Handguns and Torque, for you physicists....

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)

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Easy. The bullet starts at 0fps, and leaves the barrel at (since you used a bullet weight of 115gr, let's use the data from a Federal 9mm out of a 3.25" barrel) 1019fps. That's an average velocity of 510fps, covering that 3.25" distance in .266ms.

This particular barrel has a twist rate of 1 in 9.84" (they have to be difficult, right), so the bullet rotates .33 times in that .266ms, for a rotational speed of 74,436 RPM. Of course, that's at the muzzle. So, the bullet went from 0RPM to 74K RPM in .266ms.

^ That's assuming constant acceleration though, which isn't what happens in the gun.

I also think the average velocity thing might be the wrong way to do it even with constant acceleration... I remember doing a physics problem like that my sophomore year. I think there was a formula relating distance, time, and acceleration and that was the correct way. I can't remember exactly though..

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.

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No, the torque on the gun does not equal the torque on the bullet. It's the angular momentum of the two that are equal (and opposite). You can calculate the angular momentum simply using the bullet's muzzle velocity and the barrel's twist rate, but to convert that to torque on the gun you need to know the gun's moment of inertia, which won't be easy to compute because of its asymmetry.

I took a whack at solving for the rotation induced in a real gun by the barrel's rifling. I assumed a Springfield 1911 in .45ACP firing a 230 grain wadcutter bullet at 830 fps. The gun weighs 39 oz and has 1-in-16" rifling.

At 830 fps and one rev per 16", the bullet will be rotating at 622.5 revs/sec, or 3911 radians/sec as it leaves the barrel.

Assuming it's a solid cylinder, the bullet has a moment of inertia of (0.5)m(r^2), which for .45 caliber comes out to 2.433 * 10^-7 kg-m^2.

Angular momentum is just the product of the rotation speed and moment of inertia, so for this bullet it's 9.516 * 10^-4 newton-meter-seconds.

Angular momentum of the gun+bullet system is conserved and it was zero before the trigger was pulled. Therefore, the gun's angular momentum after the bullet leaves the barrel must be equal and opposite that of the bullet. (I'm assuming that the gun is just hanging in space when the trigger is pulled; if you attach a person to it the calculation gets really complex.)

Here's the big handwaving. I have no idea what the moment of inertia of a real gun is, because it's not a convenient shape like a cylinder or sphere. So for the purpose of calculating its moment of inertia, I assume the gun is compressed down to a solid cube of steel. For this gun, that cube would be about 5.2 cm (about 2") on each side. The moment of inertia for that cube rotating about its center of mass is 4.98 * 10^-4 kg-m^2.

So the rate of rotation of the gun after the bullet leaves the barrel would just be its angular momentum divided by its moment of inertia, which gives 1.91 radians/sec or 18.24 RPM. Since the rifling of this gun is left-handed, the gun will rotate clockwise as viewed from behind.

But the compressed cube assumption underestimates the gun's moment of inertia and so overestimates its rotation rate. I'm guessing that the rotation rate of the real gun would be about 5 to 10 RPM.

If a pistol twists under recoil, it's probably more of a function of each individual's geometry/bone structure. My son noticed my hand/arm twisting under recoil, and it's the same whether it's a left or right hand twist in the barrel.

My hand/arm rotates counterclockwise as viewed from behind even with the 1911's left hand rifling. Though the gun itself might have a tendency to counteract what my hand/arm does, the recoil far outweighs any effect the torque might have. But, that's just me. Your results might be different.

Woody

No, the torque on the gun does not equal the torque on the bullet. It's the angular momentum of the two that are equal (and opposite). You can calculate the angular momentum simply using the bullet's muzzle velocity and the barrel's twist rate, but to convert that to torque on the gun you need to know the gun's moment of inertia, which won't be easy to compute because of its asymmetry.

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I didn't have the torques set equally, I had the gun's torque reduced by proportion to the distance the guy's hand is out there? The torque that a bullet of that mass and radius spinning at that speed would exert onto the gun, which reduces the further the measuring point is from the center of the bullet.

I'm sick of talking about physics on break..

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).

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Energy applied/time=force. Besides, very few people are familiar with dyne, kilopond and poundal.

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.

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Most people don't really understand foot-pound/pound-foot either.

On that note, we should all know the conversion factors for these common units of measurement from SI to SAE and back; ft/lb to Nm, inch to mm, mile to Km, ounce to gram, etc., and be able to ballbark the conversion in our heads.

From http://en.wikipedia.org/wiki/Torque


\vec{\tau} = \vec{r}\times \vec{F}
\mathbf{\tau} = rF\sin \theta

where

* \vec{\tau} is the torque vector and \mathbf{\tau} is the magnitude of the torque,
* \vec{r}\, is the lever arm vector (vector from the axis to the point of force application), and r\mathbf{ } is the length (or magnitude) of the lever arm vector,
* \vec{F} is the force vector, and F\mathbf{ } is the magnitude of the force,
* \times\, denotes the cross product,
* \theta\, is the angle between the force vector and the lever arm vector.

