115gr vs 124gr vs 147gr Recoil/Accuracy [Archive] - Glock Talk

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JDK721
01-07-2010, 19:07
......

481
01-07-2010, 20:43
Can someone explain to me the differences in recoil/accuracy in 115gr, 124gr, and 147gr 9mm FMJ?

Which has the least recoil? Which has the most recoil? Which has the least accuracy? Which has the most accuracy?

Thanks.

"Free recoil energy" for "factory spec." FMJ ammo out of a "stock" Glock 17:

115 gr. @ 1155 fps = 5.0003 fpe of recoil energy

124 gr. @ 1120 fps = 5.3018 fpe of recoil energy

147 gr. @ 975 fps = 5.3782 fpe of recoil energy

As for the "most" or "least" accurate load; that depends upon more than just the bullet type and velocity. It is also effected by the individual weapon's quality, manufacturing tolerances, general quality of the ammunition, the shooter's skills and there is not "pat" answer for those reasons.

JDK721
01-07-2010, 20:52
......

481
01-07-2010, 21:02
So the 115gr will have the least recoil? Thanks.

Mathematically, yes.

"Perceived" recoil is another matter entirely, although with standard pressure 9mm ammunition there is little real difference or significance since none of the three examples fluctuates more than ~3.75% from the median free recoil energy value.

I'd bet most folks would be hard pressed to recognize the bullet-to-bullet difference in the recoil if they fired a magazine full of mixed 115 gr., 124 gr. and 147 gr. FMJ ammunition.

It's not a big deal, really.

coal
01-07-2010, 22:11
So the 115gr will have the least recoil? Thanks.

<S>No, not in typical factory loads. Just the opposite. Generally:
1) the more powder, the more recoil
2) the lighter bullet, the more powder
3) the more powder, the more velocity

Energy favors velocity. And, energy forward is energy back. Muzzle energy then gives a simple, rudimentary scale of comparison for recoil energy I think most can grasp since we all know: Energy (http://en.wikipedia.org/wiki/Energy) = mass x velocity squared (e=mc2)

Blazer data
115 = 1145 = 150767875
124 = 1090 = 147324400
147 = 950 = 132667500
Note: All not +p loads

Gold Dot Data
115* = 1210 = 168371500
124 +p = 1220 = 184561600
147 = 985 = 142623075
* Not +p

The 147gr will have the least recoil energy. 481's velocity is watered down for the 115gr, and if I get the gist the math s/he used is relative to momentum. Generally, the velocity for typical factory 9mm is:
115 +p = +100 to +150fps over 124gr +p
124 +p = +200 to +250fps over 147gr

"Experienced" or "felt" recoil can be subjective. And, different gun designs handle recoil differently. Regardless, comparing a 147gr at 985fps vs. 124gr at 1220fps and you are talking a 33% difference in energy forward/back; I think almost anyone will notice that.

But, I don't need math to tell me a 165gr .40sw recoils noticably more than 180gr. Or, that a 124gr +p recoils noticably more than 147gr. Both are VERY easy to discern, very.

"Accuracy" will not relate to weight. Point of impact (POI) may (e.g. high/low).</S>

See my post #29 and on...

Brass Nazi
01-07-2010, 22:17
Load a 147gr bullet with a small dose of 231 and you will experience 9mm nirvana. Great accuracy with recoil that is as soft as butter.

jesse2205
01-07-2010, 23:12
Load a 147gr bullet with a small dose of 231 and you will experience 9mm nirvana. Great accuracy with recoil that is as soft as butter.

+1 What he said...

481
01-08-2010, 14:10
Energy favors velocity. And, energy forward is energy back. Muzzle energy then gives a simple, rudimentary scale of comparison for recoil energy I think most can grasp since we all know: Energy (http://en.wikipedia.org/wiki/Energy) = mass x velocity squared (e=mc2)

Couldn't be farther from the truth.

Actually, your use of the equation, e = mc², is the relativistic expression for the conversion of an object's rest mass into energy (it works the other way, too) where the variable "c" is the speed of light, and as such, this equation has absolutely nothing whatsoever to do with calculating the kinetic energy of any imaginable handgun bullet in motion.

General Relativity aside, the correct equation for that exercise is: KE = ½mv²

Example:

Correctly calculating the KE of a 9mm 115 gr. bullet at 1155 fps using the proper equation yields:

½mv² = ~340.59 fpe

Using the improper equation, e = mc², (as suggested in your post #5, above) for the same task we get:

e = mc² = as applied to a 115 gr. bullet = 6.6975 x 10<SUP>14</SUP> Joules or 4.9398 x 10<SUP>14</SUP> fpe, which is equal to ~118 kT* (kilotons equivalent TNT).

*This represents a yield of approximately 7.8 times that of the Atomic bomb ("Little Boy") that was detonated over Hiroshima on 06 August 1945.

Obviously, as one can see from the figures, unless you have something that the U.S. Government might be interested in knowing about, the first equation (½mv²) is the correct equation for the application of calculating a bullet's KE.

481's velocity is watered down for the 115gr, and if I get the gist the math s/he used is relative to momentum.

Incorrect.

The velocity quoted for the 115 gr. FMJ is indeed correct according to several commonly available factory data and is far from "watered down" in that Remington 115 FMJs run at 1135 fps, Hornady 115 FMJs run at 1155 fps and Federal 115 FMJs run at 1160 fps. My prior post had nothing to do with anything other than standard pressure loads and I never represented it as being otherwise.

But, I don't need math to tell me a 165gr .40sw recoils noticably more than 180gr. Or, that a 124gr +p recoils noticably more than 147gr. Both are VERY easy to discern, very.

While you might be able to "feel" a very slight difference between different standard pressure loads, the difference amongst these standard pressure 9mm loads' recoil is hardly what one might call "significant".

As for not needing math to tell you anything, I'd argue that you might want to seriously consider a course in something more than "general" mathematics as well as one in physics in order to bring your skills up to "par" before dismissing the entire field of study (mathematics) as being "unneeded".

JBP55
01-08-2010, 14:38
There is less felt recoil in the heavier gr. ammunition of the same type and caliber.
147 has less felt recoil than 115.
180 has less felt recoil than 155.
230 has less felt recoil than 185.
I know this from shooting, not math.

481
01-08-2010, 15:12
There is less felt recoil in the heavier gr. ammunition of the same type and caliber.
147 has less felt recoil than 115.
180 has less felt recoil than 155.
230 has less felt recoil than 185.
I know this from shooting, not math.

J-

While this is not incorrect, "felt" or "perceived" recoil is entirely a subjective matter. Individual "perception" is also influenced greatly by the amount and burning rate/characteristics (the "pressure curve" if you will) of the powder used to propel the bullet as well as the mass of the firearm employed.

