that bullet is receiving an actual push of 1,970 lbs. How much of that is needed to escape the neck grip or rifling friction?

3) If a bullet accelerate to near maximum velocity (let us say 3000 fps) in the first 24" of a 1:9 twist barrel, that means the bullet was spun up to 240,000 rpm in about .00133 seconds. How much energy that does take?

I should have been a scientist or lab tech...

Phil

 

[firearmdoc1]

If I remember correctly, I saw something about ordnance testing done at Aberdeen Proving Grounds. The subject was about barrel erosion and flame temperatures. I believe it was written by either Hatcher or Ackley.

 

[M-61]

As a curious type, I have been wondering about a few things that I don't see answers to in my limited Google searches.

1) Just how hot is the flame in the chamber when a round is fired? I expect the temperature rises with pressure, but if the chamber is say around 50,000 psi, then what might the temperature inside the chamber be at that moment?

2) I have long wondered what amount of actual pressure it takes to force a bullet from the grip of the case neck. Or how much pressure it takes to force the bullet into the engraving and continue to push it down the barrel with the friction of the rifling trying to hold it back. A .224 diameter bullet has an area of .03940814 square inches. If 50,000 psi exists behind it, that bullet is receiving an actual push of 1,970 lbs. How much of that is needed to escape the neck grip or rifling friction?

3) If a bullet accelerate to near maximum velocity (let us say 3000 fps) in the first 24" of a 1:9 twist barrel, that means the bullet was spun up to 240,000 rpm in about .00133 seconds. How much energy that does take?

I should have been a scientist or lab tech...

Phil

Question #3. If I may add this to it. If a bullet in a 1-9 twist hits 3000 fps, how much faster would it be (accuracy excluded) from say a 1-17 twist everything else being the same?

 

[Phil3]

Good question, I wish I knew. Phil

 

[M-61]

This stuff drives me nuts as I am sure there is a way to figure this out mathematically and that's where I loose it.
If the slower twist produced 10% faster velocity (3300fps) wouldn't that mean the faster twist consumed 10% more energy just driving the bullet thru the barrel therefore resulting in a lower velocity? And if were just a speed contest, a smooth bore would result is the fastest bullet?
Sad thing is I don't know which class I should have paid more attention in!

 

[Nomad47]

You are forgetting that it takes pressure to create velocity. No such thing as a free lunch.
I'm not going to say one way or another but it was stated in a topic a while ago that twist rate has no bearing on velocity.

 

[jonbearman]

I would get german salzaar to comment.He is quite knowledgable on many things dealling with rifle questions on the technical end.

 

[jonbearman]

Go to www.frfrogspad.com/intballi.htm This should answer all your questions phil. Flame temps for ball powder is 3200-3300 degrees and srtick powder burns at 3300-3400 period.For 1.1 milleseconds.Basically a cutting torch.

 

[Phil3]

Great link! Thanks! - Phil

 

[DRNewcomb]

You know, some years back I waded through a tome on internal ballistics to understand the answers to just this question and today I can't recall any of it. Amazing how you forget partial differential equations when you don't use them. ???

 

[CJ6]

#2 should equal the pressure needed to seat the bullet (without powder compression or cold welding)

 

[Lapua40X]

... A .224 diameter bullet has an area of .03940814 square inches....

Not certain if the area displayed there is correct but, if .03940814 square inches is the area of the entire surface of the bullet, you may want to recalculate to reflect the bearing surface as one drag factor and the non-bearing surface as another.

 

[Phil3]

... A .224 diameter bullet has an area of .03940814 square inches....

Not certain if the area displayed there is correct but, if .03940814 square inches is the area of the entire surface of the bullet, you may want to recalculate to reflect the bearing surface as one drag factor and the non-bearing surface as another.

The .03940814 square inches is the area of the bullet that is subject to pressure from powder gases driving the bullet down the barrel. That is, if this was a flat based bullet, the equation is( (.224/2) x (.224/2))*3.1415927.

 

[KevinThomas]

To answer just one of the questions posed in this thread, pressure differential between various fast/slow twists amounts to about one half of one percent . . . virtually nothing. It doesn't bleed off much (if any) velocity, and adds very little in the way of pressure. Perfect example is the original M16 and the 55 grain FMJ M193 at 3250 fps. Twist was originally 1x14", which was subsequently bumped to 1x12" for the M16A1. The twist was eventually kicked to exactly twice that fast (1x7") to stabilize the SS109/M855 rounds, and yet, that hotly loaded M193 continues to be used in M4s (1x7" twists) today in training environments. No pressure issues, no blown up guns, etc..

Julian Hatcher has most of the answers to the questions posed here (and a whole lot more), and they can be found in "Hatcher's Notebook." Great read, tons of fascinating, if somewhat arcane information about a wide range of ordnance topics.