I was addressing those who argue that the force is the same in each caliber and that it shows that the recoil is the same, which is incorrect. There is, of course, some subjectivity in the observation of the recoil differences, but the recoils themselves are also very different, proven by the difference in impulses.
caliber vs. recoil
- Thread starter Flyboy_451
- Start date
I am not a physics major, and my college physics was a few decades ago. I would assert that impulse is force.
I found this:
Consider that the energy generated by firing a cartridge in a semi-auto handgun is confined to the barrel, chamber, the frame by connection to the barrel via the barrel link, and the slide.
The time that it takes for both the .40S&W and the .45ACP bullets to travel the length of the barrel should be almost identical and their muzzle velocities are similar.
I think what is needed is an objective measure of the energy exerted on the grip of the two guns as they are fired and the time it takes for that energy to be exerted. Is there an instrument which can do this. Maybe an accelerometer?
I found this:
Consider that the energy generated by firing a cartridge in a semi-auto handgun is confined to the barrel, chamber, the frame by connection to the barrel via the barrel link, and the slide.
The time that it takes for both the .40S&W and the .45ACP bullets to travel the length of the barrel should be almost identical and their muzzle velocities are similar.
I think what is needed is an objective measure of the energy exerted on the grip of the two guns as they are fired and the time it takes for that energy to be exerted. Is there an instrument which can do this. Maybe an accelerometer?
Lost Sheep
New member
Effect of the recoil spring?
You could also consider that the pressure against the breechface is not felt by the shooter at all. The breechface is part of the slide. It is the recoil spring that transfers the force to the frame, then to the shooter, then, later, the force of the slide hitting the stop.
If you went to heavier and heavier recoil springs until the slide no longer went to it's full travel against the stop, but still traveled enough to cycle (meaning the spring absorbed ALL the recoil as it compressed - over what length of travel?), this might be informative.
.
At that point, observing the difference in the spring weight of the recoil springs would give a quantitative measure. Of what, I don't quite know, but it would be a measure of something.
Yeah, we got a mystery here. The only thing actually measured by Flyboy_451 (by his intent and successfully, I would say) was perception of the sample population of shooters where simple physics tells us there should be none. That PROVED that more than simple (or simplistic) physics is involved. But I think that you guys are on the right track and I look forward to the investigation. From what I have read so far, if you combine forces, you will make real progress and I thank you both for it, as well as the other contributors here.
Now, I want to know why the felt recoil of 480 Ruger vs 454 Casull (loaded down, even) both fired in Super Redhawks is said to be so much less.
Lost Sheep
You could also consider that the pressure against the breechface is not felt by the shooter at all. The breechface is part of the slide. It is the recoil spring that transfers the force to the frame, then to the shooter, then, later, the force of the slide hitting the stop.
If you went to heavier and heavier recoil springs until the slide no longer went to it's full travel against the stop, but still traveled enough to cycle (meaning the spring absorbed ALL the recoil as it compressed - over what length of travel?), this might be informative.
.
At that point, observing the difference in the spring weight of the recoil springs would give a quantitative measure. Of what, I don't quite know, but it would be a measure of something.
Yeah, we got a mystery here. The only thing actually measured by Flyboy_451 (by his intent and successfully, I would say) was perception of the sample population of shooters where simple physics tells us there should be none. That PROVED that more than simple (or simplistic) physics is involved. But I think that you guys are on the right track and I look forward to the investigation. From what I have read so far, if you combine forces, you will make real progress and I thank you both for it, as well as the other contributors here.
Now, I want to know why the felt recoil of 480 Ruger vs 454 Casull (loaded down, even) both fired in Super Redhawks is said to be so much less.
Lost Sheep
Force = PSI * area
.40 Force = 35,000 * pi * .2^2 = 4398
.45 Force = 19,000 * pi * .225^2 = 3021
Velocity in both cases = 1000 fps
F = m*a -> a = F/m
180 grains = 0.000799224433 slugs
185 grains = 0.000821425112 slugs (to keep some semblance of real units)
.40 accel = 4398 / .000799224433 = 5502834 fps/s
.45 accel = 3021 / .000821425112 = 3677754 fps/s
a = ΔV / Δt -> Δt = ΔV / a
.40 Δt = 1000 / 5502834 = 1.817E-4 s
.45 Δt = 1000 / 3677754 = 2.719E-4 s
The .40 takes only 66% of the time to reach 1000 fps as the .45 does, that's not insignificant.
