200 vs. 230gr .45 ACP's

So now that we know how long the bullet is in the barrel, how far does the slide move in that amount of time? I do not know how fast the slide moves.

Asl far as the original question,
I accept my target that shows in this one case the difference aint squat.
I will do it again with my LW officers ACP. Same ammo, same distance.

David
 
All information and tests are useful. If memory serves, the 3 inchers owned had double recoil springs. This mitigates straight back recoil, but the pistols appear to have an increase of "lift", similar to a lot of the small lightweights available today. And even that may just be own adaptation to managing/guiding the recoil from them.

There are always other considerations. Have experienced lighter bullets needing considerable more elevation than larger heavier ones from 44 mag revolver. That was comparing mild 250 grain loads to hot rodded 180 gn jhp's.
 
Yes Zeke, my revolver would show a large difference .

I have a 5” 45 colt revolver. Thant shoots 255 grain excellent. I don’t have enough sight adjustment to shoot 275 grain no matter how much powder I put in them.

Stay tuned for more targets.

David
 
I do not know how fast the slide moves.
Multiply the bullet weight (grains) by the muzzle velocity.

Multiply the powder weight in grains by the muzzle velocity and then by 1.5

Sum the two products and divide the result by 437.5

Call that M.

M = ((Bullet Weight x muzzle velocity) + (powder weight x muzzle velocity x 1.5) )

Sum the weight of the slide (and bushing and recoil spring plug, if applicable) and barrel, all measured in ounces.

Weigh the recoil spring (ounces) and divide that weight by 2.

Sum that result with the combined weight of the slide and barrel (and bushing & recoil spring plug if applicable).

Call that W.

W = slide weight + barrel weight+ bushing weight + recoil spring plug weight+ (recoil spring weight /2)

Divide M by W and that should give you a good estimate of the velocity of the slide/barrel combo at the point that the bullet leaves the muzzle.

Multiply the slide/barrel velocity by the time that the bullet is in the bore and multiply by 12. That will give you a good estimate of the distance that the slide/barrel combo moves while the bullet is in the bore, measured in inches.
 
That math is WAY over my head but may I ask how it works with out pressure amounts ? I get that bullet weights and slide weights all mean something but doesn't how much pressure is created matter more ? Isn't everything the gun needs to do to operate dependent on the amount of pressure created .


Or does it matter if there's enough pressure to blow up the gun or create a squib , the slide will move the same amount at the same timing regardless of the pressures created ?

or how far down the barrel the bullet is when said pressures are high enough to start causing the slide and other parts to move ? Thinking burn rates and does a very slow powder for cartridge cause different timing of the parts movements then a really fast for cartridge powder would ?

EDIT

oops the velocity is basically the pressure reading correct ?
 
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the velocity is basically the pressure reading correct ?

no.

Is the rpm reading on your tacometer "basically" the vehicle speed? No. There is a relationship, but its a bit more complex.

the relationship involves not just the amount of pressure, but also the amount of time that pressure is applied. The details get complicated, but a general truth is that slower burning powders "push" a little longer than faster ones, even while the actual pressure value (in psi, cup, or however you measure it) might be the same, the "longer" push imparts more energy to the moving parts (bullet on one end, pistol on the other) so those parts get moving faster.

Generally.
 
Isn't everything the gun needs to do to operate dependent on the amount of pressure created .

No. What operates the (semi-auto recoil operated --- ) gun is recoil force. Peak chamber pressure itself does not predict if the load will produce enough recoil force to cycle the gun.

