200 vs. 230gr .45 ACP's

[polyphemus]

ok,USSR get us out of the paralysis
What have you got?

[USSR]

poly,

You know I will be doing an actual test with actual results and not surmising anything. The results will speak for themselves.

Don

[JohnKSa]

John I was disappointed to read you hold Browning to be wrong.
He was the greatest american gunsmith,bar none and he knew firearms and their workings better than most of us will ever do his patent descriptions were written by his attorneys but they are true to the facts and to this day stand to
any objective scrutiny.

I think if you read my statement more carefully you will see that I believe he was simplifying the explanation and that is what resulted in the discrepancy that is present if the statement is taken literally.

Clearly the movement of the slide and barrel is not delayed until after the bullet leaves the bore--that's not possible, it contradicts the laws of physics, and we have ultra-slow motion video to prove that the parts do move. But the movement is slowed sufficiently by the design to delay unlocking until after the bullet leaves the bore.

The slide needs V to have momentum and this is gained after it travels all of 1/10".Browning knew.

All the momentum is gained while the bullet is in the bore. Something that would be impossible if the movement of the parts were literally delayed until the bullet exits.

And one factor for you to include, the barrel is dragged along by the bullet and held fastened to the crosspin while it exits,this also contributes to the delay.

This is all factored into the system by taking conservation of momentum into account. The momentum of the ejecta (which is affected by the friction between the bullet and the barrel) is all you need to know to determine the momentum of the recoiling mass. NOT because the bullet/barrel friction can be neglected, but because it is already taken into account when one looks at the momentum figures.

The results will speak for themselves.

Some people are satisfied knowing results. Others want to know exactly why the results turn out the way they do.

There's nothing wrong with either point of view. A car doesn't get a person from point A to point B any better if the person understands the intricate details of internal combustion engine operation, nor will failing to fully comprehend the design of an automatic transmission make a person unable to drive.

That said, it is important that at least some people do have that knowledge.

[44 AMP]

I'm pretty sure that Browning knew what he was talking about. I am less sure that the wording he used describing these things cannot be dis-assembled and re-assembled by "barracks lawyers" to make it appear that Browning didn't know what he was talking about, or to make it seem he said something he didn't, or didn't intend.

We've been all over in this thread, but the original question was about how to get the lighter 200gr slugs to impact the same point as the 230gr ones at the given range.

My idea was to match the 200gr and 230 gr velocities, so that they both had the same amount of "time in the barrel".

Can we all agree that recoil, as a force, begins at the moment of bullet movement? And that since it is happening as the bullet travels down the barrel, that it has an effect on the position of the muzzle at the instant the bullet exits?? That the barrel of a handgun is rising as the bullet accelerates down the bore?

I think we can discuss how much movement there is, and how important, or not, it is, but only after we agree that it is there.

and that, therefore, two bullets spending the same amount of time in the rising barrel, and that barrel rising at the same approximate rate for both bullets, should have the same approximate point of impact at close range before factors outside the barrel exert their effect. Does this seem reasonable to you?

I think discussion of what else the barrel, slide and gun are doing at the same time or immediately afterwards doesn't address the original question.
or are we just arguing about how many angels can dance on the head of a pin?

[JohnKSa]

Can we all agree that recoil, as a force, begins at the moment of bullet movement?

Whether we all agree or not, there is video to support that assertion and it is consistent with the laws of physics.

And that since it is happening as the bullet travels down the barrel, that it has an effect on the position of the muzzle at the instant the bullet exits??

It certainly has some effect in real-world guns, but it seems clear that in some designs the effect is not significant--that is, the muzzle moves, but the only significant motion of the muzzle while the bullet is in the bore is straight backwards. And that it therefore does not result in any significant change in the point of impact.

That the barrel of a handgun is rising as the bullet accelerates down the bore?

That's really the crux of the whole discussion at this point.

I think we can discuss how much movement there is, and how important, or not, it is, but only after we agree that it is there.

That's pretty much what's happening.

[Metal god]

that's not possible, it contradicts the laws of physics,

LMAO That's what many of us have been saying from the get go . I think Poly no-think you master anymore , you just normal joe now

I crack me up

[Jim Watson]

Browning's original 1911 design required excessive cocking effort. The firing pin stop had a square bottom.

Not really.
First run 1911 firing pin stops had a 5/64" bottom corner radius, Army redo was 7/32".
The square bottom was introduced by EGW "so the customer can adjust" and probably not coincidentally, cheaper to make.
Use of the square bottom part as delivered appears to be in accordance with that fine Internet Logic, if a smaller radius is better, no radius must be best.

[JohnKSa]

That's what many of us have been saying from the get go .

Including me--here are two excerpts from my first post on this thread.

Recoil DEFINITELY begins when the bullet starts moving. There is no room for debate on that point.
...
Recoil is an unavoidable consequence of bullet movement.