With the .40 you see a higher force with a shorter time, hence the recoil is, as most people describe, a sharper kick. With the .45 you get a lower force with a longer time, so you get a recoil described by most as a longer or smoother push.
.40 Force = 35,000 * pi * .2^2 = 4398
.45 Force = 19,000 * pi * .225^2 = 3021
Velocity in both cases = 1000 fps
F = m*a -> a = F/m
180 grains = 0.000799224433 slugs
185 grains = 0.000821425112 slugs (to keep some semblance of real units)
.40 accel = 4398 / .000799224433 = 5502834 fps/s
.45 accel = 3021 / .000821425112 = 3677754 fps/s
a = ΔV / Δt -> Δt = ΔV / a
.40 Δt = 1000 / 5502834 = 1.817E-4 s
.45 Δt = 1000 / 3677754 = 2.719E-4 s
The .40 takes only 66% of the time to reach 1000 fps as the .45 does, that's not insignificant.
With the .40 you see a higher force with a shorter time, hence the recoil is, as most people describe, a sharper kick. With the .45 you get a lower force with a longer time, so you get a recoil described by most as a longer or smoother push.
Flyboy_451
New member
Ralgha,
Thanks for the input! The formulas for acceleration are above my head, so I can only assume that you are correct. It would seem that the pressure and the time in which it acts is where the difference exists. This was my theory to begin with, thus the questions posed to those that participated in the shooting. Thanks to all who have participated in this discussion! We may not have solved the mysteries of felt recoil, but I think a lot of good information was brought to light! This is what discussion forums are all about. Once again, a hearty thank you to all who participated!
JW
Thanks for the input! The formulas for acceleration are above my head, so I can only assume that you are correct. It would seem that the pressure and the time in which it acts is where the difference exists. This was my theory to begin with, thus the questions posed to those that participated in the shooting. Thanks to all who have participated in this discussion! We may not have solved the mysteries of felt recoil, but I think a lot of good information was brought to light! This is what discussion forums are all about. Once again, a hearty thank you to all who participated!
JW
"My point was simply that the forward velocity of the bullet does not generate a rearward thrust or force."
Wrong, it is precisely the forward movement of the bullet that generates recoil. The gas pressure at any given point is pushing equally in all directions so it cannot move the gun backwards. If the bullet is confined (the barrel blocked) in, say, a bolt action rifle, or in a recoil operated pistol like the M1911, and the round fired, nothing happens. The rifle just sits there or the pistol slide does not move.
(Now in a blowback pistol, the breech will open rather violently, but there will still be no recoil as the bullet does not move.)
Jim
Wrong, it is precisely the forward movement of the bullet that generates recoil. The gas pressure at any given point is pushing equally in all directions so it cannot move the gun backwards. If the bullet is confined (the barrel blocked) in, say, a bolt action rifle, or in a recoil operated pistol like the M1911, and the round fired, nothing happens. The rifle just sits there or the pistol slide does not move.
(Now in a blowback pistol, the breech will open rather violently, but there will still be no recoil as the bullet does not move.)
Jim
Rhalga:
You are assuming that everything is linear -- e.g., the pressure is constant, the acceleration is therefore constant, etc. In general, this is probably fine, but if we're talking about accelerations, we should also consider friction, and other forces that are acting upon the bullet. The friction differential between the .40 and the .45 will be significant, since the .40 has a larger bearing surface (since it's a smaller diameter, the bullet will be longer to reach ~180 grains).