Perception differs greatly from person to person and what might "feel" like heavier or lighter recoil to you may not be perceived in the same way for me. That is why I said above in post #4:

"Perceived" recoil is another matter entirely...


Still, the physics (as well as my math) above holds true and I prefer not to waste my time debating the "intangible" since our perceptions will differ anyway and certainly in an "unquantifiable" manner.

bootsx
01-08-2010, 15:12
Load a 147gr bullet with a small dose of 231 and you will experience 9mm nirvana. Great accuracy with recoil that is as soft as butter.

PM Sent

HAMMERHEAD
01-08-2010, 15:20
Generally the typical 115 fmj from Blazer or Winchester has pretty light recoil. If you buy Speer Lawman 115, the power level goes up along with recoil. Lawman usually clocks 80 -100 fps faster than WWB.
Winchester USA 147 is also pretty light, probably the lightest felt recoil in my experience, again, Speer Lawman is hotter, you can feel the difference.
Accuracy?
With Glocks it seems like anything with a reverse jacket like a hollowpoint or Winchester's Winclean BEB rounds are more accurate than fmj's. Winchester Winclean 115 BEB has ben a tack driver in all my 9mm's.
Win. USA 147 jhp has very soft recoil and excellent accuracy as well as being affordable enough to practice with.
The most accurate round I've ever tested in my G-17L is the SBR 90 grain frangible round SBR 90 at Midway (http://www.midwayusa.com/viewProduct/?productNumber=379046)
It has very light recoil due to the lead free (sintered copper) bullet. There not as cheap as discount fmj's, but they are cheaper than most hollowpoints and will drill the centers out of the targets at 50 yards.

Molon
01-08-2010, 17:16
The 147gr will have the least recoil energy.


Not necessarily. Applying the basic kinetic energy equation to the bullet weight and velocity is not the correct method for calculating recoil energy. It does not account for the weight of the powder charge, nor does it account for the portion of recoil due to the “push” of the expanding gases as they leave the muzzle. It also does not account for the weight of the firearm. The correct equations for calculating recoil energy can be found in Hatcher’s Notebook, chapter 12.<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" /><o:p></o:p>
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As examples of correct calculations of recoil energy, the following three bullet weights in standard pressure loads fired from a SIG Sauer P229 are used below: 115 grain, 124 grain and 147 grain<o:p></o:p>
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When fired from the SIG P229, Federal’s 115 grain JHP 9BP load, with an actual muzzle velocity of 1122 fps, has a recoil energy of 3.7 ft-lbs.<o:p></o:p>
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When fired from the SIG P229, Federal’s 124 grain HST load, with an actual muzzle velocity of 1098 fps, has a recoil energy of 3.9 ft-lbs.<o:p></o:p>
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When fired from the SIG P229, Federal’s 147 grain HST load, with an actual muzzle velocity of 962 fps, has a recoil energy of 4.1 ft-lbs.<o:p></o:p>
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Additional examples.<o:p></o:p>
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http://www.box.net/shared/static/a3cl0itdwg.jpg<o:p></o:p>
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http://www.box.net/shared/static/7x1yoadsyj.jpg<o:p></o:p>
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Molon
01-08-2010, 17:36
Which has the most accuracy?



As long as the proper twist rate is used, the construction/quality of the bullet will be far more important to accuracy than the actual weight. I've had great results with 9mm bullets as light as 90 grains and of course many people report great results with 147 grain loads.

Generally speaking, a quality JHP will be more accurate than most other designs. The single most accurate 9mm bullet that I have tested so far, is the Hornady 125 grain HAP. The 10-shot group pictured below was fired from 25 yards using 125 grain HAP bullets.


http://www.box.net/shared/static/q1fznfmm8c.jpg

481
01-08-2010, 18:13
Molon,

Well done.

I've come to expect no less from you and your last two posts do not disappoint.

As for your 10 shot 0.413" group at 25 yards...

I hate you.



:)

Molon
01-08-2010, 18:52
Molon,

Well done.

I've come to expect no less from you and your last two posts do not disappoint.

As for your 10 shot 0.413" group at 25 yards...

I hate you.



:)


Muchas gracias.http://www.box.net/shared/static/ig2mmpy7g3.gif
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Molon
01-08-2010, 18:56
481's velocity is watered down for the 115gr . .



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http://www.box.net/shared/static/773xj0g16b.jpg

JBP55
01-08-2010, 19:01
That 10 shot group is amazing.

coal
01-08-2010, 23:31
Couldn't be farther from the truth....
<S>We remain in disagreement on this matter. Energy, which velocity greatly favors (because it's squared), keeps it very simple and reliably states the order of recoil energy w/in caliber by weight when comparing full-power loads.

I find the more folks want to show how smart they are, the more complicated explanations get unnecessarily. Sounding smart, and being right are not the same thing.

This all can get really complicated with all the math. However, the answer, comparing typical full-power factory loads w/in caliber, is really quite simple: more powder = more velocity = more recoil. Having shot 115gr +p+, 124gr +p and 147gr (which are the ideal defensive loads in each weight IMO) from the same gun, actual and "felt" recoil of these loads are very different.

While you consistently try to make your argument in this post (and those of yours I have read along these same topics) with lots and lots and lots and lots and lots and lots and lots and lots and lots and lots and lots of math, I base mine on more simple math and experience, and still get the correct answer. And, least recoil is the 147gr when comparing typical full power factory loads w/in caliber for the reasons I mentioned.

Same goes for in the .40sw for 155gr, 165gr and 180gr. Least recoil in typical full-power loads is 180gr.

The .357sig fires 9mm .355" bullets. The 10mm fires .40" bullets. Are we really going to debate which has greater recoil, "felt" or otherwise: 9mm or .357sig ... 10mm or .40sw? Since bullet size is identical, the differences in these loadings are clearly 1) powder and 2) velocity sequel to it. And, the real difference is powder since more of it generates more velocity all other things being equal (bullet, OAL, crimp, primer and gun). Hence, more powder = more velocity = more recoil. So, so simple.

Seasoned reloaders are very, very concerned about energy. Though it provides little opportunity to display math wizardy on your part, this Reload Calculator (http://www.handloads.com/calc/recoil.asp) and this Hogdon data (http://www.hodgdon.com/PDF/Hodgdon%20Basic%20Manual.pdf) should help you wrap you head around it. Just plug these Hogdon numbers into the calculator:
124gr
Titegroup
4.4gr
1136 fps
Firearm = 1lb

147gr
Titegroup
3.6gr
929 fps
Firearm = 1lb

Let us know what you get.

The above example may be reloading data, but the principles are the same with factory loads. Comparing typical factory full-power loads w/in caliber, it's a sliding scale: any speed a heavier bullet can be pushed to, a lighter bullet can safely be pushed faster. And, velocity is the critical element. Meaning, in simple terms sans math, the relative scale of comparison (w/in caliber) remains the same: More powder = more velocity = more recoil. Again, so, so simple, and complex math not required. </S>

....
As for not needing math to tell you anything, I'd argue that you might want to seriously consider a course in something more than "general" mathematics as well as one in physics in order to bring your skills up to "par" before dismissing the entire field of study (mathematics) as being "unneeded".

