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Thread: Please help me understand MOA clicks...

  1. #1
    eicarsto
    Guest

    Please help me understand MOA clicks...


    New to the sport, so please be patient. I have a NF 12-42x56 BR Scope that has 1/8 MOA clicks.

    At the same time I purchased NightForce Ballistics Calculator which tells me how many clicks I need at certain ranges.

    Now, here is my question - When i get to 1000 yds, the program suggests 285 clicks of elevation and 81 clicks of wind.

    285 clicks Is that correct? Do I need to turn the Elev turret 285 times or am I getting confused with the whole thing.

    I am sighted at 100 yds if that helps.

    So confused... ???

  2. #2
    pphreed
    Guest

    Re: Please help me understand MOA clicks...

    Each click move the point of impact 1/8 of an inch at 100 ydsor very close an moa is not an in but close If you are jst starting in the shooting game there is a heap to learn then new stuff all the time those figures sound accurate but would need balliatic coefficient velocity and bullet wieght to double check I don"t know about nf but most scopes have numbers on the turrets the distance between those numbers in your case is 8 clicks which equals 1 inch at 100 yds and 10 at 1000 yds I would stronly suggest that you strt with shorter ranges now that you are at 100 move to 2or 3 hundundred and work your way out you will get familiar with your equiptment and your own ability as you go a lot of and a lot to learn I truly enjoy putting together my loads and seeing what they will do and what I can do Hope you enjoy shooting as much as I do Good luck Fred

  3. #3
    Elkbane
    Guest

    Re: Please help me understand MOA clicks...

    Look on your elevation turret. You should see some smaller lines and some larger lines. The smaller lines are one "click" or 1/8 MOA. There are eight of them that equal the span between the larger lines, which represent 1 MOA. It will be common to have between 6 and eight "major lines" per complete revolution of the dial. You'll have to look at your socpe and see how it's laid out. So, doing the math assuming there are 6 MOA per revolution and 8 "clicks" per MOA, then one complete revolution of the dial makes 48 clicks.

    Your "come-up" of 285 "clicks" would equate to a 35.6MOA of adjustment and 5.94 revolutions of the knob. Under this assumption, you'd make 5 complete turns + 45 "clicks".

    Recalculate the above based on how many MOA there is for one complete revolution of the dial.
    Elkbane

    You can do a search on MOA to get a better understanding. It is a "minute of angle", which equates to about 1.05 inches at 100 yards. Think about a triangle with your eye being the tip and the other end (the base) spreading with distance. The farther the distance, the wider the base of the triangle, so one MOA at 100 yards = 1.05" but that same 1 MOA at 1000 yards is 10 times as big - 10.5"...1.05 is not exact, but close = which is why we usually think of a MOA being an "inch at 100 yards".

  4. #4
    defoxer
    Guest

    Re: Please help me understand MOA clicks...

    100 yard sight-in would be very short and limiting of range, (assuming your cartridge is centrefire of reasonable potency) once you get the used to the handling of the rifle sight in at say 250yd for a reasonable point blank range - which means the envelope of distance both ahead of and beyond the sight in distance where the projectile will not raise or fall more than a set number of inches depending on the size of your quarry. This allows you to hold dead-on at any distance within the PBR and expect to be within +/-x inches on the target. Btw in this example point blank is 250yds (not holding the rifle at the deer)…

    To further elaborate on minute of angle; if you take the horizontal plane to be 0degrees and the vertical to be 90 - you break up the arc travelled from 0 to 1 degree into 60 parts then each is a minute of angle. You can then break up that minute (1/60th degree) into 60 seconds of angle if you desired. so in 1 degree (of angle) there are 60 minutes (of angle) and 3600 seconds (of angle). You can easily calculate the vertical distance with good old Pythagorean theory....

    1 MoA equates to a vertical movement ‘Y’ at distance ‘X’ as follows and the unit of ‘X’ must be the same unit ‘Y’ i.e. both must be in inches or yards (for the 100yd zero you may want to use yards!!)