A good way to see why this is problematic is the equation for distance, given a known acceleration and time. Your numbers are not consistent with known starting conditions. Let me explain -- here are the numbers from your calculations:
.40: a = 5.5E6 fps/s, Δt = 1.8E-4 seconds
.45: a = 3.7E6 fps/s, Δt = 2.7E-4 seconds
The equation for distance given a and dt, given an initial velocity of zero, is as follows:
d = (at^2)/2
If we plug your numbers in, we get resultant distances as follows:
.40: d = 5.5E6 * 1.8E-4 * 1.8E-4 / 2.0 = 8.9E-2 ft = 1 inch
.45: d = 3.7E6 * 2.7E-4 * 2.7E-4 / 2.0 = 13.5E-2 ft = 1.6 inches
Given your figures, it would appear that one of the barrels is 60% longer than the other
. Obviously, this is not the case, given the starting condition that the guns are visually indistinguishable. The reason your figures are off is, of course, as stated above, due to the fact that you are calculating things using force equations. You must account for counteracting forces, and friction will be very significant.
Mello2u is absolutely correct -- impulse, it turns out, is equivalent to the change in momentum (this is because physics really works!). Mello2u quoted an algebraic expression for the Impulse (note the use of "average acceleration"), but impulse is really the integral of a force over time. The force equation for this system is a Classic Hard Physics Problem™, and beyond any normal human's skill. The reason everyone uses momentum is that you can ignore energy, opposing forces, etc., if you have the resulting conditions (e.g., muzzle velocity). The time should be the same.
You are assuming that everything is linear -- e.g., the pressure is constant, the acceleration is therefore constant, etc. In general, this is probably fine, but if we're talking about accelerations, we should also consider friction, and other forces that are acting upon the bullet. The friction differential between the .40 and the .45 will be significant, since the .40 has a larger bearing surface (since it's a smaller diameter, the bullet will be longer to reach ~180 grains).
A good way to see why this is problematic is the equation for distance, given a known acceleration and time. Your numbers are not consistent with known starting conditions. Let me explain -- here are the numbers from your calculations:
.40: a = 5.5E6 fps/s, Δt = 1.8E-4 seconds
.45: a = 3.7E6 fps/s, Δt = 2.7E-4 seconds
The equation for distance given a and dt, given an initial velocity of zero, is as follows:
d = (at^2)/2
If we plug your numbers in, we get resultant distances as follows:
.40: d = 5.5E6 * 1.8E-4 * 1.8E-4 / 2.0 = 8.9E-2 ft = 1 inch
.45: d = 3.7E6 * 2.7E-4 * 2.7E-4 / 2.0 = 13.5E-2 ft = 1.6 inches
Given your figures, it would appear that one of the barrels is 60% longer than the other
. Obviously, this is not the case, given the starting condition that the guns are visually indistinguishable. The reason your figures are off is, of course, as stated above, due to the fact that you are calculating things using force equations. You must account for counteracting forces, and friction will be very significant.Mello2u is absolutely correct -- impulse, it turns out, is equivalent to the change in momentum (this is because physics really works!). Mello2u quoted an algebraic expression for the Impulse (note the use of "average acceleration"), but impulse is really the integral of a force over time. The force equation for this system is a Classic Hard Physics Problem™, and beyond any normal human's skill. The reason everyone uses momentum is that you can ignore energy, opposing forces, etc., if you have the resulting conditions (e.g., muzzle velocity). The time should be the same.
I really don't want to spend forever on physics stuff. I am extremely confident that momentum / impulse is the correct way to examine the question of "actual recoil." I feel I've expressed what I know well enough that if you're curious about more, I'd much rather recommend ballistics textbooks or a study of Newtonian mechanics rather than arguing about it . "Felt recoil" is its own thing, and might even have to include things like muzzle flash or sound levels.
You never stated the gun weights. I'm guessing that switching from .40 to .45 (or vice versa), with resultant changes in mags, springs, etc., will result in a weight difference. I suppose it's small, but I'd like to know what it is. If it's more than a few percent of the weight of the gun, that might well explain the difference in felt recoil.
Also, did you randomize the order in which the guns were shot? Did everyone shoot the .40 first, for example? Ideally, you would ensure a 50-50 split between subjects who shot the .40 first and those who shot the .45 first. If you randomized some, but it was more like 87-13, then that could explain your skew. Your initial post is not clear about how the samples were taken.
