1911Tuner
Moderator Emeritus
Excellent example of what I'm talkin' about!
*smack*
Why didn't I think of that?
The effect of a heavier weight bullet?
- Thread starter BruceRDucer
- Start date
- Status
Excellent example of what I'm talkin' about!
*smack*
Why didn't I think of that?
Here's one way to think of it: Recoil is caused by the pressure of the propellent gas on the bolt of the gun. All other things being equal, a heavier bullet accelerates more slowly and so takes longer to emerge from the barrel, giving the gas more time to push backward on the gun. So a heavier bullet should cause more recoil.
Sylvan-Forge
Member
Tuner, ya gone and done melted my brain!
Thank you for all the explaining .. I'll try and go through it tommorrow after some sleep.
One thing is for sure, this thread has surely shown that
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Best!
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Thank you for all the explaining .. I'll try and go through it tommorrow after some sleep.
One thing is for sure, this thread has surely shown that
.
Best!
.
1911Tuner
Moderator Emeritus
Dave...Kinda-sorta, but again...the dwell time difference is so small that we couldn't detect it. It doesn't apply the same way in a self-shicker as it does in a fixed breech gun anyway. With an autpistol...if there was no spring between the slide and frame, the "gun" will recoil while the frame stands still...until it hits the impact abutment, at least. If the frame rails were 30 feet long, you could fire it and feel almost nothing in the way of recoil. The recoil is all in the slide...which transfers its momentum through the recoil spring to the frame, and ultimately...your hand. The slide hitting the impact abutment then generates the second impulse that causes the greatest percentage of muzzle flip.
Then, the spring accelerates the slide forward...giving us another impulse as it imposes equal /opposite force against the frame.
It's known as sensory overload because it's all coming so fast that our brain can't separate the three different impulses, nor tell the difference.
With a fixed breech weapon...it's all over in one quick punch. We feel the result of the actual ballistic event...the action/reaction between bullet and bolt.
Incidentally...all the things that are acting on the slide are known as "Outside Forces." Momentum will be conserved forever in the absence of outside force. A bullet fired in the vacuum of outer space will continue at the muzzle velocity forever unless and until it either hits something or comes under the influence of a gravitational field. Anything that has the opportunity to influence a moving object will do just that.
In firing a rifle, the action/reaction event involves the bullet, bolt, and force vector provided by the expanding gas. Everything else constitutes outside force. The rifle barrel drags on the bullet. The receiver adds to the rifle's mass. The stock...the scope...and finally...the shooter. All these things are outside forces that oppose the objects in motion and work to bring them back to a stop...to restore equilibrium...the state of no motion, where compelling and resistive forces are balanced.
So, yes Sylvan. There's more to it. A lot more.
Then, the spring accelerates the slide forward...giving us another impulse as it imposes equal /opposite force against the frame.
It's known as sensory overload because it's all coming so fast that our brain can't separate the three different impulses, nor tell the difference.
With a fixed breech weapon...it's all over in one quick punch. We feel the result of the actual ballistic event...the action/reaction between bullet and bolt.
Incidentally...all the things that are acting on the slide are known as "Outside Forces." Momentum will be conserved forever in the absence of outside force. A bullet fired in the vacuum of outer space will continue at the muzzle velocity forever unless and until it either hits something or comes under the influence of a gravitational field. Anything that has the opportunity to influence a moving object will do just that.
In firing a rifle, the action/reaction event involves the bullet, bolt, and force vector provided by the expanding gas. Everything else constitutes outside force. The rifle barrel drags on the bullet. The receiver adds to the rifle's mass. The stock...the scope...and finally...the shooter. All these things are outside forces that oppose the objects in motion and work to bring them back to a stop...to restore equilibrium...the state of no motion, where compelling and resistive forces are balanced.
So, yes Sylvan. There's more to it. A lot more.
Elm Creek Smith
Member
All other things being equal, the heavier bullet results in more recoil.
Oh, yeah: Light bullets shoot low.
ECS
Oh, yeah: Light bullets shoot low.
ECS
Sylvan-Forge
Member
I think this is where we deviate.
The 180@940 event takes longer to complete than the 135@1180.
Yes, but not neccessarily more perceived recoil.
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Both events plotted out:
180@940 has more recoil energy, but longer period.
135@1180 has less recoil energy, but shorter period.
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IIRC, unlocking won't fully occur until the case lets go of the chamber walls after the bullet leaves the muzzle.