<S>:upeyes: Your closing comment was insulting; as coy and sly as you may think you were being. Few are as smart, or clever as they think. Given your math love and excitement to showcase it, I suspect an advanced Sciences Degree, so suggest an additional course in etiquette and we'll call it good.</S>

See my Post #29 on...

481
01-09-2010, 12:31
We remain in disagreement on this matter. Energy, which velocity greatly favors (because it's squared), keeps it very simple and reliably states the order of recoil energy w/in caliber by weight when comparing full-power loads.

I find the more folks want to show how smart they are, the more complicated explanations get unnecessarily. Sounding smart, and being right are not the same thing.

This all can get really complicated with all the math. However, the answer, comparing typical full-power factory loads w/in caliber, is really quite simple: more powder = more velocity = more recoil. Having shot 115gr +p+, 124gr +p and 147gr (which are the ideal defensive loads in each weight IMO) from the same gun, actual and "felt" recoil of these loads are very different.

While you consistently try to make your argument in this post (and those of yours I have read along these same topics) with lots and lots and lots and lots and lots and lots and lots and lots and lots and lots and lots of math, I base mine on more simple math and experience, and still get the correct answer. And, least recoil is the 147gr when comparing typical full power factory loads w/in caliber for the reasons I mentioned.

Same goes for in the .40sw for 155gr, 165gr and 180gr. Least recoil in typical full-power loads is 180gr.

The .357sig fires 9mm .355" bullets. The 10mm fires .40" bullets. Are we really going to debate which has greater recoil, "felt" or otherwise: 9mm or .357sig ... 10mm or .40sw? Since bullet size is identical, the differences in these loadings are clearly 1) powder and 2) velocity sequel to it. And, the real difference is powder since more of it generates more velocity all other things being equal (bullet, OAL, crimp, primer and gun). Hence, more powder = more velocity = more recoil. So, so simple.

Seasoned reloaders are very, very concerned about energy. Though it provides little opportunity to display math wizardy on your part, this Reload Calculator (http://www.handloads.com/calc/recoil.asp) and this Hogdon data (http://www.hodgdon.com/PDF/Hodgdon%20Basic%20Manual.pdf) should help you wrap you head around it. Just plug these Hogdon numbers into the calculator:
124gr
Titegroup
4.4gr
1136 fps
Firearm = 1lb

147gr
Titegroup
3.6gr
929 fps
Firearm = 1lb

Let us know what you get.

The above example may be reloading data, but the principles are the same with factory loads. Comparing typical factory full-power loads w/in caliber, it's a sliding scale: any speed a heavier bullet can be pushed to, a lighter bullet can safely be pushed faster. And, velocity is the critical element. Meaning, in simple terms sans math, the relative scale of comparison (w/in caliber) remains the same: More powder = more velocity = more recoil. Again, so, so simple, and complex math not required.



:upeyes: Your closing comment was insulting; as coy and sly as you may think you were being. Few are as smart, or clever as they think. Given your math love and excitement to showcase it, I suspect an advanced Sciences Degree, so suggest an additional course in etiquette and we'll call it good.



The vast majority of what you have said above (in both of your posts) suggests an abyssmal (at best) understanding of basic physics.

The simple math involved in the expressions of these very basic physical concepts is nowhere near as complex as you portray it, this too confirming my suspicion that you haven't any real comprehension of the "information" that you "regurgitate", oftentimes incorrectly. This is exhibited, by way of one glaring example (there are many more), in your suggested use of the relativistic equation (E=mc²), where it most certainly does not apply, for the simple Newtonian calculation of kinetic energy.

For these reasons, I cannot take seriously anything that you have to offer in either of your two posts and it is my hope that you will endeavor to increase both your education as well as your knowledge level so that you will no longer risk leading others so far astray with such fictitious "material" and misinformation.

Blitzer
01-09-2010, 12:59
Can someone explain to me the differences in recoil/accuracy in 115gr, 124gr, and 147gr 9mm FMJ?

Which has the least recoil? Which has the most recoil? Which has the least accuracy? Which has the most accuracy?

Thanks.

Considering the weapon was designed around the lighter rounds I would give the 115 & 124gr the vote for best in both areas. :cool:

sciolist
01-09-2010, 18:40
That 10 shot group is amazing.

Really. I guess I need to buy a lot more ammo. Holy moly.

coal
01-09-2010, 19:01
... The vast majority of what you have said above (in both of your posts) suggests an abyssmal (at best) understanding of basic physics.

The simple math involved in the expressions of these very basic physical concepts is nowhere near as complex as you portray it, this too confirming my suspicion that you haven't any real comprehension of the "information" that you "regurgitate", oftentimes incorrectly. This is exhibited, by way of one glaring example (there are many more), in your suggested use of the relativistic equation (E=mc²), where it most certainly does not apply, for the simple Newtonian calculation of kinetic energy.

For these reasons, I cannot take seriously anything that you have to offer in either of your two posts and it is my hope that you will endeavor to increase both your education as well as your knowledge level so that you will no longer risk leading others so far astray with such fictitious "material" and misinformation.

<s>We can compare IQ and wiener size all you want; I'm sure you'll win in one, maybe both. Regardless, being smart is not the same as being right.

So, same question you've stilll not answered: What did you get from the recoil calculator relative to recoil running those Hodgon numbers?

I already know the answer and it does not match up with your claims here. Nor, do you claims match up with reality because heavier full-power loads w/in caliber recoil less.

All along I've stated that the simple energy equation (e=mc2) is sufficient to provide a ranking of recoil w/in caliber, and is pretty sufficient between caliber much of the time. That's it. I never stated it gives the exact numerical differences, or that it's the correct formula to determine exact recoil, just the raking. If folks prefer, anyone can Google to learn all they could ever want about recoil in firearms (http://en.wikipedia.org/wiki/Recoil).

However, maybe you spend more time at Mensa potlucks than most, because, for the average Joe and Jane, I think much of your "explanation" above may as well be in Greek. You need to tailor your message to the audience IMO, even if you have to sacrfice a bit of the minutia.</s>

See my Post #29 on...

481
01-09-2010, 21:07
We can compare IQ and wiener size all you want; I'm sure you'll win in one, maybe both. Regardless, being smart is not the same as being right.

So, same question you've stilll not answered: What did you get from the recoil calculator relative to recoil running those Hodgon numbers?

I already know the answer and it does not match up with your claims here. Nor, do you claims match up with reality because heavier full-power loads w/in caliber recoil less.

