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Reduced Angles @ Yardage??

Discussion in 'Uncategorized Threads' started by fssberson, Jul 31, 2007.

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  1. fssberson

    fssberson Active Member

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    I have heard this many times, that the angles are less or reduced when shooting from yardage. Not being a math major, actually trying to avoid it whenever possible, please explain... I did pass geometry with a B... but that was over 45 years ago and I think I begged... So convince me. Fred
     
  2. lumper

    lumper TS Member

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    They are perceived as being less but they are still the same set angle from the house though the actual angle from the 27yd line is less than the actual angle from the 16yd line ... I think that might have something to do with Newtons theory of quntumler physics or possibly just some wild brained thoughts.

    Sit down and draw it on paper and you'll understand what I am thinking ... I hope.
     
  3. fssberson

    fssberson Active Member

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    Lumper: I think I will need my old geometry teacher. If the angle is less from the 27 yard line how far back do you have to go until the target becomes a straight away? Is this an urban legend of the trap field variety? Fred
     
  4. jptrapshooter

    jptrapshooter Member

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    The actual angle would be the same, but gun movement is reduced. This creates a perception of reduced angle.
     
  5. Rod Ritter

    Rod Ritter TS Member

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    For breaking the target in the same location, the angle is less. I've effectively lengthened the base leg of the triangle.
    Rod
     
  6. Hap MecTweaks

    Hap MecTweaks Well-Known Member

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    Lumper, they look less AND are less. The angle degree lessens the farther back it's measured on a trap lay-out.

    That's why two hole targets have less angle degree from the max than three or four hole settings. Hap
     
  7. lumper

    lumper TS Member

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    Hap MecTweaks ... we are correct that they not only look less and are less of an angle the farther you go back but as I said they are still the same set angle from the house.
     
  8. OhioBob

    OhioBob TS Member

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    The angle, the angle...which angle? Let's take a hard right angle from the trap.

    Actually everyone is partly correct...

    The angle between the front edge of the traphouse and the target remains the same regardless of where the shooter is standing.

    However, if we are talking about the angle that a shooter must swing his gun through to break a target at the SAME distance from the trap...then, yes, it is a smaller angle of swing for a longer yardage shooter.

    But, and here's the rub, that small angle of swing means there is less room for pointing error. That's why artllery pieces have micrometer dials and such.

    If you went back far, far enough it would begin to look like a straightaway to our eyes, but it never would be truly straight.

    We don't have shoulder fired guns capable of this anyway.

    Bob
     
  9. phirel

    phirel TS Member

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    Nice answer Bob, but if the shooter were moved back to infinity, would the angle become zero?

    Pat Ireland
     
  10. lumper

    lumper TS Member

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    What choke should you use at infinity? Would #7.5 or #8 shot be better?

    Even I would concede that 7/8oz of shot would be to little at infinity.
     
  11. Hap MecTweaks

    Hap MecTweaks Well-Known Member

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    Bob, does this mean 27 yard targets shot at the two hole settings are tougher to make/break than ,than say, four hole angles would be?

    "But, and here's the rub, that small angle of swing means there is less room for pointing error."

    For some reason I was under the impression that longer distance and timing of the shot to get there also played some part in that equasion. Another possible "rub"?

    Pat, what yard line is that located on? :) Hap
     
  12. Harold

    Harold TS Member

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    There are two angles to consider. Say A is post 5 at the 16 yard line, B is the trap house, C is the point where the target is broken, and D is post 5 at handicap yardage. If the shooter is holding over the traphouse he has to swing through angle BAC from the 16 yard line and BDC from handicap position. BAC>BDC.

    The other angle is the crossing angle between the flight path of the target and the shot. This is BCA (16 yd) or BCD (handicap) and is greater for handicap, assuming the target is broken at the same spot.
     
  13. Gilly

    Gilly TS Member

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    I'm having a existential meltdown.

    Whatever happened to point and pull the trigger.

    Bob Gilchrist


    West Virginia
     
  14. Pocatello

    Pocatello Active Member

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    It's true, but those of you who don't appreciate mathematics can stop reading now. The "proof" will involve trigonometry.

