1290 vs 1145 ?
Ok , I know all the guru's out there will jump all over my butt for this ?
I normally break targets @ 30-32 yds. When I change to 1 oz loads @ 1290 my hits seem to be a little lower--big pieces going up. Should I decrease the POI or is this just my imagination"
Phil Berkowitz
I normally break targets @ 30-32 yds. When I change to 1 oz loads @ 1290 my hits seem to be a little lower--big pieces going up. Should I decrease the POI or is this just my imagination"
Phil Berkowitz
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Another question for Phil&jhoward.
If you were to bore sight your shotguns ((exact bore centerline)), use that pre-determined mark at 10-20 or 30 yards, where would the center of the patterns strike in relation to that new aiming spot? Both fast and slow loads?
Hap
If you were to bore sight your shotguns ((exact bore centerline)), use that pre-determined mark at 10-20 or 30 yards, where would the center of the patterns strike in relation to that new aiming spot? Both fast and slow loads?
Hap
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You're shooting a faster shell at a rising target. If you are using the same sight picture, it should be hitting the target a little higher on the target. I would aim about half an inch lower on the target. Or maybe just not worry about it. I heard someone once say, you should be able to break 16 yard targets with rocksalt. Mark
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Put it on the pattern board.
All you math masters, correct me if I am wrong.
Average of 30-32 yards is 31 yards or 93 feet.
A 1290 fps load is traveling at 1,290 feet per second
An 1145 fps load is traveling at 1,145 feet per second
If you divide 93 (feet) by 1,190 you get .072 seconds for the shot to travel the 93 feet.
If you divide 93 by 1145 you get .081 seconds for the shot to travel the 93 feet.
.081 seconds - .072 seconds = .009 seconds between the speeds of the 2 loads to travel the 93 feet.
I find it hard to believe that 9/1000 of a second makes much difference.
Average of 30-32 yards is 31 yards or 93 feet.
A 1290 fps load is traveling at 1,290 feet per second
An 1145 fps load is traveling at 1,145 feet per second
If you divide 93 (feet) by 1,190 you get .072 seconds for the shot to travel the 93 feet.
If you divide 93 by 1145 you get .081 seconds for the shot to travel the 93 feet.
.081 seconds - .072 seconds = .009 seconds between the speeds of the 2 loads to travel the 93 feet.
I find it hard to believe that 9/1000 of a second makes much difference.
Phil, try and consider this possibility. Due to increased recoil to face, your pulling the gun down with forearm hand. Just something to think on. If so you, theoretically you would raise not lower poi. But you may just jerk harder and negate the change.
BigBad, wouldn't wind resistance play a role?
BigBad, wouldn't wind resistance play a role?
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Anyone here can say yes or no, but it's a fact... if you're shooting a pistol, say a .357 magnum, or even a 45 acp 1911, a top velocity load will ALWAYS shoot a little lower than a slow load at pistol distances, which for the sake of argument, may be 25 yards. It's just that way.
If your shooting at 25 yards and think... "my current load is a little high. I'll go slower and the bullet will hit lower" doesn't work. The slower shot will be even higher. A high velocity round will impact somewhat lower. Strange, but true.
Does that apply to shotguns? I really don't know. But it may be so at a certain distance.
If your shooting at 25 yards and think... "my current load is a little high. I'll go slower and the bullet will hit lower" doesn't work. The slower shot will be even higher. A high velocity round will impact somewhat lower. Strange, but true.
Does that apply to shotguns? I really don't know. But it may be so at a certain distance.
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Set the trap on straight away, go to post 3, adjust POI until you are centering targets to your satisfaction. Tom
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What Tom said.
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The 1290fps load will shoot a little higher than the 1145 due to the shot getting to the target quicker.
Gary Bryant
Dr.longshot
Gary Bryant
Dr.longshot
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Is anyone paying attention to the original post about why Phil thinks the faster shot is shooting a little lower? Do you really think that what he has cited as the reason for his belief makes any sense after the Target Break Reading Cyber Challenge and the demonstration of everyone's total inability to actually read target breaks for information about where the main cloud of the shot is?
