Rifle Scopes Mil Dot Reticle
- Thread starter hollowpoint
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Re: Mil Dot Reticle
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: hollowpoint</div><div class="ubbcode-body">Any suggested reading to get the most out of a mil dot reticle? </div></div>
http://www.mil-dot.com/
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: hollowpoint</div><div class="ubbcode-body">Any suggested reading to get the most out of a mil dot reticle? </div></div>
http://www.mil-dot.com/
Re: Mil Dot Reticle
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: hollowpoint</div><div class="ubbcode-body">awesome hamstur, ive been at it for hours lol
</div></div>
Just in case ... as long as you're playing with ranging, here's a note on windage. That demo has wind fixed at perfectly southwest. Most people instintually think that cuts wind in half, but it still has 70% effect.
If you took them, remember Pythagorean's theorem from geometry, and vectors from physics:
A^2 + B^2 = C^2
If you have a perfectly southwest 10MPH wind...
- Perfectly southwest means A = B
- C = 10
- A^2 + B^2 = C^2
- 2 A^2 = 10^2
- 2 A^2 = 100
- A^2 = 50
- A = 7.07
Which means a perfectly southwest wind at 10MPH provides a 7MPH effective crosswind.
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: hollowpoint</div><div class="ubbcode-body">awesome hamstur, ive been at it for hours lol
</div></div>
Just in case ... as long as you're playing with ranging, here's a note on windage. That demo has wind fixed at perfectly southwest. Most people instintually think that cuts wind in half, but it still has 70% effect.
If you took them, remember Pythagorean's theorem from geometry, and vectors from physics:
A^2 + B^2 = C^2
If you have a perfectly southwest 10MPH wind...
- Perfectly southwest means A = B
- C = 10
- A^2 + B^2 = C^2
- 2 A^2 = 10^2
- 2 A^2 = 100
- A^2 = 50
- A = 7.07
Which means a perfectly southwest wind at 10MPH provides a 7MPH effective crosswind.
Re: Mil Dot Reticle
What would be more advantageous would for you to constuct that in MOA's or MIL's.
I understand geometry, trig, calculus, euclidian, bayesian inferential statistics, but a lot of the guys here don't.
Simple is good.
What would be more advantageous would for you to constuct that in MOA's or MIL's.
I understand geometry, trig, calculus, euclidian, bayesian inferential statistics, but a lot of the guys here don't.
Simple is good.
Re: Mil Dot Reticle
Not sure I could convert it into MOA or Mil, cause wind calculations are dependant on drift of load & wind strength ... and all that is assuming a perfect world scenario of constant wind. Best I can think off on a pinch is a brief visual on the % crosswind effect of a wind at N, NNE, NE, ENE, and E:
From a model perspective of that ShooterReady demo, if your target is 700 yards, drift is 3.9" per MPH, wind is 8MPH NE:
(1) Calc crosswind: 8 MPH * 70% (due to NE) = 5.6 MPH crosswind
(2) Calc drift: 5.6 MPH * 3.9" per MPH = 21.84" at 700 yards
(3) Scale to 100 yards: 21.84" at 700 yards = 3.12" at 100 yards
(4) Wind holdover in Mil: 3.12" / 3.6" per Mil = 0.866 Mil holdover
If anybody is bored and really wants to discuss, feel free to PM.
Not sure I could convert it into MOA or Mil, cause wind calculations are dependant on drift of load & wind strength ... and all that is assuming a perfect world scenario of constant wind. Best I can think off on a pinch is a brief visual on the % crosswind effect of a wind at N, NNE, NE, ENE, and E:
From a model perspective of that ShooterReady demo, if your target is 700 yards, drift is 3.9" per MPH, wind is 8MPH NE:
(1) Calc crosswind: 8 MPH * 70% (due to NE) = 5.6 MPH crosswind
(2) Calc drift: 5.6 MPH * 3.9" per MPH = 21.84" at 700 yards
(3) Scale to 100 yards: 21.84" at 700 yards = 3.12" at 100 yards
(4) Wind holdover in Mil: 3.12" / 3.6" per Mil = 0.866 Mil holdover
If anybody is bored and really wants to discuss, feel free to PM.
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