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Discussion Starter #1
Has anyone ever figured this out?

In the time it takes a load of light 8's to travel 33 yards (approx. .120 sec.), how many VERTICAL inches does a properly thrown straightaway target rise?
 

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Roger,

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
 

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Easystreet- Your calculations seem accurate but leave out two important factors. One is the deceleration of the target. A target leaves the house at 43 MPH and slows down to around 35 MPH at thirty five yards. Next, is target fall due to gravity. Neither of these things are linear and I have no idea how to put them into a calculation of the rate of target rise.

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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Pat,

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
 

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Easystreet, your calculations are right on the numbers that the accomplished trapshooters know and the donaters either don't or won't believe! A trap gun that shoots 15 inches high would be considered a neutral gun which is touching the target while shooting, target rise via spinning dimpled clays, shot dropping approx. 4 inches due to gravity at these distances all point to the need for built in height. Trying to "float a target is folly as all you do is shoot under it all the while you believe you are over it! The shooters that shoot quickly have to have their guns impacting even higher and the area 0f 22 to 24 inches is very popular, even higher for the fast boys. incinerate em!
 

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I can accept the 31 MPH target speed. My number of 35 MPH is just sort of an average I read on my radar gun of the last speed before a shooter reduces the target to dust. What did you use for average shot velocity between the end of the barrel and the target? For a 1200 ft/sec shell, I think an average velocity of 1100 would be about right (1200 ft.sec at 3 feet-1000 ft/sec at 35 yards). I agree that the best we can do is come up with a close approximation to the answer. I am certainly not challenging your estimate. I just want to make sure I understand it.

Pat Ireland
 

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Easystreet, support for your guess of 15 degrees is on page 3 of the linked file. A bit of trig leads to a number like yours.

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
 

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Discussion Starter #10
Damn you guys are good. Exactly what I wanted to know. I realize there are a hundred variables in actually making the shot but I wanted to know approximately how far the bird actually rose during the shot travel time. Many thanks.

Roger
 

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Roger, stick around. That number is going to stay, but what it "means" in terms of a gun's "correct" POI (mathematically, not practically) is going to be cut to six inches, more or less. We are just going to get there step-by-step so everyone can see how we got there.

Neil
 

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don't forget to figure in the weight of the trap shooters standing next to you, and what station you are standing on! :)

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


tony
 

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Pat,

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
 

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Good question!
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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We aren't shooting rifles are we?

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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Wolfram, what can that velocity be, a couple of feet per second? In a tenth os a second shot-flight, that can't amount to much. Add to that, my frame-by-frame VCR's and CD player show that one of the instructional stars slows his gun dramatically just an instant before the shot goes off. You can see this in a shootoff. There's no followthrough at the Clay Target resolution, the gun is just (about) pointing as it was when the shot was fired. That means there can't have been much gun movement at muzzle-exit.

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
 

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when I got my new gun, I spent some time thinking this out too and offer at least one additional variable. One is the point on the arc of the target that you want to shoot it. You pointed out a distance of 33 ft., but as Bigben pointed out, different places in the arc require different setups. My estimate at the start was that with an assumption of a launch height of 2 ft. and a height at 10 yds of 9ft., the target is climbing at a rate of around 14fps, neglecting air friction, gravity, etc. with a launch speed of around 20 yards per second. It flattens out some thereafter.

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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d, sure. We'll never get it right. But we can play "let's pretend" and at least winnow out some of the more popular calculations and make us think more critically about many of our comforting assumptions and conclusions. How much of what we think we know passes the math test and how much are we going to have to let go of because it doesn't?

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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Well Neil,

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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Though I never disagree with you, Wolfram, I have to just this once. What we are looking at is the speed of the muzzle at shot exit and we agree that it is some small positive number. However, that speed stays with the shot until it bleeds off. It is just that speed, not 30x, since once it is out the the muzzle, there's no force to accelerate it up. It just continues with that same vertical velocity component (attributable to muzzle movement.)

Neil
 
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