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Wind effect

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19 June 2017, 19:25
Instructor
Wind effect
At 20mph wind from either 3 or 9 o'clock @ 600yds would tell me that the correction would be as follows: Range 600 x V wind velocity of 20mph = 12,000/1000 = 12moa adjustment either right or left. This is for full value wind and if not able to determine if it is truly a full value, would reduce adjustment by some half of the 12 moa. I fudge this a bit and take some 60% of the 12 or 7.5 moa. Not likely would end up in the X ring, but would be on the target face and easy and quick to correct for second shot, providing the wind does not switch on you or the mirage is boiling, etc., etc. Placing that small projectile in the desired spot is a science, but also and art to master.