    As a representative case let's set up a triangle. One vertex is at the point where the target is released from the trap machine. Call that vertex A. Let's assume that the target thrown is an extreme left angle, and the point where the target is broken is vertex B. Vertex C is at the shooter's location, post 3, 27 yardline for this example. We are interested in angle C, the angle the shooter swung through to break this extreme target. Following the usual convention side a is opposite angle A, and is the distance the shot traveled to break the target, side b is opposite angle B, and is the distance of the shooter to the release point (i.e.27 yards), and side c is opposite angle C and is the distance the target flew from the trap house before breaking. (Neil - do you like this so far? Targets fly!).

    The Law of Sines applies: [sin(A)/a]=[sin(B)/b]=[sin(C)/c], and since C is an acute angle, C=arcsin(c*sin(A)/a)

    Now set up a second triangle with the shooter at the 16 yardline, post 3. Again A is at the release point and angle A is the same as before if the trap throws the same extreme left angle. B' is the new point where the target is broken, probably slightly closer to the house than where the 27 yard target. C' is at the shooter, 16 yards from the release point. Again let a', b' (=16 yards), and c' be the respective opposite side lengths. Applying the Law of Sines we get C'=arcsin(c'*sin(A)/a'). Note that angle A is the same as before. Since arcsin is a monotonically increasing function, we can compare the size of the respective angle C and C' by comparing the size of the respective arguments c*sin(A)/a versus c'*sin(A)/a'. Since sin(A) is present on both sides of the equation, divide by it and compare c/a to c'/a'. Once again, c and c' are the respective distances the target travels from release until break, while a and a' are the respective distances the shot travels until break. a and a' are roughly proportional to b and b' respectively (27 yards and 16 yards respectively), while c and c', in my experience are not. That is, it does not appear to me that a target shot from the 27 yard line is roughly 1.7 times as far from the house when it breaks as is a target shot from the 16 yard line. To me it appears that the multiplier is no more than 1.1, certainly less than 1.2. So, for the sake of argument, say c is approximately 1.15*c', while a is approximately 1.7*a' (1.7 is approximately 27/16). Then c/a is about 0.68 times as big as c'/a' (0.68 is approximately 1.15/1.7), i.e. it is smaller, hence the angle C is smaller than angle C'.
     
  15. KelleyPLK

    KelleyPLK TS Member

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    Further back you go its all visual ! Its merely more disceptive you have to give better leads passing targets back 27 than short yardage or 16 yds u can shoot headon mostly . and you cannot remove your eye from bird bead check or you stop the gun shoot behind targets . 16 doesnt show your mistakes as does 27

    Pat Kelley
     
  16. wolfram

    wolfram Well-Known Member

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    Pat Ireland hit the nail on the head, as the distance from the trap house aproaches infinity, the angle from the shooter to the target approaches zero. 27 yards is not infinity, although it seems like it at times, but the angle of the line of sight to the target relative to the trap house is less than if the given target were sighted from the 16 yard line. Most people don't get too hung up on what those angles actually measure but rather they imprint a sight picture in their brain and make shooting decisions relative to that.
     
  17. Hap MecTweaks

    Hap MecTweaks Well-Known Member

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    Gilly, I'm learning somethin new here, my poke-n-hope method hasta be differentn yer point and shoot method. You back to whuppin the guys again, hopefully? Hap
     
  18. smartass

    smartass TS Member

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    Thanks for the simplification, Sam. That made a world of sense on some planet, I'm sure.
     
  19. JBrooks

    JBrooks TS Member

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    Good lord, jptrap is the only one close to correct. The angle of the target in relationship to the sidewalk remains the same. It is the length of the ARC that you swing the gun through to cover the same length of the target path that is reduced as you move back.

    Stand 5 feet from the center of your garage door and point your finger at one edge and move your finger across the full width. Now, move back to the street and do the same thing and see how much less you had to move your arm.
     
  20. fssberson

    fssberson Active Member

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    Jim; So it is not the angle of the target that changes, but the arc of the swing of the gun that is reduced... I almost understand this explanation. With some of these expanations, no wonder I like singles better than 27 yards. Fred
     
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