Energy Ft. Lbs. relationship to Recoil and Pressure

^^^^^^^
I saw a video stating that .460 rowland is close to 1000 ft. lbs.
.460 rowland conversion in a .45 acp glock 21.
Comparable to .44 mag. according to the video.
 
Ron is correct. I did look at a few different online calculators yesterday.
The results were that one pretty nearly agreed with what I posted about 1911
Free recoil of the gun = RG in ft.lbs.
Recoil velocity of the gun = VG in fps
1. RG = 5.17 VG = 10.5 as above for the 1911.
2. RG = 4.58 VG no data
3. RG = 4.3 VG no data
4. RG = 5.1 VG = 10.4 (*)
5. RG = 4.43 VG = 9.75
6. RG = 4.9 VG = 10.3 (*)
 
In the same gun, you can get a rough idea of relative recoil by multiplying the weight of the projectile times the muzzle velocity of the projectile and comparing that for various loadings.

I'm not claiming that's the exact answer, nor that it's a formula for 'felt recoil' but it will give you a pretty good way of comparing multiple loads in one gun.

Felt recoil is much trickier. Some people process the noise of the report as part of the felt recoil. The grip of the gun can affect felt recoil--with a good fit reducing it and a poor grip fit making it worse. The only way you're going to know how a particular load feels in a particular gun is to actually shoot it.

But it sounds like you're looking for a quick and dirty comparison that you can calculate--for that you can just compare mass velocity products.
 
JohnKSa said:
In the same gun, you can get a rough idea of relative recoil by multiplying the weight of the projectile times the muzzle velocity of the projectile and comparing that for various loadings.
Click to expand...

Also known as momentum or power factor.
 
BondoBob said:
Can someone explain this relationship between energy ft. lbs and recoil? I'm trying to decide on a defense load for my 357 revolvers. Critical Defense is listed as a low recoil round but has 624 ft lbs. Critical Duty is listed as more powerful and penetrating but less energy at 487 ft. lbs. How can this be? Is there an other component to this formula I'm missing like pressure? I would think the higher the ft lbs the more felt recoil in the same gun. I'm a bit recoil sensitive I guess.
Click to expand...
Power and Energy are 2 different things, so it's possible one factor can be more and the other less.
The differentiator is Time.

Energy = Power x Time.

As to penetration, if one bullet A is more aerodynamic (and hydrodynamic hitting a water based target) than another bullet B, then A may penetrate further than B into the target.

As to the recoil, most revolvers don't have muzzle brakes and have no recoil mitigation mechanisms as some semi handgun (and rifles do). So the shooter is going to feel the full recoil as a fairly sharp pulse containing all the energy over a one to a few milliseconds. Heavier revolvers take longer to put in motion from the recoil force, so a proportional increase in time. The heavier the revolver, the slower the recoil.

A semi-spreads that out even more as multiple movements / actions are involved, so the felt recoil is less even though the muzzle energy might be the same.

Here's a slo mo of those reactions on a semi-rifle:
12,000 ft-lbs of bullet (and recoil) energy in (almost) slow motion. The bullet exits the barrel before any appreciable movement in the weapon. Then the barrel starts moving back against 2 stout springs. This action is followed by the bolt carrier moving back against a single large spring (similar to an AR-15). Just after that you can see the shooter starting to feel the delay and spread out recoil.

The 12,000 ft lbs of energy 'felt' by the shooter is the same amount of energy felt by a target at pointblank range that fully absorbs the bullet energy..... except the shooter feels it spread out over about 1/4 second, where the target is absorbing the energy in well under 1 millisecond.

rEnAOCR.gif


So you might be recoil sensitive to 'sharp' (fast) recoil, but less so to recoil that is more of a 'push' as seen in this video.

Does that help any?

(P.S. The 12,000 ft lbs of recoil from the Barrett is effectively a push and much less unpleasant than the sharp 280 ft lbs from a Bersa .380.)
 
Folks, this ain't hard. Simply use one of the online recoil calculation programs. There are several to choose from, this is just one.


http://www.shooterscalculator.com/recoil-calculator.php

You need to know 4 things 1st.

Weight of the bullet
Weight of the powder charge
Weight of the firearm
Muzzle velocity

You can go to any of the online reloading sites to get a rough idea of powder charge. Most firearms websites list weight. On rifles you need to also figure the weight of optics, mounts, and anything else on the rifle.

It is best to actually weigh the gun and shoot ammo over a chronograph, but using published numbers will get you pretty close.

The weight of the powder charge makes a significant difference, especially with rifles, and is often overlooked. This is why it is possible for a 308 to shoot the same bullet weight as a 30-06, and shoot it to the same velocity, and still have significantly less recoil. The 30-06 needs 8-10 gr more powder to achieve the same speed. 10-13 gr more powder to get 100 fps more than 308.

This is actual recoil. Some guys like to talk about "felt" recoil somehow being different. That difference is mostly due to differences in stock shape. That makes it a rifle issue, not a cartridge issue.

The rest is between the ears. The mind is easily fooled and if someone "believes" a certain firearm, or cartridge will have excessive recoil they will feel it. That doesn't make it so.

I don't deal with abstracts. Figure the real recoil and you have a baseline. The rest is a firearm issue or mental issue.
 
TXAZ said:
The 12,000 ft lbs of energy 'felt' by the shooter is the same amount of energy felt by a target at pointblank range that fully absorbs the bullet energy..... except the shooter feels it spread out over about 1/4 second, where the target is absorbing the energy in well under 1 millisecond.
Click to expand...

