Dwell Time in barrel

[Archie]

Or, How long is the bullet in the barrel?

In another thread the subject of dwell time occurred to me.

It is a fairly simple matter of arithmetic. One figures the average velocity of the bullet as it leaves the muzzle (which is the initial velocity and the final velocity in the barrel - that is zero [bullet in case] and muzzle velocity so it's half the muzzle velocity); then divide this into the ballistic length of the barrel (distance from base of bullet to muzzle) IN FEET (because of how velocity is measured).

The length of the muzzle in feet is (length in inches in decimal notation) divided by twelve (inches per foot.

One can assemble a formula table on the math spreadsheet program in one's computer. (I have an Apple and it's called 'Numbers' for us simple folk. I believe it's called "Excel" on Windows.)

Just for fun a .30-06 with 22 inch barrel using a bullet fired at 2700 fps is out of the barrel in .00123 seconds. Still enough time to buck and throw the shot off.

 

[WrongHanded]

It is a fairly simple matter of arithmetic. One figures the average velocity of the bullet as it leaves the muzzle (which is the initial velocity and the final velocity in the barrel - that is zero [bullet in case] and muzzle velocity so it's half the muzzle velocity); then divide this into the ballistic length of the barrel (distance from base of bullet to muzzle) IN FEET (because of how velocity is measured).

At a glance that would appear to be correct. However, acceleration is not necessarily linear, or the same between bullet weights.

Consider that if you have a 150gr bullet and a 200gr bullet, which both leave the same barrel at the same velocity. The heavier bullet has more inertia to overcome, so it will likely get a slower start "off the line". That has to be made up somewhere farther down the barrel to reach the same velocity at the muzzle.

 

[edwardware]

It is a fairly simple matter of arithmetic. One figures the average velocity. . .

That's a rough approximation, but you can do better with QuickLoad. You'll find that acceleration is almost all accomplished in the first couple inches, so transit time is lower than your estimate.

 

[ExtremeSquared]

You've got to break out the calculus, sorry. You can't get average velocity that way.
F=ma
a=F/m
F(s)=some function more or less proportional to pressure
Once these are known,
integrate ( F(s)/m ) [from s=0 to s=dwelltime] = known muzzle velocity
Find indefinite integral. Solve for dwelltime.

 

[GHinNH]

First couple inches? I see a projectile that has moved 2.657" (dotted) at Pmax, FPS (solid blue) increases all the way to the end of the barrel. Otherwise the bullet would slow from drag. There are just far too many variables involved to pin any one number on it.

 

[ExtremeSquared]

First couple inches? I see a projectile that has moved 2.657" (dotted) at Pmax, FPS (solid blue) increases all the way to the end of the barrel. Otherwise the bullet would slow significantly from drag. There are just far too many variables to pin a number on it.

Note that the x axis on here is position, not time. A velocity/time graph should be even less linear. Trying to approximate dwell time, that graph will be more valuable.
Edit: Looking again, I'm wrong about that.

 

[denton]

Yes, it is a calculus problem, and not a simple one. You have powder burning, trying to increase pressure, and the bullet moving down the barrel, which increases volume/decreases pressure.

Still, the estimate of 1.23 milliseconds is not far off the mark, both by QuickLoad and by PressureTrace measurements.

You can take it a couple of steps farther and calculate that a 30-06 has an average acceleration of about 500g, and the rifle only moves about 1/16-1/8" before it stops accelerating back toward your shoulder. By then, it has all the momentum it is going to have.

 

[lysanderxiii]

Some actual clocked times:

Cartridge, Ball, 7.62mm NATO, M59 (Vm = 2,765 fps) - 1.30 msec (22 inch Barrel)
Cartridge, Ball, 9mm Parabellum, Canadian Dominion Production (Vm = 1,386 fps) - 0.70 msec (8 inch barrel)
Cartridge, Ball, Cal. .30, M2 (Vm = 2,838 fps) - 1.25 msec (24 inch barrel)
Cartridge, AP, Cal. .30, M2 (Vm = 2,750 fps) - 1.40 msec (24 inch barrel)
Cartridge , Ball, 5.56mm, M193 (Vm = 3,249 fps) - 1.24 msec (20 inch barrel)

Of course, this varies from lot to lot.

For example the M193 data is a average, the lots varied in time from as low as 1.15 msec to as high as 1.36 msec

 

[Archie]

That's a rough approximation, but you can do better with QuickLoad. You'll find that acceleration is almost all accomplished in the first couple inches, so transit time is lower than your estimate.

Right. So much more accurate to accept a generic estimate from a mass program than to measure your own velocity yourself. That probably sounds sarcastic.

