Handguns and Torque, for you physicists....

I'll bite,

Yes they torque but I don't think it would be nearly enough to be noticeable to anybody. The 3 components you're looking for in a torque are the force exerted, the distance from the axis of rotation and the portion of force exerted on a moment arm. The force is fairly large for sure but the distance from the axis of rotation is small and the angle is small as well. I think that conceivably it could affect the path of the bullet and for most handguns it would tend raise the point of impact.

http://en.wikipedia.org/wiki/Torque#Force_at_an_angle

Not a physicist, but common sense tells me there is not enough rotating mass over a long enough period of time. The bullet is small in diameter, only rotates about 1/3 to perhaps 2.5 turns at most (depending on rifling twist and length of barrel) during its travel down the barrel, and has exited the muzzle before the much heavier (in comparison to the weight of the bullet) gun has had any chance to twist. Whatever tiny amount of torque is generated is too small to feel when shooting, or affect the bullet's point of impact.

I worked up the numbers on this once, using a 1911 with a hardball load. Any rotating mass generates torque, so yes, torque is appied to the firearm in the oppositire direction of the bullet's rotation. However, while I don't really remember the number I came up with, it was ridiculously small, like .013 joules (0.1 in/lbs) or somesuch.

I've been asked this question a number of times by folks who notice that a handgun recoils not only upward, but also to one side. Had to point out in each one of those instances that rotational tendency of the handgun under recoil is a product of our skeletal structure, not torque generated by the accelerating projectile. Some didn't believe me, so I let them shoot two of my guns and then showed them that the rifling was LH in one and RH in the other.

there is also torque from the recoil right? I'm thinking of my wrist or my shoulder as the axis of rotation?

The twist of the bullet down the barrel probably contributes negligible torque, especially compared to forces of recoil.

Well, you guys know more about math than I do, but when I watch other people shoot my S&W 500 Mag with 440g balls, it sure looks like it twists all over the place in the hands of not-so-strong-gripped people.

Question:

bottom shelf said:

Well, you guys know more about math than I do, but when I watch other people shoot my S&W 500 Mag with 440g balls, it sure looks like it twists all over the place in the hands of not-so-strong-gripped people

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This is most likely the answer:

Grizfire said:

there is also torque from the recoil right? I'm thinking of my wrist or my shoulder as the axis of rotation?

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there is also torque from the recoil right? I'm thinking of my wrist or my shoulder as the axis of rotation?

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That's what I was trying to explain to people who thought is was the bullet's rotation causing the twisting motion of the gun as it recoiled.

When a handgun recoils, it causes the radius to roll toward the body's centerline as the wrist bends backward (since the flexion of the wrist toward the radius is very limited). This means for righty's, the gun twists up and right. For Southpaws, up and left.

Well, you guys know more about math than I do, but when I watch other people shoot my S&W 500 Mag with 440g balls, it sure looks like it twists all over the place in the hands of not-so-strong-gripped people.

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You're still only talking tenths of inch pounds.

Torque is measured in newton-meters. The diameter of a bullet is virtually nothing. something like 9/1000th of a meter. Because of the bullet's tiny mass, the force required to spin the bullet is also virtually nothing. Not even a 1/1000th of a ft-lb. So the lateral torque can be considered nonexistent. There's significant torque in recoil though, but not to the sides of the gun. There's torque in your hand since the handle is always below the axis of the bore.

Torque is measured in newton-meters.

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Nm is just one unit of force measurement.

joules (1 Nm)

Kilogram meter (.102 Nm)

foot pound (1.36 Nm)

Inch pound (.113 Nm)

Other units such as poundals and Inch ounce are used, and Nm (joule) can also be converted into brake horsepower or Kilowatt-seconds, Kilowatt-hours, dyne centimeters, etc.

The torque of the gun rotating upwards since the pivot point, (your wrist) is below the axis of the barrel. Yes, perceived as recoil. If the grip was directly behind the barrel, there would be no upward movement. (just rearward)
The torque of the bullet being spun by the rifling? Yes, but not perceived by the shooter, because the torque is trying to rotate the barrel in the opposite direction, but the barrel is locked in the gun and that mass is large compared to the torque of the bullet.
(I think)

Nm is just one unit of force measurement.

joules (1 Nm)

Kilogram meter (.102 Nm)

foot pound (1.36 Nm)

Inch pound (.113 Nm)

Other units such as poundals and Inch ounce are used, and Nm (joule) can also be converted into brake horsepower or Kilowatt-seconds, Kilowatt-hours, dyne centimeters, etc.

