Wesquire
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Posts posted by Wesquire
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1 hour ago, AcePrater said:
Very interesting. I've got a PCC and some czs to load for. Are you running it in competition? If so you may be limited to the certian weights of bullets if you load for major.
I would think that you can either try to load hotter with slower powder to work the comp - like a 9mm open gun,OR
Try a light PF load with 147s - but note that wont work the comp as much.
Normally I run my CZ 124 minor loads through the PCC, but I just worked up to open major with a comp so now I'm wondering what that would be like in the Ruger PCC.
The difference in the effect on the comp is almost completely negligible. It is a completely different beast than open guns (like 3-4 times less pressure at the muzzle than in a pistol). I'm pretty sure Max even suggests to tune your load without the comp on the gun. The comp just lets you relax a tad more. -
The lighter bullet, the better. 124gr was a significant improvement over 147ge. 105gr was a significant improvement over 124gr. It isn't about gas working the comp, it is about a quicker impulse.
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On 9/1/2020 at 7:41 PM, Cuz said:
Bumping this to see if there’s any new info bin looking to upgrade the trigger in my Ruger PCC and notice there seem to be a few more options available now. For those of you Shooting the Ruger, what are you doing about a trigger upgrade?
Sent from my iPhone using TapatalkGet the Mcarbo metal trigger, polish everything really well, and get the wolff reduced power sear spring and return spring. This gets it as nice as it will probably get until/if volq improves their trigger
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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I'm curious what effect I should expect when experimenting with different recoil spring lengths/weights in my Ruger PCC. Also, how analogous would this be to AR buffer springs?
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This isn't going anywhere. Recoil calculators are readily available online if anyone is actually curious.
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14 minutes ago, Umbrarian said:
f= ma, m is constant. Thus when a is peak, f is peak. It cannot go up after reaching peak, as that then would be peak.
I do not see how momentum matters. You have Newtons' Second Law -> f=ma, and you have his Third Law -> equal and opposite. So the f number is what it is. Now felt recoil, that is a different and things like gun mass, spring weight, force duration, etc. all factor in.
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.
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2 minutes ago, Umbrarian said:
Since f=ma, and m is constant, f cannot get any higher then when acceleration is at peak. Since peak acceleration occurs early, Longer BBL lengths after that do not matter wrt to f.
Except peak force is just one aspect. The total force still goes up after peak force. And again, force isn't even the most relevant metric here, momentum is.
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Just now, Dan Sierpina said:
I would look at the lubricity of the two, oh, if the extensions need a bit of tweeking to match your individual mag for smooth feeding, I'll take Delrin every day.
That and the weight are why I ended up getting a delrin extension for my big stick. However, I went with a brass basepad for the mag I use when I need a mag change due to it being heavier (for ejection) and stronger.
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1 minute ago, Dan Sierpina said:
Delrin is a very tough plastic. Maybe if you dropped a loaded mag from a 2nd story onto concrete, it may.
I don't know what the best metric would be, but I'm seeing a tensile strength of 11k psi for delrin and 27k for aluminum
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Would delrin not break more easily when dropped?
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38 minutes ago, Cuz said:
If the maximum force is applied before the bullet is half way down the barrel, wouldn’t the longer barrel produce lower velocities?
The bullet has to still be increasing speed during the last half inch of the longer barrel.
Sent from my iPhone using TapatalkNo, he is right that maximum force happens relatively early. It is just that total force is still higher with a longer barrel (assuming everything else is constant).
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13 minutes ago, Umbrarian said:
Yes it is true.
Peak acceleration, and thus peak force, is achieved within the first few centimeters of bullet movement. After that, pressure is dropping fast, and while it is still accelerating, it is accelerating at a decreasing rate.
Except you didn't just say peak force.
"So shortening the BBL will lower the PF, but not the force."
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I haven't see anything crazy at a match. However, one time I was at a public range and an obviously inexperienced lady shooter had brass eject into her shirt. She jumped up and down in a circle and swept everyone while her finger was still on the trigger. She was even laughing about it and had no idea she was doing anything dangerous.
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Given the same weight, the longer barrel/faster bullet gun will have more recoil because momentum is conserved.
3 hours ago, Umbrarian said:From a physics point of view (f=ma), the maximum f has already been achieved before the bullet is halfway down the barrel. So shortening the BBL will lower the PF, but not the force.
This is not true because the bullet is obviously still accelerating, otherwise the muzzle velocity would be identical. And beyond that, force is not what is relevant here.
Load development for Ruger PC Carbine / Charger
in 9mm/38 Caliber
Posted
Also, a 16" barrel has 4 times the volume of a 4" barrel. This means the gas will already be dispersed over 4 times the area, meaning 4 times less pressure before even factoring in mechanical differences.