 

[Archie]

'Average'. "(adjective) {C}onstituting the result obtained by adding together several quantities and then dividing this total by the number of quantities" from a web search.

No, it's not anymore complicated than that. One knows the starting velocity - zero - and the final (in the barrel) velocity - muzzle velocity. This actually varies from the chronometer reading, but not enough to matter. The intermediate velocities at various distances from the beginning position are essentially smushed together in the average. Pressure is not of interest for this exercise. (That is a different discussion.) Weight has no bearing on this. This only considers the time from first movement to departure.

LysanderXIII: Good information.

 

[bangswitch]

You'll find that acceleration is almost all accomplished in the first couple inches,

Acceleration continues the full length of the barrel. The "almost" doesn't quite qualify your statement. If that were true, barrel length would make no difference to muzzle velocity, and as we all know, longer barrel = greater MV (to a practical maximum) for a given cartridge. For purposes of steady aim, I don't think the differences in dwell time play as much a role as powder load and bullet weight, but it does affect things like cycling in gas impingement systems.

I can barely balance my checkbook, so I'll leave the higher math to others to figure out specifics. But I do know dwell time is pretty short.

 

[F-111 John]

Acceleration continues the full length of the barrel. The "almost" doesn't quite qualify your statement. If that were true, barrel length would make no difference to muzzle velocity, and as we all know, longer barrel = greater MV (to a practical maximum) for a given cartridge.

The acceleration of a bullet is not even remotely close to being linear for the entire length of the barrel. It will be a function of powder burn rate and total volume behind the bullet.

Example: Does a bullet exiting a 20" barrel have exactly double the velocity of the same bullet exiting a 10" barrel? No, of course not. Therefore the acceleration rate reduces as the bullet travels down the barrel as it is being pushed by burning powder gasses filling a larger and larger volume behind the bullet, providing less total pressure (thus force) on the base of the bullet.

Acceleration in the last few inches of the barrel may approach zero for a very long barrel, and the bullet will in fact decelerate if the barrel is too long for the powder charge being burned, or if the burn profile of the powder is wrong for the barrel length in question. (A pistol powder designed to burn very quickly vs. a rifle powder designed to burn more slowly. Or consider a squib load where the bullet gets lodged in the barrel. The bullet accelerated, then decelerated back to zero velocity all within the length of the barrel.)

However, I think it is very fair to say that that the majority of the bullet's final velocity is gained in the first few inches of barrel travel, with the rest of the barrel travel providing additional velocity, but by an ever lower amount the further along the bullet travels.

 

[F-111 John]

No, it's not anymore complicated than that. One knows the starting velocity - zero - and the final (in the barrel) velocity - muzzle velocity. This actually varies from the chronometer reading, but not enough to matter. The intermediate velocities at various distances from the beginning position are essentially smushed together in the average.

Think about a different sport all together, drag racing. A vehicle that quickly reaches its terminal velocity vs. a vehicle that experiences a constant acceleration for the entire quarter mile.

Assume that both cars have the same trap speed at the end of the quarter mile (a.k.a. muzzle velocity.) The car that accelerated to terminal velocity faster then held that velocity for the rest of the track will have a lower elapsed time (a.k.a. your "dwell time") than the car that had constant acceleration for the entire length of the track, even though both cars hit the end of the track at the exact same final speed.

Therefore you must take varying acceleration as the bullet travels down the barrel into account in order to calculate a meaningful '"dwell time."

 

[tark]

This is an interesting study in ballistics. There are four stages in the study of ballistics. Interior, transitional, exterior and terminal. Most people aren't aware of the second stage. The transitional stage begins the instant the base of the bullet leaves the muzzle until the pressure behind the bullet is equalized. To simplify, until the gasses from combustion are no longer accelerating the bullet.

The transitional stage is, obviously, the shortest and quickest stage, lasting only milliseconds. It is over by the time the bullet only an inch or two out of the barrel. I'm damned if I know how this is all measured.

F-111 John gave a very good analogy. The two cars won't hit the finish line at the same time, but they will have the same top speed.

Nice Aardvark, John! Did you pilot one?

 

[F-111 John]

Nice Aardvark, John! Did you pilot one?

Nope, just helped fix them for a short time.

 

[bangswitch]

The acceleration of a bullet is not even remotely close to being linear for the entire length of the barrel. It will be a function of powder burn rate and total volume behind the bullet.

Re-read my comment. Nowhere did I say that acceleration was linear, only that it continues. I agree that powder type, as well as bullet weight, affect the rate of acceleration, but as long as there is gas pressure behind the bullet in the barrel, it will continue to accelerate until it leaves the muzzle.

