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Post by TMan on Jun 11, 2005 19:27:23 GMT -5
There have been a lot of discussions about the stopping power of the .45 ACP vs. the 9 mm Luger. That is outside to scope of this thread i.e. please don't go there. My shooting instructor encouraged me to get a 9mm vs. my .45 because it wouldn't have as much recoil. Really? When I look at www.ammoguide.com they show the 9mm Luger with a weight of 124 gr 1110 fps and 340 ft-lbs of muzzle energy. The .45 ACP is 230 gr 835 fps and 356 ft-lbs of muzzle energy. The is roughly a 5 % increase in muzzle energy. It is my belief (correct me if I'm wrong) the the muzzle energy is the contributing factor to felt recoil. If you had two guns of the same weight, same recoil springs, same barrel length and one was a 9mm Luger and the other the .45 ACP, would you notice a difference in recoil? True the projectile is almost twice as heavy in the .45, but isn't that reflected in the muzzle energy figure? I'm not noticing much difference in shooting 9mm vs. .45 ACP. Am I alone? |
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Post by TA on Jun 11, 2005 22:58:03 GMT -5
Here, I'll confuse you even more. ;D I looked up the specs on Sellier & Bellot ammo. 9mm 124 gr - 1181 fps , 382 ft. lbs. .45 230 gr. - 853 fps , 371 ft. lbs. (less than 9mm) 357 Sig 140 gr. - 1352 fps , 563 ft. lbs. I don't have anything in 9mm so I don't have a personal comparison. I think there are many factors affecting felt recoil. The size of the bullet, action/reaction. The design and weight of the gun also greatly affect felt recoil. All things being equal, such as being able to measure recoil on test barrels and charging the rounds to all achieve the 563 ft lbs. of the 357 Sig, the .45 230 gr would be the highest recoil, I suspect. (don't try this at home kids  ) In equal size/weight guns would the .45 have more felt recoil than 9mm with factory ammo? Probably. I would say it is a contributing factor but I think there is much more to it all. Also understand, this is just pulled from my head applying basic physics. There are others here that can probably give a more scientific explanation. All that being said, I don't find the .45 acp uncomfortable at all to shoot, especially in a full size 1911. There are too many people out there that will tell you "she really kicks" because 45 is one number higher than .44 magnum. |
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Post by Mountaineer on Jun 12, 2005 11:21:58 GMT -5
TMan, Everything else being equal, it seems bullet weight has a lot to do with felt recoil. I normally shoot 230gr. LRN in my 45's. One day I loaded some 200gr. SWC bullets and when I went to the range and shot them after the 230gr. bullets, it felt like the recoil was cut in half. I'm not sure about the Velocity's or muzzle energy of these rounds, but I could definitely tell the difference in felt recoil of the two rounds.
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Post by klmhq on Jun 12, 2005 13:33:10 GMT -5
Newton's Laws. Number 2 is "For every action there is an equal an opposite reaction." The bullet is pushing on the gun just as much as the gun is pushing on the bullet.
A heavier bullet pushes harder and the gun has to push harder on it.
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Post by "DoubleAction" on Jun 12, 2005 14:20:48 GMT -5
This is from a previous thread I made on the 38 Casull. Casull CA 3800 / 1911/ .38 Casull
Post by DoubleAction on Jun 1, 2005, 11:17pm Even though it's been well over four years since it's introduction, Dick Casull hurdled a big fence in putting a .355" diameter bullet on a .45 acp cartridge case.This did not come without the pistol to launch a 124 gr. FMJ bullet into velosities of 1863 FPS,and some beyond.The pistol is a Casull CA 3800, built from the 1911 design using a six inch barrel & slide with a 30 lb. recoil spring. For some who has a question regarding the recoil of a .357 sig verses the .45 acp, this is what happens when the same similar bullet is mounted on a .45 acp cartridge case. www.casullarms.com/handguns1.htm |
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Post by TMan on Jun 12, 2005 17:42:48 GMT -5
Things to ponder on a Sunday afternoon... I could be doing something exciting like reading a passage from Leviticus or being outside by the pool getting my hide fried further. The sunburn I got on my back last week when company was here was worst than I thought. My back was itching like crazy in church this morning. I kept squirming in my seat trying to scratch it. The preacher kept looking at me. I could read it in his eyes: "oh, you sinner - I'm really making you squirm".
