9mm bullets, guns, and Physics question...

[Cuz]

The fact that I wanted to make PF in 3 different pistols is what got this thought process started.


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[Wesquire]

On 8/4/2020 at 4:27 PM, MikeBurgess said:

real physics answer is all the above are wrong because the barrel and slide assembly move, so your felt recoil is a combination of the recoil spring pushing the gun back into your hand and the barrel stopping on the locking block and the slide bottoming out on the frame and the recoil spring pushing the slide forward and the slide loosing energy as it strips a round from the magazine and the barrel going into battery and stopping the slide. 

 

the chances that you could fire a round from one gun with its combination of springs and weights and losses and then fire a round in a different gun with a different combination of the above and feel a meaningful difference that is directly related to a 5.5% change in PF is as close to zero as you can get

 

We are talking about if everything else was kept equal. You will get more recoil out of the same ammo in a longer barrel than a short one.

[Wesquire]

5 hours ago, belus said:

 

I'm not a physicist or mechanical engineer, and never even took a dynamics class in college which would have given me the real tools to discuss this.

But aren't you confusing Force with Energy and Impulse, ie the integral of Force over Distance and Time respectively?

 

No. I'm just not ignoring the additional force that happens after the first half of the barrel. The rate of acceleration slows, but the total force still goes up.

[belus]

3 minutes ago, Wesquire said:

No. I'm just not ignoring the additional force that happens after the first half of the barrel. The rate of acceleration slows, but the total force still goes up.

In that case where does this additional force come from?

The pressure is decreasing as the bullet moves down the barrel and the area of bullet is constant, but F = P*A so the force decreases once the bullet starts moving.  That's two separate lines of reasoning: peak acceleration and peak pressure that imply the force is at a maximum shortly after the bullet leaves the case but long before it leaves the barrel.

I still think you're confused about the difference between an accumulated quantity, either Energy (ft-lbs or whatever units you want) and Impluse, and an instantaneous one like Force.

[Cuz]

For the record, with a timer, I'm always just a little faster with the longer slide on the G34.

 

[George16]

On 8/3/2020 at 8:52 PM, Wesquire said:

 

You are misunderstanding this. For example, let's say a 100gr bullet gets up to 750fps when it is half way down the barrel. Then in the other half it still accelerates to 1000fps. The peak happened in the first half, but the TOTAL force is still higher by the time the bullet exits the muzzle than it was at the half way point.

 

Force is redundant and is derived by the rate of change in momentum. It is also not conserved like momentum is. Newton's second law is about the rate of change in momentum. Additionally, Newton's second has also been supplanted by the momentum principle  because Newton's second breaks down when an object approaches the speed of light

 

It doesn’t work that way when you’re calculating for force. Acceleration is the rate of change of velocity over a set period of time. As such, you first have to determine the acceleration by subtracting  the initial velocity from the final velocity  divided by the initial time from the final time as stated by this formula:

a=v(f)−v(i) divided by t(f)−t(i) wherein a is acceleration, vf is final velocity, vi is initial velocity, tf is final time and ti is initial time.  

 

750 going through the barrel is not even part of your velocity equation (if you say that your 1000 FPS is your final velocity). Peak happens at the highest speed the bullet achieved. Peak doesn’t happen at 750 or first half or any increase in speed. As seen on the formula I wrote above, you have to subtract your initial velocity (zero FPS with bullet in chamber) and final velocity (1000 FPS or whatever you got from your chrono). Once you have the difference of the two, you divide it with the difference on final and initial time it took to accelerate from zero seconds to however it took to reach 1000. 

So if it takes 5 seconds to go from zero to 1000 FPS, then acceleration is 1000/5 which equals to 200 feet/second square ( in truth, everything should be converted to metric system since  Acceleration is measured in meters/second square, but for simplicity, I’ll stay with English the  system) Then you take this 200 ft/second square and multiple it with your mass (100 Gr) to get the force.

 Remember, this equation is only for straight line acceleration. There are other equations for angular and centripetal acceleration.

 

Now back to our regular programming  

[BoyGlock]

This is my un-engineering take: whenever I see the reddish flame spouted out of the muzzle while firing, and see my iron sights still aligned  and the flame in the background, it says recoil starts/occurs after the bullet exits the muzzle. So analizing recoil w/ bullet movement inside the barrel does not agree w/ my range experience.  