There are at least two force vectors that can result in torque in firing a handgun. The most noticeable is what is felt as recoil. The second less forceful torque is the energy imparted by the bullet contacting the lands and grooves of the barrel.

"Energy applied/time=force"

Energy per unit time is power, not force.

Tim

Energy per unit time is power, not force.

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Would that not be energy used (work done) per time unit is power? As in horsepower-hour or kilowatt-hour?

Perhaps I'm just not wording things right, since I never had a physics class.

Force=mass*acceleration, acceleration being a linear measurement in time units, right?

We agree that Energy applied per time unit causes a certain rate of acceleration, as in the measure of foot-pound-force, where you have mass, distance and time used to calculate, correct? Using the poundal as the common measurement in that system, we have a unit that describes the force required to move 1 pound one foot per second with a rate of accelleration of one foot per second. As a linear equation, then, one poundal (which is one half ounce acted on with the energy of earth's gravitational pull, correct?) of force will accelerate a mass of one pound to 5 FPS after 5 feet traveled and 5 seconds.

In the same, then does one foot-pound of energy in action (now a pound-force, as it is causing acceleration) applied not have the force of 32 poundals and making it able to accelerate a one pound mass to 32 FPS or a 32 pound mass to 1 FPS in one second?

I really am asking here, since this is not a calculation I use, but should be able to better explain it when asked.

"Would that not be energy used (work done) per time unit is power? As in horsepower-hour or kilowatt-hour?"

Horsepower and watt are units of power, which is energy (or work) per unit time. Power times time, then, as in horsepower.hour or kilowatt.hour is energy (or work).

"We agree that Energy applied per time unit causes a certain rate of acceleration..."

We don't agree on that. Applying energy to a mass does not necessarily accelerate it. Consider, for example, frictional losses in an automobile that require energy input just to keep speed constant.

"Force=mass*acceleration, acceleration being a linear measurement in time units, right?"

Acceleration has units of distance per unit time per unit time. For example, meters per second(squared).

Tim

Would that not be energy used (work done) per time unit is power? As in horsepower-hour or kilowatt-hour?

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Yes, power is work done per unit time. Work is just a transfer of energy, so the terms energy and work are considered synonymous. Power can be measured in horsepowers or watts. The units you mention are measures of energy, not power, because they are power multiplied by time.

Force=mass*acceleration, acceleration being a linear measurement in time units, right?

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F=ma is correct. In the English system, acceleration is measured in feet per second per second, or ft/sec^2.

We agree that Energy applied per time unit causes a certain rate of acceleration, as in the measure of foot-pound-force, where you have mass, distance and time used to calculate, correct?

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It would be more correct to say that applying a force to a mass causes a certain rate of acceleration, imparting energy to the mass.

Using the poundal as the common measurement in that system, we have a unit that describes the force required to move 1 pound one foot per second with a rate of accelleration of one foot per second. As a linear equation, then, one poundal (which is one half ounce acted on with the energy of earth's gravitational pull, correct?) of force will accelerate a mass of one pound to 5 FPS after 5 feet traveled and 5 seconds.

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Yes, one poundal is the gravitational force felt by a mass weighing one half ounce. If a force of one poundal acts on a mass weighing one pound, the mass will accelerate at one foot per second per second. So after 5 seconds, the mass will be moving at 5 FPS. However, its average speed over that 5 seconds is 2.5 FPS, so in 5 seconds it will have moved 5 * 2.5 = 12.5 feet.

In the same, then does one foot-pound of energy in action (now a pound-force, as it is causing acceleration) applied not have the force of 32 poundals and making it able to accelerate a one pound mass to 32 FPS or a 32 pound mass to 1 FPS in one second?

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You're confusing force and energy here. If you apply a force of 32 poundals to a mass weighing one pound for one second, the mass will get accelerated to 32 FPS. If the same force is applied to a mass weighing 32 pounds, it will be accelerated to 1 FPS in one second. But the energy imparted to the masses is different, because kinetic energy is proportional to speed squared. In fact, the lighter mass will end up with 32 times as much energy as the heavier mass.

Yes heavier bullets cause more torque. Check out this link
http://www.470mbogo.com/BigBoreCompendium/
specifically this:
The 600 Overkill has the ability to push a 900 grain bullet at more than 2400 f.p.s. if the shooter can withstand the recoil and maintain his grip on the gun. At that level of power, as the bullet spins down the barrel, the rifle is torqued in the opposite direction, tending to twist the forearm of the rifle out of the shooter's hand. Brass for the 600 Overkill is available on a special order basis from Dieter Horneber (www.huelsen-horneber.com) of Germany. Rifles may be ordered from American Hunting Rifles for not much more than US$3000. For the big bore fan who needs the biggest affordable big bore, the 600 Overkill based on a CZ550 action is for him.