DA, I had heard of the .454 Casull, but not the .38 Casull. With the muzzle energy approaching that of the .44 Magnum, and the price that of a good Wilson CQB, I don't think I'll be buying one soon. However, scanning down the page I saw the CA2000. I want one.
KLMHQ, I think the Newton's law you quoted is why it stuck in my mind that using Muzzle Energy was the best way to judge the recoil. How the gun absorbs that recoil is another topic, which we have discussed e.g. a heavier recoil spring will spread out the energy over a longer period of time, and that we you don't feel it as much.
I went looking through some books to find out how muzzle energy was calculated to no avail. Then I remembered Google and did a search. Wow, why didn't I think of that in the first place. My favorite formula gives the results in ft-lbs and is: KE=mv2/450400. There are others like KE = .5mv2 but that gives the results in funky units of measure.
What I find most interesting is that my impression, and that of others, that the larger the bullet the larger the recoil is true, but velocity has a much more effect because it is the velocity2. So if you double the weight of the bullet, you double the recoil, but if you double the velocity it will be four times as much recoil.
TA, I'm shooting the S&B 115gr FMJ at 1280 fps, which is giving me a KE of 421 ft-lbs. vs. 356 ft-lbs on the Winchester WB for the .45 ACP. No wonder I don't notice any more recoil shooting a .45 than with the 9mm.
I wish I had a Kimber in .45 since I have one in 9mm. It would be interesting shooting them with the same weight recoil springs. Hmmm, the weight of the ammo is different so I'd have to load less rounds in the .45. Even with that if the final weight was the same, the weight distribution (like front to back in a car) would be different.
What really got me back thinking about this was shooting the Accelerator with CCI Maxi Mags and a ME of 312 ft-lbs. This is a heavy gun and there was barely a perceptible recoil. Note: the recoil spring is massive. I can't rack the slide with my left hand. I then shot the little Taurus 85 with S&B .38 Specials and a ME of 278. The recoil on the Taurus is the worst of any gun I own.
One last example again using S&B data: the .40 S&W is 180 gr. 968 fps and ME of 375 ft-lbs. Take the same weight bullet, 180 gr in .30/06 Government and with a fps of 2673 fps. you get a whopping 2866 ft-lbs of muzzle energy. No wonder my brother complains about it. (My reply: leave the deer alone!!!)
Based on what I've felt in shooting 9mm and .45 ACP, I think I was misinformed. Now I just need to reshape my head. The difference in recoil on the .45 is in my wallet.
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Post by MLB on Jun 12, 2005 20:19:17 GMT -5
TMan, I like the way you think.
The recoil is a product of the energy in the powder. I think ME is a good measure of basic recoil. I say basic because (as a structural engineer) I look not only on the force to be resisted, but also how it is applied.
It seems to me that if you could put a 9mm and a 45 through the same gun, you'd feel similar recoil when they are loaded to produce the same ME. Bore axis and handgun mass seem to be the major factors here. I'd imagine that the springs would have to be similarly tuned for each load too to get the same rate of absorbtion.
Unscientifically, I don't see a direct coorelation between caliber and recoil either. My .40 Walther is worse than any .45 I've fired (steel 1911s and a Glock).
Just my 0.02. Shoot me down (figuratively of course) where appropriate...
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Post by klmhq on Jun 12, 2005 23:46:10 GMT -5
You know, a less painful test may be available. DA, I believe, has a Sig with both the .357 Sig and .40S&W barrels. I don't know if the recoil springs change with this... ??
If the springs don't change, then we have the perfect test bed. I would still be subjective, but a lot more informative. Heck, maybe find a cheap accelerometer and tape it to the bottom of the frame and check that too.
Anyway, I'm just thinking too...
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Post by TMan on Jun 13, 2005 8:32:32 GMT -5
Klmhq, unless we could find ammo in both calibers that had the same muzzle energy, we wouldn't be proving anything.
I think our basic question is: the formula of ME = mv2 can it be proven? I think we all tend to think that the larger the bullet size is coming out of the barrel the greater the recoil will be, and that is true, but I think what is equally true is the exponential increase in recoil that you get from increasing the velocity.
Perhaps some scientists have proven it; I don't know. I've often wondered about ole Albert's E=mc2. We know there is a tremendous amount of energy holding the protons, which hate each other, together in the atom, but how did he come up with that formula. I've often thought that he pulled it out of his butt and then challenged anyone to prove him wrong. Suppose the real formula was E=mc1.873
MLB, you are certainly right about handgun mass and the bore axis. The bore axis for the two barrels on my derringer are different. When I fire the top barrel, I find far more muzzle flip than I do when I fire the bottom barrel. Which raises something else that I find interesting.
