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MOA/MOA or MIL/MIL

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Easy2 View Drop Down
Optics GrassHopper
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Easy2 Quote  Post ReplyReply Direct Link To This Post Posted: January/09/2011 at 11:34
Originally posted by Jon A Jon A wrote:

[QUOTE=Bitterroot Bulls]
Say you range something and look up in your chart your 10 MPH wind correction for that range and it's .7 Mils, for example.  You measure with your Kestral, observe foliage and mirage all the way to the target and you estimate your total average wind value to be 7 MPH.  What's your hold?  .49 (.5) Mils .  How about 3 MPH?  .21 (.2) Mils.  How about 11 MPH?  .77 (.8) Mils.  How about 15 MPH?  1 Mil.  See how that works?  Now try the same for 2.5 MOA.  If you're good at math, it can certainly be done, just not as quickly and easily (for me, anyway).

I am a lost here, if my windage says 2.5 moa I dial 2.5 moa, there is no math?
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Easy2 Quote  Post ReplyReply Direct Link To This Post Posted: January/09/2011 at 11:36
Originally posted by Easy2 Easy2 wrote:

Originally posted by Jon A Jon A wrote:

[QUOTE=Bitterroot Bulls]
Say you range something and look up in your chart your 10 MPH wind correction for that range and it's .7 Mils, for example.  You measure with your Kestral, observe foliage and mirage all the way to the target and you estimate your total average wind value to be 7 MPH.  What's your hold?  .49 (.5) Mils .  How about 3 MPH?  .21 (.2) Mils.  How about 11 MPH?  .77 (.8) Mils.  How about 15 MPH?  1 Mil.  See how that works?  Now try the same for 2.5 MOA.  If you're good at math, it can certainly be done, just not as quickly and easily (for me, anyway).

I am a lost here, if my windage (windage chart) says 2.5 moa I dial 2.5 moa, there is no math?
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Jon A View Drop Down
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Jon A Quote  Post ReplyReply Direct Link To This Post Posted: January/09/2011 at 16:25
In the interest of saving space and being simple, many or even most simple drop charts people make list only the 10 MPH wind value.  So the chart says 2.5 MOA for 10 MPH wind.

So if the wind is only blowing 7 MPH, how much do you dial?
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338LAPUASLAP View Drop Down
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Post Options Post Options   Thanks (0) Thanks(0)   Quote 338LAPUASLAP Quote  Post ReplyReply Direct Link To This Post Posted: January/09/2011 at 17:31
After the PM's here is my explanation at the bottom.

The cover up of Dots is an entire other debate or explanation but mils vs MOA is easy IMO.

Skip next paragraph I misinterpreted the question.

What is the temp, humidity and what is the altitude, barometric pressure, angle of wind,  bullet weight, velocity, range? = 62.832/60 = 1 min so 1 min = 1.0472~ Wind drift (in MOA / INCH = 12 x (88 x s /60) x sin a x (t - (3 x r) / v)  NO CHART is going to do it for you or no scope that I know of regardless of reticle math is necessary.  Not able to do math not able to hit 1st time...  No rock or ground shooting...

Mil = miliradian. A radian is approx. 57.295645 degrees, or there are 6.2832 radians in a circle. This makes 6283.2 miliradians in a circle and makes 3.438 moa in a mil. this is often rounded to 3.44 and we use 3.5 for ease of math. With our math this would make 35 inches at 1000 yards in stead of 36 inches.

Mils can range a target by using is height in meters X 1000 / mil height.  METERS

Can MOA do this?

Is it not the most important thing of shooting a target?

Knowing its (target or animal) approx. distance.

Conversion of mm to inches?

3.1416 x 2 equals 6.2832 (number of radians in a circle) with 360 divided by 6.2832 equals 57.295645 degrees. A miliradian is 1000th of that or 6283.2 in a circle. Thus 1 mil equals 21600 (number of moa in a circle) divided by 6283.2 equals 3.4377387 or 3.438 moa in a mil. Or 100 x 2 x 3.1416 – 628.32 yrds then 628.32 x 36 – 22619.52 inches 22619.52 / 360 – 1 degree 1 degree = 62.832 inches – 62.832 / 60 – 1 minute 1 minute – 1.0472 or =100*2*3.1416*36/360/60=1.0472

