20 MOA Base

[Gundoktr]

I have a long range rifle, 6.5 x 284, in the shop that needs an elevated/tapered base. It is a sleeved action, with a full dovetail on top and need to make a picatinny style base with at least 20 MOA taper to fit the dovetail.

My question is this: How much taper, in thousanths of an inch would I need in 8 inches of base? Could someone measure a base and tell me the taper and length of the base they have

 

[Toolbreaker]

You'll need .0419" of rise at the rear of the base compared to the front for 20 MOA. Sine(angle) x length, so sine(.3) x 8. = .04188".

 

[Gundoktr]

That was quick, thanks, but what are you using for an angle and how did you determine it?

 

[ED3]

I calculated it to be: 0.0465". But I am not real strong in math/ geometry, so I did it a 'long' way:


8" base = 8" circle radius= 16" diameter.
16" Circle circumference = 50.2656" ( pi x diam)
50.2656/ 360 = 1° arc = 0.13962666666667"
0.13962666666667/ 60 = 0.00232711111111" = 1 MOA
0.00232711111111" x 20 = 0.0465422222222" = 20 MOA for 8" radius.

If this is incorrect, I'd like to learn the error of my methods.
Regards,
ED

 

[dalewoolumjr.]

Both answers are correct depending on how many decimal places you use for sine. 1 place (.3) = .04188" 8 places (.33333333) = 0.04654264"

Just a second thought decimal degrees to minutes
.3 = 18.0
.33 = 19.8
.333 = 19.98
.3333 = 19.998
.33333 = 19.9998
.333333 = 19.99998
.3333333 = 19.999998
.33333333 = 20.0

Dale

 

[Toolbreaker]

ED3 did have the answer that was closer than mine. I rounded 20 minutes of angle to .3 degrees, where I should have rounded it to .3333 degrees. It's actually .3333 ad infinitum. In use, since one angle is known (.3333 degrees) and since one length is known (8" for the length of the base), the remaining dimensions can be calculated using trigonometry. In this particular case, the base length can be used as the hypotenuse of a right triangle, and the .3333 degrees as the angle adjacent. After that, it's a straightforward calculation. Here's a link to a calculator that might make it clearer...

http://www.csgnetwork.com/righttricalc.html

 

[ED3]

ED3 did have the answer that was closer than mine. I rounded 20 minutes of angle to .3 degrees, where I should have rounded it to .3333 degrees. It's actually .3333 ad infinitum. In use, since one angle is known (.3333 degrees) and since one length is known (8" for the length of the base), the remaining dimensions can be calculated using trigonometry. In this particular case, the base length can be used as the hypotenuse of a right triangle, and the .3333 degrees as the angle adjacent. After that, it's a straightforward calculation. Here's a link to a calculator that might make it clearer...

http://www.csgnetwork.com/righttricalc.html

Thanks for the additional information. I should probably now try to learn that which I should have 40+ years ago.
The precision of my numbers was quite accidental. Those were the decimal places my computer's 'pocket calculator' were kicking out that I simply copied and pasted.

 

[ebb]

Use Burris signature Z rings and get the insets to do the work for you.

 

[Gundoktr]

The way it turned out was not enough; just short of elevation at 1K. I ended up making another with .046: taper in the 8" rail. Thanks for all the info though.

Jim

 

[Ggmac]

Precision shooting magazine had a great article on the amount needed . I' m at the Dr .now but the # and method are great for future builds . It was a little diff than what was posted . I used the formula for a 22 lr at 300 yds and it came close . I will post it when I get home .
Gary

 

[Ggmac]

It is Precision Shooting ,May 2003 . It works out to .00582 per inch , 6.5 " long center to center (front ,rear mount screw) a 20 moa = .03783

Remember its the mounting point that matters , a 1" overhang at the front should not count .

 

[carlsbad]

Guys. Don't make this difficult. I think all the answers here are right but here is a simple way to think of it if you know basic trigonometry:

20 moa is 1/3 of a degree is .33333 deg

8" long rail raised by 1/3 of a degree becomes the hypotenuse.

Definition of sine is opposite over hypotenuse.

so x/8=sin.3333 x= sin .33333x8 = .0465" raise

--Jerry

 

[jscandale]

Guys. Don't make this difficult. I think all the answers here are right but here is a simple way to think of it if you know basic trigonometry:

20 moa is 1/3 of a degree is .33333 deg

8" long rail raised by 1/3 of a degree becomes the hypotenuse.

Definition of sine is opposite over hypotenuse.

so x/8=sin.3333 x= sin .33333x8 = .0465" raise

--Jerry

Jerry,
You have landed on the correct answer by virtue of the hypotenuse being arbritrarily close to the adjacent side of the triangle (8"). You are correct that sin does equal opposite over hypotenuse, however your equation does not match that assertion. The only way to solve for the hypotenuse is to know some info that is not stated. The correct answer is to use the tangent function;

Tan (angle theta) .3333 = opposite(x) / adjacent(8"),

so... x = 8 Tan .3333, therefore, x = .046"

In your equation, the greater the angle becomes, the longer the hypotenuse gets in relation to the adjacent side. Additionally, once the 8" rail is raised by .046" and becomes the new hypotenuse of the triangle, the adjacent side becomes a length smaller than 8". As the angle becomes greater, the adjacent side gets smaller and smaller.

JS