I have always considered myself "recoil sensitive". However, I've found that I'm more "big noise" sensitive than I am to the actual handgun movement. Wearing both foam inserts and muffs reduces the noise quite a bit. The second part is that there are two aspects of recoil: the vertical movement and the horizontal movement.
For me, the vertical movement is the part about recoil that bothers me the most, which in retrospect is why two of my favorite guns are the Sig 220 Sport and the Taurus 608, both of which are compensated.
Having said that, it brings to mind that there is another aspect of recoil, which is the absorption of energy by components inside the gun and transferred to me. This happens to me mostly with the little Taurus 85 when the trigger stings my shooting finger. (Also in the Sig P210 when the hammer comes back and embeds itself in the flesh of my hand.)
TA, unless it is a mis-print. I found something interesting about the S&B .40 S&W ammo. Ammoguide.com lists the nominal muzzle energy for the cartridge as 412 ft-lbs. S&B lists for their 180gr FMJ only 375 ft-lbs. This is the first time I've seen S&B be lower than the nominal value.
Well, that is enough thinking for me for now.
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Post by klmhq on Jun 13, 2005 10:48:52 GMT -5
**Disclaimer: I have tried to organize my thoughts in an effective path, but this discussion is pretty technical from a Physics standpoint. If you just want the summary, stop when you get to the example and skip to the conclusion. Klmhq, unless we could find ammo in both calibers that had the same muzzle energy, we wouldn't be proving anything. I disagree, the point is proving which ammunition has the greater recoil. Technically, we're talking about acceleration of the gun. All we would have to do is find bullets in the two calibers with the same mass. Don't go there TMan  . The formula for Kinetic Energy is actually 1/2 the mass times the velocity squared. Greater minds than ours have shown those formulas (even Einstein's) to be true many thousands of times. Upon further reflection, the Muzzle Energy (or Kinetic Energy at the muzzle) doesn't have anything to do with your felt recoil. OK, what is recoil? -- the acceleration of the gun after the bullet is fired. Force is equal to the mass of the object multiplied by its acceleration. F = m x a So a larger object requires more force to be acclerated at the same rate as a smaller object. Makes sense, a F-350 requires much more engine to accelerate quickly than a Miata. This also explains why a little bitty .22 can go way faster than a .45 with a lot less powder. When you pull the trigger and the powder explodes, the bullet is pushed forward by rapidly expanding gas. That gas is also pushing against the gun itself (recoil and working the slide). This is Newton's 2nd Law (Every action has an equal and opposite reaction, or each force has an equal and opposite force). So the mass times the acceleration of the bullet is the same as the mass times the acceleration of the gun. a bullet x m bullet = a gun x m gunDoes this make sense? A polymer, compact .45 will have much more recoil than a steel, commander .45. The small gun has less mass so it must have more acceleration. If we know what the mass of the bullet is, the mass of the gun, and the acceleration of the bullet, we can find out the acceleration of the gun. If we have two guns with the same mass and two bullets with the same mass, then the difference in acceleration of the two guns will be based entirely on the acceleration of the two bullets. That's why I say that the Kinetic Energy isn't important to the subject of recoil. The energy of the bullet is only a sub issue of the acceleration of the bullet when the gun is fired. Example: I'm using a CZ-75 to fire a 90 grain 9mm bullet and a 150 grain 9mm bullet. I have to convert everything to metric just because I'm used to it.
90 grain 9mm
mass = .0058319 kg
1212 fps = 369.4176 m/s (meters per second)
150 grain 9mm
mass = .0097198 kg
997 fps = 303.8856 m/s
CZ 75B
mass = 1 kg
barrel length = 4.7 inches = .11938 meters
Acceleration of 90 grain bullet
a = Vfinal2 - Vinitial2
----------------------------------------------
2 x distance
a = 369.4179 m/s2 - 0 m/s2
---------------------------
2 x .11938 m
a = 571576 m/s2
(Now that's a big freaking acceleration!!)
Force of a 90 grain bullet
F = m x a
F = .0058319 kg x 571576 m/s2
F = 3333 N (Newtons)
Acceleration of CZ 75 with 90 grain bullet
a = F / m
a = 3333 N / 1 kg
a = 3333 m/s2
** Please note, that this is the acceleration of the gun without anyone or anything holding it. This also does NOT include working the action of the gun which eats up some big chunk of this energy.