All along I've stated that the simple energy equation (e=mc2) is sufficient to provide a ranking of recoil w/in caliber, and is pretty sufficient between caliber much of the time. That's it. I never stated it gives the exact numerical differences, or that it's the correct formula to determine exact recoil, just the raking. If folks prefer, anyone can Google to learn all they could ever want about recoil in firearms (http://en.wikipedia.org/wiki/Recoil).

However, maybe you spend more time at Mensa potlucks than most, because, for the average Joe and Jane, I think much of your "explanation" above may as well be in Greek. You need to tailor your message to the audience IMO, even if you have to sacrfice a bit of the minutia.




Clearly there is nothing to be gained by debating this further with you.

You'll notice that I've not bothered to answer any of your questions.

These are the reasons:

Since you, even now, have failed to grasp even the simplest notion; that the correct formula for calculating KE is KE=½mv², not E=mc² (as you so incorrectly persist in suggesting) despite the fact that anyone with even the most cursory understanding of basic physics would know to be fact; tells me all that I need to know regarding you and your understanding of basic physics and the simplest of mathematics.

Additionally, your failure to understand that my initial comments in (my) posts #2 and #4 (I am assuming that you read them of course) of this thread relates only to standard pressure ammunition, yet persist in inferring that I addressed an aspect of the topic that I did not, suggests that you are far beyond reason and rationality.

In that light, I have no desire to waste my time attempting to explain the plainly obvious to the hopelessly ineducable.


At what point you became convinced that you were knowledgeable in this topic matter, I cannot say. However you became convinced of this or whomever might've told you so; I can tell you right now that it was quite clearly in error.

I've repeated myself enough here. I hope that you get "the gist".




Regardless, being smart is not the same as being right.

In this case, it is. Better luck next time.

Molon
01-09-2010, 21:27
We can compare IQ and wiener size all you want; I'm sure you'll win in one, maybe both. . . bla, bla bla. . .

What did you get from the recoil calculator relative to recoil running those Hodgon numbers? . . .

124gr
Titegroup
4.4gr
1136 fps
Firearm = 1lb

147gr
Titegroup
3.6gr
929 fps
Firearm = 1lb


bla, bla, bla . . .I already know the answer and it does not match up with your claims here. Nor, do you claims match up with reality because heavier full-power loads w/in caliber recoil less.

All along I've stated that the simple energy equation (e=mc2) is sufficient to provide a ranking of recoil w/in caliber, . . . bla, bla, bla.



I say the following with all due respect; STFU already and pay attention to what 481 is telling you. You might actually learn something. Your ignorant posts demonstrate your complete lack of understanding of the subject of recoil energy. What part of "the speed of light" has absolutely nothing to do with calculating and ranking recoil energies don't you understand?<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" /><o:p></o:p>
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The correct equations used to calculate and rank recoil energies are not a matter of your personal opinion versus another's opinion. They're a matter of non-debatable scientific facts; facts which you have obviously failed to grasp.<o:p></o:p>
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As for the examples you’ve listed for comparison in your previous post; the only thing that they demonstrate is your intellectual dishonesty and lack of integrity. For all of your carrying-on about comparing “full power loads”, instead of choosing apples to apples examples for comparison, you chose to compare a 147 grain load that runs at more than 3,000 PSI lower than your 124 grain load. Pathetic. (Hodgdon’s manual, pages 61 &62) A stunt like this would earn you a censure at a more reputable location. <o:p></o:p>
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Using a much more apples to apples comparison, we obtain the following results. The following loads are from Lyman’s 48<SUP>th</SUP> edition reloading manual. (page 323) The loads are within 700 C.U.P.s of each other; the closest example I could find on short notice.<o:p></o:p>
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Bullet weight: 125 grains<o:p></o:p>
Powder charge: 4.2 grains of Titegroup<o:p></o:p>
Listed velocity: 1045 fps<o:p></o:p>
Pressure: 31,200 C.U.P.<o:p></o:p>
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Bullet weight: 147 grains<o:p></o:p>
Powder charge: 3.6 grains of Titegroup<o:p></o:p>
Listed velocity: 983 fps<o:p></o:p>
Pressure: 31,900 C.U.P.<o:p></o:p>
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Firing the above loads from a pistol with a weight of 1.9875 pounds (such as a SIG Sauer P229) and assuming the same velocities, the 125 grain load will have a recoil energy of 3.47 ft-lbs. The 147 grain load will have a recoil energy of 4.03 ft-lbs; the exact opposite of your BS claims and just what 481 has been trying to explain to you. In short, 481 knows what he is talking about and you don’t. He’s right and you’re not. Try to learn from that, for it is posts like yours that have earned this website a very poor reputation in the firearms community as a source for factual information. <o:p></o:p>
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coal
01-09-2010, 22:03
Clearly there is nothing to be gained by debating this further with you.

You'll notice that I've not bothered to answer any of your questions.

These are the reasons:

Since you, even now, have failed to grasp even the simplest notion; that the correct formula for calculating KE is KE=½mv², not E=mc² (as you so incorrectly persist in suggesting) despite the fact that anyone with even the most cursory understanding of basic physics would know to be fact; tells me all that I need to know regarding you and your understanding of basic physics and the simplest of mathematics.

Additionally, your failure to understand that my initial comments in (my) posts #2 and #4 (I am assuming that you read them of course) of this thread relates only to standard pressure ammunition, yet persist in inferring that I addressed an aspect of the topic that I did not, suggests that you are far beyond reason and rationality.

In that light, I have no desire to waste my time attempting to explain the plainly obvious to the hopelessly ineducable.


At what point you became convinced that you were knowledgeable in this topic matter, I cannot say. However you became convinced of this or whomever might've told you so; I can tell you right now that it was quite clearly in error.

I've repeated myself enough here. I hope that you get "the gist".

In this case, it is. Better luck next time.

