Does porting make you Lose Velocity? And do you think Porting hurts your Patterns? Thanks
Question for Neil Winston
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Mike, it's the math that explains the .50 BMG result you cite.
Using the same format as we did the shotgun example above,
The momentum of the bullet: 700 x 2800 = about two million. Again, no units, just two million.
The momentum of the powder 200 x 1400 = about a quarter milliion
The "jet effect" uses a multiplier of about 1.75 for rifles, except super high speed. So that gives us: 200 x 2800 x 1.75 = about a million.
To the accuracy we are working with here, that puts the powder momentum about 2/3 as large as the bullet momentum. And together, about 3 and a quarter milliion.
Going back to our ideal universe, the one where the powder goes in the opposite direction from the bullet. First we subtract a million and a quarter, since it's not recoiling toward the shooter, leaving just the 2 million, as we expect. Then we take off an additional one and a quarter, since it's going the other way, leaving us about three-quarters of a million. So the recoil reduction we've gotten in this ideal world where we can redirect all the powder force in the opposite direction is from 3.25 million to .75 million, or more than 75%.
And that's why a muzzle brake works so well with the .50 BMG; its math is ideal for the purpose.
When (if) I finally get my recoil- on- the -shoulder setup working right, I would indeed like to borrow your device. Maybe sooner, if I just settle on the the less-than-ideal one which seems to work some too. But if I took it now I wouldn't have any way to measure it, and so would never answer the question. But I will certainly keep you in mind and hope someday to be able to take you up on your fair offer.
Neil
Using the same format as we did the shotgun example above,
The momentum of the bullet: 700 x 2800 = about two million. Again, no units, just two million.
The momentum of the powder 200 x 1400 = about a quarter milliion
The "jet effect" uses a multiplier of about 1.75 for rifles, except super high speed. So that gives us: 200 x 2800 x 1.75 = about a million.
To the accuracy we are working with here, that puts the powder momentum about 2/3 as large as the bullet momentum. And together, about 3 and a quarter milliion.
Going back to our ideal universe, the one where the powder goes in the opposite direction from the bullet. First we subtract a million and a quarter, since it's not recoiling toward the shooter, leaving just the 2 million, as we expect. Then we take off an additional one and a quarter, since it's going the other way, leaving us about three-quarters of a million. So the recoil reduction we've gotten in this ideal world where we can redirect all the powder force in the opposite direction is from 3.25 million to .75 million, or more than 75%.
And that's why a muzzle brake works so well with the .50 BMG; its math is ideal for the purpose.
When (if) I finally get my recoil- on- the -shoulder setup working right, I would indeed like to borrow your device. Maybe sooner, if I just settle on the the less-than-ideal one which seems to work some too. But if I took it now I wouldn't have any way to measure it, and so would never answer the question. But I will certainly keep you in mind and hope someday to be able to take you up on your fair offer.
Neil
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Dutch, that's an extremely interesting point.
I think the fact that the muzzle is moving up and so imparting a vertical component to the horizontal movement we all mostly consider must happen, and - to throw out some guesses at numbers - a foot a second movement at the moment of exit will end up almost two inches at 40 yards. And sure, any force, such as ports, which slow that upward speed will lower the POI, just as you say.
Neil
I think the fact that the muzzle is moving up and so imparting a vertical component to the horizontal movement we all mostly consider must happen, and - to throw out some guesses at numbers - a foot a second movement at the moment of exit will end up almost two inches at 40 yards. And sure, any force, such as ports, which slow that upward speed will lower the POI, just as you say.
Neil
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AA, it'll take more than marketing - more even, if I dare say it, than Tron and Foggy.
Neil
Neil
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Andrew, I made several determintations of the POI of the MX-8 referred to above before I had it ported. The top barrel shot flat to 1/2 inch high, on the average, from 13 yards, off a rest, shooting light 7 1/2 Gold Medals.