. "Felt recoil" is its own thing, and might even have to include things like muzzle flash or sound levels. You never stated the gun weights. I'm guessing that switching from .40 to .45 (or vice versa), with resultant changes in mags, springs, etc., will result in a weight difference. I suppose it's small, but I'd like to know what it is. If it's more than a few percent of the weight of the gun, that might well explain the difference in felt recoil.
Also, did you randomize the order in which the guns were shot? Did everyone shoot the .40 first, for example? Ideally, you would ensure a 50-50 split between subjects who shot the .40 first and those who shot the .45 first. If you randomized some, but it was more like 87-13, then that could explain your skew
. Your initial post is not clear about how the samples were taken.Felt recoil is an insufficient term for recoil. Muzzle blast is also important. So it is really more of a perceived recoil, a combination of felt recoil and how loud it is. I think what you done is measured peoples sensitivity to muzzle blast. The .40 is higher pressure and louder.
Not an engineer but I do have 3 year of physics.
Pressure is involved but it's not the main element. The bullet momentum is the main factor. The second factor is the weight and velocity, multiplied together also equal momentum, of the powder gasses. A higher pressure will give you higher gas velocity and increase recoil. In this example the heavier bullet weight and heavier powder charge of the .45 will probably offset the difference. So the total momentum is
Mb + Mp = Mt
Mb = momentum of bullet
Mp = momentum of the powder gasses
Mt = total momentum
The momentum of the slide will also equal Mt. So the slides will have the same velocity and the recoil will be the same.
Your dwell times are way off for several reasons. First the pressure is not even close to a constant. It's continuously variable. The .40 will probably have a faster rise to max pressure but it will also fall off faster than the .45.
Second you don't calculate the drag from the frictional forces. Last you don't know what the pressures are for these rounds.
Some software, quik load I think, will give you the dwell times. From what I remember I think for service type pistols it's usually about .001 seconds. The differance for the loads we are looking at will be at most .0002 seconds. If ya'll think you can feel that difference you must have spidy senses.
Someone brought Impulse into the argument. They are correct in that the longer the time over which the momentum change occurs, the lower the average force and therefore the lower the felt recoil. It explains why a revolver will have more felt recoil than a semi-auto if they are the same weight and load. In the revolver the momentum is almost completely transfered to the gun by the time the bullet leaves the barrel. The reason it's almost is because the powder gasses do impart some momentum change after the bullets leaves the barrel. For the semi-auto, the slide gets the recoil in the same time as the revolver. But the slide transfers that to the frame, hence your hand, over the entire duration of the slide travel.
In the examples I think time difference is to small to make a difference in felt recoil. This is because the bullet in both cases will leave the barrel before the slide has moved .1 inch. The difference in movement will be something on the order of .01 or .02 inch. This difference is such a small part of the total slide travel I don't think you can feel it. See how much pressure it takes to move your slide .1 inch. That's all the force you will feel before the bullet exits the barrel.
The .40 is considered snappy because it's usually shot in a 9mm size gun that hasn't been modified enough to account for the increased momentum of the .40 as compared to the 9mm. Take a G17 9mm vs a G22 .40. They use the same recoil spring and the slide is less than an ounce heavier. So the slide has a higher recoil velocity and hits the frame harder. Most people think the snappiness of the .40 is due to the faster acceleration. It's not, the slap is at the end when the slide hits the frame. Look at some slow motion videos and you'll see most of the muzzle flip occurs when the slide hits the frame. I have a G21 .45 and a G20 10mm with a .40 conversion barrel. In these guns the .40 and .45 have no noticeable difference in recoil and the split times don't show any difference as well. I like using the split times because its not subjective.
Look at the firing a handgun under water thread and watch when the muzzle rise occurs.
http://www.youtube.com/watch?v=_eUlpPY96Ok
Here is another video that shows when the main muzzle rise occurs.
http://www.trippresearch.com/tech/video.html#
Not an engineer but I do have 3 year of physics.
