Geometry question.... Angular movement

[Thumper580]

I'm old and my geometry is long gone. Weird question I cannot answer.
For sake of argument lets assume a rifle will put every round though the same hole......
What amount of movement at the bench would cause 1/4" impact movement at 100 yards? 1/16", 1/8"??
I've long forgotten how to calculate this. Maybe some of you big brain, smarter guys can tell me. Thanks.

 

[Makinchips208]

Define movement and pivot point better.

For example, is the bipod fixed, and 24” behind that at the buttstock movement amount? In this example, Movement the thickness of a human hair will be enough to cause more than 1/4” at 100 yards.

 

[Wannashootit]

Which still amazes me when punching a bullet through the same hole, or tiny groups in the 2's.

 

[Aftermath]

Consider the 1/4" as the base of an Isosceles triangle. That is a triangle with 2 sides the same length.

When the base is divided in 2 and then a line drawn from that midpoint thru the apex, you create 2 right triangles.

In this case, the new line is 100 yards long (3600 inches) and the short side is 1/8" long.

Some Triginometry...the angle at the pivot point is found by use of the Tangent function. Tan A = Opposite Side/Adjacent side
Tan A= 1/8"/3600"= 0.00003472

To find the angle A, you can look it up in a table (probably a real old one since that is a tiny angle) or you can use a calculator.
Arctan 0.00003472 (or tan^-1 0.00003472) is ABOUT 0.002 degrees

EDIT: https://www.rapidtables.com/calc/math/Arctan_Calculator.html

EDIT EDIT: If you can imagine, the closer you get to the "point", the closer you get to zero. We know that at 100 yards, that 0.002 degrees results in a 1/8" right triangle base. How big is that base at 1"? How big (small?) is it at 1/1000"? Food for thought if you are nerd.

 

[cro789]

Half a frog hair.

 

[Aftermath]

Consider the 1/4" as the base of an Isosceles triangle. That is a triangle with 2 sides the same length.

When the base is divided in 2 and then a line drawn from that midpoint thru the apex, you create 2 right triangles.

In this case, the new line is 100 yards long (3600 inches) and the short side is 1/8" long.

Some Triginometry...the angle at the pivot point is found by use of the Tangent function. Tan A = Opposite Side/Adjacent side
Tan A= 1/8"/3600"= 0.00003472

To find the angle A, you can look it up in a table (probably a real old one since that is a tiny angle) or you can use a calculator.
Arctan 0.00003472 (or tan^-1 0.00003472) is ABOUT 0.002 degrees

EDIT: https://www.rapidtables.com/calc/math/Arctan_Calculator.html

EDIT EDIT: If you can imagine, the closer you get to the "point", the closer you get to zero. We know that at 100 yards, that 0.002 degrees results in a 1/8" right triangle base. How big is that base at 1"? How big (small?) is it at 1/1000"? Food for thought if you are nerd.

Yep....NERD here. Just ask @Makinchips208

Consider the rifle from muzzle to buttplate as the length to consider. I think this is useful because it's the movement at the muzzle that matters...right? Well, let's agree that it is for this discussion.

So, for shits and giggles, a 24" barrel and say 14 more inches to the buttplate. 38"

0.00003472 = X/38"
38" times (because the * was sort of lost) 0.00003472 = X
0.00131936" = X

That much movement at the muzzle will result in 1/8" movement of point of impact at 100 yards.

 

[Thumper580]

Thanks guys....

 

[Makinchips208]

Consider the 1/4" as the base of an Isosceles triangle. That is a triangle with 2 sides the same length.

When the base is divided in 2 and then a line drawn from that midpoint thru the apex, you create 2 right triangles.

In this case, the new line is 100 yards long (3600 inches) and the short side is 1/8" long.

Some Triginometry...the angle at the pivot point is found by use of the Tangent function. Tan A = Opposite Side/Adjacent side
Tan A= 1/8"/3600"= 0.00003472

To find the angle A, you can look it up in a table (probably a real old one since that is a tiny angle) or you can use a calculator.
Arctan 0.00003472 (or tan^-1 0.00003472) is ABOUT 0.002 degrees

EDIT: https://www.rapidtables.com/calc/math/Arctan_Calculator.html

EDIT EDIT: If you can imagine, the closer you get to the "point", the closer you get to zero. We know that at 100 yards, that 0.002 degrees results in a 1/8" right triangle base. How big is that base at 1"? How big (small?) is it at 1/1000"? Food for thought if you are nerd.

You sound like a math teacher or somethin… lol

 

[Aftermath]

If you really want to be blown away, do the calcs for point of impact change for a shitty group of say 2".

If the rifle is some sort of laser where atmospherics and gravity (ballistics) do not have anything to do with the reality, you might be surprised at how little movement it takes to make even that large (?) of a difference in point of impact (or point of aim, if you will).

 

[308pirate]

Half a frog cunt hair.

FIFY

 

[cro789]

FIFY

Red one at that

 

[LRI]

Morning OP.

Here's my attempt to answer your question. I made the drawing using a 26" barrel. It does not factor in a receiver length, nor a specific location of the sight(s) or optic. I used the muzzle as the datum to calculate the movement required to create a 1/4" POI deviation.

.0018" is what I came up with. Just to be clear, .0018" is a linear value, not angular.
The angular dimension is .004 degrees.

Research from a long time ago revealed that a healthy human eye has a resolution of about .002" (2 thousandths). A 20x power optic should increase this to .0001" (1/10th of a thousandth) resolution. It starts to paint a picture of how bad things actually look when a shot goes "off the grid."

In 2003, I ran around a bit with the US Palma Team. It was impressive to me to watch folks pound a 3-4 inch shot plot from the 1,000 with iron sights. (When conditions were nice) While still impressive, dissecting the math, and a better understanding of how the eye works, helps to explain some of this.


Hope this helps.