Acceleration of 150 grain bullet
a = Vfinal2 - Vinitial2
----------------------------------------------
2 x distance
a = 303.8856 m/s2 - 0 m/s2
---------------------------
2 x .11938 m
a = 386775 m/s2
Force of a 90 grain bullet
F = m x a
F = .0097198 kg x 386775 m/s2
F = 3759 N
Acceleration of CZ 75 with 90 grain bullet
a = F / m
a = 3759 N / 1 kg
a = 3759 m/s2
Conclusion
1) without knowing the amount of energy involved in working the action of the gun, all this is 2 hours of wasted effort... sigh.
2) It is still safe to conclude that a bigger gun will have less recoil.
3) From the same gun, the larger bullet will probably have the larger felt recoil... though I'm still curious about the difference between a .40 and .357 Sig in the same gun.
4) Muzzle energy only affects what the bullet does to the target, not what the recoil does to you.
Sources:
www.ammoguide.com
www.cz-usa.com
www.onlineconversion.com/
You might also look at the following:
www.chuckhawks.com/handgun_recoil_table.htm
(I do not know how the Physics works on his formula. I'd like to see how that formula came into being. I suspect he may have some correction for blowback operations or something like that. I can't view the other articles he mentions.) |
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Post by MLB on Jun 13, 2005 13:35:05 GMT -5
This is good stuff...  OK TMan, at one time long ago, someone gave me a good explantion for the formula for kinetic energy and how it was derived. Unfortunately, I don't remember how they did it and am not ambitious enough to rejustify it in my head. So I'll just have to list that as an assumption for now. ;D (As for Einstein's theories, I haven't managed to get my head around them enough to say one way or another. Fortunately, I couldn't afford a fusion powered handgun anyway ;D) KLM, them's some fancy formulae slingin' ya done there. It'll take a bit more than the time I've got now to do justice to it. I'll dig into it a bit more tonight. However, I'll throw this your way for now: Say I loaded 2 45 cal cartridges, one hot and one light, and fired them out of the same handgun. Now you have no variables but the ME to deal with. I think it should be clear that the ME is directly related to the recoil at this point. I'm not saying that acceleration is irrelavent, just that it is a by-product of the total energy in the system rather than its cause and is affected by too many variables to calculate simply with statics (as your conclusion bears.) It remains to be determined however, if we can generalize the ME in this situation to different calibers (I think it can) or to differing handguns (I think not - more variables) whew! |
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Post by TMan on Jun 13, 2005 18:27:28 GMT -5
Pandora's box comes to mind. Before I go any further, I'd like to personally thank those that have replied to this thread and in particular klmhq for the amount of time and effort.
The formula for muzzle energy (aka kinetic energy) that I previously gave was: KE=mv2/450400, which will give the results in familiar ft-lbs. The .5 times the mass is factored into the constant so I ignore it when I say that KE basically is mv2.
From klmhq:
Very good. I like it in those terms, but what it is to me as a shooter is any discomfort that I feel as a result of firing the gun. Energy that is absorbed by the gun can be transmitted to me, but primarily it is the acceleration of the gun that I'm objecting to.
Oh, this is my favorite part, and I never really thought of it in those terms, but I've experienced it because the Sig P226 Sport isn't compensated, but is a big heavy gun with negligible recoil. My ultra-lite Taurus 85 only shoots .38 Specials and has a horrid recoil. I should have expected that before I bought it when I noticed that it was ported. Why would they have a compensator on a .38 Special? For a good reason.
Well, this is where we part company. I think we are in agreement that the acceleration of the bullet is the major factor in the amount of felt recoil, but I think since the KE factors in both the size of the bullet and the acceleration it is a better factor than just acceleration alone. If I just looked at acceleration, I could easily conclude that a .22 would have the same recoil as a .50 given the same acceleration, and it wouldn't be true.
Okay, I'm totally lost here. Don't we have the same amount of energy coming back at us? Actually, I would think we would have more because some of the bullet's energy was lost in friction while traveling through the bore. (I wonder if the bore getting hot is primarily because of the powder being burned or from friction?)
Keeping in my that my little Bond Arms Derringer has interchangeable barrels that weigh approximately the same, I still maintain that when thinking about getting a new caliber barrel, looking at the muzzle energy of the candidate rounds will give me a good indication of the amount of recoil that I'm going to experience.
