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Ballistic trajectory of a projectile (ballistic loophole)

  1. Jun 12, 2014 #1
    I am trying to find the formula for the so called "ballistic loophole" which is a technique where one will cut a small hole in a wall or structure (no bigger than 3 inches wide or tall) and shoot through it at a target without placing himself to close to the wall.

    I know I need the height from bore to center of optic which is 2.5inches
    Distance to target 300yards
    100yards zero so its 3"low at muzzle to 10yards, 1"low at 50, 0"low from 70yards to 130yards, 0.3"low at 150yards, 2"low at 200yards, 5" low 250yards,and 10"low at 300 yards.
    weight of projectile 62grains (4.017grams) velocity-2980fps at the muzzle

    I'm just going by eye and I can shoot through a hole 1.5inches wide and 2 inches tall and hit targets 470yard away but I would like an actual formula. Thank you for the help. I apologize if this isnt the place to ask this question
  2. jcsd
  3. Jun 13, 2014 #2


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  4. Jun 13, 2014 #3


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    My grandson is a sniper in the US Army. He told me that the answer to the OP's question is highly secret.

    Of course basic ballistics physics is well known and hardly secret; a rule-of-thumb formula that people in the field can execute in their heads is something else. The suggestion is that would-be terrorists would like to know the answer to that question.
  5. Jun 13, 2014 #4


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    I guess the key is that the hole must be both on the line of sight to the target and on the ballistic path of the bullet. The maths can't be too hard but since a lot of web posts claim it's secret I don't plan on doing the calculations.
  6. Jun 13, 2014 #5
    reply to anorlunda

    No, its not highly secret thats whats going around and it just simply isn't so. Terrorists don't want to "snipe" you thats a myth circulated by movies they want to cause as much damage as possible as quickly as possible. There are commercial pda devices available that will calculate the formula for you. Besides anyone with 30minds some cardboard and a rifle can do it well enough without the formula. I'm asking for a simple math formula so I can understand in depth whats actually going on.

    EDIT-I've found the formula but man oh man am I in over my head. I'd really appreciate some help

    BD=Bullet diameter (.224inches)
    SH=sight height (2.5inches)
    LD=distance to loophole (120inches)
    LH=Loophole Height (3inches
    NZ=near zero (50yards)
    FZ=far zero (200 yards)
    HA=Holdover angle (5.3 degree)

    H1=SH(NZ-LD)/NZ for LD<=NZ
    LH=H0+the greater of (H1+H2orH0)~=H0+H1+H2~=H1+H2

    H1 is the simple triangular interpolation of the sight radius onto the Loophole (we can consider the trajectory linear at such short distances). H2 is an amount you can shrink the loophole at the top for holdover (you don't need the whole scope field of view to see) or the amount you have to add to the keyhole height at the top to be able to see the target in a holdunder situation. H1 is 0 at the near zero and far zero

    Can someone spell this out for me? I guess I'm gonna have to go to the local library because alot of this is over my head. I will certainly be glad once I start college classes.
    Last edited: Jun 13, 2014
  7. Jun 13, 2014 #6
    To put people at ease

    Look its a simple technique if you want to just get the raw jist of it. If it makes you feel better I'm not asking you to make anything public thats not available in classes or from ballistics software. I'll even show you. http://longshotsoftware.com/eshop.php?view=productListPage&category=3 pc software to calculate it (I have a version similar to this) and here http://www.horusvision.com/pda.php either one of these 3 will do the math for you and much much more.

    I have the formula I just need help understanding it.
  8. Jun 14, 2014 #7


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    I would start by putting all that data onto a drawing.

    It's quite possible you know more about the problem than we do. Many people here aren't shooters but will understand the ballistics side. If you have to take into account things like the field of view of telescopic sights it sounds like it's going to be pretty complicated.
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