Maybe this hitch is what changes things just enough to improve felt recoil (for me anyways).
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Sylvan-Forge
Member
In my log books, I've 10 of 31 results detailing the same deal - 9 revolvers and 1 bolt action with heavier projectiles giving higher recoil energies but somehow softer felt recoil than lighter/faster comparison loads.
It's got to be the time factor.
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Or maybe I'm just crazy
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It's got to be the time factor.
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Or maybe I'm just crazy
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1911Tuner
Moderator Emeritus
But how much longer...and can you perceive the difference in time that's measured in just a few milliseconds? You still don't seem to be grasping that you don't feel recoil in an auto from the ballistic event. You feel the slide compressing the spring and the impact with the frame...and by the time that you actually start to feel it, the bullet is gone.
Yeah. Those are fixed-breech guns. The perception of recoil is different than a self-shucker. All you have is the recoil impetus. With autoloaders, you've got all that slam-bang action and a moving slide.
An example is an old 1937 Brazillian contract Smith & Wesson revolver in .45 ACP that my step-father now owns. When I had it...about 20 years ago...a friend was with me at the range, and we shot it a little along with my 1911s. He asked why it "Kicked less" with the same ammunition. I didn't have a good, plausible answer for him, and that's what got me really studyin' on it. If you can find a Smith and Wesson 10MM revolver, and fire the two .40 loads back to back...you'll also notice that they both feel diferent than when they're shot in an autopistol.
1911Tuner
Moderator Emeritus
A little long...so read carefully.
Let's try this again with a brief description of how a short recoil pistol works. Since I'm most familiar with the 1911, I'll use it. There's very little difference anyway.
Bang! The bullet starts its trip through the barrel. Assuming zero headspace, the slide starts rearward at the same instant. The slide pulls the barrel backward with it via the upper barrel lugs.
At 1/10th inch of slide travel, the barrel reaches the beginning of the linkdown point...and the bullet exits...and the barrel starts to link down. The bullet MUST be gone at that point, because if the breech opens before the pressure drops, the case will burst. Once the bullet is gone, it has no more influence.
1/10th inch of slide travel means that the spring has compressed 1/10th inch further than its static preload. The static spring preload in a 5-inch pistol with a 16 pound spring is about 8.5-9 pounds. Add a tenth inch of extra compression to that preload...and that's what you feel pushing against the frame. Go to the chart and find the gun/ammo combination that gives 9 pounds of free recoil...and you can tell how much recoil that pistol is providing at the point of bullet exit. Once the bullet has left the barrel, the action side of the primary action/reaction event is gone, and is no longer part of the equation. From that point, all recoil that you feel is up to the slide and the spring.
Now, the slide is compressing the spring. The spring becomes a force vector in a separate closed system. That force vector is between the slide and the frame. As the spring compresses, it fights to push the slide forward. As it pushes the slide in one direction, it pushes the frame in the opposite direction with equal force. Just prior to the point of slide to frame impact...you're feeling 16 pounds of free recoil.
Wham! The slide hits the frame at speed, and torques the gun upward in what we call "Muzzle Flip" Before that impact, the muzzle has "flipped" very little. Slow-motion videos will bear this out...and they will also show that at the instant of bullet exit, the gun has barely moved.
Let's try this again with a brief description of how a short recoil pistol works. Since I'm most familiar with the 1911, I'll use it. There's very little difference anyway.
Bang! The bullet starts its trip through the barrel. Assuming zero headspace, the slide starts rearward at the same instant. The slide pulls the barrel backward with it via the upper barrel lugs.
At 1/10th inch of slide travel, the barrel reaches the beginning of the linkdown point...and the bullet exits...and the barrel starts to link down. The bullet MUST be gone at that point, because if the breech opens before the pressure drops, the case will burst. Once the bullet is gone, it has no more influence.
1/10th inch of slide travel means that the spring has compressed 1/10th inch further than its static preload. The static spring preload in a 5-inch pistol with a 16 pound spring is about 8.5-9 pounds. Add a tenth inch of extra compression to that preload...and that's what you feel pushing against the frame. Go to the chart and find the gun/ammo combination that gives 9 pounds of free recoil...and you can tell how much recoil that pistol is providing at the point of bullet exit. Once the bullet has left the barrel, the action side of the primary action/reaction event is gone, and is no longer part of the equation. From that point, all recoil that you feel is up to the slide and the spring.