Then I had it ported - more and bigger holes than many drill - and now, again in several tests, that same barrel shoots one to one-half inch low, off the same rest, with the same shells. I call that an inch change down, which is a 3-inch change at 40 yards.
I've not done the calculations, I take it they come out with smaller numbers. I'm pretty sure of my effect, and unless the pictures of the effect of porting are all fixed, then there is _something_ going on when you drill holes in barrels.
(aside here: what people do is put an LED on the end of the gun, then use a strobe and longer - 1/25 sec? - shutter speed in quite low light. The flash gets the shooter and gun, and so a scale, and the LED traces out the movement of the end of the gun. I saw one set of photos in Sproting Clays which looked - well - at least questionable, if you get my drift. But there have been others and they all tell besically the same story, that the muzzle lifts less with ports in it. Shooting I can't tell. )
By the way, do my powder-effect replies to Mike make any sense to you? Am I saying things in a way that anyone can understand?
Neil
Then I had it ported - more and bigger holes than many drill - and now, again in several tests, that same barrel shoots one to one-half inch low, off the same rest, with the same shells. I call that an inch change down, which is a 3-inch change at 40 yards.
I've not done the calculations, I take it they come out with smaller numbers. I'm pretty sure of my effect, and unless the pictures of the effect of porting are all fixed, then there is _something_ going on when you drill holes in barrels.
(aside here: what people do is put an LED on the end of the gun, then use a strobe and longer - 1/25 sec? - shutter speed in quite low light. The flash gets the shooter and gun, and so a scale, and the LED traces out the movement of the end of the gun. I saw one set of photos in Sproting Clays which looked - well - at least questionable, if you get my drift. But there have been others and they all tell besically the same story, that the muzzle lifts less with ports in it. Shooting I can't tell. )
By the way, do my powder-effect replies to Mike make any sense to you? Am I saying things in a way that anyone can understand?
Neil
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Not, Southpark, I didn't. I do all this mostly in late spring, summer and autumn late-afternoons. Just warm weather, mostly. I think that at 13 yards air density and so on are of no measureable effect. I know there is a poster here you says you can change POI with shot speed. When I tested Extra-lights and Handicaps at 25 yards, where you can still tell with some accuracy where the shot's center is, I was unable to show any reliable difference, the POI's just generally overlapped. So I don't worry about it. And besides, I redo everything at least twice, often more.
Neil
Neil
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Andrew, you have to factor the powder in twice. Once for its speed and mass, and the second time for its jet effect. As far as the rest, I can't see anything to explain the difference. It looks, if anything, like the effect should be less. We both agree, I suppose, that if you turn out to be right, then porting is totally ineffective at controlling muzzle jump. I have to say I just can't tell.
I've been wanting to do that photo test for years; looks like it's time. In the meanwhile, I'll work on the math another way - the unbalanced force method I wrote to Jimmy Borum about and see how that comes out.
And of couse, the acid test. 25 shots with one barrel, then port it, then retest. It's getting warm here, so that will probably be first.
Neil
I've been wanting to do that photo test for years; looks like it's time. In the meanwhile, I'll work on the math another way - the unbalanced force method I wrote to Jimmy Borum about and see how that comes out.
And of couse, the acid test. 25 shots with one barrel, then port it, then retest. It's getting warm here, so that will probably be first.
Neil
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(thinking)
Neil
Neil
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Interesting deal you have there Mike. How do you keep it on the gun? And here's the critical question - have you ever tried it backwards?
You've got me interested in a test now. A comparison of the results frontwards and backwards would let you nail down the effect exactly by just halving the difference. What a great thing you have made there, Mike!
Neil
You've got me interested in a test now. A comparison of the results frontwards and backwards would let you nail down the effect exactly by just halving the difference. What a great thing you have made there, Mike!
Neil
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And to think all I ask was does porting effect speed and pattern!!!!!
