stevelong: Sir; without a chrono. aint know way to know.
Trust this POINT. Sir; I'll not shoot anyones reloads. Sir; my firearms are to valuable to me. My risk with factory ammo is minimal by comparison.
These fine folks that are selling reloads are probably the most honorable individuals around. My firearms will miss out on said reloader's. [redundant for sure]
Powder burn rate=pressure=push
Winchester super x; good quality factory loading= trust each will be very close to the next.
44's by nature with hotter loads heavier lead can beat you up. Without Proper information I could very easily ramble on for the rest of the night. NOT GOOD.

Originally Posted by neophyte

stevelong: Sir; without a chrono. aint know way to know.
Trust this POINT. Sir; I'll not shoot anyones reloads. Sir; my firearms are to valuable to me. My risk with factory ammo is minimal by comparison.
These fine folks that are selling reloads are probably the most honorable individuals around. My firearms will miss out on said reloader's. [redundant for sure]
Powder burn rate=pressure=push
Winchester super x; good quality factory loading= trust each will be very close to the next.
44's by nature with hotter loads heavier lead can beat you up. Without Proper information I could very easily ramble on for the rest of the night. NOT GOOD.

I couldn't agree more. Never shoot someone elses reloads.Your putting a lot of trust into someone you don't even know. Get yourself a press and reload your own and you will know what you have. Good Luck.

First let me say that I'm not a reloader - I bought a press some time back, but it hasn't seen action yet. That said, I don't see how two bullets of the same mass with the same muzzle velocity would recoil that much differently. Pressure is a function of temperature (and Volume) by the ideal gas law PV=mRT. Anyway, your final (combustion) temperature is determined mainly by the constituents in the propellant. The pressure is simply a force per unit area, and it's that force that persuades the bullet down the barrel as well as pushing backward on the gun that we perceive as recoil. Extraneous info aside, pressure is what accelerates the bullet and what causes recoil. I don't see how it's possible, but I sure don't know everything.

Snowman: Sir; you are absolutely correct. Now part of your answered thinking takes into account; correct reloading practices.
Sir; just a simple variable of any consequence very easily change the physical chemical reaction to catastrophic ending.
My reason for not using someone else's reload? Trust.
Two reloads mathematically correct in all ways: I expect them to be very close in reaction to the action.
Two reloads with different 'lead' captive pressures: =caution
Two reloads with different "lengths" = caution
either or will cause with all being the same; pressure differences.
A grossly simplified thought.

Originally Posted by neophyte

Snowman: Sir; you are absolutely correct. Now part of your answered thinking takes into account; correct reloading practices.
Sir; just a simple variable of any consequence very easily change the physical chemical reaction to catastrophic ending.
My reason for not using someone else's reload? Trust.
Two reloads mathematically correct in all ways: I expect them to be very close in reaction to the action.
Two reloads with different 'lead' captive pressures: =caution
Two reloads with different "lengths" = caution
either or will cause with all being the same; pressure differences.
A grossly simplified thought.

Hi Neophyte. You are also correct. I agree with everything you said here. I hope it didn't appear that I was disagreeing with you; rather, I was attempting to answer the OP's question which was (I think) "Could the two loads recoil so differently, and still shoot at the same velocity?" Haha, as I reread my post it's not really clear what the heck I was trying to say, so hopefully this will clear it up:

I don't think it's possible for the two loads of largely different recoil to both shoot at 1200 ft/s. Also, don't shoot anyone's reloads that you don't trust dearly.

My own handload is a .44 Magnum that throws a 240gr. cast bullet at 1200 f.p.s. which is very accurate and pleasant to shoot. I think the Winchester 240gr. JHP factory load is much faster, probably pushing the 1400 f.p.s. range. My own heavy JHP hunting loads are right at 1400 f.p.s. and very nearly duplicate factory loads.

And, what has been point out, same weight bulets at identical velocities in the same gun will produce the same amount of recoil.

The slight difference in powder burning rates will make no noticable difference.

And, incidentally, the heaviest loads use slower burning powders, lighter loads the faster burning.