Pressure is involved but it's not the main element. The bullet momentum is the main factor. The second factor is the weight and velocity, multiplied together also equal momentum, of the powder gasses. A higher pressure will give you higher gas velocity and increase recoil. In this example the heavier bullet weight and heavier powder charge of the .45 will probably offset the difference. So the total momentum is
Mb + Mp = Mt
Mb = momentum of bullet
Mp = momentum of the powder gasses
Mt = total momentum
The momentum of the slide will also equal Mt. So the slides will have the same velocity and the recoil will be the same.
Your dwell times are way off for several reasons. First the pressure is not even close to a constant. It's continuously variable. The .40 will probably have a faster rise to max pressure but it will also fall off faster than the .45.
Second you don't calculate the drag from the frictional forces. Last you don't know what the pressures are for these rounds.
Some software, quik load I think, will give you the dwell times. From what I remember I think for service type pistols it's usually about .001 seconds. The differance for the loads we are looking at will be at most .0002 seconds. If ya'll think you can feel that difference you must have spidy senses.
Someone brought Impulse into the argument. They are correct in that the longer the time over which the momentum change occurs, the lower the average force and therefore the lower the felt recoil. It explains why a revolver will have more felt recoil than a semi-auto if they are the same weight and load. In the revolver the momentum is almost completely transfered to the gun by the time the bullet leaves the barrel. The reason it's almost is because the powder gasses do impart some momentum change after the bullets leaves the barrel. For the semi-auto, the slide gets the recoil in the same time as the revolver. But the slide transfers that to the frame, hence your hand, over the entire duration of the slide travel.
In the examples I think time difference is to small to make a difference in felt recoil. This is because the bullet in both cases will leave the barrel before the slide has moved .1 inch. The difference in movement will be something on the order of .01 or .02 inch. This difference is such a small part of the total slide travel I don't think you can feel it. See how much pressure it takes to move your slide .1 inch. That's all the force you will feel before the bullet exits the barrel.
The .40 is considered snappy because it's usually shot in a 9mm size gun that hasn't been modified enough to account for the increased momentum of the .40 as compared to the 9mm. Take a G17 9mm vs a G22 .40. They use the same recoil spring and the slide is less than an ounce heavier. So the slide has a higher recoil velocity and hits the frame harder. Most people think the snappiness of the .40 is due to the faster acceleration. It's not, the slap is at the end when the slide hits the frame. Look at some slow motion videos and you'll see most of the muzzle flip occurs when the slide hits the frame. I have a G21 .45 and a G20 10mm with a .40 conversion barrel. In these guns the .40 and .45 have no noticeable difference in recoil and the split times don't show any difference as well. I like using the split times because its not subjective.
Look at the firing a handgun under water thread and watch when the muzzle rise occurs.
http://www.youtube.com/watch?v=_eUlpPY96Ok
Here is another video that shows when the main muzzle rise occurs.
http://www.trippresearch.com/tech/video.html#
Last edited:
I did oversimplify a bit, by ignoring the mass of the powder, which contributes to recoil. (Note that as the powder burns and is converted to gas, its volume increases, but its mass does not change.)
Powder mass is not too significant in a handgun when we have a 230 grain bullet, say, and a 7 grain powder charge. But in a rifle, with, say, a 150 grain bullet and a 50 grain powder charge, the momentum of the powder and powder gas contributes significantly to recoil.
But pressure does not. In a closed container (which the cartridge case-barrel-bullet is, pressure acts equally in all directions. At any given time and any given bullet position, pressure is equal in all directions within that container. Nor is friction involved. Recoil can and does occur even when the bullet is so loose fitting in the barrel that there is effectively no friction at all.
Recoil is the effect of the equal and opposite motion business. m x v of the bullet and powder/gas going one way creates the M x V of the gun recoiling the other way.
If in doubt, consider a space rocket. Some folks consider the rocket like a bullet, but it is the opposite - the rocket is the "gun". As the fuel burns, it is pushed out the back at high speed and that is what causes the recoil that sends the rocket into space.
So, again, if the bullet does not move, the gun does not recoil. Period.
