Wind holdover in MILs question

[Myrradah]

Working on getting more into Precision long range shooting.

Doing a lot of research and seeing what i need to learn more of.

With a MIL scope and a xmas tree reticle - the windage holdovers are in MILS. So I was trying to find a conversion of wind speed to MILS. After watching a few videos I combined a few formulas and came up with this -

Windspeed / (Ballistic Coef x 10) x Range / 1000

Example -
Remington 30-06 165 grain has a BC of .338
Range - 600 yards
wind speed - 10 MPH
10 / 3.38 x 600 / 1000
2.96 x .6 = 1.78 MILs

im assuming it would be easier to round the BC to the nearest whole number (3 in this case).

Does this look right or is there an easier way?

New to the site - lots of interesting things here to explore

 

[KnowNothing256]

Working on getting more into Precision long range shooting.

Doing a lot of research and seeing what i need to learn more of.

With a MIL scope and a xmas tree reticle - the windage holdovers are in MILS. So I was trying to find a conversion of wind speed to MILS. After watching a few videos I combined a few formulas and came up with this -

Windspeed / (Ballistic Coef x 10) x Range / 1000

Example -
Remington 30-06 165 grain has a BC of .338
Range - 600 yards
wind speed - 10 MPH
10 / 3.38 x 600 / 1000
2.96 x .6 = 1.78 MILs

im assuming it would be easier to round the BC to the nearest whole number (3 in this case).

Does this look right or is there an easier way?

New to the site - lots of interesting things here to explore

Most people shooting competition precision rifle (PRS, etc.) use what we call the “gun number” or “gun mph.” These terms actually refer to your load, not the gun itself, but either way, here’s the idea:

Wind deflection increases fairly linearly with wind speed. So, you find the wind speed in a direction perpendicular to your bullet flight (so from 3 o’clock or 9 o’clock, referred to as a “full value wind”) that equates to about one tenth of a mil of deflection per 100 yards (or meters). The estimate works best at around 600 yards for most centerfire loads.

The way you find it is to set your wind to 3 o’clock, then look at your windage at 600 yards. Add wind speed until that windage reaches 0.6 mil. That speed is your “gun number.”

Then, to use it, I take the actual wind speed, divide that by my gun number to get what I call the number of “wind units” that are blowing, then adjust for the wind angle (there are charts to reduce the wind speed value based on angle). Finally, I take that number, divide by ten, then multiply that times my target distance in hundreds of yards. Sounds complicated, but you catch on quick.

Example: I have a 6 mph gun number. Wind speed is 15 mph from 11 o’clock. Target is at 400 yards. That’s 2.5 wind units, then I apply the correction factor for the angle, which is 50%; so now we’re at 1.25 wind units. I divide by ten (0.125) then multiply that by 4 for the target distance, to arrive at my final wind hold of 0.5 mils.

Lots of videos and posts out there about this method. Good luck!

 

[KnowNothing256]

Btw, mentally I don’t actually divide the final wind units by ten, I just know that I’m counting tenths of a mil. Same story for the target distance; I’m counting hundreds of yards. Feels less like math and more like counting on my fingers when I do it that way, and seems to go way faster for me.

So in my original example, my thought process is the same as what I originally said until I get to that 1.25 “full value wind units.” My target is at distance 4 (“distance units,” if you will), and so I have 4 x 1.25 tenths of hold, or five tenths. So then I hold 0.5.

 

[Krob95]

Look up wind rose. Your answers will be there

 

[Krob95]

Or use a tremor 3. That’s what I did.

 

[6.5SH]

Gun MPH Wind Bracket Rose

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[Huskydriver]

Or use a tremor 3. That’s what I did.

Gross

 

[Krob95]

Gross

Quiet down over there, Mr. MSR2