Velocity, barrel length and bore volume

AI Thread Summary
Bullet propulsion in a bore is influenced by factors such as barrel length, bullet mass, and chamber pressure. A longer barrel allows black powder to burn more completely, increasing velocity, while smokeless powder burns faster, enabling shorter barrels in modern handguns. The discussion explores whether bore volume or barrel length is more critical for achieving higher velocities, concluding that barrel length is essential for optimal combustion time. Increasing caliber can compensate for shorter barrels, but this requires a proportional increase in powder charge to maintain pressure and velocity. Overall, the interplay of these factors is complex and crucial for understanding interior ballistics.
kalkyl
Messages
3
Reaction score
0
Hello.

How does a bullet propell inside a bore? What determine its velocity? I read that a bullet in cal .44 propelled by black powder from a 3” barrel is as powerful as a .25 ACP, however with a longer barrel, the velocity increase significant. With a 8” barrel a .44 black powder bullet is as powerful as a modern 9mm.

The reason that handguns today has shorter barrels compared to the black powder guns of the old days, is because the smokeless powder burn faster. Simply put, black powder burn to slow to create high enough velocity inside a short barrel.

However, is it really a matter of the length of the barrel or in fact a matter of the volume? Let suppose that a .57 barrel that’s 3” long has the same bore volume as a .44 barrel that’s 8” long. With equal volume, the black powder should be able to burn to the same extent, or? Will the .57 bullet in the 3” barrel have the same velocity as the .44 bullet in the 8” barrel provided that the same charge of black powder is used and that the bullets has the same weight?

If we take into account that the .57 ball will weigh more than the .44 ball, how does that change the ballistics and the energy outcome?
 
Physics news on Phys.org
kalkyl said:
How does a bullet propell inside a bore? What determine its velocity?

There are many factors, including bullet mass, chamber pressure, barrel length, and more.

kalkyl said:
However, is it really a matter of the length of the barrel or in fact a matter of the volume? Let suppose that a .57 barrel that’s 3” long has the same bore volume as a .44 barrel that’s 8” long. With equal volume, the black powder should be able to burn to the same extent, or? Will the .57 bullet in the 3” barrel have the same velocity as the .44 bullet in the 8” barrel provided that the same charge of black powder is used and that the bullets has the same weight?

It's not really about the volume, it's about having enough barrel length for the black powder to burn more completely before the bullet leaves the barrel. In other words, you need more time, not volume. An 'ideal' scenario might involve some sort of mechanism that prevents the bullet from accelerating until the pressure in the chamber reaches a certain point. That way you can reach a higher chamber/barrel pressure which will accelerate the bullet faster and require a shorter barrel for equal velocities. But since real firearms don't have such a mechanisms, we have to use a longer barrel instead.
 
There is more than that, too much to cover here. Cartridge case shape, bullet weight, powder burn rate, and more are all important.

Search terms interior ballistics will get you a lot of good information. The same search terms in Amazon brings up several books on the subject.
 
It’s a complex subject for sure. Let’s keep it as basic as possible and hold on to black powder. I thought that more volume gave more time for the powder to burn? If we have a 3” barrel in .44 and gradually increase the size of the caliber (and the weight of the bullet too, let’s assume that we use roundballs), to a certain point the bullet shot from the 3” barrel will be big enough to generate an equal amount of ft/lbs as the .44 ball shot from the 8” barrel - is that correct? With other words, a bigger caliber will compensate a shorter barrel?

If so, is it more a matter of heavier ball/more mass than increased volume of the bore?
 
kalkyl said:
I thought that more volume gave more time for the powder to burn?

In this case, yes, because a longer barrel has a larger volume, so it just happens to come out that way. Think about what happens if you increase the chamber volume instead. You'll actually have reduced pressure because the larger volume takes more gas (from the vaporized powder) to get to the same pressure.

kalkyl said:
If we have a 3” barrel in .44 and gradually increase the size of the caliber (and the weight of the bullet too, let’s assume that we use roundballs), to a certain point the bullet shot from the 3” barrel will be big enough to generate an equal amount of ft/lbs as the .44 ball shot from the 8” barrel - is that correct? With other words, a bigger caliber will compensate a shorter barrel?

It will as long as you also increase the amount of powder as you increase the caliber. So it's not really a fair comparison since you aren't use the same amount of powder in both cases.
 
Hi there, im studying nanoscience at the university in Basel. Today I looked at the topic of intertial and non-inertial reference frames and the existence of fictitious forces. I understand that you call forces real in physics if they appear in interplay. Meaning that a force is real when there is the "actio" partner to the "reactio" partner. If this condition is not satisfied the force is not real. I also understand that if you specifically look at non-inertial reference frames you can...
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...

Similar threads

Back
Top