A slow powder that requires a large charge weight can produce more recoil force and lower pressure than a fast powder that requires a much smaller charge weight even when pushing the same bullet to the same speed. http://www.shootingtimes.com/editorial/compensators-pressure-gas/99170
 
David,

The details may seem complex to add up, but the easy thing to remember is the momentum (velocity × mass) of the bullet is equal and oppositely directed from the total momentum imparted to moving parts in the gun. So, simply put, if your moving gun parts weigh 100 times more than the bullet, they will be going 100 times slower than the bullet. If they weigh 300 times more (common in hunting rifles), they will be going 300 times slower than the bullet. This applies not only to the final velocity at the muzzle but to the velocity of the bullet at any point along its way down the barrel. Just don't forget, both the gun and the bullet start at zero, so if you need a velocity to figure out how far the moving gun parts travel during barrel time, it will average half the final velocity, and you want to use that average value as an approximation. It won't be exactly true because of the uneven shape of the pressure curve, but it'll usually get you within 10% or so.

Once you understand that basic principle, you can start tweaking precision by getting into more detail. In addition to the bullet mass, add the mass of the portion of the powder and gas that gets blown down the bore with the bullet when finding your weight ratio. Then add to the bullet momentum the momentum from expelling the powder gas and unburned powder when the bullet base uncorks the muzzle and lets that stuff rocket out. Lots of accounting fun to be garnered here.
 
So, simply put, if your moving gun parts weigh 100 times more than the bullet, they will be going 100 times slower than the bullet. If they weigh 300 times more (common in hunting rifles), they will be going 300 times slower than the bullet.

Ok so now this is getting interesting , well it's been interesting . I think I may finally be understanding Poly's point . :eek: .This momentum and movement ratio would seem to indicate a fixed object/firearm . Meaning a bolt action or single shot firearm or revolver . That ratio can't be accurate if there is a slide being resisted by a spring which is being held by a not so stable hand , correct ? There would seem to be lots of give/extra allowable movement in the equation when using a semi auto ? You'd need to know how much the spring is resisting the movement as well as how much the hand hold actually freely lets the frame move . Maybe not freely but your hand is no vise so the frame is going to move hundreds if not tenths of an inch with relatively no resistance based on the soft tissue of the hand .

So If I'm understanding my own point there . Because the semiauto has all these allowable ( micro movements ) the over all movement ( muzzle rise ) is less then if it was a fixed single shot firearm because these( micro movements ) are nullifying/canceling out some of the momentum ???

I'm not even sure that makes any sense . Man I wish I was smart enough to put into words what my brain is thinking . :o
 
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...how it works with out pressure amounts ? I get that bullet weights and slide weights all mean something but doesn't how much pressure is created matter more ? Isn't everything the gun needs to do to operate dependent on the amount of pressure created .
That's the beauty of conservation of momentum. Once you know the momentum of what comes out of the barrel, you also know the recoil momentum.

Barrel friction, pressure, force, all of that wraps into one single number--the muzzle momentum.

NOT because those factors (pressure, force, barrel friction) are unimportant, but because they are already taken into account when one considers muzzle momentum.
Or does it matter if there's enough pressure to blow up the gun or create a squib , the slide will move the same amount at the same timing regardless of the pressures created ?
The momentum of the recoiling mass (in this case the slide, barrel, bushing, recoil spring plug, and half the recoil spring) will be equal to the momentum of what comes out of the muzzle--gases and bullet.

Once you know the momentum of the recoiling mass, you can calculate its velocity by dividing the momentum by the combined weight of the recoiling mass. Then with the velocity you can figure how much the slide moves in a given amount of time.
or how far down the barrel the bullet is when said pressures are high enough to start causing the slide and other parts to move ?
The slide/barrel begins to move when the bullet begins to move. If the bullet has moved part of the way down the barrel then the slide has already begun moving.
Because the semiauto has all these allowable ( micro movements ) the over all movement ( muzzle rise ) is less then if it was a fixed single shot firearm because these( micro movements ) are nullifying/canceling out some of the momentum ???
Very close.

Muzzle rise happens when recoil momentum drives a recoiling mass AND the resistance to the recoiling mass is below, not directly behind the muzzle. That unbalanced resistance results in the muzzle rising as the gun torques around the center of resistance.