Yes they torque but I don't think it would be nearly enough to be noticeable to anybody.

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Wrong...and I've got a .45-70 Sharps Cavalry carbine that'll prove it. Full-bore loads in a Super Blackhawk will too.

Alright, so I know I'm straying from the topic a bit here, but I want to check my math with the experts.

So to figure out how much force is required to accelerate a 230 gr. bullet to 850 FPS in a 5 inch barrel, and for the purpose of this excercise, assuming the rate of acceleration is constant, rather than curved, we must first turn bullet weight in grains into a usable measurement, as 23/700ths of one pound. We then figure out how long it takes for the bullet to reach that speed, which is roughly .49 ms. This now gives us the rate of acceleration of 1,734,694 FPS per second.

To figure out how many poundals are needed for this rate of acceleration, we now multiply the mass of 23/700ths of one pound by the rate of acceleration at 1,734,694, right? So if my math works, and if bullet acceleration was constant and environmental factors notwithstanding , the manstopper .45 hardball fired from an M1911 pistol needs 57k poundals of force to do it's job.

Now to go one step further, and still assuming that the rate of acceleration is constant, to figure out the pressure we would convert the poundal back into pound-force and then divide it by the bullets's area that is being acted upon. So 57,000 poundals becomes 1,781.25 pounds force, divided by the surface area of the bullet (~.16 sq.in.), which would tell us that 57,000 poundals exerted on a .4515" diameter surface generates 11,133 PSI, yes?

So to figure out how much force is required to accelerate a 230 gr. bullet to 850 FPS in a 5 inch barrel, and for the purpose of this excercise, assuming the rate of acceleration is constant, rather than curved, we must first turn bullet weight in grains into a usable measurement, as 23/700ths of one pound. We then figure out how long it takes for the bullet to reach that speed, which is roughly .49 ms. This now gives us the rate of acceleration of 1,734,694 FPS per second.

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Yep, but you're off by a factor of two because you used the bullet's muzzle velocity rather than its average speed in the barrel. Since we're assuming that its acceleration was constant, its average speed was 850/2 = 425 FPS and it takes .98 msec to reach the muzzle. So the acceleration is 867694 FPS per second.

To figure out how many poundals are needed for this rate of acceleration, we now multiply the mass of 23/700ths of one pound by the rate of acceleration at 1,734,694, right? So if my math works, and if bullet acceleration was constant and environmental factors notwithstanding , the manstopper .45 hardball fired from an M1911 pistol needs 57k poundals of force to do it's job.

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You've got the concept right. Since the acceleration is really half of yours, the force on the bullet must be 28509 poundals.

Now to go one step further, and still assuming that the rate of acceleration is constant, to figure out the pressure we would convert the poundal back into pound-force and then divide it by the bullets's area that is being acted upon. So 57,000 poundals becomes 1,781.25 pounds force, divided by the surface area of the bullet (~.16 sq.in.), which would tell us that 57,000 poundals exerted on a .4515" diameter surface generates 11,133 PSI, yes?

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Yep, except that you're still off by two. It's 5537 PSI.

That may sound like an unreasonably small number, but consider two things. First, the pressures that are usually cited for cartridges are peak, not average. And our assumption that the acceleration, and therefore the pressure, was constant is unrealistic. In a real gun, the pressure reaches a peak very quickly and then drops quickly, because the chamber volume increases as the bullet moves down the barrel. So the average pressure is considerably less than the peak.

Thanks, Dave. I understand the averaging now. Force calculations are not something I've ever really used in hobby or career. Knew what the units were, but applying them always seemed more theoretical than practical, since you have sooooo many variables in real-world acceleration of objects. It would probably take me longer to calculate the pound-force required to run 12's in the quarter than it would to tune the car and drive it to the race track.

That may sound like an unreasonably small number, but consider two things. First, the pressures that are usually cited for cartridges are peak, not average. And our assumption that the acceleration, and therefore the pressure, was constant is unrealistic. In a real gun, the pressure reaches a peak very quickly and then drops quickly, because the chamber volume increases as the bullet moves down the barrel. So the average pressure is considerably less than the peak.

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I chose the .45 hardball to keep it gun related; I'm pretty well versed in internal ballistics and the pressure curves, which is why I was careful to denote that we were assuming constant acceleration only for the purpose of my little excercise.

I'm know if we were to examine the pressure curve and the true rate of acceleration using the first few microseconds of bullet travel, the rate of acceleration would be many, many times faster.

Hmmmmm..........I wonder what the acceleration looks like for a magnetic rail gun. I'll bet it's more of a straight line.

I fired some .45-70 out of a Thompson Center handgun and you could most certainly feel the torque on that bad boy. Ouch.


-Mark.