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Hmm, Andrew. Using a kind of flaky momentum analysis, I get just what you did less than a tenth of the predicted 1/10 inch rise. It appears that using the weights, speeds, corrections and so on we've used so far, gasses will never get the job done. Time for "real world" testing, that's for sure. We'll just have to wait and see how that comes out. I'm pretty certain of replication, especially based on the photos I've seem.
But now, looking into a possible future, let's say it all falls apart, that the muzzle is in essentially the same place when the shot exits, porting or no. I've always thought "If porting works, the shot must be placed lower, based on the idea that if the muzzle-rise is not (partially) counteracted by the time the shot exits, there's no reason it should happen later." If, in fact, it turns out that POI depression doesn't happen or haqppens, as both our maths are telling us, only to the degree of 1/100 inch, doesn't this necessarily mean that porting is a complete zero, muzzle-lift-wise?
Neil
But now, looking into a possible future, let's say it all falls apart, that the muzzle is in essentially the same place when the shot exits, porting or no. I've always thought "If porting works, the shot must be placed lower, based on the idea that if the muzzle-rise is not (partially) counteracted by the time the shot exits, there's no reason it should happen later." If, in fact, it turns out that POI depression doesn't happen or haqppens, as both our maths are telling us, only to the degree of 1/100 inch, doesn't this necessarily mean that porting is a complete zero, muzzle-lift-wise?
Neil
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Andrew, you wrote
"Neil, Regarding your second paragraph suggesting that if porting does not affect POI then this indicates "no useful muzzle lift reduction", I don't necessarily agree with this reasoning. Think of the main recoil backwards (or muzzle lift upwards) sans porting; the recoil SPEED is initiated and increased while the shot goes down the bore, but recoil MOVEMENT continues for some (relatively long) time after the shot leaves the bore until it is fully arrested by the shooter."
Is this what you had in mind?
(edit here. There was a graph confirming Andrew's speculation, but I have to to take it down now. If you just read Andrew's quoted post, you'll know what it showed anyway.)
Neil
"Neil, Regarding your second paragraph suggesting that if porting does not affect POI then this indicates "no useful muzzle lift reduction", I don't necessarily agree with this reasoning. Think of the main recoil backwards (or muzzle lift upwards) sans porting; the recoil SPEED is initiated and increased while the shot goes down the bore, but recoil MOVEMENT continues for some (relatively long) time after the shot leaves the bore until it is fully arrested by the shooter."
Is this what you had in mind?
(edit here. There was a graph confirming Andrew's speculation, but I have to to take it down now. If you just read Andrew's quoted post, you'll know what it showed anyway.)
Neil
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Tim, figure between milliseconds three and four, favoring four. You see what a graph like that does to slow powder theories of recoil, don't you?
Neil
Neil
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I'll email one this evening, Andrew. No, I've never felt any difference with ports v. no ports.
Remember, the transducer is one inch forward of the shellhead, and I'm told if I want to measure pressure near the muzzle, that's where I have to measure it.
Neil
Remember, the transducer is one inch forward of the shellhead, and I'm told if I want to measure pressure near the muzzle, that's where I have to measure it.
Neil
Okay, I wrote this up before I read today's postings, but I'll throw it out anyway, just to add to what Neil and Andrew seem to have more or less settled on.
I am going to take a stab at this problem using a little different approach. The porting effect must be less than the muzzle jump; otherwise it would not just cancel out the muzzle jump but would force the muzzle down below horizontal. So how much is the muzzle jump?
I will calculate the angular momentum imparted by a 1 ounce shot charge leaving the barrel at 1200 fps, then figure out the rotational velocity of the gun using conservation of angular momentum.
If the path of the projectile is directly in line with the center of mass of the gun, there is no rotation. The angular momentum is proportional to the distance from the center of mass to the bore centerline.
L=r*m*v
Where L is the angular momentum, r is the distance from the center of mass, m is the mass of the shot charge, and v is the muzzle velocity.
My 870 has a center of gravity maybe about an inch below the bore centerline so I will use 1/12 of a foot for r. This gives an angular momentum of 1 oz. *1200 fps/12 = 100 oz-ft^2 per second.