I say the following with all due respect; STFU already and pay attention to what 481 is telling you. You might actually learn something. Your ignorant posts demonstrate your complete lack of understanding of the subject of recoil energy. What part of "the speed of light" has absolutely nothing to do with calculating and ranking recoil energies don't you understand?<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" /><o:p></o:p>
<o:p></o:p>
The correct equations used to calculate and rank recoil energies are not a matter of your personal opinion versus another's opinion. They're a matter of non-debatable scientific facts; facts which you have obviously failed to grasp.<o:p></o:p>
<o:p></o:p>
As for the examples you’ve listed for comparison in your previous post; the only thing that they demonstrate is your intellectual dishonesty and lack of integrity. For all of your carrying-on about comparing “full power loads”, instead of choosing apples to apples examples for comparison, you chose to compare a 147 grain load that runs at more than 3,000 PSI lower than your 124 grain load. Pathetic. (Hodgdon’s manual, pages 61 &62) A stunt like this would earn you a censure at a more reputable location. <o:p></o:p>
<o:p></o:p>
Using a much more apples to apples comparison, we obtain the following results. The following loads are from Lyman’s 48<SUP>th</SUP> edition reloading manual. (page 323) The loads are within 700 C.U.P.s of each other; the closest example I could find on short notice.<o:p></o:p>
<o:p></o:p>
Bullet weight: 125 grains<o:p></o:p>
Powder charge: 4.2 grains of Titegroup<o:p></o:p>
Listed velocity: 1045 fps<o:p></o:p>
Pressure: 31,200 C.U.P.<o:p></o:p>
<o:p></o:p>
<o:p></o:p>
Bullet weight: 147 grains<o:p></o:p>
Powder charge: 3.6 grains of Titegroup<o:p></o:p>
Listed velocity: 983 fps<o:p></o:p>
Pressure: 31,900 C.U.P.<o:p></o:p>
<o:p></o:p>
<o:p></o:p>
Firing the above loads from a pistol with a weight of 1.9875 pounds (such as a SIG Sauer P229) and assuming the same velocities, the 125 grain load will have a recoil energy of 3.47 ft-lbs. The 147 grain load will have a recoil energy of 4.03 ft-lbs; the exact opposite of your BS claims and just what 481 has been trying to explain to you. In short, 481 knows what he is talking about and you don’t. He’s right and you’re not. Try to learn from that, for it is posts like yours that have earned this website a very poor reputation in the firearms community as a source for factual information. <o:p></o:p>
<o:p></o:p>
<o:p></o:p>
<o:p></o:p>

<S>I AM TRULY NOT TRYING TO BE OBTUSE. So, let's ignore me and my apparent ignorance: Neither of you has explained why recoil calculators, using manufacturer published reloaing data, do not support your claim.

I think the above is an example of finding numbers that support a view. 1045fps is a slow velocity 125gr (lead) load, whereas 945fps is a typical velocity 147gr (FMJ) load. And, in energy, velocity is highly favored. 1150fps-1200fps is a more typical 124gr FMJ load, and this naturally changes the numbers.

Hogdon data is Hogdon's data, and Hogdon makes the powder. And, you both appear to dismiss Handloads.com's recoil calculator (http://www.handloads.com/calc/recoil.asp), which is a very well-regarded resource among reloaders. Which, gives results that are pretty much consistent with other reload calculators for reloaders.

Much posted here looks very impressive (e.g. the math) and clearly shows a keen mind. However, again, Neither of you has explained why recoil calculators, using published reloaing data, do not support your claim. That's what I'd most like to hear/read. If I am wrong, so are the reload calcutors I have used over the years in reloading, or I am misinterpretting them. So, I could really learn something here, and so could many others, especially reloaders. In all seriousness, truly, please explain why the reload calculators are all wrong using published manufacturer powder data because, as you say, I am really not getting it. And, whether you believe it or not, I very much would like to understand why all the reload calulators are all wrong.

Those reloader's resources are the resources I trust because I reload, and have for many years. And, they have matched reality (i.e. heavier bullets recoil less). So, again, please explain to me why the recoil calculators are all wrong calculating recoil (because they all show heavier bullets recoiling least).

This is the question I should have asked initially. In not doing so, I apologize.</S>


See my Post #29 and on...

481
01-09-2010, 23:30
I AM TRULY NOT TRYING TO BE OBTUSE. So, let's ignore me and my apparent ignorance: Neither of you has explained why recoil calculators, using manufacturer published reloaing data, do not support your claim.

I think the above is an example of finding numbers that support a view. 1045fps is a slow velocity 125gr (lead) load, whereas 945fps is a typical velocity 147gr (FMJ) load. And, in energy, velocity is highly favored. 1150fps-1200fps is a more typical 124gr FMJ load, and this naturally changes the numbers.

Hogdon data is Hogdon's data, and Hogdon makes the powder. And, you both appear to dismiss Handloads.com's recoil calculator (http://www.handloads.com/calc/recoil.asp), which is a very well-regarded resource among reloaders. Which, gives results that are pretty much consistent with other reload calculators for reloaders.

Much posted here looks very impressive (e.g. the math) and clearly shows a keen mind. However, again, Neither of you has explained why recoil calculators, using published reloaing data, do not support your claim. That's what I'd most like to hear/read. If I am wrong, so are the reload calcutors I have used over the years in reloading, or I am misinterpretting them. So, I could really learn something here, and so could many others, especially reloaders. In all seriousness, truly, please explain why the reload calculators are all wrong using published manufacturer powder data because, as you say, I am really not getting it. And, whether you believe it or not, I very much would like to understand why all the reload calulators are all wrong.

Those reloader's resources are the resources I trust because I reload, and have for many years. And, they have matched reality (i.e. heavier bullets recoil less). So, again, please explain to me why the recoil calculators are all wrong calculating recoil (because they all show heavier bullets recoiling least).

This is the question I should have asked initially. In not doing so, I apologize.

Fair enough, coal. :)

First of all, please accept that science is not a "democratic" or "communal" process or a "popularity contest" and the laws (mathematical) of physics (in this case, "mechanics") are, for all practical purposes, "set in stone" and not subject to "emotion" or "opinion".

Second, please accept that the following are the accepted, proven expressions for calculating the following principles in mechanics and that the related information is not my opinion; it is scientific fact.

OK?

1. mv = mv, this is the oft-quoted mathematic expression for the conservation of momentum (ρ) as stated in Newton's Third Law of Motion: "To every action, an equal and opposite reaction."

2. Forget, for the time being, the equation, E=mc². This equation has no place here and if you'd like, I'd be happy to explain the significance of this "not-so-simple" equation and its rather vast implications to you via PM at a later date.

3. Instead, for the calculation of a bullet's kinetic energy (K.E.), please use the following equation: KE = ½mv². It is the proper "form" for the purpose, but it does require some modification for unitary agreement so that it will produce a numerical answer that makes sense.

4. I have, for your convenience, made those "modifications" to the equation so that you can calculate KE to an unnecessary (and insane) level of precision should you want to do so.

Here it is:

KE = (3.24 x 10<SUP>-5 </SUP>M) (.3048 V)² (.737562149277)<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" /><o:p></o:p>

You can enter the value of M in grains, V in feet per second and the result will be expressed in the ever common "foot pounds" of energy (fpe). (the "English" system, not to be confused with the GT member of the same name :cool:)

Get rid of the conversion "factor", .737562149277, and your result will be expressed in Joules (J). (metric system; Kg-M-S)

5. As for calculating the "free recoil energy" of any given firearm weight/cartridge/bullet weight/propellant charge weight/velocity arrangement, I have no need of any such "calculators" provided by anyone or any company.

The mathematic expression of physical laws that dictate mechanics are the exclusive property of no one person or entity and since I have no idea as to how such programing might be written by any one party or entity, I do not trouble myself with them.

