Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Clar on 16/12/2013 22:52:26
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There is some way to measure the force of bullet lunching by windcheater? I want to use a some kind of this force meters to it
But question is where I should strapped this force meters? Is it real to measure it by this device?
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Please append a version of this question in your native language as the English version seems to make no sense
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There is some way to measure the force of bullet lunching by windcheater? I want to use a some kind of this force meters to it
But question is where I should strapped this force meters? Is it real to measure it by this device?
I'm having a little difficulty myself understanding, is it possible you're referring to the make of rife: Winchester?
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The key to measure either recoil of a gun, or the force of impact of a bullet is to have high speed measurements. So, choose a scale system that can be computerized, and also includes recording software to capture high speed dynamic changes in force.
If you are measuring the recoil of a rifle, perhaps mount the rifle from above using swinging wires. Mount your scale or sensor to something solid such as a concrete wall. I would assume you would select some kind of a compression sensor rather than a hanging sensor.
Pistol recoil may be more complex as leverage may tend to cause the pistol to twist.
Obviously make sure your bullet trajectory is well controlled, protected from ricochet, and has a level of safety in the design in case all doesn't go as planned.
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There is some way to measure the force of bullet lunching by windcheater?
I don't understand this question at all. E.g. what is a windcheater?
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There is some way to measure the force of bullet lunching by windcheater?
I don't understand this question at all. E.g. what is a windcheater?
I think he meant Winchester, as in rifle.
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I don't understand this question at all. E.g. what is a windcheater?
It's a trademark for a windbreaker jacket.
I suspect that Clar meant "launching" rather than "lunching". I got a mental image of a jacket dining on bullets.
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Clifford almost got the answer. Assuming we mean "launched from a Winchester" and not eaten by a raincoat, the trick is to suspend the rifle with the barrel horizontal, from two wires of equal length (the longer the better at least 2 or 3 meters) and add a substantial mass, say 20 kg of lead. A long photographic exposure, or a tape measure, will show you how high the rifle rises as the recoil turns into a swinging motion.
Then if the total mass of the rifle plus ballast = M and the maximum height of the swing is h, the recoil energy was Mgh = 0.5 MV2 where V was the initial speed of the recoil.
Now if the mass of the bullet is m and its muzzle velocity was v we know that MV = mv.
and since the bullet accelerated from rest, v2 = 2as where s is the length of the barrel and a is the average acceleration
so a = (Mgh/m)2/2s
thus F = m x (Mgh/m)2/2s = (Mgh)2/2ms
with no sophisticated instrumentation required.
If you aren't happy about measuring the upswing of the ballasted rifle, you can get a rough estimate of F by ignoring air resistance and finding where the bullet hits the ground from a horizontal shot. I'll leave the arithmetic to you!
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Clifford almost got the answer. Assuming we mean "launched from a Winchester" and not eaten by a raincoat, the trick is to suspend the rifle with the barrel horizontal, from two wires of equal length (the longer the better at least 2 or 3 meters) and add a substantial mass, say 20 kg of lead. A long photographic exposure, or a tape measure, will show you how high the rifle rises as the recoil turns into a swinging motion.
Then if the total mass of the rifle plus ballast = M and the maximum height of the swing is h, the recoil energy was Mgh = 0.5 MV2 where V was the initial speed of the recoil.
Now if the mass of the bullet is m and its muzzle velocity was v we know that MV = mv.
and since the bullet accelerated from rest, v2 = 2as where s is the length of the barrel and a is the average acceleration
so a = (Mgh/m)2/2s
thus F = m x (Mgh/m)2/2s = (Mgh)2/2ms
with no sophisticated instrumentation required.
If you aren't happy about measuring the upswing of the ballasted rifle, you can get a rough estimate of F by ignoring air resistance and finding where the bullet hits the ground from a horizontal shot. I'll leave the arithmetic to you!
That method measures the momentum of the bullet and exhaust gas. Most ballistic pendulums do the same thing, except they measures how high a wood block pendulum swings after capturing the bullet. It's important to block the exhaust gas from the muzzle from pushing the wood block, and the pendulum must capture the bullet, not deflect it.
A more modern device to measure bullet velocity is the ballistic chronograph (http://"https://www.google.com/#q=ballistic+chronograph"). From the velocity and bullet mass, you can calculate momentum and energy.
The OP asked about force; that is not usually a concern. You can calculate the average force if you know the velocity and the time in the barrel. However, there is a time lag from the time the firing pin hits the primer and the time when the bullet begins to accelerate, so the measure is not accurate. You can assume constant acceleration/deceleration and calculate force based on the distance traveled during the acceleration or deceleration.
