How do I determine distances with Mil-Dot rectile

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Bell Cowboy

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I just bought a Bushnell 6x24 with mil dots and I'm about to sight it in at the range. My question to those who know more than I is... "I understand the Mil dots do not represent exactly 300, 400, and 500yds (if sighted in to 200yds) but the local range only goes to 200yds so how can I figure out the distance those mil dots represent if I center them on the target. Is there a equation to figure out the approx distance they (mil dots) represent? Such as if 200yds is zero than 1st dot down is 275yds? I know the ammo used affects this but I will be using the same ammo afield.

Thanks
 
Using th emetric system makes it much easier than using the Imperial system, so I'll give you the Metric way of doing things;

First, you have to set the power to the RED number (usually around 10X), for calibration sake. Unless you are very, very familiar with the scope and the way to use the dots, always use the RED number.

The distance between two dots (center-to-center) represents (1mills) or 1 meter @ 1000 m.
So, at 500 meters, if something is filling up the distance between two dots (C-to-C), you now know it measures 5 meters.

There is a small formula to calculate the distance from a know target size;

Target real height (m) * 1000 / number of Mill(s) it Covers = Distance

Example;
You know a man measures average of 1.80m. The man covers 4 Mills when the scope is at the nominal power (red number).

1.80 X 1000 / 4 = 450 meters.


You can use the dots for compensating for the drop, but usually, it's easier to estimate the distance then adjust the zero (with the Target Knobs) to match the impact point.

For the instructions for using the Milldot of your Bushnell scope, go here; http://www.bushnell.com/customer_service/manuals/riflescopes/MIL_DOT_generic.pdf
 
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I know very little about the mil dot system but the internet has tons of information on it.

From what I understand, the mid dot system was basically designed to estimate the distance to a target by knowing approximately the height of the target in metric or whatever and then using mil dots to estimates the distance to it. Once you know the distance to the target, you can adjust your elevation accordingly.

I suppose that you could use the mildots to compensate for distance having determine their equivalent MOA (similar approach used with some archery scopes).

http://www.mil-dot.com/index.htm

Duke1
 
1 mil @ 100 meters = 10 cm
1 mil @ 200 meters = 20 cm

0.1 mil @ 100 meters = 1 cm
0.1 mil @ 200 meters = 2 cm
and so on.
Usually top quality tactical scopes with mill dot usually have a .1 mil adjustment option and a first focal plane retical.
If its a second focal plane scope your mils will only work at max power.
If its a 1/4 moa click on your elevation and windage then good luck figuring that out.
 
The distance between the center of dot A and and the center of dot B should be 3.6 minutes of angle at 100 yards/meters.
So the distance between the exact same dots is 7.2 moa at 200 yards, 10.8 moa at 300 yards and so on.


The trick to start ranging is to be as accurate as possible with the number of mils covering the intended target. Is that 1.5 mils or is it 1.3 mils. To start out break the ranging into 1/2 mil increments. Then work out to 1/4 mil increments. Then right down to 1/10th mil increments. Make sure you know the size of the target or be damn close to its approximate size.


Best example I've found is as follows:

Lets say you are out varminting and you spot a fat groundhog in the field. You estimate him to be 10 inches. You put your rifle to your shoulder and mil Mr. groundhog at .6 mils.........NOW these are the two math formulas that you can use. We will range in yards. There are 36 inches in a yard....thus the 3.6 inches between mils.
The objects size in inches devided by 36 X 1000 devided by mils read OR The objects size in inches X 27.77 devided by mils read.

10 inches devided by 36 x 1000 devided by .6 = 462 yards OR
10 inches X 27.77 devided by .6 = 462 yards.



The 27.77 is a math constant for distance in yards and 25.4 is the math constant for meters. It is certainly fun to practice ranging in the field, but I ended up investing in a rangefinder to eliminate the math work involved.



Hope this isn't too confusing.
 
Unless you have a mil dot master or a calculator and enough time, I would suggest getting a laser rangefinder. I use my mildots for different hold over shots but then there is the wind to take into consideration.
Long shots take Me a bit of time to set up and shooting at a stationary target is all fine and good but hitting a moving target at an unknown distance can be very hard to do without a lot of practice.
 
Bell Cowboy said:
I just bought a Bushnell 6x24 with mil dots and I'm about to sight it in at the range. My question to those who know more than I is... "I understand the Mil dots do not represent exactly 300, 400, and 500yds (if sighted in to 200yds) but the local range only goes to 200yds so how can I figure out the distance those mil dots represent if I center them on the target. Is there a equation to figure out the approx distance they (mil dots) represent? Such as if 200yds is zero than 1st dot down is 275yds? I know the ammo used affects this but I will be using the same ammo afield.

Thanks
Click to expand...

I see a number of very interesting replies from very knowledgeable shooters and from those, I have learned a lot about the subject.

I can now say that I'm starting to understand a wee bit about the mil dot system. With this in mind, I went back and read the original post very carefully and now I believe I know what Bell Cowby wants to do using this system.

He basically wants to set his sights at 200 yd and then use the mil dots to compensate for distance. This can be done by the following:

1. Obtain from ballistics tables the bullet drop in inches at the various distances up to whatever distance you want to shoot. For example, a 308, 150 gr caliber bullet with a muzzle velocity of 2820 fps will have the following trajectory when zeroed at 200 yd (this is for a particular type and shape of a bullet):

100 yd: +2.4"
150 yd: +2.0
200 yd: 0.0"
250 yd: -3.8"
300 yd: -9.8"
400 yd: -29.3"
500 yd: -62.0"

2. Then convert these drops in inches to mils using the relationships

1 mil = 3.6" at 100 yd, 7.2" at 200 yd, 10.8" at 300 yd, etc. (distance in yd/100yd X 3.6")

So, for example, at 300 yd, a mil is equal to 10.8" so with a bullet drop of 9.8", this would equivalent to 0.9 mils. (bullet drop in inches/distance of 1 mil at 300 yd)

And, for example at 500 yd, a mil is equal to 18" so with a bullet drop of 62", this would be equivalent to 3.44 mils

3.0 All that you have to do now is draw up a table using what ever distances you need and calculating the equivalent mils as explained above using the trajectory information for the caliber you are using. Ballistics information can be obtained from various sources including the Internet, reference manuals, etc.

In the example used here we would have:

Yards ..........Mil
100 ------ 0.67 above
200 ------ 0.00 zero
250 ------ 0.42 below
300 ------ 0.90 below
400 ------ 2.03 below
500 ------ 3.44 below

4.0 All this could be programmed in an Excel spread sheet with the bullet drop and the distance independent variables (input) and mils as output (dependant variable).

5.0 To go further and use each mil dot to represent the correction for a specific distance in yards using a specific caliber, etc, would require considerably more complex calculations. This is because the distance in inches covered by 1 mil changes as the shooting distance to the target changes. This also applies to the bullet drop with the drop not being a linear change. For example to come up with something like 1 mil dot is the correction when using caliber/bullet X when shooting at Y yards. If we look at the situation above, we will note that 1 mil dot would be a correction for a shooting distance of slightly greater than 300 yards.

In the end, a laser range finder would be of great help for hunting although theoretically, you could use the mil dot scope to estimated the distance. The main problem with using this approach would be to find an object that you can estimate the height/width fairly accurately.

Please let us know how you make out.

Duke1
 
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I thank everyone for all the insight... I tried to go to the range twice but up here in the North the range was -32 and -28 thanks to windchill but once it gets warm enough I will test all of this out!

Thanks again for all your help
 
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