Assume the gun is 4 feet long and has a moment of inertia approximately like a uniform cylindrical rod. The formula for the moment of inertia of a uniform cylindrical rod rotating about the mid point is
I = M l^2/12
Where l is the length of the gun (4 feet) and M is 8 lb = 128 oz. This gives a moment of inertia I= 128*16/12 = 170.6 oz-ft^2
The formula for angular momentume is
L = I* omega
Where L is the angular momentum, I is the moment of inertia, and omega is the rotational speed in radians per second. Pi radians is equal to 180 degrees.
Setting the angular momentum of the gun equal to that of the projectile,
I * omega = 100
Omega = 100 oz-ft^2 per second/170.6 oz ft^2 = 0.6 radians per second.
This is the final rotational velocity. Now to find out the angle the gun has rotated when the barrel is at the end of the muzzle, we need to find out how long the projectile takes to leave the barrel. If the projectile were going 1200 fps the majority of the time it is in the barrel, and the barrel is 30 inches (2.5 feet) long the time is 2.5/1200 = 0.002 seconds. Then the angle of rotation at the moment of exit is 0.6 radians per second times 0.002 = 0.0012 radians per second *180 degree/pi radians = .068 degrees which is equal to 60*.068 = 4 minutes of angle. As we all know 4 minutes of angle is 4 inches at 100 yards, or 2 inches at 50 yards. This is the increase in POI due to muzzle jump. I would expect the porting effect to be significantly less.
Conclusion: Neil needs to redo his test.
I am going to take a stab at this problem using a little different approach. The porting effect must be less than the muzzle jump; otherwise it would not just cancel out the muzzle jump but would force the muzzle down below horizontal. So how much is the muzzle jump?
I will calculate the angular momentum imparted by a 1 ounce shot charge leaving the barrel at 1200 fps, then figure out the rotational velocity of the gun using conservation of angular momentum.
If the path of the projectile is directly in line with the center of mass of the gun, there is no rotation. The angular momentum is proportional to the distance from the center of mass to the bore centerline.
L=r*m*v
Where L is the angular momentum, r is the distance from the center of mass, m is the mass of the shot charge, and v is the muzzle velocity.
My 870 has a center of gravity maybe about an inch below the bore centerline so I will use 1/12 of a foot for r. This gives an angular momentum of 1 oz. *1200 fps/12 = 100 oz-ft^2 per second.
Assume the gun is 4 feet long and has a moment of inertia approximately like a uniform cylindrical rod. The formula for the moment of inertia of a uniform cylindrical rod rotating about the mid point is
I = M l^2/12
Where l is the length of the gun (4 feet) and M is 8 lb = 128 oz. This gives a moment of inertia I= 128*16/12 = 170.6 oz-ft^2
The formula for angular momentume is
L = I* omega
Where L is the angular momentum, I is the moment of inertia, and omega is the rotational speed in radians per second. Pi radians is equal to 180 degrees.
Setting the angular momentum of the gun equal to that of the projectile,
I * omega = 100
Omega = 100 oz-ft^2 per second/170.6 oz ft^2 = 0.6 radians per second.
This is the final rotational velocity. Now to find out the angle the gun has rotated when the barrel is at the end of the muzzle, we need to find out how long the projectile takes to leave the barrel. If the projectile were going 1200 fps the majority of the time it is in the barrel, and the barrel is 30 inches (2.5 feet) long the time is 2.5/1200 = 0.002 seconds. Then the angle of rotation at the moment of exit is 0.6 radians per second times 0.002 = 0.0012 radians per second *180 degree/pi radians = .068 degrees which is equal to 60*.068 = 4 minutes of angle. As we all know 4 minutes of angle is 4 inches at 100 yards, or 2 inches at 50 yards. This is the increase in POI due to muzzle jump. I would expect the porting effect to be significantly less.
Conclusion: Neil needs to redo his test.
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