confusion on dialing in range

Discussion in 'The Basics, Starting Out' started by foreign, Jun 28, 2008.

  1. foreign

    foreign Well-Known Member

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    hi all im new here and to long range shooting, been reading heaps. i think too much, getting confused. im going to talk in inches because thats what my scope works in. so one inch at 100yrds is 10 inchs at 1000(roughly).so if i move the scope up one inch then will it hit 10 inchs high at 1000?????(if sited in at 100). now lets say that sited in at 100yrds the drop of the bullet is 220inchs at 1000.would i have to move the scope 22inchs or 88 clicks since the scope is in 1/4inchs.
    anyway. in going a little crazy thinking about it. help would be great.cheers
     
  2. Shawn Carlock

    Shawn Carlock Sponsor

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    You are headed down the right road. Your scope is probably in MOA (Minutes Of Angle) as opposed to inches of angle. Check on this it should say. If it is in fact MOA then it is 1.047" for every 100 yards. At 1000 yards if you need 220" of correction divide 220" by 10.47" thatis 1 moa at 1000 yards and you get 21.25 MOA ( to the nearest 1/4) forget thinking in clicks. The whole numbers on your scope are MOA go to 21 MOA + 1/4. If you let us know what scope you have this would help. My explanation is based on MOA and it is most likley what you have.
     

  3. johnnyk

    johnnyk Well-Known Member

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    foreign,
    You are correct in your computations, Shawn is trying to make it "easier" for you. Using scope manufactorers data; 1 click = 1/4" @ 100yds, 1.25" @ 500yds and 2.50" @ 1000yds. When sighted in dead on at 100yds and the bullet drops 220" @ 1000yds, divide 220" by 2.50" = 88 clicks.
    What Shawn is trying to get you to see (I think) is instead of counting each "click" look at your scopes dial and the value to/from each whole number is 4 clicks or an MOA. An MOA is roughly an inch, more precisely it's 1.047" at 100yds and 10.47" @ 1000yds. So with that in mind divide your drop (220") by what an MOA represents at 1000yds (10.47") and you more precisely get 21.012416 (21 1/4 MOA). If you reset your scopes turret back to "0" after rifle/load was zeroed at 100yds, this would have you go from "0" on your elevation turret to "21" and one more little "click".
    Gets you there quicker and more precisely. I learned it the way you did, along time ago and it was hard for me to grasp the MOA version.Hope this all helps. JohnnyK.
     
  4. foreign

    foreign Well-Known Member

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    hey shawn i have a burris fullfield 2. doesnt have turrets.i wont be shooting past 500m any time soon so i thought it un needed. plus all i could afford. stuffs real expensive in nz...so johnnyk now lets say theres a target at 500yrds and drop for that is 50inchs. you would divide 50/1.25=40 clicks or 10 inchs worth on my scope sited in at 100. then in MOA. so 50/5.235=9.55109. so 9.6 MOA. so up to 9 then 2 more clicks (.25moa per click).. that right???if target was at 255yrds then i would go 1.047*2.55=2.66985 MOA. then drop is 7 inchs lets say so 7/2.66985=2.621870143. so 2.6moa adjustment to be zeroed at 255yrds?
     
  5. foreign

    foreign Well-Known Member

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    can all this be done using meters for the calculations. and are there any ballistic calculators on the net which i could down load for free that would give me corrections in moa and inchs
     
  6. WildcatB

    WildcatB Well-Known Member

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    Here's Two:

    JBM - Calculations - Trajectory It's mostly based on Robert L. McCoy formulas. He was a US Army's ballistician and researcher for 30 years. While this is free, it's web-based so if you're not connected you can't use it. You can make printable trajectory card with it though.


    Here's another free one that runs on ti-83/84/89 calculators. It's based on Art Pejsa's formula's.Killzone
     
  7. foreign

    foreign Well-Known Member

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    so once i have a balistic calculator to find the drop then i have to work out the corrections to dial in my scope. whether in inchs or moa. depending on the scope. eg. .223 drop at 500 yrds. 55.35 inchs. so in moa. 5*1.047=5.235. 55.35/5.235=10.506moa adjustment if sighted in at 100. is that right anyone. and in inchs just use 1 instead of 1.047 but it will be less accurate and compounded over distance.
     
  8. tlk

    tlk Well-Known Member

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    OK, So a living example question

    I have the same question. If I understand Shawn correctly, this is the process:

    1. Determine distance to target. (x)
    2. Determine 1 mil for that distance. [(x/100)*1.047”]
    3. Determine drop (in inches) for bullet at that distance for a given speed out of the barrel.
    4. Divide drop by 1 mil at the given distance [drop”/((x/100)*1.047”)], rounding to nearest ¼ MOA (adjustment can be either up or down).
    5. Adjust scope to number of MOA plus ¼ MOA required.


    Make it easy: say I use this on a .22 that is zeroed at 50 yards. Scope has 1/4 MOA adjustment. Shot to be taken at 100 yards:

    1 mil @ 100 yards = 1.047”
    bullet drop at 100 yards = -5.7”
    adjustment: 5.7/1.047 = 5.444 moa = 5.5 MOA adjustment.

    Is this the correct way to look at this? Is it really this easy? If it is, I interpret this to mean that all I need is a rangefinder and a trajectory table (AT THE MOST BASIC LEVEL - i.e., no concern for humidity, wind, etc) for a given load and this basic equation.

    Any elaboration would be GREATLY appreciated.
     
  9. KRP

    KRP Well-Known Member

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    Your formula is correct but I get 10.57MOA, maybe that was a typo.



    A mil(milliradian) and MOA(minute of angle) are two different things. 1 milliradian at 100 yards is 3.6", 1 MOA at 100 yards is 1.047".
     
  10. tlk

    tlk Well-Known Member

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    KRP, thanks for the correction. I have made the adjustments in my notes.

    Other than this, is it correct? Anything else that I need to adjust?
     
  11. KRP

    KRP Well-Known Member

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    You seem to have the concept figured out, everything else looks good.
     
  12. tlk

    tlk Well-Known Member

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    OK, Now the next one: This equation assumes a perfectly flat world. How does angle work into this equation, specifically? In other words, is there another equation or is this equation altered, and what is it?

    I am trying to take this apart piece by piece...
     
  13. KRP

    KRP Well-Known Member

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  14. foreign

    foreign Well-Known Member

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    hey krp. yea i must have done some thing cause just re calculated and got 10.57 too. human error. im glad that the concept is that simple. feel like iunderstand it. now just to apply it. that will be the interesting part.