 

[kennedy]

here is a real time example, bought a sig .40 226, police trade in, supposed to be zeroed in, but shot 6 inches high with my reloads, figured my reloads were too light and the bullet was still in the barrel when the muzzles started climbing. Sooo increased my load and now shoots to point of aim.

 

[Slamfire]

The more you shoot, the more you find out how important followthrough is. Bud's who have electronic devices on their rim rifles tell me, the device can see their movement before, during, and after trigger fall. I can say, if either the pistol or rifle recoils in a certain direction, other than up, than I have pushed or pulled on the thing. And the bullet goes in the direction of the flinch.

Hard to believe that I can move fast enough to influence an event as quick as bullet travel through a barrel, but I can. And I am not Superman either, I am much slower than a speeding bullet.

 

[Archie]

Think about a different sport all together, drag racing.

Yes. I am most familiar with the endeavor.

Assume that both cars have the same trap speed at the end of the quarter mile (a.k.a. muzzle velocity.) The car that accelerated to terminal velocity faster then held that velocity for the rest of the track will have a lower elapsed time (a.k.a. your "dwell time") than the car that had constant acceleration for the entire length of the track, even though both cars hit the end of the track at the exact same final speed.

Not exactly a valid comparison.
By the way, dragsters are ideally geared to accelerate all the way to the finish line. Not too low a ratio as the top speed would be too low; not too high as the acceleration would be sluggish (comparatively).
So if the two cars have the same et, it means both will have the same average speed for the quarter mile. Even if one accelerates faster for one segment of the race.

Therefore you must take varying acceleration as the bullet travels down the barrel into account in order to calculate a meaningful '"dwell time."

No. What you are discussing is regulating the type and amount of powder to produce a desired result. Which does not enter in to this exercise at all.

Tell you what: If you don't like this, make up your own plan. Just ignore me.

 

[ExtremeSquared]

Yes. I am most familiar with the endeavor.

Not exactly a valid comparison.
By the way, dragsters are ideally geared to accelerate all the way to the finish line. Not too low a ratio as the top speed would be too low; not too high as the acceleration would be sluggish (comparatively).
So if the two cars have the same et, it means both will have the same average speed for the quarter mile. Even if one accelerates faster for one segment of the race.
No. What you are discussing is regulating the type and amount of powder to produce a desired result. Which does not enter in to this exercise at all.

Tell you what: If you don't like this, make up your own plan. Just ignore me.

You posted incorrect math in a public forum. What occurred after is the entire point of forums.

 

[denton]

Archie, no offense intended, but your physics and math need some adjustment. What you have said is not correct, at all.

A full and correct explanation requires integral calculus, or the numerical methods approximation, which is beyond what is possible to explain in this forum.

 

[ExtremeSquared]

Also, to disprove a hypothetical model, it helps to look at absurd extremities, and see if the model fails.

Hypothesis: Knowing exit velocity, and distance, you can derive dwell time.

Absurd extremities:
Car A spends 5 minutes idling for 10 feet, stomps the gas, and crosses the line at 200mph. Dwell time: 310 seconds.
Car A stomps the gas, and crosses the line at 190mph. Dwell time 12 seconds.

Why is there a higher dwell time here with a faster exit velocity?
Because acceleration a is a function of time in these scenarios. In the absurd car, a(seconds) is decided arbitrarily by the driver.
In a barrel, a(seconds) is going to vary based on pressure curve. Maybe other things -- air column compression / barrel friction / etc.

Maybe the model works if acceleration is a known constant. But this is replacing an entire variable in the calculus I posted above -- F(s) becomes a constant. One errant variable is all it takes to ruin the solvability of an equation.

 

[Atavar]

Note that the x axis on here is position, not time. A velocity/time graph should be even less linear. Trying to approximate dwell time, that graph will be more valuable.
Edit: Looking again, I'm wrong about that.

Umm, isn’t ms time?

 

[tark]

Uh-oh... Here we go....

 

[lysanderxiii]

Actually, it does not require calculus, even if you want numbers (quantitative information), if you have a plot of the pressure time curve.

You break the P-T curve down into small time intervals, say 0.01 msec, so for each point on the P-T curve you can calculate:

Average Pressure (from time tx to time tx+1) x area (constant) = Average Force (over time interval tx+1)

(Force - friction*) x mass (constant) = acceleration (over time interval tx+1)

And with acceleration and given a small time interval you can calculate delta V and thus delta S over the time interval, and the sum of all the delta V = total velocity and delta S = total travel

The acceleration is the slope of the velocity curve, the closer the line is to vertical, the higher the acceleration. (Qualitatively, it is shaped like the pressure curve)



______________
* found by simply pushing a bullet down the barrel, you can measure the engraving force too, so the friction is a function of travel S.