Now, the slide is compressing the spring. The spring becomes a force vector in a separate closed system. That force vector is between the slide and the frame. As the spring compresses, it fights to push the slide forward. As it pushes the slide in one direction, it pushes the frame in the opposite direction with equal force. Just prior to the point of slide to frame impact...you're feeling 16 pounds of free recoil.
Wham! The slide hits the frame at speed, and torques the gun upward in what we call "Muzzle Flip" Before that impact, the muzzle has "flipped" very little. Slow-motion videos will bear this out...and they will also show that at the instant of bullet exit, the gun has barely moved.
Using the 1911 as an example, IPSIC/IDPA as the sport, many people feel the heavier 230 grain at a lower velocity compared to the 200 grain at higher velocity, feels "softer" to shoot. Both at the same power factor.
I'm not saying I can feel it, or everyone can, but it seems many can. That could be the slide speed, different "feel", muzzle coming back to point, etc.
I have no point to make on this,, simply observing. Of course, changing recoil spring weights, mainspring combo's etc can/do "affect" how you want your "recoil" sensation.
I'm not saying I can feel it, or everyone can, but it seems many can. That could be the slide speed, different "feel", muzzle coming back to point, etc.
I have no point to make on this,
, simply observing. Of course, changing recoil spring weights, mainspring combo's etc can/do "affect" how you want your "recoil" sensation.1911Tuner
Moderator Emeritus
Found what I was lookin' for! Play this on High Res, and pay particular attention to the pistol that stars in two clips back to back. In one, it's tethered by a spring. In the other, it's not.
Then watch the last 1911 just before the revolver. It's an excellent stop-action sequence that reveals many things. First will be a light gas blow-by...followed by the bullet nose peekin' out...followed by the blast of gas after the bullet exit. The slide is moving the whole time, but take note of how little it's moved at the point of bullet exit.
Most interesting is the bolt bounce in the AR15 sequence.
Watch and learn...
http://www.trippresearch.com/media/movement/hispeedgateway.html
Then watch the last 1911 just before the revolver. It's an excellent stop-action sequence that reveals many things. First will be a light gas blow-by...followed by the bullet nose peekin' out...followed by the blast of gas after the bullet exit. The slide is moving the whole time, but take note of how little it's moved at the point of bullet exit.
Most interesting is the bolt bounce in the AR15 sequence.
Watch and learn...
http://www.trippresearch.com/media/movement/hispeedgateway.html
Vern Humphrey
Member
Or just look at a fired case -- clearly, the gun is still locked before the bullet exits the muzzle (if it were not, the case wouldn't be in any shape to be reloaded!) Then note how much rearward motion of the slide is needed to unlock the gun.
Sylvan-Forge
Member
1911Tuner .. I've learned a lot of this stuff from you and your references and do appreciate your helpfulness. Please don't take any of this the wrong way .. Not trying to win, I'm trying to understand .. there has to be some kind of explaination for my and others counter-intuitive felt-recoil experiences from these loads.
Mirror plot. 180 top, 135 bottom.
If it takes longer to get the bullet out of the muzzle, it should take longer for the slide to hit the abutement due to the counteracting (slowing) forces of the barrel being pushed forward by the bullet, the case remaining obturated due to pressure in the bore still sealed by the bullet and possibly spring dynamics .. effectively delaying things fractionally (perhaps enough) at the link-down point. Over doing it of course would let the extractor slip out of the groove, rip the case head off or rupture the case.
?
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Thanks for the vid link .. I'm on dial-up, so which one to get?
Can't see extreme left of page either ..
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I honestly don't know, perhaps it is the differing magnitudes of force delivered over time that counts .. did some unscientific doodling and came up with this:
Mirror plot. 180 top, 135 bottom.
Around 16lbs. you say, prior to abutement..
Abrupt stop! The red lines in the doodle graph.
If it takes longer to get the bullet out of the muzzle, it should take longer for the slide to hit the abutement due to the counteracting (slowing) forces of the barrel being pushed forward by the bullet, the case remaining obturated due to pressure in the bore still sealed by the bullet and possibly spring dynamics .. effectively delaying things fractionally (perhaps enough) at the link-down point. Over doing it of course would let the extractor slip out of the groove, rip the case head off or rupture the case.
?
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Thanks for the vid link .. I'm on dial-up, so which one to get?
Can't see extreme left of page either ..
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Attachments
Vern Humphrey
Member
Not necessarily.
The motion of the slide is governed by the momentum of the ejecta, not the dwell time.