You're thinking of momentum, a conserved vector quantity. The gun has the same amount of momentum coming back at you as the bullet has going away from you (ignoring the momentum of the gas, which creates a small correction).

But... the bullet has over 99% of the kinetic energy of the bullet + gun system. Kinetic energy isn't a vector. The gun has some, the bullet has some. But because the bullet is so much lighter, it has much more of the energy -- almost all of it. That's why your shoulder gets a slight bruise, while the deer has its guts blown out.
 
jmr40 said:
The weight of the powder charge makes a significant difference, especially with rifles, and is often overlooked.
Click to expand...

If you make the bullet light enough and the powder charge big enough, eventually you can't ignore the powder any more. So, e.g. with extreme varmit loads it's unavoidable that recoil is going to get more complicated.

At the opposite end is handgun loads. You might typically be firing a 147gr 9mm, using ~4gr of powder. That 4gr of powder is there, sure enough, but it's not going to ruin basic conclusions if you ignore it.
 
To answer the OP's question directly.

The relationship between energy and recoil is that energy divided by the muzzle velocity of the projectile will provide a number that will allow a comparison between the recoil of various loadings when fired from the same handgun.

The reason it works pretty well is that muzzle energy is mass times velocity times velocity times a scaling factor. When you divide that by velocity you just end up with mass times velocity times a scaling factor which is just muzzle momentum times a scaling factor. Since momentum is directly related to recoil, you can compare the resulting numbers to each other and get an idea of how the loadings will compare in terms of recoil.

This wouldn't work so well in high-velocity rifle rounds, but for handguns it's going to work pretty well. The reason it won't work so well in high-velocity rifle rounds is because the powder mass velocity product can get to be fairly significant compared to the mass velocity product of the projectile and so you need to take it into effect since recoil is really related to the momentum of EVERYTHING that comes out the muzzle. But in handgun rounds, you can ignore it without causing too much trouble.

The relationship between energy and pressure is complicated and exploring that avenue in the pursuit of getting something that will provide you a simple way to compare recoil for various loadings is not going to be productive.
 
Looking back at the original first post:
Energy Ft. Lbs. relationship to Recoil and Pressure
Can someone explain this relationship between energy ft. lbs and recoil? I'm trying to decide on a defense load for my 357 revolvers. Critical Defense is listed as a low recoil round but has 624 ft lbs. Critical Duty is listed as more powerful and penetrating but less energy at 487 ft. lbs. How can this be? Is there an other component to this formula I'm missing like pressure? I would think the higher the ft lbs the more felt recoil in the same gun. I'm a bit recoil sensitive I guess.
Click to expand...

Looking at the two mentioned cartridges, The Hornady Critical Defense and Hornady Critical Duty. The numbers given of 624 (Critical Defense) and 487 (Critical Duty) are both expressed in LbFt or Food Pounds of force.

Reading the boxes Hornady claims a velocity of 1500 FPS (Foot Per Second) for their Critical Defense ammunition 125 grain bullet and 1275 FPS for their 135 grain bullet. Now keep in mind at this point those are Hornady numbers for their test gun on the given day they fired the gun. The same ammo in the same gun will slightly vary shot to shot. Those numbers will produce the muzzle energy numbers mentioned. Calculating muzzle energy only requires two numbers. bullet weight and muzzle velocity. Velocity and Bullet Weight.

In keeping with the US system of measure this simplified would be :

Velocity X Velocity X Bullet Weight / 450,240 = Energy expressed in Ft/Lb

So Hornady Critical Defense

1500 * 1500 = 2250000 * 125 = 2.8125^8 / 450,240 = 624.6 FtLb. Which is their claim.

So Hornady Critical Duty

1275 * 1275 = 1,625,625 * 135 = 2.1946^8 / 450,240 = 487.4 FtLb. Which is their claim.

What is important to note is that while Bullet Weight and Muzzle Velocity are used to calculate ME (Muzzle Energy) they are only two factures which are used to calculate FRE (Free Recoil Energy) and that ME and FRE are two different animals. While both are units of energy and expressed (here in the US) as the unit of measure FtLb not the same.

Ron
 
Metric said:
You're thinking of momentum, a conserved vector quantity. The gun has the same amount of momentum coming back at you as the bullet has going away from you (ignoring the momentum of the gas, which creates a small correction).

But... the bullet has over 99% of the kinetic energy of the bullet + gun system. Kinetic energy isn't a vector. The gun has some, the bullet has some. But because the bullet is so much lighter, it has much more of the energy -- almost all of it. That's why your shoulder gets a slight bruise, while the deer has its guts blown out.
Click to expand...
We could argue dot and cross products, and we'll get 2 different (both correct) answers depending on if we're in an engineering or Physics environment... But...
Last time I checked, a vector is a "quantity having direction and magnitude". In this instance, the kinetic energy clearly has a direction and magnitude.
 
KE definitely isn't a vector, and it's not controversial. Take two freight trains of equal mass and speed, heading toward each other (i.e. in opposite directions). The net momentum is zero, since momentum is a vector and the vectors cancel out, due to the opposite direction. But the kinetic energy is definitely not zero -- there is twice as much kinetic energy as a single train. You can add up kinetic energy, and direction does not matter.
 
Metric said:
KE definitely isn't a vector, and it's not controversial. Take two freight trains of equal mass and speed, heading toward each other (i.e. in opposite directions). The net momentum is zero, since momentum is a vector and the vectors cancel out, due to the opposite direction. But the kinetic energy is definitely not zero -- there is twice as much kinetic energy as a single train. You can add up kinetic energy, and direction does not matter.
Click to expand...
Poor example.
For a definitive definition: https://www.dictionary.com/browse/vector?s=t
 
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