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Nm is the international STANDARD unit for torque.

Obviously there's an infinite number of ways you could measure it, as long as you have a unit of force and a unit of distance. But the SI unit is used for calculations and other forms of measurement are usually converted to SI units for anything.

Nm is the international STANDARD unit for torque.

Obviously there's an infinite number of ways you could measure it, as long as you have a unit of force and a unit of distance. But the SI unit is used for calculations and other forms of measurement are usually converted to SI units for anything.

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If you use SI units this statement is true. The standard unit in English units in usually ft-lbs.

Are you from the US?

Everyone uses SI units for science, even americans. Nobody would accept a physics report using anything else, and you cant do anything with ft-lbs without also either converting it to SI or converting everything else in the formulas to the same system. In any case, the lateral torque on the gun from the bullet rotating down the barrel is infinitesimally small.

ericyp, for the less scientifically inclined here, foot pounds are a unit of measure they understand, they aren't going to know the relative measure of a newton meter.

annyywayy,

if you are thinking of torque, then think about the comparison of a bullet to an engine. A low powered VW aircooled engine generates an average of 75 ft-lbs at 2500 rpm or so. Consider the weight of the crankshaft, connecting rods and cylinders, and the flywheel.

now think about the bullet spinning at a few thousand rpm that weighs 23 grams. negligible? mostly, present? yes.

The design of the S&W X-frame would seem to counter the position that the torque is negligible. From the original Shooting Times review of the 500:

However, the fore/aft frame alignment of the Model 500 installation actually works with recoil, pressing the latch more firmly into its notch, rather than opening it. Plus the Model 500's direction of rifling twist causes the frame to torque more firmly against the yoke, rather than away from it, so the total effect of the 500's design is to actually strengthen the yoke-frame interface when firing, rather than stress it.

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http://www.shootingtimes.com/handgun_reviews/monster_1103/

Note also the comment on gain twist rifling as used in the 460 S&W:

Gain-twist rifling essentially starts out with a very slow twist and then “spins up” to the desired rate for the specific caliber/bullet configuration as the bullet passes down the bore. The gain-twist benefit for a high-velocity revolver cartridge is that it allows the bullet to transition more slowly from a non-rotating state when it crosses the barrel/cylinder gap under extreme pressure and slams into the rifling at the rear of the barrel. This allows a much more concentric and consistent alignment of the bullet with the bore axis, a more positive and non-distorted land/groove engagement, consequent enhanced accuracy, and less wear and tear on the bore itself. Plus, “peak torque” is reduced, lessening the abrupt sideways twist a shooter feels when firing other high-pressure big-bore handguns.

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http://www.shootingtimes.com/handgun_reviews/ST_magrevstateoftheart_200805/index1.html

The above comments correspond to my own experiences with the S&W 500, in that the torque is very real and very noticeable.

ericyp said:

Everyone uses SI units for science, even americans. Nobody would accept a physics report using anything else, and you cant do anything with ft-lbs without also either converting it to SI or converting everything else in the formulas to the same system. In any case, the lateral torque on the gun from the bullet rotating down the barrel is infinitesimally small.

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In the engineering world, particularly civil engineering, it remains fairly commonplace for English units to be used. At least that was the case when I was working on my ME degree a few years back... about a third or so of our work was done using English units, and we had to know our way around both systems.

Also, physics threads on THR tend to make my head hurt.

Inertia is going to fix the position of the gun until the bullet gets out the end of the barrel.

Recoil does affect bullet path

Take a 357 Magnum revolver.

Bench rest it with light 38Spl loads.
Bench rest it (to the same sight picture and target) with 38Spl +P loads.
Bench rest it (to the same sight picture and target) with 357 Mag loads.

You will find as the power of the bullet goes up, so does the impact point on the target. Not feet, mind you, but the more powerful bullets begins to raise the barrel before the bullet exits the barrel. It can easily be a few inches at 10-15 yards.

Upward torque from recoil will affact the impact point. My SP101 is dead on with 357 loads and shoots 38Spls lower using the same sight picture.

I suspect torque from bullet rotation through the barrel is basically irrelevent (within the accuracy variation of handguns).

Nm is just one unit of force measurement.

joules (1 Nm)

Kilogram meter (.102 Nm)

foot pound (1.36 Nm)

Inch pound (.113 Nm)

Other units such as poundals and Inch ounce are used, and Nm (joule) can also be converted into brake horsepower or Kilowatt-seconds, Kilowatt-hours, dyne centimeters, etc.