In case I haven't mentioned it: this little gun is a real hoot to shoot.
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Post by 5ontarget on Jun 13, 2005 21:29:08 GMT -5
Bah, It's all ball bearings these days. Man, my brain hasn't had to process numbers and formulas like that for a number of years. Reminds me why I only took the minimun 2 semesters of physics in college for my degree. I still wonder why I needed that much . MLB, if you take the same bullet and load one light, and one hot, (why is it not hot and cool/cold, or heavy and light. agrh, another tangent. what's a tanget? a line perpendicular....help I can't stop.) ME will of course be less with the lighter load, because the velocity is lower. Why, because it didn't accelerate as much as the hot load. So, it can still be linked back to aceleration as the starting point, if you want. KMLQ - why not just make your equation for revolvers, no energy syphoned off to work the action on them. Then your equation would still stand, right? So, according to your theory, we could do something along what MLB was stating, and find a hot, and a light load and could shoot them as our test? 9mm and .22 come in a huge variety of +P, hi velocity, and sub sonic loads. Hey, Tman did you ever find/fire those .22 CBs, if they are the same weight as the cci mini mags or stingers? put an accelerometer on the end of a revolver bbl and your good to go. I'm too tired to think this hard. This probably doesn't make any sense at all, to anyone but me, right now. maybe I should just forget it and try again tomorrow. Why didn't I pay better attention in class? Oh yeah, this is the first time I've ever attempted to use anything from those classes. There is some math and engineering in my genetic past, but I think my father and sister took it all out of the gene pool. |
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Post by TMan on Jun 13, 2005 21:39:57 GMT -5
5ontarget, yes I did get a box of the CCI CB's. Unfortunately the gun to shoot them in is currently unavailable. Sportsman's Warehouse in Lewisville has just about any round that you could possibly want. Their gun prices vary from 7% to 14% over their cost. Why the variance? I didn't ask - never look a gift horse in the mouth. I appreciated the information. They also have a wide selection of handguns too. Can't ever have enough - never know when you might lose the key to your safe. &^#@u%$  |
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Post by klmhq on Jun 13, 2005 22:11:50 GMT -5
This is good stuff... I like it too. ME is still a side issue. The muzzle energy is related to the speed of the bullet as it leaves the barrel. It's impact energy is related to its speed at the time of impact. However, the recoil is determined by how hard the bullet had to accelerate to get to that velocity... I think Too true for words. Darn reality, always messing with good simple math! TMan But the force required to accelerate the .22 is way, way less. A .500 S&W weighs 440 grains. A .22 LR might weigh 30 grains. Less than one tenth. Sure we have that smae energy coming back at us. That's why people who are shot don't fly backwards through windows... cause the shooter would too. In the case of semi-autos we also have to consider that a lot of the energy must be used to load the next round. I think that it would be a pretty good indicator. But you should also consider the mass of the gun, recoil spring, etc. As an example, my dad tried to make a sabot round for a 7.62 pistol. The round fired OK, but it didn't have enough energy to work the action. Recoil, what recoil? Even though the bullet was fast as heck, it just didn't have the mass. 5ontarget: Revolvers... there's an idea... say, someone shoot a 38 spl and a .357 a tell us which has greater felt recoil... oh wait, I know the answer to that one. I suspect the .357 accelerates much, much harder than the .38 for the same bullet mass. More accel, more recoil. More speed out the barrel, more muzzle energy... way more because velocity is squared. Finally: It would be difficult, but the best way to test this would be to make a load for each of two different mass bullets (like .40 and .357 Sig) where the muzzle velocity would be the same. Then the larger bullet would have more recoil because it has more mass to accelerate... hence more force. 2nd experiment: Have two guns with two different length barrels (like a .38 spl 2in and 4 in). Create loads (same mass bullets) for each such that the muzzle velocity is the same. This means that the acceleration for the shorter barrel is higher. I predict that the shorter barrel gun would have more felt recoil... you would also have to add weights to the smaller gun so that the mass of the two was equal. I wonder if my dad would teach me to use his reloading gear... |
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Post by 5ontarget on Jun 14, 2005 13:55:46 GMT -5
KMLQ,
In your second experiment, shouldn't the guns have similar mass distribution too. One would have to add weight to the smaller gun so that it is distributed the same way as the long gun. A front heavy gun feels very different than a butt heavy gun. (insert bad joke/pun here about butt heavy) I'd imagine that would play into it as well.