It is possible measure the instantaneous force pushing the bullet thru the barrel, using a special laser ruler (http://"https://www.google.com/#q=laser+ruler")which measures the position of the bullet every few microseconds in the barrel. This can be done by reflecting the laser off of a sacrificial mirror so it looks down the barrel. This kind of measure is helpful for optimizing the powder load when reloading ammo.
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Good point about the exhaust gas momentum. In my calculation m should be the combined mass of the projectile and projectile charge, since nearly all the gas leaves the barrel.
I still like the simplicity of estimating recoil force by simply measuring how far the bullet travels from a horizontal launch, and it occurs to me that you could instead fire it vertically and time its return to earth.
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time its return to earth
The speed of the bullet when falling will be much lower than the velocity when rising, so this would involve some careful measurement and modelling of wind resistance (as would measuring the horizontal range of the bullet).
Maybe we should just use "Maths Rules" and assume that the experiment is done in a vacuum?
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Or, you can just do what the OP asked and Clifford hinted.
Put the back end of the gun against a block fitted with strain gauges and measure the gauge output.
You will need a fairly fast data logger.
The net force on the bullet is going to be pretty much the same as the kick.
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I wonder if Clar who seems to have difficulty writing English if in fact that is his or her native language has sufficient skill in arithmetic to apply the formulae suggested.
F = m x (Mgh/m)2/2s = (Mgh)2/2ms
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Or, you can just do what the OP asked and Clifford hinted.
Put the back end of the gun against a block fitted with strain gauges and measure the gauge output.
You will need a fairly fast data logger.
The net force on the bullet is going to be pretty much the same as the kick.
You would need an extremely rigid linkage between the back of the barrel and the strain gauge, and the strain gauge itself would have to be extremely rigid. If any part of the gun accelerates backward at all, its mass times acceleration would take away from the force at the strain gauge. It's much easier to measure momentum than force in ballistics.
A nanosecond light pulse is 300 mm in length. To get 1 mm accuracy, your laser ruler will need to time the reflection with an accuracy of 2/300 ns; that's 6.67 ps. To get the necessary measurements, you will need a laser ruler that pulses for a few picoseconds once ever microsecond.
Rather than f = ma, it might be better to use the definition of force, f = dp/dt. That's the time rate of change of momentum.
Using the laser ruler, you get the distance traveled inside the barrel by the bullet at many instants of time. The change of distance between two measurements divided by the time interval is the average velocity for that interval. The change of velocity between two intervals times the mass the bullet is the change of the bullet's momentum. Change of momentum between successive intervals divided by the time interval is the force pushing against the bullet.
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Good point about the exhaust gas momentum. In my calculation m should be the combined mass of the projectile and projectile charge, since nearly all the gas leaves the barrel.
While the bullet is in the barrel, the gas and unburned powder expand in such a way that the front part is going as fast as the bullet and the rear part is going back as fast as the barrel. The momentum of the gas is therefor equal to ½(velocity of bullet + the velocity of barrel) times the mass of the gas. (Since the barrel's velocity is negative, it subtracts when you add it.)
The mass of the gas is equal to the mass of the burned powder before it burned. So if all the powder burns before the bullet reaches the muzzle, the exhaust gas will have the same mass as the powder load. Some ammo is loaded in such a way that the powder is still burning after the bullet has gone down range. It's hard to say whether the unburned powder stays at the back of barrel until it burns.
As soon as the bullet leaves the muzzle, the gas explodes in different ways, depending on the type of muzzle. A muzzle brake may actually send the gas rearward. Without a brake, the gas goes out ahead of the bullet.
So there are good reasons to fire the bullet into a block of wood hanging on strings, rather than measure the backward momentum of the gun. To get only the momentum of the bullet, fire it thru a small hole in a sheet of metal. That will block the gas from pushing the block.
Maybe Clar is wondering how much force the gun butt exerts against his shoulder (or hand). That could be measured by a force transduce between the butt and the shoulder. The quicker your shoulder stops the gun, the greater the force.
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You would need an extremely rigid linkage between the back of the barrel and the strain gauge, and the strain gauge itself would have to be extremely rigid. If any part of the gun accelerates backward at all,...
...then it would be pretty much the same as usually happens when someone fires a gun
The block holding the gun would, ideally, be as rigid as the shoulder that normally holds it. That way the measured value would correspond to the practical one.
You could do that with a small force gauge between the gun and the shoulder.
Just as an excercise in thought, imagine the gun starts off lying on a frictionless table in a vacuum chamber.
The gun fires , the bullet comes out one way, and the gun flies off in the opposite direction.
There's no friction and no air resistance.
So, is the force on the gun zero?
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This discussion brings up an important point: it's a lot more natural to talk about momentum and energy than it is to talk about force when dealing with problems like this. Energy and momentum are conserved, whereas the applied force can vary depending on the time/distance over which it is applied.