Consider the formula: M1 X V1 = M2 X V2.
The mass of the slide (M2) is the fixed component in this equation. It isn't changed by ejecta mass or velocity. So if you change ejecta mass or velocity, you automatically change slide velocity.
If a given round has more momentum, the slide will come back faster -- even if that bullet has less velocity than a lighter bullet.
Sylvan-Forge
Member
Ahh ok, that makes sense Mr. Humphrey .. equal and opposite ..
Doh!
Do you agree on "case" obturation slowing it a bit at link-down?
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Edit: sorry, forgot to specify 'case' obturation
Doh!
Do you agree on "case" obturation slowing it a bit at link-down?
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Edit: sorry, forgot to specify 'case' obturation
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Vern Humphrey
Member
Friction or rifling resistance may have an effect -- rifling can raise pressures and retard velocity somewhat. But any difference between different weight bullets in the same barrel would be very small.
Sylvan-Forge
Member
BruceRDucer
Member
/
Sylvan Forge, in your original post, you showed the calculation for a pistol bullet, and indicated:
3.7409 ft-lbs of recoil
So roughly speaking (and a guestimate is as good enough for me) does that mean if a rifle fired the same, the impact on the shoulder would be 3.7 POUNDS?
Does ft-lbs equate to pounds of impact force exactly, or is the shoulder dimension part of the calculation?
If a 30-06 rifle fires a 150 versus a 180 grain bullet, what might the difference in recoil impact be on the shoulder?
(also, this would be using standard, off the shelf factory loads....nothing exotic, custom or fancy)
/
Sylvan Forge, in your original post, you showed the calculation for a pistol bullet, and indicated:
3.7409 ft-lbs of recoil
So roughly speaking (and a guestimate is as good enough for me) does that mean if a rifle fired the same, the impact on the shoulder would be 3.7 POUNDS?
Does ft-lbs equate to pounds of impact force exactly, or is the shoulder dimension part of the calculation?
If a 30-06 rifle fires a 150 versus a 180 grain bullet, what might the difference in recoil impact be on the shoulder?
(also, this would be using standard, off the shelf factory loads....nothing exotic, custom or fancy)
/
Sylvan-Forge
Member
I meant case obturation, not bullet .. sorry bout that.
Vern Humphrey
Member
Case obturation depends on internal case pressure. When the bullet exits the bore, the internal pressure drops dramatically. In a properly timed gun, unlocking occurs as the bullet exits. There are exceptions to this -- blowback and delayed blowback weapons are examples.
So in a locked breech weapon, with proper timing, case obturation should have little effect.
So in a locked breech weapon, with proper timing, case obturation should have little effect.
Sylvan-Forge
Member
BruceRDucer,
Yes it would be 3.7lbs delivered to the shoulder assuming muzzle vel and rifle weight are the same as factored for the pistol.
Here's the calculator:
http://www.handloads.com/calc/recoil.asp
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My pitch is that with heavier bullets the powercurve is shaped differently and drawn out over a longer period of time giving some folks the sense that it recoils less even if the formulas say more.
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YMMV!
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Yes it would be 3.7lbs delivered to the shoulder assuming muzzle vel and rifle weight are the same as factored for the pistol.
Here's the calculator:
http://www.handloads.com/calc/recoil.asp
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My pitch is that with heavier bullets the powercurve is shaped differently and drawn out over a longer period of time giving some folks the sense that it recoils less even if the formulas say more.
.
YMMV!
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1911Tuner
Moderator Emeritus
I understand, Sylvan...but you still haven't wrapped your head around the fact that...while the 180 may indeed have more momentum at exit velocity...it may not necessarily be the case at the beginning of the event, where the majority of recoil occurs. Muzzle velocity has very little to do with what you feel as recoil, and...in an autopistol...you feel nothing from the actual internal ballistic event. By the time you feel it, the action and reaction between the bullet and the breechblock is over. Done. Finito.
What you feel comes from the slide compressing the spring and its impact with the frame. If you've watched the videos, you've seen it.
In order for the slide to impact the frame harder, it must have more momentum. In order for it to have more momentum, it must be moving faster at impact. Must be. Momentum is a function of mass and velocity.
In order for it to have more velocity at impact, it must have started out faster...during the peak pressure and force imposed on it. That occurs at about a half-inch of bullet travel, or less...just as the bullet hits the rifling.
