AR Sponsor
Posted: 3/31/2005 8:59:51 PM EDT
| I'm shooting a 62 grain bullet out of my 1/9 RRA 16" how much effect does wind actually have at distances under 250 yds? |
|
MrScaramanga, To put you in the ballpark, with a 10mph full value(90Degree to bullet path) wind, with your load you will have about 4" of wind effect. Wind will affect the bullet directly proportionate to the velocity of the wind. 5mph will have half the effect. 20 mph will have twice the effect. The angle of the wind relative to your bullet path has an effect also. The full value has the most effect. A wind coming straight at you or straight away can be ignored, for all practical purposes. A 45 degree wind will have an effect of appx.7/10 that of a full value wind. (Alot of shooters consider almost any quartering wind 1/2 value. For the purposes of your example ,at limited ranges,that would work.) Example. you have a target at 200 yards with a 20 mph 45 degree wind. Think 4(original 10 mph effect)x2(because a 20mph wind twice as strong as 10mph)=8 , Then 8 x .7 because it is a quartering wind=5.6" of drift. This is meant to be a field use formula, not an exact scientific answer. For other ranges, you can be real close if you use the formula r squared= inches of wind drift , where r equals your range in yards expressed as a single digit. (based on a 10 mph wind for the base calculation) Example- A 10 mph wind will blow a 223 bullet ; 200 yds- 2x2=4" 300yds - 3x3=9" 400yds - 4x4=16" 500yds- 5x5=25 " Then correct the base calculation to account for 1)Actual wind velocity if not 10mph, and 2) Actual wind value if not Full Value. Again, this is a field formula, not a definitive answer. Jeff |
AR Sponsor