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Not enough to get away from me!! LOL

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It depends on what the target's angle of climb is at that distance. I don't know what that angle is, but I would guess that it's about 15 degrees. If that's correct, then the vertical climb amount would be:

Vertical Climb = Distance along the arc the target travels in .120 seconds (times) Sin 15*

VC = 5.4' x 0.2588

VC = 1.40 feet

Keep in mind that this is the approximate amount that the target climbs during the .120 seconds that the shot takes to get there. However, if your gun barrel is angled upward while shooting (and certainly it is), then that would have to be taken into condideration if you are trying to figure out how high the gun needs to pattern to hit the target.

Easystreet

Then, if we really want to get controversial, we could add in a third factor; possible target lift due to the dome shape. Remington made up a diagram many years ago showing the target height at different distances. I used the words "made up" intentionally.

Pat Ireland

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I already took those factors into consideration in my calculations in figuring how far the target would travel and at what vertical angle. I make no claims about my figures being exact, but they are a pretty close approximation. BTW, I estimated the target speed at that point at about 31 mph or 45 ft/sec.

Easystreet

Pat Ireland

You forgot to put the gravitational pull of the moon into your equation. HMB

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I'd like you to expand on

"if your gun barrel is angled upward while shooting (and certainly it is), then that would have to be taken into condideration if you are trying to figure out how high the gun needs to pattern to hit the target."

since it changes the expected requirement for high-shooting considerably. Let's assume the muzzle of the gun is five feet off the ground when fired. I'd like to give credit here to zzt for the initial clue that shooter-height factors into this.

My standard disclaimer: while the math can be done and results in a number, none of this has much of anything to do with the actual shooting of a trap target, since nervous system delays make all of this no more than a math, rather than a shooting, enterprise. Still, it's informative and fun. Especially as it drifts into handicap.

Neil

Roger

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Neil

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along with the position and phase of the moon, or time of day!

tony

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I didn't figure the velocity. I simply used the time of flight (0.120 sec) as provided by Roger and confirmed in a table which I got from Shotgun Sports Magazine.

Then I multiplied the velocity of 45 ft/sec times 0.12 sec and got 5.4 feet. For short distances like this, the length of the arc approximates a straight line, so that is close enough.

Then, as stated above, the vertical climb is 5.4 ft times the sin of 15*. This gives us 1.4 feet.

BUT, this doesn't mean we can shoot 1.4 feet above the target and expect to break it because the muzzle of the gun will be elevated at almost as steep an angle as the angle of climb of the target. I believe that the difference in angles would be less than 1 degree, but I have no way to measure it nor to estimate it accurately. My best GUESS (based on experience) is that the amount a shooter has to shoot over the target to break it at this distance is only a few inches (perhaps 3" to 5" would be my guess). For that matter, due to pattern spread, the shooter could shoot right at it and break the target with the top part of his/her pattern.

Easystreet

Take it to a local college and ask for a physics teacher to get the right formula. equation would use speed of target, target lbs (momentum), angle of rise, gravity, coeffecient drag, Atmospheric conditions (or perfect Colorado day), I think that is it. Never thought of it but it is some thing that we should know.

Michael Sharkey, DC

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Remember, the end of the shotgun usually has an upward velocity when the shot goes off. The human computer tends to do a lot of fancy manipulations with that factor that makes these other details seem trivial.

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However, I agree, that the human computer does make this all just so much talk, but it's winter and what's better to do?

Neil

I agree with the additional premise that the arc is not quite like one would predict as the dome shape seems to make the target "frisbee" and after the first 10 or so yds from the house, it seems to straight line for quite a distance. Try standing at the side of the house and look how far the bird seems to climb on a seemingly straigt path.

If you were shooting right out of the house you might need 0.12 X 14= 1.7 ft.

The 12-15" expectation is what I have always thought was right slightly more downrange.

In the end, Neil is right (of course) that by the time you add in all the affects of rate of climb of the barrel, neurological reaction time, etc., the number of variables far exceeds the equations, so just go out and shoot until you get it right.

d

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Here's where we are.

Bird speed about 45 fps.

Bird flight and height and so on according to page 3 of the linked Remington trap Guide.

Shot speed at the bird about 700 fps

Time to the bird for the shot 0.12 seconds for singles, 0.14 seconds for handicap

Muzzle height 6 feet from the ground when fired.

So where should the gun shoot if we were shooting .22's? Who will take this on?

Show your work.

Neil

. . . and, for extra credit, adjust your result using "Wolfram's correction," below.

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one inch of muzzle rise with a 30" tube ewuates to 1 yard of pattern rise at 30 yards, thats quite a bit. Its also the reason we like to keep our guns moving when we fire the shot.

You tell me.....

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Neil

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