It's at that point in the sequence that peak force is imposed, and it's at that point that about 95% of the recoil impulse is generated...and you don't even feel it yet. The slide and bullet do...but all you get is a little push from the spring...and it's at that point that the speedy 135's momentum is greater...even though it may not be at exit.
Thank you, Vern!
Unlock begins at about 1/10th inch of slide travel. The bullet must be gone, or at the point of leaving before that can happen. By 1/8th inch, the breech is open, and even the residual gasses are gone, and can't impose any more thrust on the slide. The slide can't accelerate any faster after that happens any more than a bullet can accelerate after it leaves the muzzle.
Sylvan...Read that again. Once the bullet exits the muzzle, there is no more recoil. The action/reaction event is over. The bullet and the slide are moving because of the momentum that they conserved during the event.
Now, back to the bullet for a minute.
The bullet is resisted by the frictional forces imposed by the barrel. An object of lesser mass loses momentum at a faster rate than one of greater mass...so the 135 bullet loses momentum at a faster rate. The frictional resistance is constant.
Now for the slide and spring.
The slide's mass is constant. The spring's rate is constant. The slide's loss of momentum is controlled by the spring.
In firing the 135/1180 load, the slide is moving faster at the moment of peak force...pressure. It has to be in order to match the bullet's momentum.
Now for an analogy that may help.
Fire a .30 caliber bullet at 3,000 fps muzzle velocity and a 180 grain bullet at 2800 fps MV. At 100 yards, the lighter, faster bullet has lost a greater percentage of its initial velocity than the heavier one. By 300 yards, the 180's speed is roughly equal to the 150's. By 400 yards, they're neck and neck...the 180 is closing the gap. At 500 yards, the 180 overtakes the 150 and passes it.
It's actually moving faster, even though it started slower.
Why?
Two reasons. Momentum is the obvious answer...even if they started out at equal momentum. The faster the bullet is moving when it hits the air, the harder the air fights it...offering it greater resistance than on the slower bullet...even though the atmospheric pressure is identical.
The faster any object is moving when it encounters an outside force, the faster that force will slow it down.
So, forget momentum calculations at muzzle velocity, and consider what happens while the bullet and breechblock are compelled by peak pressure and force.
To recap:
In order for a given load to produce more felt recoil than another one in an autopistol, the slide must hit the frame with more momentum. Because momentum is a function of mass and velocity, it must be moving faster. If it's moving faster, it's compressing the spring faster. When the spring compresses faster, it pushes backward on the frame faster.
What you feel comes from the slide compressing the spring and its impact with the frame. If you've watched the videos, you've seen it.
In order for the slide to impact the frame harder, it must have more momentum. In order for it to have more momentum, it must be moving faster at impact. Must be. Momentum is a function of mass and velocity.
In order for it to have more velocity at impact, it must have started out faster...during the peak pressure and force imposed on it. That occurs at about a half-inch of bullet travel, or less...just as the bullet hits the rifling.
It's at that point in the sequence that peak force is imposed, and it's at that point that about 95% of the recoil impulse is generated...and you don't even feel it yet. The slide and bullet do...but all you get is a little push from the spring...and it's at that point that the speedy 135's momentum is greater...even though it may not be at exit.
Thank you, Vern!
Unlock begins at about 1/10th inch of slide travel. The bullet must be gone, or at the point of leaving before that can happen. By 1/8th inch, the breech is open, and even the residual gasses are gone, and can't impose any more thrust on the slide. The slide can't accelerate any faster after that happens any more than a bullet can accelerate after it leaves the muzzle.
Sylvan...Read that again. Once the bullet exits the muzzle, there is no more recoil. The action/reaction event is over. The bullet and the slide are moving because of the momentum that they conserved during the event.
Now, back to the bullet for a minute.
The bullet is resisted by the frictional forces imposed by the barrel. An object of lesser mass loses momentum at a faster rate than one of greater mass...so the 135 bullet loses momentum at a faster rate. The frictional resistance is constant.
Now for the slide and spring.
The slide's mass is constant. The spring's rate is constant. The slide's loss of momentum is controlled by the spring.
In firing the 135/1180 load, the slide is moving faster at the moment of peak force...pressure. It has to be in order to match the bullet's momentum.
Now for an analogy that may help.
Fire a .30 caliber bullet at 3,000 fps muzzle velocity and a 180 grain bullet at 2800 fps MV. At 100 yards, the lighter, faster bullet has lost a greater percentage of its initial velocity than the heavier one. By 300 yards, the 180's speed is roughly equal to the 150's. By 400 yards, they're neck and neck...the 180 is closing the gap. At 500 yards, the 180 overtakes the 150 and passes it.