I have never commented on the aforementioned "recoil calculators", simply because I have no need of them. I can determine the results for myself through a relatively simple equation that takes into account all of the pertinent variables requisite to such a finding.

That equation is simple enough.

Here it is:

"Free Recoil Energy" = [(Mbullet x Vbullet + 4700 x Mpropellant) ÷ 7000]² ÷ [64.34809711] ÷ [Mfirearm]


<o:p>Please:</o:p>
<o:p></o:p>
<o:p>- enter the mass of the bullet (Mbullet) in grains</o:p>
<o:p>- enter the velocity of the bullet (Vbullet) in feet per second</o:p>
- enter the mass of the propellant (Mpropellant) in grains
- enter the mass of the firearm (Mfirearm) in decimal pounds (for example 1.55, 2.25, 4.50 7.75, 11.67 etc.)

*Note: the term 64.34809711 is a term of conversion for the English unitary weight (pounds) into English unitary mass (slugs) and encompasses the conversion of "g" (the acceleration of gravity) in English units as ~32.17404856 feet/second²

*Additionally, the co-efficient "4700" in the above equation is the commonly accepted "guesstimation" :supergrin: for the propellant gas velocity immediately as it exits the bore of the firearm. This number most likely varies per application, but serves as a "fairly reasonable" value for the purpose of calculating recoil energy of small arms.

The result will be expressed in foot pounds of energy.

Now you have no need of those "recoil calculators" and the (possibly questionable) math behind them. You have the answers because you know have the ability to figure them out for yourself.


So, there you go, coal. This oughtta keep you busy for awhile. I hope that you enjoy these equations and the understanding (of the concepts addressed above) that they bring to you.


Good night,

:)

Tarowah
01-10-2010, 08:48
481 and Molon I hate you both for digging up so many bad memories from my youth, my brain hurts now =P

All kidding aside those are some of the best posts and explanations I have seen to date on the board, well done.

coal
01-10-2010, 10:59
Fair enough, coal. :)

...

Here it is:

"Free Recoil Energy" = [(Mbullet x Vbullet + 4700 x Mpropellant) ÷ 7000]² ÷ [64.34809711] ÷ [Mfirearm]

<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" /><o:p>...</o:p>

The result will be expressed in foot pounds of energy.

Now you have no need of those "recoil calculators" and the (possibly questionable) math behind them. You have the answers because you know have the ability to figure them out for yourself.

...

First, I do apologize for any 'tude I had. I appreciate your time and insight.

Thanks for that information. The "free energy" formula was the same methodology I was using when I double-checked the recoil calculator's math again after this post, and before when I first started using it*.

And, not to beat a dead horse, I did find that KE=.5mc2 and E=mc2 pretty much gets you the same ranking "shorthand" (as each other and the "free recoil" formula) comparing full-power loads. So, I'm still not clear because full-power loads recoil less and this is what I always understood the OP to be asking.*

Where I've been lost this whole time is the contention that the heavier bullet has more recoil comparing typical full-power loads. And, I still am honestly*.

Here's what I have run:
((124*1150+4700*4.4)/7000)
23.325714285714285714285714285714
23.325714285714285714285714285714*23.325714285714285714285714285714
544.08894693877551020408163265305
544.08894693877551020408163265305/64.34809711
8.4554007247282142700191749718867/1
Handloads.com = 8.13 (free recoil energy using 124gr @ 1150 w/ 4.4gr and 1lb firearm)

((147*950+4700*3.6)/7000)
22.367142857142857142857142857143
22.367142857142857142857142857143*22.367142857142857142857142857143
500.28907959183673469387755102041
500.28907959183673469387755102041/64.34809711
7.7747299774322220776215514575674/1
Handloads.com = 7.52 (free recoil energy using 147gr @ 950 w/ 3.6gr and 1lb firearm)

Those are pretty close, and "off" by roughly the same amount. However, both still show the heavier bullet, when comparing typical full-power velocity loads, as having less recoil energy. WHERE ARE WE MISSING EACH OTHER?*

Basically, as the formula clearly illustrates a 23gr change in weight is easily offset by a 200fps change in velocity or .8gr change in powder due to multiplication. This is why I've always stated that energy favors velocity. However, clearly, if you "weaken" the 124gr load to 1050fps, the the 147gr has more recoil (because the combined effect of bullet weight and powder charge can then offset velocity).*

*It appears we (and recoil calculators) are using the same method. It would be silly if our whole disconnect was/is that you (and others) were talking about "weak" 124gr vs. full-power 147gr, whereas I was comparing full-power loads in each (115gr +p = 1250+fps, 124gr +p = 1150+fps and 147gr = ~950fps). That appears the case to me now. I have, and always have gotten, the same results running a "weak" 1050fps 124gr (i.e. less recoil) vs. a full-power 950fps 147gr (i.e. more recoil). But, I always think apples-to-apples making these comparisons.


p.s. I clearly over-simplified with the "energy only" formula (because it's an easy "shortcut" comparing full-power loads I thought most could best grasp). KE is the (still simplified) correct "physics", simply dividing the result by 2. Both formulas sqaure velocity - because it's all energy afterall - meaning velocity is favored, having the same impact on the product/result. Point noted to not simplify too much because there are those that "get it". This is one of those "tailor the message to the audience" deals IMO. I'll just direct folks to the Handloads.com's Recoil Calculator (http://www.handloads.com/calc/recoil.asp) in the future because my "simplified" approach appeared to complicate this post.

481
01-10-2010, 12:05
First, I do apologize for any 'tude I had. I appreciate your time and insight.

Thanks for that information. The "free energy" formula was the same methodology I was using when I double-checked the recoil calculator's math again after this post, and before when I first started using it*.

And, not to beat a dead horse, I did find that KE=.5mc2 and E=mc2 pretty much gets you the same ranking "shorthand" (as each other and the "free recoil" formula) comparing full-power loads. So, I'm still not clear because full-power loads recoil less and this is what I always understood the OP to be asking.*

Where I've been lost this whole time is the contention that the heavier bullet has more recoil comparing typical full-power loads. And, I still am honestly*.