[superdude]

1 hour ago, BoyGlock said:

This is my un-engineering take: whenever I see the reddish flame spouted out of the muzzle while firing, and see my iron sights still aligned  and the flame in the background, it says recoil starts/occurs after the bullet exits the muzzle.  

 

It might look that way, but it doesn't work that way. Recoil starts as soon as the bullet starts moving forward.

[Wesquire]

3 hours ago, belus said:

In that case where does this additional force come from?

The pressure is decreasing as the bullet moves down the barrel and the area of bullet is constant, but F = P*A so the force decreases once the bullet starts moving.  That's two separate lines of reasoning: peak acceleration and peak pressure that imply the force is at a maximum shortly after the bullet leaves the case but long before it leaves the barrel.

I still think you're confused about the difference between an accumulated quantity, either Energy (ft-lbs or whatever units you want) and Impluse, and an instantaneous one like Force.

 

Force is not instantaneous. Say the 100gr bullet gets to 1000 fps in the first half of the barrel and it takes 1 second. The F would be equal to 100gr x 1000fps^2. However, in the second half of the barrel it accelerates further to 1200 fps and it takes .5 seconds (simplifying). That's an additional F equal to 100gr x 400fps^2. 

 

You don't just look at what the maximum interval of force was. The powder is still burning for the full duration the bullet is in the barrel. IDK why you would think the force would be instantaneous.

Edited August 8, 2020 by Wesquire

[Wesquire]

2 hours ago, George16 said:

 

It doesn’t work that way when you’re calculating for force. Acceleration is the rate of change of velocity over a set period of time. As such, you first have to determine the acceleration by subtracting  the initial velocity from the final velocity  divided by the initial time from the final time as stated by this formula:

a=v(f)−v(i) divided by t(f)−t(i) wherein a is acceleration, vf is final velocity, vi is initial velocity, tf is final time and ti is initial time.  

 

750 going through the barrel is not even part of your velocity equation (if you say that your 1000 FPS is your final velocity). Peak happens at the highest speed the bullet achieved. Peak doesn’t happen at 750 or first half or any increase in speed. As seen on the formula I wrote above, you have to subtract your initial velocity (zero FPS with bullet in chamber) and final velocity (1000 FPS or whatever you got from your chrono). Once you have the difference of the two, you divide it with the difference on final and initial time it took to accelerate from zero seconds to however it took to reach 1000. 

So if it takes 5 seconds to go from zero to 1000 FPS, then acceleration is 1000/5 which equals to 200 feet/second square ( in truth, everything should be converted to metric system since  Acceleration is measured in meters/second square, but for simplicity, I’ll stay with English the  system) Then you take this 200 ft/second square and multiple it with your mass (100 Gr) to get the force.

 Remember, this equation is only for straight line acceleration. There are other equations for angular and centripetal acceleration.

 

Now back to our regular programming  

 

This is the point I've been making. Other people are trying to break it down further from just the muzzle velocity and look at just the segment within the barrel in which the bullet accelerates fastest.

[Wesquire]

Bottom line, F = m*a. The bullet still has mass and is still accelerating for the entire length of the barrel. Ergo, the total force increases for the full duration the bullet is in the barrel. This should not be controversial.

[BoyGlock]

9 minutes ago, superdude said:

 

It might look that way, but it doesn't work that way. Recoil starts as soon as the bullet starts moving forward.

True. But our recoil as felt and as seen can matter more than physics. Specially in the way we shoot. Im also an engr. and appreciate the technicalities above. But I prefer the more practical view than the physics of it contrary to the Op. 

[belus]

10 hours ago, Wesquire said:

 

Force is not instantaneous. Say the 100gr bullet gets to 1000 fps in the first half of the barrel and it takes 1 second. The F would be equal to 100gr x 1000fps^2. However, in the second half of the barrel it accelerates further to 1200 fps and it takes .5 seconds (simplifying). That's an additional F equal to 100gr x 400fps^2. 

 

You don't just look at what the maximum interval of force was. The powder is still burning for the full duration the bullet is in the barrel. IDK why you would think the force would be instantaneous.

 

Its tough to explain things to a lawyer who is already convinced of their position. I know because I'm married to an attorney.
Instantaneous was maybe a confusing word choice, but I was trying to draw some contrast between the concepts of Energy and Force without invoking calculus.

If F=ma, why are you squaring velocity? m*V^2 gives you an energy not a force. It might look like you have mass * length / seconds^2 (force) in your example, but you actually have mass * length^2 / seconds^2 which is energy, and those are not the same. Energy increases as the bullet continues to gain velocity in the later half of the barrel, but the force/pressure/acceleration does not.