            1 Milliradian = 1/1000th of a radian, 1 radian = 2 PI

            1 Milliradian = .0573 degrees or 6283 parts of a circle

USAR: 360 degrees = 1 circle
   6400 mils = 1 circle, 360
    17.8 mils = 1 degree
   360 degrees divided by 6400 = .0563 multiplied by 60 = 3.375 MOA or
    1 mil = 3.375 moa  

 

USMC: 360 degrees = 1 circle
   6283 mils = 1 circle, 360
   17.5 mils = 1 degree
   360 degrees divided by 6283 = .0573 multiplied by 60 = 3.438 MOA or
    1 mil = 3.438 moa

USAR: 3.375 moa X 1.047” = 3.533625” @ 100yrds or 35.33625” @ 1000yrds
USMC: 3.438 moa X 1.047” = 3.599586” @ 100yrds or 35.99586” @ 1000yrds

Since a radian is approx. 57.295645 degrees, or there are 6.2832 radians in a circle. This makes 6283.2 miliradians in a circle and makes 3.438 moa in a mil. this is often rounded to 3.44 and we use 3.5 for ease of math. With our math this would make 35 inches at 1000 yards in stead of 36 inches.


MIL/MIL


Original Question...


For hunting and long range target shooting, would the MIL/MIL system or MOA/MOA system be more beneficial.  I will typically be dialing for elevation and holding for wind.

Any experience and/or preference out there?


MIL/MIL

 

I haven't had much trouble with a MIL reticle/ MOA turrets either, but it just makes plain sense to match them up.  Sighting in is quick and easy, and one less column on the dope chart.

I guess my precise questions are:

1.  Would you prefer the slightly finer MOA adjustments or the slightly faster MIL adjustments? (especially when using the reticle to hold for wind) 

1/8 is very nice for precise MIL is nice for ease.

2.  Do the reticles look very different?  Is the MIL reticle less cluttered?  Does the MOA reticle take up less of the field of view?

MIL


MIL is the only way to do this IMVHO...

DOTS/CIRCLES/OVALS are another subject that gets me goin
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jonoMT View Drop Down
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Post Options Post Options   Thanks (0) Thanks(0)   Quote jonoMT Quote  Post ReplyReply Direct Link To This Post Posted: January/09/2011 at 17:32
Originally posted by Jon A Jon A wrote:

In the interest of saving space and being simple, many or even most simple drop charts people make list only the 10 MPH wind value.  So the chart says 2.5 MOA for 10 MPH wind.

So if the wind is only blowing 7 MPH, how much do you dial?


Multiplying .7 * 2.5 = 1.75 MOA. Of course, looking for the easier math, I thought "25 * 7 = 175" then divided that by 100. You'll be close enough at a fair ways, but if you run your load(s) through JBM, you'll see that the results vary more as you get out there.
Reaction time is a factor...
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Jon A View Drop Down
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Jon A Quote  Post ReplyReply Direct Link To This Post Posted: January/09/2011 at 19:12
Originally posted by jonoMT jonoMT wrote:

Multiplying .7 * 2.5 = 1.75 MOA. Of course, looking for the easier math, I thought...

Correct.  But most everybody will be faster answering 7 times 7 and less likely to make a mistake.  .7 * .7 = .49 Mils.  Easy.
Originally posted by 338LAPUASLAP 338LAPUASLAP wrote:

What is the temp, humidity and what is the altitude, barometric pressure, angle of wind,  bullet weight, velocity, range? = 62.832/60 = 1 min so 1 min = 1.0472~ Wind drift (in MOA / INCH = 12 x (88 x s /60) x sin a x (t - (3 x r) / v)  NO CHART is going to do it for you or no scope that I know of regardless of reticle math is necessary.  Not able to do math not able to hit 1st time...  No rock or ground shooting...

You're missing the point.  All that math is done when you make up the chart.  Everybody realizes that at extreme ranges a simple chart doesn't do the job and individual solutions with real-time weather input is required.  This is a small percentage of most people's shooting.  Any time your 10 MPH wind correction is only .7 mils, you are close enough a chart does just fine and changes in the weather won't affect the answer enough to worry about (they'll be smaller than your ability to judge the wind is accurate).
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Easy2 View Drop Down
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Easy2 Quote  Post ReplyReply Direct Link To This Post Posted: January/10/2011 at 00:51
  I follow you now. I miss understood the question. Yes , you are right the math is easier for the 
wind correction from that standpoint. But not overly difficult to do in moa.
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