Here's what I have run:
((124*1150+4700*4.4)/7000)
23.325714285714285714285714285714
23.325714285714285714285714285714*23.325714285714285714285714285714
544.08894693877551020408163265305
544.08894693877551020408163265305/64.34809711
8.4554007247282142700191749718867/1
Handloads.com = 8.13 (free recoil energy using 124gr @ 1150 w/ 4.4gr and 1lb firearm)

((147*950+4700*3.6)/7000)
22.367142857142857142857142857143
22.367142857142857142857142857143*22.367142857142857142857142857143
500.28907959183673469387755102041
500.28907959183673469387755102041/64.34809711
7.7747299774322220776215514575674/1
Handloads.com = 7.52 (free recoil energy using 147gr @ 950 w/ 3.6gr and 1lb firearm)

Those are pretty close, and "off" by roughly the same amount. However, both still show the heavier bullet, when comparing typical full-power velocity, as having less recoil energy. WHERE ARE WE MISSING EACH OTHER?*

Basically, as the formula clearly illustrates a 23gr change in weight is easily offset by a 200fps change in velocity or .8gr chage in powder due to multiplication. If you "weaken" the 124gr load to 1050fps, the the 147gr has more recoil.*

*It appears we (and recoil calculators) are using the same method. It would be silly if our whole disconnect was/is that you (and others) were talking about "weak" 124gr vs. full-power 147gr, whereas I was comparing full-power loads in each (115gr +p = 1250+fps, 124gr +p = 1150+fps and 147gr = ~950fps). That appears the case to me now. I have, and always have gotten, the same results running a 1050fps 124gr (i.e. less recoil) vs. a 950fps 147gr (i.e. more recoil). But, I always think apples-to-apples making these comparisons.


p.s. I may have over-simplified with "energy only" formula (because it's an easy "shortcut" comparing full-power loads I thought most could best grasp). KE is the (still simplified) correct "physics", simply dividing the result by 2. Both formulas sqaure velocity - because it's all energy afterall - meaning velocity is favored, having the same impact on the product/result. Point noted to not simplify too much because there are those that "get it". This is one of those "tailor the message to the audience" deals IMO.

coal,

Thanks. Glad that we are "cool" now.

A couple of points that hit me after reading your post:

Seems that you have been using the incorrect variable, but the correct values when calculating KE. The variable "c" in, E=mc², is taken to always represent the speed of light in physics equations, whereas the variable "v", in ½mv² is always used to represent the velocity of an object whose momentum, kinetic energy or trajectory is being calculated.

As for the differences given by the online calculator and our equations, I suspect that the answer is rather simple and easily explained.

Where you have obtained the results:

Free recoil energy calculated using 124gr @ 1150 w/ 4.4gr and 1lb firearm
From the equation: 8.45
From Handloads.com: 8.13 (3.936% less than the equation yield)

Free recoil energy using 147gr @ 950 w/ 3.6gr and 1lb firearm
From the equation: 7.77
Handloads.com: 7.52 (3.3345% less than the equation yield)

I have no ability to review the software and programming of these sorts of 'sites, let lone the math that they are employing so it is hard to say what is causing the difference especially since their proportions (%) vary with each result. If the proportional variance was the same for each result I could say with near 100% certainty that the effect was being caused by their use of a different coefficient (like a lower propellant velocity which would produce a lower number like you are seeing) but, the difference the two results is not constantly proportional which suggests some sort of programming or software induced effect.

This issue is why I opined in my prior post:

Now you have no need of those "recoil calculators" and the (possibly questionable) math behind them. You have the answers because you know have the ability to figure them out for yourself.


Rest assured that the equation provided, which is based upon solid physics and algebraic expression, is the soundest and most accurate method for calculating the free recoil energy of a firearm. The coefficient for the propellent velocity (4700) is the only "uncertainty" that I can see that might exist between the actual equation and that of the software on the 'site. I know what I know and I know little of how the 'site obtains their result (other than that it is "close" to what we are getting) and I always defer to the "hands on" control of equations whenever I can.

The choice is yours, since I have no stake in what option you prefer to employ.

The unexpected difference bewteen the recoil of "heavy" (147 gr.) and "light" (124 gr. and 115 gr.) loads likely comes from the amount of powder being used in order to propel the bullet. Powder weight has alot to do with recoil generated, more than most realize, and my suspicion is that the "lighter" (124 and 115) loads utiltize more propellant weight than the "heavy" (147) which causes the seemingly contradictory result.

Have a look at this:

9mm handloads from my bench.

Both projectiles have very similar momentum and similar (but a different powder: Bullseye under the 124 and Unique under the 147) propellant weight:

124 gr. @ 1125 fps over 4.50 grs. in a (Glock 17)/1.55 lbs. = 5.280778761 fpe

147 gr. @ 965 fps over 4.30 grs. in a (Glock 17)/1.55 lbs. = 5.374214302 fpe

These are the numbers that the equation produces.

"Felt" recoil is another story altogether...and one that I am not going to get into. :)

Dandapani
01-10-2010, 12:28
Recoil is simply momentum conservation. The simple M x V should work. The round with the higher MV should recoil more.

481
01-10-2010, 13:07
Recoil is simply momentum conservation.

It is.

The simple M x V should work.

It does.

The round with the higher MV should recoil more.

Of course it does.


Once one (having an inquiring mind) finds the recoil velocity of a particular cartridge/firearm combination, its free recoil energy can then be calculated by using the equation:

"Free Recoil Energy" = [(Mbullet x Vbullet + 4700 x Mpropellant) ÷ 7000]² ÷ [64.34809711] ÷ [Mfirearm]

481
01-10-2010, 13:16
481 and Molon I hate you both for digging up so many bad memories from my youth, my brain hurts now =P

All kidding aside those are some of the best posts and explanations I have seen to date on the board, well done.

Tarowah,

So sorry. :sorry:

This one is on me. :winkie:


http://www.excedrin.com/images/coupons/migraine_save_1.jpg (http://bricks.coupons.com/bstart.asp?o=56232&c=EX&p=uP4BRQLY)

coal
01-10-2010, 13:38
Recoil is simply momentum conservation. The simple M x V should work. The round with the higher MV should recoil more.

Agreed. It's all forms of energy, and in loads velocity is the greater extreme in the formula (especially considering when "energy" values are squared). So, velocity generally trumps weight and powder grain differences.

Considering full-power loads, I attempted to "simplify" a topic with "complicated' physics. 481 called me on that, and introduced the more complex math and concepts of why (http://en.wikipedia.org/wiki/Recoil). So much for "shortcuts" (though MxV is a good one, too).

For the OP, and all, to get the exact answer, you have to compare the exact loads. There are "rules of thumb" when comparing full-power (e.g. defensive loads), but clearly those had little bearing in this discussion.

I reload, and have for quite some time, and so my loads are all pretty consistent. But, target factory loads are generally "weak", watered-down loads in 115gr and 124gr (e.g. "lawyer-proof"), whereas 147gr loads are pretty consistent (i.e. full-power) in target and defensive offerings (as the data below shows). However, I did not realize just how "weak" current 9mm target factory loadings had become for 115gr and 124gr. Folks shoot that "weak" factory 115gr (~1150fps) or 124gr (~1050fps) target load thinking "Sweet, light recoil", then load up that 124gr +p (~1250fps) defensive load and have a slightly different experience. This does not occur w/ 147gr. I learned something here: Current factory 9mm 115gr and 124gr loads are very "weak". But, I digress.

It really all depends mostly on how fast each relative object being compared is moving. As I became very curious about current factory loadings, I took some time to get the "free recoil energy" of a few loads. I'll leave the results up to personal interpretation.