One more piece of the puzzle:

Velocities listed on charts and those produced in the field can differ significantly. The only way you can trust velocitiy numbers is if they were obtained with a chronograph. IMHO

Snowman: Sir; you were absolutely spot on. My addition to your thoughts was brought about from other thoughts.
Most "new to understanding bullets" have too much to do learning about there new hobby. It takes time and quality answers like yours.
Sometimes when Quote [hobbiest] know something; conveying the thought leaves out some of the necessary. You adding to mine; mine adding to others; others adding to others is a proper learning tool
I appreciate good thinking. I enjoy THOUGHT provocation.
I appreciate that most information coming from this group isn't 'condescending' with respect to one and another

Hello, thanks for the (few) straight replies to the point of the question.
I'm new to this particular forum, just signed up in starting this post.
I began loading trap loads with my Dad around 1962, then I taught him centerfire handgun & rifle reloading processes beginning around 1970, and in the last few years I began loading & shooting match centerfire ammo for centerfire varmint matches.
Not a rookie reloader.
I just do not have a chrony.
The gun show reloader guy has his own business, sells tens of thousands - heck maybe hundreds of thousands of rounds each year, for both rifle & handgun, at gun shows & mail order. I've only used his reloads when I buy a gun at a show, and know that I don't have anything loaded at home to quickly function-fire the newly acquired treasure.
I'm guessing the reloader guys' labels no longer reflect his load recipe, and that he's just using it for referencing caliber and bullet weight.
?
Steve.

Originally Posted by TOF

One more piece of the puzzle:

Velocities listed on charts and those produced in the field can differ significantly. The only way you can trust velocitiy numbers is if they were obtained with a chronograph. IMHO

You're mighty right about that! While I could chronograph my handloads, I set up to revolvers for each cartridge, one with either a six inch or seven and a half inch barrel, the other with four inch barrel or 4 5/8" barrel. I also chronographed factory ammunition to compare with my results.

There were a few surprises, especially in .357 Magnum, where the 4" barrel often produced higher velocities than the longer barrel. And, the old powder, #2400, easily outstripped the newer H110 and Winchester #296 powders. The ball powders were hard pressed to push a 250 gr. cast SWC 1400 fps in my .44 Magnums, yet #2400 exceeded 1500 fps in some cases.

And, the closest I could get to the magical 1900 fps with a 180 gr bullet was 1715 fps using #2400.

Made for an interesting period in my studies.

Bob Wright

I agree with the don't trust relaods part. I also agree with less recoil equals less volocity. I would also think it affects the muzzle energy.

Originally Posted by Capt. Mike

I agree with the don't trust relaods part. I also agree with less recoil equals less volocity. I would also think it affects the muzzle energy.

Bingo. As several of us alluded to, energy is directly influenced by velocity by the kinetic energy equation:

KE = 1/2 m*v^2 (KE = one half times mass times velocity squared)

So in fact it affects muzzle energy exponentially.

Mmmmmm...

Is it more useful to us to reason in terms of kinetic energy [Joules] or Force needed to stop the gun's acceleration F = ma [kN] ??

I'd probably reason in terms of Force, or momentum-torque [kNm] which my hand-arm system, in example, has to bear to slow down and stop the gun accordin to his main acceleration vector component when it recoils...

There is a clear example of that in the following link:
http://en.allexperts.com/q/Physics-1358/Gun-Recoil.htm

the example appears to me simpler than the handgun's case, since it is viewed as motion along a single axis, whereas the hangun's motion is more complex...

Also Wikipedia illustrates teh basic concepts of guns physics
http://en.wikipedia.org/wiki/Physics_of_firearms

I recvently saw an article expressing a gun's recoil in energy - Joules units. Probably that's less intuitive though than force units, although involving less calculations (energy being a state variable if I get it right).

They are invariably related. Energy being the ability to do work, and force which will accelerate a mass. You're correct; it is force that is perceived as recoil as well as torque which is force times the distance between the applied force normal to your hand.

As you said, it is much easier to calculate energy than instantaneous force upon a body F = dp/dt. Momentum, hence force, is most useful in macroscopic applications like this since it is conserved. Energy is conserved of course, but it becomes impossible to employ the conservation of energy on something like this having to account for losses like heat, sound and light.

However, this is a topic that has little practical meaning. The question was 'will rounds that recoil differently have the same energies' - which is really the point of interest.

I've been mulling over a few of teh above issues.

As a matter of fact, teh case of teh handgun is only sglightly more complex than teh case of the rifle.