Jim
Powder mass is not too significant in a handgun when we have a 230 grain bullet, say, and a 7 grain powder charge. But in a rifle, with, say, a 150 grain bullet and a 50 grain powder charge, the momentum of the powder and powder gas contributes significantly to recoil.
But pressure does not. In a closed container (which the cartridge case-barrel-bullet is, pressure acts equally in all directions. At any given time and any given bullet position, pressure is equal in all directions within that container. Nor is friction involved. Recoil can and does occur even when the bullet is so loose fitting in the barrel that there is effectively no friction at all.
Recoil is the effect of the equal and opposite motion business. m x v of the bullet and powder/gas going one way creates the M x V of the gun recoiling the other way.
If in doubt, consider a space rocket. Some folks consider the rocket like a bullet, but it is the opposite - the rocket is the "gun". As the fuel burns, it is pushed out the back at high speed and that is what causes the recoil that sends the rocket into space.
So, again, if the bullet does not move, the gun does not recoil. Period.
Jim
Wiki has a very good technical discussion of how to calculate free recoil. There is no term in the equation for either pressure or time. Of course, it's the pressure that moves the bullet and leads to the recoil effect, but the important thing is the mass * velocity of the ejecta, not whether they are accelerated to that velocity by high pressure over a short period of time or lower pressure over a longer period.
I think the point being missed is this: The two masses were nearly identical and moving at the same velocities. This means that the energy or work done by the expanding gases in the barrel are virtually identical, and that the recoil energy would be the same. This is not really up for argument, since it's so obvious. If you want an equation for this, I suggest KE = 0.5mv^2, or kinetic energy is equal to one half of the square of the bullet velocity multiplied by the mass of the bullet. These numbers should come out to be pretty darn similar for the two custom loads that the OP presented to us.
The difference, as the OP has pointed out, was in the pressure of the rounds. I suggest that the higher pressure of the .40 means that the gases in this cartridge did their work more quickly than the lower-pressure gases in the .45. Essentially we're talking about F (force, or the pressure exerted) = m (mass of the bullets, which is the same) * a (acceleration, or how rapidly the bullet changes velocity). If the force is higher in one cartridge thanks to the higher pressure and the mass is the same, then the acceleration must increase to keep the equation balanced. Keep in mind that the force of the expanding gases are applied to the entire sealed system of the chamber, case, barrel, and bullet. Since the bullet moves away, the gases are going to push the case and bolt-face toward the shooter to compensate, thus recoil.
So, if we go back to our first equation, we have equal final kinetic energies but one bullet accelerates to that energy level quicker, meaning that the expanding gases are going to push toward the shooter in compensation faster or an imbalance in the system will result. This is why, even if they have the same total recoil energy, a high-pressure round like the .40 is going to have "sharper," that is really "quicker" recoil. It's not really subjective - if you put identical test barrels on a machine that graphed force over time, the 40 would have a higher peak recoil force than the .45, despite the fact that the two rounds have equal energy. However, the .45's force pulse would be longer. It has been a long time since I took calculus, but I imagine that if the energies are the same, then the integral of the force over time curves would be the same - that is, the area under the force x time curves would be equal.
If we were able to exaggerate the differences in pressure and diameter with a large-diameter projectile made of lighter material and a smaller, higher-pressure round with normal bullets, the difference in recoil speed or "snappiness" would be exaggerated still further - while the muzzle energies remained the same.