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Post by MLB on Jun 14, 2005 16:04:56 GMT -5
... I suspect the .357 accelerates much, much harder than the .38 for the same bullet mass. More accel, more recoil. More speed out the barrel, more muzzle energy... way more because velocity is squared. ... Yes, a greater ME. We're getting to the same point I think. The acceleration of the bullet and the ME are both determined by the muzzle velocity. The velocity of the bullet is determined by the powder charge (i.e. total energy available). If you accept that acceleration of the bullet is a good measure of "felt recoil", you must also accept that ME is equally as good a measure.The difference is that we can easily measure the ME with a chronograph and a scale and it is a published quantity. The tough part is that the relationship between the acceleration of the bullet (or the ME) and the acceleration of the handgun is difficult to relate. |
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Post by klmhq on Jun 14, 2005 18:03:12 GMT -5
... I suspect the .357 accelerates much, much harder than the .38 for the same bullet mass. More accel, more recoil. More speed out the barrel, more muzzle energy... way more because velocity is squared. ... Yes, a greater ME. We're getting to the same point I think. The acceleration of the bullet and the ME are both determined by the muzzle velocity. The velocity of the bullet is determined by the powder charge (i.e. total energy available). If you accept that acceleration of the bullet is a good measure of "felt recoil", you must also accept that ME is equally as good a measure.I agree that we're pretty much on the same point. My problem with Muzzle energy is that it comes into play after everything else involved with recoil is already over. Once the bullet leaves the barrel, the gases can expand without interference from a bullet, case, chamber, or barrel and the force that pushes the gun back into your hand ends. True. Again, we've got the same ideas. Once you measure the velocity of the bullet as it exits, it is simple math to figure its acceleration too. I agree that it is trickier to determine an actual number for the recoil on the gun, but it's also more exact than something like 'felt recoil'. I'm attempting to show an actual number. Now if I can only figure out how much energy the blowback system uses. 5OnTarget: As I went to bed last night, I realized the same thing you brought up and the image of 14 38. spl bullets duck-taped to the barrel of a revolver wouldn't go away. |
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Post by TMan on Jun 14, 2005 18:26:22 GMT -5
klmhqThis is a derringer i.e. a mini shotgun and the difference in mass would be very small and would amount to the thickness of the metal in the barrel after it has been bored. The frames are heavy in order to withstand the forces of some of the rounds. There are no recoil springs.
I'm coming around to your way of thinking in terms of F = ma, and I see where you put the emphasis on acceleration. However, the end result of the acceleration is a bullet leaving the muzzle at a certain velocity. To leave at a higher velocity requires more acceleration. Consequently, if you measure the velocity when the bullet leaves, and you know the mass of the bullet, the greater either of these number are the more energy is required in obtaining the acceleration. Therefore, I'm stuck on the fact that muzzle energy or mass of the bullet and the velocity are a good indication of how much energy is going to be reflected back to the shooter in terms of recoil; some of which may be absorbed by the gun.
I could say more, but no one really wants me too; besides I have to go mow the lawn. ;D
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Post by MLB on Jun 15, 2005 9:37:16 GMT -5
Aw come on TMan, KLM and I are beating it to death. The more the merrier...
I'm pretty convinced that you're right on TMan. If you know the ME, you know the bullet acceleration too. I think we all agree that the bullet acceleration is relavant (although KLM is getting there through Newtons's second and I'm taking the conservation of energy path).
Getting back to your original goal of determining "recoil" in any particular handgun; say we find a way to determine the acceleration of the frame after the action has absorbed up whatever energy it will. This acceleration will need to be a vector (magnitude and direction) rather than a simple scalar (magnitude only). That is, the resulting acceleration won't be straight back in the opposite direction of the bullet.
The applied force will of course (equal and opposite right?), but since the applied force is not at the center of mass of the frame (it's at the bore axis height) we will have a torque (or a moment). This will result in the handgun accelerating back and up as well.
Complicated stuff, this recoil business. So, I think any numerical index of "recoil" in an uncompensated handgun will need to include the following three items (at a minimum):
The applied average force (calculated from either ME or bullet acceleration), handgun mass & CG, and bore axis. If you know these three things, I can't think of any other factor that would affect the recoil.
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) In equal size/weight guns would the .45 have more felt recoil than 9mm with factory ammo? Probably.
. The formula for Kinetic Energy is actually 1/2 the mass times the velocity squared. Greater minds than ours have shown those formulas (even Einstein's) to be true many thousands of times. 