Edited August 8, 2020 by belus

[Wesquire]

1 hour ago, belus said:

 

Its tough to explain things to a lawyer who is already convinced of their position. I know because I'm married to an attorney.
Instantaneous was maybe a confusing word choice, but I was trying to draw some contrast between the concepts of Energy and Force without invoking calculus.

If F=ma, why are you squaring velocity? m*V^2 gives you an energy not a force. It might look like you have mass * length / seconds^2 (force) in your example, but you actually have mass * length^2 / seconds^2 which is energy, and those are not the same. Energy increases as the bullet continues to gain velocity in the later half of the barrel, but the force/pressure/acceleration does not.

 

My bad, I was trying to notate acceleration as fps per second. Not to square the velocity. 

 

I'm not saying that the force in the second half of the barrel is higher than the first half. I'm saying that only looking at the first half because it is highest in that interval is wrong. The bullet still accelerates (albeit at a slower rate) in the second half, ergo additional force to what happened previously. Furthermore, my point is that force is not the relevant measurement here. I think you'd agree with that. Momentum is what dictates recoil.

Edited August 8, 2020 by Wesquire

[George16]

8 minutes ago, Wesquire said:

 

My bad, I was trying to annotate acceleration as fps per second. Not to square the velocity. 

 

I'm not saying that the force in the second half of the barrel is higher than the first half. I'm saying that only looking at the first half because it is highest in that interval is wrong. The bullet still accelerates (albeit at a slower rate) in the second half, ergo additional to what happened previously. Furthermore, my point is that force is not the relevant measurement here. I think you'd agree with that. Momentum is what dictates recoil.

Actually, force is greater in the second half of the barrel because the acceleration is greater. The velocity of the bullet is higher after a given time from its initial position which is zero. 

 

Now, you’re agreeing to what majority of the posters is talking that the higher PF, the greater the recoil when you say momentum is what dictates recoil. 
 

Here is the formula for momentum:

P = MV where P is momentum, M is mass and V is velocity

 

Here’s the formula for power factor:

PF = weight of bullet (mass) x velocity (FPS) divided by 1000

 

If you look at those two equations, they are the same except for dividing 1000 from the product of mass and velocity.

 

With that said, the higher the PF, the higher the momentum which will result in a stronger recoil compared to a lower PF ammo as long as everything else is equal (gun weight, length etc).

 

 

[Wesquire]

5 minutes ago, George16 said:

Actually, force is greater in the second half of the barrel because the acceleration is greater. The velocity of the bullet is higher after a given time from its initial position which is zero. 

 

Now, you’re agreeing to what majority of the posters is talking that the higher PF, the greater the recoil when you say momentum is what dictates recoil. 
 

Here is the formula for momentum:

P = MV where P is momentum, M is mass and V is velocity

 

Here’s the formula for power factor:

PF = weight of bullet (mass) x velocity (FPS) divided by 1000

 

If you look at those two equations, they are the same except for dividing 1000 from the product of mass and velocity.

 

With that said, the higher the PF, the higher the momentum which will result in a stronger recoil compared to a lower PF ammo as long as everything else is equal (gun weight, length etc).

 

 

 

If you are wanting to segment into the first half and second half like some posters are doing, then the force in the second half is lower because you start from the velocity at the end of the first half, not from zero. However, I don't see any reason to segment it like this. 

 

I have said momentum is the relevant metric from the beginning. I'm not who brought force up. I've only been discussing force because someone made the claim that the force is equal between long and short barrels because peak acceleration happens early.

Edited August 8, 2020 by Wesquire

[George16]

8 minutes ago, Wesquire said:

 

If you are wanting to segment into the first half and second half like some posters are doing, then the force in the second half is lower because you start from the velocity at the end of the first half, not from zero. However, I don't see any reason to segment it like this. 

 

I have said momentum is the relevant metric from the beginning. I'm not who brought force up. I've only been discussing force because someone made the claim that the force is equal between long and short barrels because peak acceleration happens early.

I segmented it like that based on your post. Otherwise, acceleration will always be higher after the bullet leaves the barrel because of the increase in velocity. Velocity will always be higher outside the barrel due to the friction encounter by the bullet as it travels trough the barrel’s rifling. 

 

The force will be different between short and long barrels due to the different rates of acceleration over a given period of time.