It's actually moving faster, even though it started slower.
Why?
Two reasons. Momentum is the obvious answer...even if they started out at equal momentum. The faster the bullet is moving when it hits the air, the harder the air fights it...offering it greater resistance than on the slower bullet...even though the atmospheric pressure is identical.
The faster any object is moving when it encounters an outside force, the faster that force will slow it down.
So, forget momentum calculations at muzzle velocity, and consider what happens while the bullet and breechblock are compelled by peak pressure and force.
To recap:
In order for a given load to produce more felt recoil than another one in an autopistol, the slide must hit the frame with more momentum. Because momentum is a function of mass and velocity, it must be moving faster. If it's moving faster, it's compressing the spring faster. When the spring compresses faster, it pushes backward on the frame faster.
Sylvan-Forge
Member
1911Tuner, thanks! I'm pretty sure I've got the mechanics down, mostly.
And I think I might now understand what I have been experiencing:.
The powercurve leading up to that point was broader and smoother or whatever, giving the sensation that things are not abrupt as they seem to be with the lighter faster load.
More prepped or 'momentumized' leading up to the finale rather than a such a recurved/rapid rise to finale.
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Sound good?
If not, that's cool .. my eggs are fried.
Good exersize though! Thanks for it!
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And I think I might now understand what I have been experiencing:.
The powercurve leading up to that point was broader and smoother or whatever, giving the sensation that things are not abrupt as they seem to be with the lighter faster load.
More prepped or 'momentumized' leading up to the finale
rather than a such a recurved/rapid rise to finale..
Sound good?
If not, that's cool .. my eggs are fried.
Good exersize though! Thanks for it!
.
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1911Tuner
Moderator Emeritus
I dunno. Sounds a bit too complicated...
"Power Curve" suggests that the actual recoil impulse occurs over a longer period of time..which even though it probably does...wouldn't make any practical difference with the powder burn rates in an autopistol cartridge. Studying the pressure charts with quick powders shows a rapid peak instead of a slow climb, with very little area under the "curve" Once it peaks, it drops about as fast as it took to get there.
Anyway...
It seems like you've seen that what you feel as recoil doesn't come from the ballistic event...in an autopistol.
The greatest percentage of the felt recoil...the gun's movement...comes from the slide impacting the frame.
If the bullet is gone at 1/10th to 1/8th inch of slide travel, by the time the slide hits the frame...the bullet has already hit a 25-yard target and likely come to a stop in the berm.
Because recoil is the reaction side of an action/reaction event...once the action side of the equation is missing...the reaction is over. IOW, by the time the slide hits the frame, and you feel the effects, there is no more recoil because the bullet isn't there. Recoil occurs while the bullet is in the barrel and being driven forward. No bullet/no force/no recoil.
"Power Curve" suggests that the actual recoil impulse occurs over a longer period of time..which even though it probably does...wouldn't make any practical difference with the powder burn rates in an autopistol cartridge. Studying the pressure charts with quick powders shows a rapid peak instead of a slow climb, with very little area under the "curve" Once it peaks, it drops about as fast as it took to get there.
Anyway...
It seems like you've seen that what you feel as recoil doesn't come from the ballistic event...in an autopistol.
The greatest percentage of the felt recoil...the gun's movement...comes from the slide impacting the frame.
If the bullet is gone at 1/10th to 1/8th inch of slide travel, by the time the slide hits the frame...the bullet has already hit a 25-yard target and likely come to a stop in the berm.
Because recoil is the reaction side of an action/reaction event...once the action side of the equation is missing...the reaction is over. IOW, by the time the slide hits the frame, and you feel the effects, there is no more recoil because the bullet isn't there. Recoil occurs while the bullet is in the barrel and being driven forward. No bullet/no force/no recoil.
BruceRDucer
Member
I have a box of Remington Express Bronze Point in 30-06 Springfield 180 grain
Is there an online chart that will show:
(1) Powder Charge in Grains
(2) Velocity in fps
My rifle's weight is 6 3/4 lbs
I would like to use your calculator or formula to obtain the recoil data. Thanks.
/
Is there an online chart that will show:
(1) Powder Charge in Grains
(2) Velocity in fps
My rifle's weight is 6 3/4 lbs
I would like to use your calculator or formula to obtain the recoil data. Thanks.
/
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