Click to expand...

Nm is the international STANDARD unit for torque.

Click to expand...

You guys are confusing two units. The newton-meter and the other units referred to by the first poster are units of energy (not force). A Nm is the energy required to raise a mass weighing one newton a distance of one meter. The newton itself is a unit of force.

The newton-meter of torque is the torque generated by exerting one newton of force on the end of a lever (a wrench, for instance) that's one meter long.

Same units, but torque is a vector. Work and energy are scalars.

Tim

Shooting a rifled firearm will create a small amount of torque around the gun's bore axis, because angular momentum is always conserved. Before you pull the trigger, there is no angular momentum. That is, nothing is rotating. After you pull the trigger the bullet is rotating, so the gun and possibly part of the attached person must rotate in the opposite direction to maintain zero total angular momentum. The direction of the rifling determines the direction of the torque on the gun and person; the gun will always rotate in the opposite direction as the bullet. However, the effect on the gun/person is minor, because the bullet's mass is much less than that of the gun and person.

If gun is not held at a position that's in line with the bore axis, the firing will also recreate a torque around the hold point, but this torque is around a left-to-right axis, not the bore's axis. That's why the recoil of a handgun causes the muzzle to rise.

Single Action at the OK Corral

Torque is the product; multiplication, of an applied force; which is, mass of the bullet:grains divided by 7000 to convert to pounds, divided by the acceleration of gravity, 32.2 fpsps to arrive at mass -times the acceleration of the bullet; angular. The acceleration of the bullet is induced by the rifling and the linear acceleration along the barrel.

The more rifling -twist, more muzzle velocity, the more the acceleration.
Both reactions: the linear; recoil and rotational. Both are simultaneous. At once.

Just by consideration, it is obvious that the mass of the bullet is miniscule, while the rotational acceleration is large as is the linear acceleration.

However, your hand on that grip gives you a large lever; wrench, to resist that rotational twist; applied at the center of the bore through your palm and into the grip.

The other rotational twist; "recoil" is at the same distance; bore to palm, but in this case the center of rotation is your wrist!

The bore lies somewhere beyond, that is, above your palm; grip, and the recoil; force is applied at that distance to your palm, causing the gun to rotate in the up and down plane; the sighting plane, in your hand.

You should see and feel it in my Ruger Blackhawk firing some stout loads!

That thing bucks up and around at the muzzle and if I would release my fingers, except for the trigger finger {and could manage not to get struck in the teeth with it}, it I could let it spin or twirl around my finger like Val Kilmer a.k.a. "Doc Holiday" can twirl a tin cup.

Units of torque are customarily given in force x distance units, so as not to confuse them with work units which are given in distance x force.
Example of torque units: pound x feet or pound x inches. These are pounds -force.

Inertia is going to fix the position of the gun until the bullet gets out the end of the barrel.

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Not true. Take a look at this video, especially around the 2:15 mark. You'll see the pistol start to recoil before anything leaves the barrel:

http://www.youtube.com/watch?v=XXZUMGRxPM4

Note that the entire pistol begins to recoil backwards, even before the slide starts to actuate (watch the end of the guide rod).

And as far as the torque imparted on the firearm by means of the bullet being infinitessimal - well, that's certainly relative to the particular firearm. The manual of my .50AE Desert Eagle warns to use a very tight grip, lest the recoiling firearm will torque in the shooter's hand, causing the case to eject into the shooter's face. I experienced this very thing when I didn't use enough grip... it fired, rose up, twisted left, and the ejected case bounced off my forehead.

In the engineering world, particularly civil engineering, it remains fairly commonplace for English units to be used. At least that was the case when I was working on my ME degree a few years back... about a third or so of our work was done using English units, and we had to know our way around both systems.

Also, physics threads on THR tend to make my head hurt.

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My best friend is completing his masters in mechanical engineering, metric and standard units are what's used now.

while the rotational acceleration is large as is the linear acceleration

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Why? It would depend on the rate of twist.

Someone can calculate the torque force on the bullet.

A bullet mass of .0075 kilograms. Radius of .0045 meters. So the bullets moment of inertia is .0075*.0045^2*.5

Multiply this by the bullets rotational acceleration (which I blatantly admit I can't figure out. I have no idea how to figure out how long it took for the bullet to reach the end of the barrel to get time)


And that's the torque force being exerted. Then consider the vertical distance between the center of the bullet and the gun hand, and we have our final number.