Think about it. How is the target supposed to know where the pellets that don't hit it are and so be able to respond "in conformance to theory?"
Neil
Think about it. How is the target supposed to know where the pellets that don't hit it are and so be able to respond "in conformance to theory?"
Neil
OK, I will take the bait. A faster load shoots lower because it is out of the barrel quicker and is less affected by recoil motion. Recoil motion begins at detonation and the gun starts to rotate upward around your shoulder because of the design of the stock. ShooterR is 100% correct and I believe this handgun fact applies to any gun.
BigBad Bob, while you have your calculator warmed up, how far did the bird move vertically in that 9/1000"s of a second?
BigBad Bob, while you have your calculator warmed up, how far did the bird move vertically in that 9/1000"s of a second?
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wrd, there are two phases of recoil according to John Brindle. In the first phase, the gun rotates around its center of gravity. The one you are talking about is the second phase and that's a lot later, though he doesn't say how much. Probably it doesn’t amount to much until long after the shot is gone.
Though BBB didn't account for the drop in speed of both loads, his estimate of the difference in time to get to the bird at 30 yards is very close. Shotgun Ballistics for Windows says 0.0087 seconds but we will use his 0.01 for convenience.
Pat Ireland and I, using different radar guns, both estimated the speed of targets when they were smoked (and so the radar gun stopped seeing them) was close to 30-plus miles an hour, call it 45 feet per second.
So in a tenth of a second the bird flies 4.5 feet and in 0.01 second it flies 0.4 feet. But that's the speed it leaves the shooter and not the rise, which
we will say is based on a flight-angle of ten degrees, half-way between the start of 20 degrees and the eventually flattening out. The rise is the opposite/adjacent of the resulting triangle and that's the tangent and the tangent of 10 degrees is about 0.2 so the rise is about 0.2 times 0.4 feet giving a linear rise difference of about 8/100 of a foot, call it an inch.
That’s more of a difference than really occurs, of course, since the bird is not the only thing that's rising, the shot is too. The effect is to reduce differences like this, but not much and it depends on the height of the shooter but let's not worry overmuch about that.
My view is that
1. You can't read target breaks for the location of the shot at all at all, not once and
2. Even if you could, the difference Phil thinks he is looking at is less than about an inch and so I think he's imagining it.
That’s the problem with target-break reading. It looks like it works though it doesn't and tricks you into seeing all kinds of other things which aren't true either. The only defense is to quit doing it. If you can't do it alone, there's probably a TBRA meeting in your area to offer support. With help from the whole TBRA community, you can beat this thing!
Neil
Though BBB didn't account for the drop in speed of both loads, his estimate of the difference in time to get to the bird at 30 yards is very close. Shotgun Ballistics for Windows says 0.0087 seconds but we will use his 0.01 for convenience.
Pat Ireland and I, using different radar guns, both estimated the speed of targets when they were smoked (and so the radar gun stopped seeing them) was close to 30-plus miles an hour, call it 45 feet per second.
So in a tenth of a second the bird flies 4.5 feet and in 0.01 second it flies 0.4 feet. But that's the speed it leaves the shooter and not the rise, which
we will say is based on a flight-angle of ten degrees, half-way between the start of 20 degrees and the eventually flattening out. The rise is the opposite/adjacent of the resulting triangle and that's the tangent and the tangent of 10 degrees is about 0.2 so the rise is about 0.2 times 0.4 feet giving a linear rise difference of about 8/100 of a foot, call it an inch.
That’s more of a difference than really occurs, of course, since the bird is not the only thing that's rising, the shot is too. The effect is to reduce differences like this, but not much and it depends on the height of the shooter but let's not worry overmuch about that.
My view is that
1. You can't read target breaks for the location of the shot at all at all, not once and
2. Even if you could, the difference Phil thinks he is looking at is less than about an inch and so I think he's imagining it.
That’s the problem with target-break reading. It looks like it works though it doesn't and tricks you into seeing all kinds of other things which aren't true either. The only defense is to quit doing it. If you can't do it alone, there's probably a TBRA meeting in your area to offer support. With help from the whole TBRA community, you can beat this thing!
Neil
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