Blazer FMJ
115 = 1145 = 7.14
124 = 1090 = 7.39
147 = 950 = 7.76

Federal FMJ
115 = 1180 = 7.53
124 = 1150 = 8.13
147 = 1000 = 8.50

Federal HST
115 = N/A
124 = 1150 = 8.13
147 = 1000 = 8.50
124 +p = 1200 = 8.77
147gr +p = 1050 = 9.28

S&B FMJ
115 = 1287 = 8.78
124 = 1188 = 8.62
147 = N/A

Remington GSHP
115 = 1180 = 7.53
124 = 1125 = 7.82
147 = 990 = 8.35

Remington Express
115 = 1250 = 8.33
124 = 1100 = 7.52
147 = 990 = 8.35

Gold Dot
115 = 1210 = 7.87
124 = 1220 = 9.04
147 = 985 = 8.28

Winchester FMJ
115 = 1190 = 7.64
124 = 1140 = 8.01
147 = 990 = 8.35

Winchester Ranger T-Series
115 = N/A
124 = 1250 = 9.44
147 = 990 = 8.35

Winchester Ranger JHP
115 = 1335 = 9.37
115 = 1225 = 8.04

Double Tap
115 +p = 1415 = 10.40
124 +p = 1310 = 10.27
147 +p = 1135 = 10.69

Fiocchi FMJ
115 = 1260 = 8.45
124 = 1180 = 8.26
147 = 1000 = 8.50

MagTech
115 = 1155 = 7.25
115gr +p = 1328 = 9.28
124 = 1109 = 7.62
147 = 990 = 8.35

Wolf
115 = 1150 = 7.19
124 = 1130 = 7.88
147 = 985 = 8.28

PMC FMJ
115 = 1160 = 7.30
124 = 1110 = 7.64
147 = N/A

Notes:Assumptions*:
Recoil Calculator (http://www.handloads.com/calc/recoil.asp)
Velocity pulled from manufacturer website today
115gr = 4.6gr powder
124gr = 4.4gr powder
147gr = 4.2gr powder
1lb = handgun weight
*Manufactures don't list powder charge weight. I suspect slight variations, and more velocity to come from more of the same powder, or less of a faster powder. Still, differences of 0.xxgr in powder amounts will not skew results significantly.

Edit:
For OP, here's what I see looking at the data above:
If the 147gr is traveling at 950fps:
- 115gr needs to be traveling at greater than 1201fps
- 124gr needs to be traveling at greater than 1120fps
for the 147gr to have less recoil.

If the 124gr is traveling at 1050fps:
- 115gr needs to be traveling at greater than 1125fps
for the 124gr to have less recoil.

coal
01-10-2010, 14:19
coal,

Thanks. Glad that we are "cool" now. Agree. Good discussion IMO. :thumbsup:

A couple of points that hit me after reading your post:

Seems that you have been using the incorrect variable, but the correct values when calculating KE. The variable "c" in, E=mc², is taken to always represent the speed of light in physics equations, whereas the variable "v", in ½mv² is always used to represent the velocity of an object whose momentum, kinetic energy or trajectory is being calculated. Correct. I use the lay-person term, but know it's the speed of light (in a vaccuum). Velocity is often replaced by the lay-person and works for lay-person objectives.

As for the differences given by the online calculator and our equations, I suspect that the answer is rather simple and easily explained. Agreed; I was thinking the same as below. The calculators are easier to use, the product close enough, and the variances (in product between each weight) fairly consistent. I run mid-power reloads, so it's close enough.

Where you have obtained the results:

Free recoil energy calculated using 124gr @ 1150 w/ 4.4gr and 1lb firearm
From the equation: 8.45
From Handloads.com: 8.13 (3.936% less than the equation yield)

Free recoil energy using 147gr @ 950 w/ 3.6gr and 1lb firearm
From the equation: 7.77
Handloads.com: 7.52 (3.3345% less than the equation yield)

I have no ability to review the software and programming of these sorts of 'sites, let lone the math that they are employing so it is hard to say what is causing the difference especially since their proportions (%) vary with each result. If the proportional variance was the same for each result I could say with near 100% certainty that the effect was being caused by their use of a different coefficient (like a lower propellant velocity which would produce a lower number like you are seeing) but, the difference the two results is not constantly proportional which suggests some sort of programming or software induced effect.

This issue is why I opined in my prior post:



Rest assured that the equation provided, which is based upon solid physics and algebraic expression, is the soundest and most accurate method for calculating the free recoil energy of a firearm. The coefficient for the propellent velocity (4700) is the only "uncertainty" that I can see that might exist between the actual equation and that of the software on the 'site. I know what I know and I know little of how the 'site obtains their result (other than that it is "close" to what we are getting) and I always defer to the "hands on" control of equations whenever I can. I agree. The 4700 is the most obvious "variance" because the rest is all known. As noted, variances are not exact each run, but, IMO, close enough for a relative comparison.

The choice is yours, since I have no stake in what option you prefer to employ.

The unexpected difference bewteen the recoil of "heavy" (147 gr.) and "light" (124 gr. and 115 gr.) loads likely comes from the amount of powder being used in order to propel the bullet. Powder weight has alot to do with recoil generated, more than most realize, and my suspicion is that the "lighter" (124 and 115) loads utiltize more propellant weight than the "heavy" (147) which causes the seemingly contradictory result. Not contradictory at all. Heavier loads use less propellant (or, ideally, a slower burning one) because they use a longer bullet that seats deeper, thereby having less case space/volume remaining, plus a greater resistance to movement, thereby increasing pressure in two ways. Simple concept, especially for reloaders.

Have a look at this:

9mm handloads from my bench.

Both projectiles have very similar momentum and similar (but a different powder: Bullseye under the 124 and Unique under the 147) propellant weight:

124 gr. @ 1125 fps over 4.50 grs. in a (Glock 17)/1.55 lbs. = 5.280778761 fpe

147 gr. @ 965 fps over 4.30 grs. in a (Glock 17)/1.55 lbs. = 5.374214302 fpe

Agree. Those velocities will give those results. I consider 950-975fps and 1150-1200fps to be more representative of each respectively. However, my recent research into current factory loads shows the 124gr has been "watered-down" even more to the ~1050fps range, whereas the 147gr remains unchanged. I reload, and I had not realized target loads had been "lawyer-proofed" and "weakeded" to that degree. This was my error (i.e. I was using ~1150+fps for 124gr and ~1250fps for 115gr from memory). It all depends on the velocity of the loads you are comparing.

These are the numbers that the equation produces.

"Felt" recoil is another story altogether...and one that I am not going to get into. Let's not even get into "felt" recoil. :)

Thoughts above. New post (http://glocktalk.com/forums/showthread.php?t=1169109)created to "crunch numbers".