James: I think you're missing the concept referred to as "work." Yes, the velocity of the bullet and the speed of the gases under a rocket are related to moving the associated items, but the velocity is not the cause. The speed of a body has no ability to directly affect the speed of another body unless the two have actually collided, and a bullet never collides with anything in the gun - it only moves down the barrel, resulting in friction. In the example of the rocket, it is the force of the excited gas molocules pushing (creating pressure) on the inside of the rocket engines and the nozzles which accelerates the rocket and creates its velocity. In the gun, it is the force of the excited gas molocules pushing (creating pressure) on the inside of the casing, bolt face, barrel, muzzle, etc. The difference is, unlike the rocket, they are restricted. In a totally closed container, constant pressure (the force of gaseous molocules on the container) does no work (which is measured in units of energy), because to have work something must be moved. In a container like the chamber/casing/barrel system of a gun, the pressure can only do work in one direction, which is to accelerate the bullet out of the barrel. Since the force is able to perform work (that is, accelerate the bullet) in only one direction, the associated "equal and opposite reaction" must have its vector in the opposite direction. Thus, it is the pressure in the chamber rather than the speed of the bullet which is actually the agent doing the work. The presence of the bullet simply allows the system to be "temporarily closed" which results in the pressure increase. As soon as the pressure overcomes friction, the bullet begins to move and increase the volume of the closed system, meaning that the pressure decreases as it moves down the barrel. It's all precisely timed, which is why it is possible to achieve very different results with different powders - they all produce pressure at different speeds and with different curves. To achieve maximum velocity in a cartridge without exceeding the maximum pressure allowable, you have to pick a powder which will exert the most permissible pressure possible on the bullet for the longest period of time achievable - (this means keeping the force of F=ma as high as possible for as long as possible, keeping the bullet accelerating quickly) thus, the most energy imparted to the bullet, and since energy and velocity are related, the more energy the more velocity. Velocity of the bullet is not the cause of anything - it is the result of everything.
DISCLAIMER: I don't have a degree in engineering, but I did take some engineering classes, and I had a great physics professor. Take out of the above what you will, but I can assure you it is correct in essentials.
Another edit: Regarding the idea that recoil is perceived as snappy because of slide movement only, this is disregarding slide velocity. If there are two different pressures working on the slide, then slides of equal weight will have different velocities, momentums, energies, etc. Since the pressure peak of the 40 is higher and shorter than that of the 45, it is likely to accelerate more quickly and therefore reach the top speed and the slide stop faster, thus pressure is related to slide-induced felt recoil.
The difference, as the OP has pointed out, was in the pressure of the rounds. I suggest that the higher pressure of the .40 means that the gases in this cartridge did their work more quickly than the lower-pressure gases in the .45. Essentially we're talking about F (force, or the pressure exerted) = m (mass of the bullets, which is the same) * a (acceleration, or how rapidly the bullet changes velocity). If the force is higher in one cartridge thanks to the higher pressure and the mass is the same, then the acceleration must increase to keep the equation balanced. Keep in mind that the force of the expanding gases are applied to the entire sealed system of the chamber, case, barrel, and bullet. Since the bullet moves away, the gases are going to push the case and bolt-face toward the shooter to compensate, thus recoil.
So, if we go back to our first equation, we have equal final kinetic energies but one bullet accelerates to that energy level quicker, meaning that the expanding gases are going to push toward the shooter in compensation faster or an imbalance in the system will result. This is why, even if they have the same total recoil energy, a high-pressure round like the .40 is going to have "sharper," that is really "quicker" recoil. It's not really subjective - if you put identical test barrels on a machine that graphed force over time, the 40 would have a higher peak recoil force than the .45, despite the fact that the two rounds have equal energy. However, the .45's force pulse would be longer. It has been a long time since I took calculus, but I imagine that if the energies are the same, then the integral of the force over time curves would be the same - that is, the area under the force x time curves would be equal.
If we were able to exaggerate the differences in pressure and diameter with a large-diameter projectile made of lighter material and a smaller, higher-pressure round with normal bullets, the difference in recoil speed or "snappiness" would be exaggerated still further - while the muzzle energies remained the same.