    100yds= 3600” -> 1MoA = Y x tan(1/60 degrees) = 3600 x 2.9E-4 = 1.047”
    1000yds= 36000” -> 1MoA = Y x tan(1/60 degrees) = 36000 x 2.9E-4 = 10.47”

    for those without a Casio 1000… the figures at the top =distance in yards, down the left are minutes of angle therefore you can view the vertical distance in inches at range from 100 to 1000 and arc of between 1 and 30 minutes of angle. e.g. 30 MoA at 1000yds equals 314inches

    moa /yds 100 200 300 400 500 600 700 800 900 1000
    1 1.047 2.094 3.142 4.189 5.236 6.283 7.330 8.378 9.425 10.472
    2 2.094 4.189 6.283 8.378 10.472 12.566 14.661 16.755 18.850 20.944
    3 3.142 6.283 9.425 12.566 15.708 18.850 21.991 25.133 28.274 31.416
    4 4.189 8.378 12.566 16.755 20.944 25.133 29.322 33.510 37.699 41.888
    5 5.236 10.472 15.708 20.944 26.180 31.416 36.652 41.888 47.124 52.360
    6 6.283 12.566 18.850 25.133 31.416 37.699 43.982 50.266 56.549 62.832
    7 7.330 14.661 21.991 29.322 36.652 43.982 51.313 58.643 65.974 73.304
    8 8.378 16.755 25.133 33.510 41.888 50.266 58.643 67.021 75.398 83.776
    9 9.425 18.850 28.274 37.699 47.124 56.549 65.974 75.398 84.823 94.248
    10 10.472 20.944 31.416 41.888 52.360 62.832 73.304 83.776 94.248 104.720
    11 11.519 23.038 34.558 46.077 57.596 69.115 80.634 92.154 103.673 115.192
    12 12.566 25.133 37.699 50.266 62.832 75.399 87.965 100.531 113.098 125.664
    13 13.614 27.227 40.841 54.455 68.068 81.682 95.295 108.909 122.523 136.136
    14 14.661 29.322 43.983 58.643 73.304 87.965 102.626 117.287 131.948 146.608
    15 15.708 31.416 47.124 62.832 78.540 94.248 109.956 125.665 141.373 157.081
    16 16.755 33.511 50.266 67.021 83.776 100.532 117.287 134.042 150.798 167.553
    17 17.803 35.605 53.408 71.210 89.013 106.815 124.618 142.420 160.223 178.025
    18 18.850 37.699 56.549 75.399 94.249 113.098 131.948 150.798 169.648 188.497
    19 19.897 39.794 59.691 79.588 99.485 119.382 139.279 159.176 179.073 198.970
    20 20.944 41.888 62.833 83.777 104.721 125.665 146.609 167.553 188.498 209.442
    21 21.991 43.983 65.974 87.966 109.957 131.949 153.940 175.931 197.923 219.914
    22 23.039 46.077 69.116 92.155 115.193 138.232 161.271 184.309 207.348 230.387
    23 24.086 48.172 72.258 96.344 120.430 144.515 168.601 192.687 216.773 240.859
    24 25.133 50.266 75.399 100.533 125.666 150.799 175.932 201.065 226.198 251.331
    25 26.180 52.361 78.541 104.722 130.902 157.082 183.263 209.443 235.624 261.804
    26 27.228 54.455 81.683 108.911 136.138 163.366 190.594 217.821 245.049 272.277
    27 28.275 56.550 84.825 113.100 141.375 169.649 197.924 226.199 254.474 282.749
    28 29.322 58.644 87.967 117.289 146.611 175.933 205.255 234.577 263.900 293.222
    29 30.369 60.739 91.108 121.478 151.847 182.217 212.586 242.956 273.325 303.694
    30 31.417 62.833 94.250 125.667 157.084 188.500 219.917 251.334 282.751 314.167


    as above demonstrates; to alter the distance you simply factor in the y as above but this does not hold strictly true for factoring the angle though the error is very small. Therefore 10MoA at 1000yds equiv to 10x10.47 = 104.7”

    The arc of the bullet which is not a circular arc at all but more so a negative exponential curve so the although the rate at which the bullet drops is absolutely the same whether a .22 bullet from a rimfire or a .30cal bullet from a 30-378weatherby; the horizontal velocity of the projectile is significantly less with the .22 allowing the bullet to drop a lot closer to the rifle therefore giving the appearance of the trajectory of a potato…

    As a side note the trajectory depends on only 2 factors; bullet velocity and ballistic coefficient, for example if a .25 and a .50 calibre projectile had the same ballistic co-efficient and left the rifle barrel at the same velocity they would exhibit equal trajectories, I digress..

    D

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