To amke it simple, let's assume all the gun's movement results in a shift along the barrel's axis. The hands do not hold the barrel though, they hold the grips, located slightly underneath it, and that distance is the momentum radius, or leverage.
Due to the degrees of freedom of the wrist's joint, and since the arm's posture tends to prevent a purely translational motion, the barrel/handle system rotates around teh fulcrum, th egrip surface. It is a clockwise rotation, so the muzzle tends to move upwards, as it is the case. Should teh grip be above the handle, teh rotation would be counterclockwise and the muzzle would go downward.

We might easily figure out the muzzle force knowing the gun's recoil speed as per link's example. we need to fit into momentum and angular velocity though...

Sure the energy calcs is easier.

In the case of the above 44 mag, 1200 fps velocity = 400 m/s
weight bullet? 10 grams = 2/100 ofa pound = 0.01 kg, so energy in joules = 0.5mV^2 = 0.05*400 = 800 joules. Projectile's weight may be wrong.

Know we should convert this to gun's acceleration, which is harder...

Snowman,
sorry, I saw your post after my latest post. I agree, frictional and thermal dissipation are hard to quantify.

The question was 'will rounds that recoil differently have the same energies' - which is really the point of interest.

I believe you mean: will rounds which cause a different recoil (in terms of force/torque magnitude) have the same energies.

well, I tried to solve teh .44mag question using teh rifle example of the link.

-i simplified the problem assuming only non-rotational motion is involved.

In this case, m1 = bullet weight = 0.01 kg, v1 = muzzle velocity = 400 m/s, m2 = gun's mass = 1.5 kg, v2 = m1v1/m2 = 2.7 m/s = gun's velocity

a = v2^2/(2d) where d = half an inch, movement of gun in the shooter's hands.
a = 284 m/s^2
F = m*a = 427 N = a little less than 100 pounds, not too bad after all....

Now, if we increase by 50% the bullet's weight >>>> F = 960 N, force doubled

We get the same result increasing velocity by 50%

In the first case (bullet's mass increased) we have Energy = 1200 J, whereas in the second case (velocity increased) we have energy = 1200 J, with the same recoil force.

Conversely, we can easily verify that m1=0.01 and v1=490 gives the same muzzle energy (1200) than teh pair m1=0.015 and v1 = 400 >>E = 1200 J.

So, the answer to the original question would be that yes, different recoil may result in same muzzle energies (if the abova calcs are right, please verify).

At the end of all this, it would appear that muzzle energy is not such an important factor as muzzle momentum : m1*v1=bullet mass times bullet's velocity.

I went on searching. A very good and practical article on recoil is fopund on the wikipedia:
http://en.wikipedia.org/wiki/Recoil

It turns out on the interent there is a free recoil calculator, which might help and answer stevelong's original answer:

http://handloads.com/calc/quick.asp

A quick view:
using stevelong's data, and a tyhpical gun weight (3 #), the only left variable is powder weight.

Powder charge gr. recoil in ft/lbs

10 11.36
15 12.76
20 14.37
25 15.30

So, even a great change in powder charge only implies a change in about 35-40% recoil.

Probably other factors at work: velocity not as stated (probably), faster detonation, you name it, I'm not a firearms expert...

What is left out of the equations; give considerations to dynamics
machine held
strong arm held
flex arm hold

Any argument is the basis, or completeness

Laws of Physics cannot account between different mounts, with a human variable

Application mathematics=reference only.

Those Wnchester loads are made to punish what you point your weapon at, deer sized game probably. The more energy the ammo produces, the more recoil you will feel. If you feel less recoil out of the same gun, your load is producing less energy.



My 5" classic on top. I handload 240gr. Nosler SPs to 1000fps and the result is a "mid range" load with moderate recoil that doesn't beat up my gun, and they have killed pigs and deer with no failures to date.

Oldscot 3,
your 5" S&W baby on top sure has beatiful grips...

The top one is a beauty.

The bottom one is a cutie.

Thanks Mccoy, I can appreciate all types of firearms and like to think of myself as broad minded, but when I plunk down my hard earned cash I ususally go home with a steel gun wearing wood grips. By the way the 44 came with rubber grips, I found the wood Smith & Wesson grips at a gun show. They're easier on my hand in my opinion, the rubber ones gave good purchase but rubbed my fingers sore after shooting for a spell.

I like to use the 5" 38 with CCI ratshot, cleaned a squirrel out of my blind this year... just point and click.

oldscot3: That is some fine looking lumber on the classic

Yes, recoil calculations should always include the "weight of ejecta", which is the total weight of the propellant charge and the projectile.

AA#2 is a pretty fast powder for a .44 magnum.