James: I think you're missing the concept referred to as "work." Yes, the velocity of the bullet and the speed of the gases under a rocket are related to moving the associated items, but the velocity is not the cause. The speed of a body has no ability to directly affect the speed of another body unless the two have actually collided, and a bullet never collides with anything in the gun - it only moves down the barrel, resulting in friction. In the example of the rocket, it is the force of the excited gas molocules pushing (creating pressure) on the inside of the rocket engines and the nozzles which accelerates the rocket and creates its velocity. In the gun, it is the force of the excited gas molocules pushing (creating pressure) on the inside of the casing, bolt face, barrel, muzzle, etc. The difference is, unlike the rocket, they are restricted. In a totally closed container, constant pressure (the force of gaseous molocules on the container) does no work (which is measured in units of energy), because to have work something must be moved. In a container like the chamber/casing/barrel system of a gun, the pressure can only do work in one direction, which is to accelerate the bullet out of the barrel. Since the force is able to perform work (that is, accelerate the bullet) in only one direction, the associated "equal and opposite reaction" must have its vector in the opposite direction. Thus, it is the pressure in the chamber rather than the speed of the bullet which is actually the agent doing the work. The presence of the bullet simply allows the system to be "temporarily closed" which results in the pressure increase. As soon as the pressure overcomes friction, the bullet begins to move and increase the volume of the closed system, meaning that the pressure decreases as it moves down the barrel. It's all precisely timed, which is why it is possible to achieve very different results with different powders - they all produce pressure at different speeds and with different curves. To achieve maximum velocity in a cartridge without exceeding the maximum pressure allowable, you have to pick a powder which will exert the most permissible pressure possible on the bullet for the longest period of time achievable - (this means keeping the force of F=ma as high as possible for as long as possible, keeping the bullet accelerating quickly) thus, the most energy imparted to the bullet, and since energy and velocity are related, the more energy the more velocity. Velocity of the bullet is not the cause of anything - it is the result of everything.
DISCLAIMER: I don't have a degree in engineering, but I did take some engineering classes, and I had a great physics professor. Take out of the above what you will, but I can assure you it is correct in essentials.
Another edit: Regarding the idea that recoil is perceived as snappy because of slide movement only, this is disregarding slide velocity. If there are two different pressures working on the slide, then slides of equal weight will have different velocities, momentums, energies, etc. Since the pressure peak of the 40 is higher and shorter than that of the 45, it is likely to accelerate more quickly and therefore reach the top speed and the slide stop faster, thus pressure is related to slide-induced felt recoil.
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" Since the bullet moves away, the gases are going to push the case and bolt-face toward the shooter to compensate, thus recoil."
Nope, go back to your physics classes. Pressure has no effect on recoil, only forward movement of matter does. As the bullet moves away, the pressure at any given point in time or distance is the same in all directions.
Jim
Nope, go back to your physics classes. Pressure has no effect on recoil, only forward movement of matter does. As the bullet moves away, the pressure at any given point in time or distance is the same in all directions.
Jim
Yes with a machine you could measure the .40 to have a higher peak recoil. The average recoil will be the same. In practice you will never feel it. The bullet is out of the barrel by the time the slide has moved about .1 inch. Take a pistol and push the slide back .1 inch and note how much force it takes. You'll never feel the difference when shooting. The difference in the force pulse is on the order of .0001 to .0002 seconds. You simply can't detect this small a difference with human senses.
This is wrong. Pressure has very little to do with slide velocity. This is a conservation of momentum problem. The slide will have the same momentum as the bullet. If the bullets have the same momentum the slides will have the same momentum. Since the slides are identical they will have the same velocity. For identical guns, as long as the bullets have the same momentum the slides will have the same momentum and the impact speed of the slide against the frame will be the same. As mentioned earlier the difference in slide cycle times is on the order of .0001 to .0002 seconds. You just can't feel that small a difference.
While the equations don't have a term for pressure, one of the terms is dependent on pressure. The velocity of the powder, really powder gasses and unburnt powder, is dependent on pressure. The higher the pressure the higher the velocity.
Time is not a factor in calculating the total free recoil energy. Time is important in felt recoil. If you increase the time over which the energy is transfered to your hand you will decrease the felt recoil.
Generally, yes, but it's the area under the pressure curve that determines muzzle velocity, not the peak pressure, which is why some powders can achieve higher velocities (same cartridge, bullet, and firearm, obviously) than others when both are loaded to not exceed the SAAMI pressure spec for the cartridge.
Yes, exactly. Some of the preceding posts seemed to not distinguish very well between free recoil, which is a matter of physics and therefore able to be calculated objectively, and felt recoil, which has a large subjective component related to time and many other factors.