[polyphemus]

Neither Browning nor I say that the slide does not start to move rearward when the bullet begins its own forward motion,the snide remarks aside.
We are talking about an extremely short event here,about a milisecond and less than a 1/8" travel length.The slide begins to move and for the first.005"only mass,friction and springs oppose it then the grooves contact the barrel ribs and
begin to drag it back here the slide meets barrel inertia and resistance from the bullet's forward impulse to the barrel,the link begins to rotate back and as the bullet exits and pressure is released the link pulls the barrel down and out of the way.

[44 AMP]

repeatedly restating how the 1911 cycles does nothing to address the original question, which was "how to get the 200 & 230gr bullets to strike the same point"

[polyphemus]

Thank you for the remainder 44
All done.

[Metal god]

Neither Browning nor I say that the slide does not start to move rearward when the bullet begins its own forward motion,the snide remarks aside.

You deserve an apology for my snide remark , sorry about that . I had to look up the definition to be sure I understood the word correctly . Although my statement was snide it was meant in jest rather then malice . I almost never mean to be rude or condescending , most of my post are in a jovial manner but don't always come across that way when written out .

Again sorry about that
MG

[David R]

Been following the whole thread.

Did some shooting today.

5" Dan wesson Vaor 45, stock form.

Shot target loads and hot loads along with one target with Remington Factory 230 Ball.


Results...........

Top left 185 Jacketed HP with minimal powder probably 700 fps. Did not eve work the action. I only had 2.

Top right my target load. 200 SWC at 675 fps.
Bottom left, 200 SWC, 7.0 Unique 900 fps
Bottom right a new load I tried today. 8.0 CFE Pistol 200 SWC, should be 1100 fps. Not chronographed, but ..... Max is 8.2 grains on Hodgun website for 1142 fps.

Center Reimington factory Ball.

Gun was rested on sand bags.
Distance 25 yards.
Sight Burris FF III NO adjustments made.

Make your own conclusion.

David


I did some quick math. Bullet goes from zero to 850 fps in 5 inches.

850 Feet * 12 inches = 9600 inches per second. Divide that by 5 inches and get 1900th of a second for barrel time.

It IS a little longer, but I dont get into math that deep to figure acceleration.

The barrel and slide do move back in that .0019 of a second. Some one else can figure the distance the slid moves in that little time. I have seen vidieo of the slide moving before the bullet exits. The slide and barrel move together, so elevation does not change.

David

[zeke]

Am thinking we need some person way smarter than i to set up a poll, based on the average poi difference between the loads in USSR's test. Something like 1 in or less, than 1 to 2 in and more than 2 inch.

1 in or less = ain't squat
1 to 2 in = little more than ain't squat, adjust sights/hold for 50 yds
over 2 in = adjust the sights for differing load, or adjust load for smaller poi

Then we need a prize?

Am in for ain't squat, and already know how far i can p_ss in the snow.

MERRY CHRISTMAS!

[Unclenick]

Part of the problem with small changes is determining the statistical significance of the group to try to discern what is real. It may need 30-shot groups (or combining 3 10-shot groups) to do it. I couldn't conclude anything from the small groups trying to distinguish real from random. We still don't know what the OP's range was. We also don't know if the bag stabilized movement of the gun.


David,

850 ft/s times 12 in is 850 ft²/s or 10,200 in²/s.

For 850 ft/s (or any other velocity) the bullet starts out at 0 ft/s, so the average velocity inside the barrel is the average of 0 ft/s and the MV. In this case 0 ft/s and 850 ft/s, or 425 ft/s. The barrel may be 5 in. long, but the bullet sits forward in the case and has bullet base travel of about 4.35 in. to the muzzle.

4.35 in/12 in/ft = 0.3625 ft

In order to get the feet to cancel out and leave seconds, you have to divide that distance by the mean velocity:

0.3625 ft / 425 ft/s = 8.53 × 10⁻⁴ s, or 0.000853 seconds.

So, a bit less than a thousandth of a second. In reality, that time is a little long. The reason is the peak pressure occurs when the bullet is only half an inch so into the throat, and the peak pressure gives it a third to half its speed by the time it has gone an inch, so it starts down the rest of the barrel with speed already approaching the average velocity and getting faster from there. It will vary with the powder burn rate because that determines how far forward the bullet is at its peak, but it looks like, for a fast powder like Bullseye, the actual barrel time would be around a tenth of a millisecond shorter than that calculation gives.

[David R]

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

[zeke]

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.

[David R]

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

[JohnKSa]

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.

[Metal god]

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 ?

[44 AMP]

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.

[74A95]

Originally Posted by Metal god

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/editori...sure-gas/99170

[Unclenick]

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.

[Metal god]

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 . .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 .

[JohnKSa]

...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.