In a revolver, the WHOLE gun is the recoiling mass and the resistance to the recoiling mass is well below the bore. So the instant that the barrel starts recoiling, the muzzle starts rising.

In a locked breech Browning type recoil operated pistol, the resistance to the recoiling mass--UNTIL THE BARREL UNLOCKS FROM THE SLIDE--is very small. So the slide and barrel move almost straight back until the barrel unlocks from the slide. There's very little muzzle rise during this phase.

The other half of the equation is that, BY DESIGN, a locked breech Browning type recoil operated pistol does not unlock the slide from the barrel until the bullet is out of the bore.

Combine those two facts and you end up with the result that there is very little muzzle rise until the bullet leaves the bore.
That math is WAY over my head...
No, it's not. Here's what's going on.

First of all, the math calculates the momentum of the bullet (bullet weight times muzzle velocity). Momentum is mass times velocity.

Next the math calculates (approximates) the momentum of the gases that escape the muzzle. That's powder weight times 1.5 the muzzle velocity of the bullet. That's a reasonable approximation for the gas momentum in a pistol.

Then the two momentums are summed into one total momentum figure of M.

NOW we know the total momentum of what comes out of the bore. But we really want to know the slide velocity.

Conservation of momentum tells us that the momentum of the recoiling mass is equal to the total muzzle momentum. So the total muzzle momentum figure (M) is ALSO the momentum figure for the recoiling mass.

Since momentum is mass times velocity, if we divide the momentum figure by the weight of the recoiling mass, we can calculate its velocity.

(Mass x velocity)/Mass = velocity

So next we figure up the total weight of the things that make up the recoiling mass. The slide, barrel, half the recoil spring weight, the bushing, and the recoil spring plug.

Now we have the weight, we can divide the momentum figure by that weight to get the velocity of the recoiling mass at the instant that the bullet exits the muzzle.

The physicists will note that I'm being sloppy by pretending that mass and weight are identical. While that is not actually true, in this case, because we're going to cancel out the weight and just be left with velocity, we don't need to worry about the correction factors since they will cancel out too.
 
To answer that, requires knowing the combined weight of the barrel and slide, and, if applicable, the bushing and recoil spring plug.
 
Assuming that the slide velocity is about 20fps (which is probably in the right ballpark); and using Unclenick's calculated barrel dwell time, it moves about 0.2" before the bullet leaves the bore.
 
Never thought such a simple question could generate so much interest and it has been a very interesting read with lots of different opinions.
At any rate I bought some .230gr. Hornady HAP bullets and loaded up 3 different boxes of 50. all with an overall length of 1.230 in new Starline brass with a taper crimp of .467.

1. 6.6gr. of power pistol.

2. 6.0gr. of universal clays

3. 5.4gr. of HP-38

All 3 shot to the point of aim elevation wise with the universal load delivering the best overall accuracy at 30ft. Both the Universal and the HP-38 loads seemed a bit hot so I'm going to reduce to 5.8 and 5.2 respectively.
next time out.

Brought 2 .45's with me a 5" with target sights and the commander. I definitely shoot the 5" better with the longer sight radius and sights but that's just old age creeping up on me.:p

Hope everyone has a happy new year.
 
74A95 - you're probably right, but do you have a slo-mo video of a Commander length 1911? I shouldn't have said "zero", but "imperceptible" instead.
 
Am not a video expert, but does it appear the hand holding those pistols is video chopped in place? If they were video chopped, would be interesting to see what was actually holding the guns in place.
 
One more time.
25 yards Light Weight Officers ACP 3.5" barrel.

Center target Remington Ball. 230 grain


Top left 200 swc 1,000 fps
Bottom left 200 SWC 676 fps
Bottom Right 200SWC 900 fps.
Velocities are for a 5" gun.

Evevation looks petty much thesame to me.

David
4d3d830d7f37914d438c4a50f421a925.jpg


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