MOA/ranging math... but really programming question

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  • jrh84

    Sharpshooter
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    It's just a proportion. In simple terms:
    1st dot : 1.4 MOA * 16 power / current magnification = current MOA

    So
    @ 4 power:
    1st mark: 1.4 MOA * 16 power / 4 power = 5.6 MOA
    2nd mark: 4.3 MOA * 16 power / 4 power = 17.2 MOA
    etc.

    @ 9 power:
    1st mark: 1.4 MOA * 16 power / 9 power = 2.48 MOA
    2nd mark: 4.3 MOA * 16 power / 9 power = 7.64 MOA
    etc.

    Simplified, it's just

    Current MOA = original MOA * 16 / current magnification

    1st mark: 22.4 / current magnification = current MOA of 1st mark below center
    2nd mark: 68.8 / current magnification = current MOA of 2nd mark below center
    3rd mark: 115.2 / current magnification = current MOA of 3rd mark below center
    4th mark: 166.4 / current magnification = current MOA of 4th mark below center
    5th mark: 240 / current magnification = current MOA of 5th mark below center


    Hope this helps. Let me know if I'm way off.
     

    T.Lex

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    It's just a proportion. In simple terms:
    1st dot : 1.4 MOA * 16 power / current magnification = current MOA

    So
    @ 4 power:
    1st mark: 1.4 MOA * 16 power / 4 power = 5.6 MOA
    2nd mark: 4.3 MOA * 16 power / 4 power = 17.2 MOA
    etc.

    @ 9 power:
    1st mark: 1.4 MOA * 16 power / 9 power = 2.48 MOA
    2nd mark: 4.3 MOA * 16 power / 9 power = 7.64 MOA
    etc.

    Simplified, it's just

    Current MOA = original MOA * 16 / current magnification

    1st mark: 22.4 / current magnification = current MOA of 1st mark below center
    2nd mark: 68.8 / current magnification = current MOA of 2nd mark below center
    3rd mark: 115.2 / current magnification = current MOA of 3rd mark below center
    4th mark: 166.4 / current magnification = current MOA of 4th mark below center
    5th mark: 240 / current magnification = current MOA of 5th mark below center


    Hope this helps. Let me know if I'm way off.
    Actually, I was just getting ready to post that to help. Across all magnifications, it is a multiplier, like you say, across the dot "level." So dot 3 at 16x is linearly related to dot 3 at 4x. But dot 2 at 16x doesn't look mathematically related to dot 3 at 16x, unless its some tan/sin/cos thing.

    The BDC dots look evenly spaced, but the differences in their "heights" does not appear uniform, so I'm not sure. Along with the owners manual came 2 sheets of stickers for the eyepiece that show the BDC drops for all sorts of calibers and bullet weights. So, I thought there was math behind it.

    I'll try to post pictures of how it explains the reticle.
     

    T.Lex

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    So, playing around with the numbers a bit more, I think I need some pi.

    If you take 16x at 1.4 and figure out what 1x would be at that same rate, you get .0875. Using that as a multiplier M, the second BDC dot is at M*(16+33.3333). (The first dot is at M*16.) The third dot is at M*(16+ (2*33.3333)). So, now I'm thinking there's a pattern. But wait. The fourth dot is at... M*(16+(3*34.3333)). The drop down to the post is M*(16+(4*39)). That drop does look a bit longer than the others.

    But, that's a bit of progress toward figuring out the progression. I think.
     

    jrh84

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    The spacing between the BDC marks increases to mimic the parabolic path of the bullet vs. distance. It might be perfect for some load, somewhere, but it's really just an average of all common rounds. Each mark corresponds to holdovers for 200 yards, 300 yards, etc. Out to 600 yards, there's not that much difference between typical FMJ/hunting loads in .223, .308, .30-06, .270, etc. The BDC won't be perfect, but will usually be within a few inches. It might be spot on at 200 and 300, 2" low at 400, 3" low at 500, and 1.5" high at 600. Close enough for most shooters and rifles.

    You can look at it like that, or you can figure out what yardages the marks ARE perfect for your rifle/load. They might match at 195, 300, 415, 525, and 630 yards for a .30-06, or 180, 290, 420, 515, and 604 for .223. (Not real numbers, just showing differences).
     

    T.Lex

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    Oh yeah, I'm familiar with BDC. And this scope came with stickers that mark out the bullet drop for my preferred round for this gun (.308, 150 gr.). So, I should be able to tell where the bullet will land.

    But, before I can use the BDC, I have to have a range to the target. :) The dots above center and horizontally are more evenly spaced. I have some known-distance/height objects out my back door that I can use to get a sense of those. Just trying to up my game (with technology) for hunting.

    With the +33.3333 thing, I do think I'll get to a "close enough" point, though. Unless someone comes along that knows the real answer. :)
     

    Ericpwp

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    I think it would be best to range at 16X only. You are just adding complexities when trying to range at any zoom level. Plus, who is to say the scope is zoomed to the level you think it is. To add to that, the BDC would only be correct at one zoom level.
     

    T.Lex

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    I think it would be best to range at 16X only. You are just adding complexities when trying to range at any zoom level. Plus, who is to say the scope is zoomed to the level you think it is. To add to that, the BDC would only be correct at one zoom level.

    Yeah, the BDC is "set" at 8x. I'll probably use that for ranging.

    Along those same lines, between the Huntstand app and a laser rangefinder for visual cues, my actual ranging needs for our usual hunting grounds is limited. :D More just an exercise in backup processes.
     

    Ericpwp

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    So using JRH's equation (props :yesway:) you get
    2.8
    8.6
    14.4
    20.8
    30
    @8x

    The BDC can be used at any zoom as long as you know the values.
     

    T.Lex

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    Yeah, I'll probably use a sharp-pointed sharpie to write in the MOA numbers on the eyepiece sticker. That'll make it somewhat easier.
     

    thunderchicken

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    I poked my nose in this thread hoping to learn something but all this math is way out of my wheelhouse. I've never shot further than 100yds but have aways been amazed by those who can. However, if it's as difficult as this thread makes it appear I don't think it's for me.
    All I know about long rage shooting is...slow your heart rate, let out half a breath, hold steady and squeeze the trigger smoothly. Otherwise as far as I know it's just windage and elevation.
    Good luck figuring this out
     

    jrh84

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    I poked my nose in this thread hoping to learn something but all this math is way out of my wheelhouse. I've never shot further than 100yds but have aways been amazed by those who can. However, if it's as difficult as this thread makes it appear I don't think it's for me.
    All I know about long rage shooting is...slow your heart rate, let out half a breath, hold steady and squeeze the trigger smoothly. Otherwise as far as I know it's just windage and elevation.
    Good luck figuring this out

    Don't let this thread discourage you at all. This was just us trying to set up a computer program to use a BDC reticle in his scope differently that how it was intended. Long (er) range shooting does require a bit of math, but nothing the average Joe isn't capable of. T Lex is some sort of programmer (I'm guessing) and I'm an engineer. This was just something for S&G's.
     

    T.Lex

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    Yeah, I'm really not a long range shooter AT ALL. This is more of an intellectual exercise. :) I'd also like to get into longer range shooting, someday. When I have more time. And patience. And a computer program. :)
     
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