[Wesquire]

5 minutes ago, George16 said:

I segmented it like that based on your post. Otherwise, acceleration will always be higher after the bullet leaves the barrel because of the increase in velocity. Velocity will always be higher outside the barrel due to the friction encounter by the bullet as it travels trough the barrel’s rifling. 

 

The force will be different between short and long barrels due to the different rates of acceleration over a given period of time.

 

And I segmented it in response to previous posts segmenting it.

 

Yes, the force will be different between short and long barrels. That's been my point.

 

Seems like we are in agreement.

[George16]

1 minute ago, Wesquire said:

 

And I segmented it in response to previous posts segmenting it.

 

Yes, the force will be different between short and long barrels. That's been my point.

 

Seems like we are in agreement.

Yes we are now that you explained yourself clearly. Your previous posts wherein @belus responded to were not succinctly explained, hence some members responded to them asking for clarifications.

[Umbrarian]

14 hours ago, Wesquire said:

 

Force is not instantaneous. Say the 100gr bullet gets to 1000 fps in the first half of the barrel and it takes 1 second. The F would be equal to 100gr x 1000fps^2. However, in the second half of the barrel it accelerates further to 1200 fps and it takes .5 seconds (simplifying). That's an additional F equal to 100gr x 400fps^2. 

 

This is the equation to use f=ma

 

This is what you are doing f = (100*1000) + (100*400)

which is f = ma1 + ma2 and is incorrect.

[Umbrarian]

2 hours ago, George16 said:

Actually, force is greater in the second half of the barrel because the acceleration is greater.

 

In almost every round made, the acceleration is greatest in the first 1/2/3 inches, after that it is declining rapidly.

 

This is explained well in paper that was posted. He even provided graphs,:

[Wesquire]

1 hour ago, Umbrarian said:

This is the equation to use f=ma

 

This is what you are doing f = (100*1000) + (100*400)

which is f = ma1 + ma2 and is incorrect.

 

No. That is what you are doing. I'm saying that it should just be f=ma and you use the muzzle velocity, not an arbitrary point inside the barrel where acceleration peaks.

[Umbrarian]

1 minute ago, Wesquire said:

 

No. That is what you are doing. I'm saying that it should just be f=ma and you use the muzzle velocity, not an arbitrary point inside the barrel where acceleration peaks.

 

Muzzle velocity is not acceleration.

 

It is not an arbitrrary point, it is the point where a is greatest, thus f must be greatest since m does not change.

 

Did you read the paper? It is a peer reviewed journal, I get not believing me, but the reputation of the journal is well established. it supports everything I have written.

 

Here is the graph again. Note Peak Force occurs early in the cycle. Note it occurs at a specific instant (arbitrary according to you). Note once Peak Force is obtained, Force does not get higher. I am not trying to be argumentative, but you are entirely wrong on this.

 

 

 

 

 

 

[Wesquire]

14 minutes ago, Umbrarian said:

 

Muzzle velocity is not acceleration.

 

It is not an arbitrrary point, it is the point where a is greatest, thus f must be greatest since m does not change.

 

Did you read the paper? It is a peer reviewed journal, I get not believing me, but the reputation of the journal is well established. it supports everything I have written.

 

Here is the graph again. Note Peak Force occurs early in the cycle. Note it occurs at a specific instant (arbitrary according to you). Note once Peak Force is obtained, Force does not get higher. I am not trying to be argumentative, but you are entirely wrong on this.

 

 

 

 

 

 

 

Yes, obviously it would be muzzle velocity divided by the time.

 

Nobody has denied that peak force happens then. The point is that the total force is not limited to just the peak force.

[MikeBurgess]

And once again everyone reverted to non reciprocating locked breach calculations.

remember as the bullet accelerates the barrel ans slide move backwards, how that recoil impulse gets driven into the barrel slide assembly makes almost no difference to what you feel as recoil.

first you feel the recoil spring being compressed then you feel the impact of the slide hitting the frame then the spring pushing the slide back 

the biggest difference you will feel is how hard the slide hits the frame if you need proof watch some slow motion video of someone shooting and you will see the movement of the frame in the hands is pretty minimal as the slide cycles to the rear until the slide bottoms out then you will see comparatively lots of movement.

 

so calculating the recoil forces that take place in the fraction of time the bullet is in the barrel is pretty close to useless for answering the question posed, all you can learn is what the slide velocity should be.