confusion on dialing in range

[tlk]

I have the same question. If I understand Shawn correctly, this is the process:

1. Determine distance to target. (x)
2. Determine 1 mil for that distance. [(x/100)*1.047”]
3. Determine drop (in inches) for bullet at that distance for a given speed out of the barrel.
4. Divide drop by 1 mil at the given distance [drop”/((x/100)*1.047”)], rounding to nearest ¼ MOA (adjustment can be either up or down).
5. Adjust scope to number of MOA plus ¼ MOA required.


Make it easy: say I use this on a .22 that is zeroed at 50 yards. Scope has 1/4 MOA adjustment. Shot to be taken at 100 yards:

1 mil @ 100 yards = 1.047”
bullet drop at 100 yards = -5.7”
adjustment: 5.7/1.047 = 5.444 moa = 5.5 MOA adjustment.

Is this the correct way to look at this? Is it really this easy? If it is, I interpret this to mean that all I need is a rangefinder and a trajectory table (AT THE MOST BASIC LEVEL - i.e., no concern for humidity, wind, etc) for a given load and this basic equation.

Any elaboration would be GREATLY appreciated.

[KRP]

Originally Posted by foreign

so once i have a balistic calculator to find the drop then i have to work out the corrections to dial in my scope. whether in inchs or moa. depending on the scope. eg. .223 drop at 500 yrds. 55.35 inchs. so in moa. 5*1.047=5.235. 55.35/5.235=10.506moa adjustment if sighted in at 100. is that right anyone. and in inchs just use 1 instead of 1.047 but it will be less accurate and compounded over distance.

Your formula is correct but I get 10.57MOA, maybe that was a typo.

OK, So a living example question

I have the same question. If I understand Shawn correctly, this is the process:

1. Determine distance to target. (x)
2. Determine 1 mil for that distance. [(x/100)*1.047”]
3. Determine drop (in inches) for bullet at that distance for a given speed out of the barrel.
4. Divide drop by 1 mil at the given distance [drop”/((x/100)*1.047”)], rounding to nearest ¼ MOA (adjustment can be either up or down).
5. Adjust scope to number of MOA plus ¼ MOA required.


Make it easy: say I use this on a .22 that is zeroed at 50 yards. Scope has 1/4 MOA adjustment. Shot to be taken at 100 yards:

1 mil @ 100 yards = 1.047”
bullet drop at 100 yards = -5.7”
adjustment: 5.7/1.047 = 5.444 moa = 5.5 MOA adjustment.

Is this the correct way to look at this? Is it really this easy? If it is, I interpret this to mean that all I need is a rangefinder and a trajectory table (AT THE MOST BASIC LEVEL - i.e., no concern for humidity, wind, etc) for a given load and this basic equation.

Any elaboration would be GREATLY appreciated.

A mil(milliradian) and MOA(minute of angle) are two different things. 1 milliradian at 100 yards is 3.6", 1 MOA at 100 yards is 1.047".

[tlk]

KRP, thanks for the correction. I have made the adjustments in my notes.

Other than this, is it correct? Anything else that I need to adjust?

[KRP]

Originally Posted by tlk

KRP, thanks for the correction. I have made the adjustments in my notes.

Other than this, is it correct? Anything else that I need to adjust?

You seem to have the concept figured out, everything else looks good.

[tlk]

OK, Now the next one: This equation assumes a perfectly flat world. How does angle work into this equation, specifically? In other words, is there another equation or is this equation altered, and what is it?

I am trying to take this apart piece by piece...

[KRP]

Read this:

Long Range Hunting - Angle Shooting

There are also several lengthy threads on the subject in the forum if you do a search.

[foreign]

hey krp. yea i must have done some thing cause just re calculated and got 10.57 too. human error. im glad that the concept is that simple. feel like iunderstand it. now just to apply it. that will be the interesting part.