What fraction of stock movement changes the impact at 100 yards? How much?

[gamewarden]

I've heard multiple different answers but trying to get the correct one to drive the point home.

Last one I jotted down was from the MDS podcast (I believe) was 5/1000 of an inch= 1/2" at 100 yards.

Is the above accurate?

Thanks

 

[Tokay444]

Do the trig dude.

 

[gamewarden]

As helpful as that was I can't remember the last time I did "trig". Anyone else?

 

[Conaso]

I think it's 0.005".

 

[JBoomhauer]

It depends…

 

[Conaso]

It depends…

on sight radius...

 

[Tokay444]

Trig is a life essential skill.
It stops you from looking like an idiot on the internet.
Let's start with, how long is the rifle?

 

[JBoomhauer]

Trig is a life essential skill.
Is stops you from looking like an idiot on the internet.
Let's start with, how long is the rifle?

Just give me an answer. I dont measure shit either!

 

[Gemsbok]

But if you don't measure stuff, why are you concerned w/change in POI?
Do you buy lottery tickets?

 

[DocRDS]

Gimme some time and I'll fo the "trig for shootings g dummies" today

I live for this shit

 

[A&8's]

Math is your friend!

Think about it like this:

If you take the specs on a Rem700 LA 20moa rail from Badger Ord. It measures 7.175" long. The difference between the front and the back is .112". That will change the POI 20" @ 100yrds. (actually 20.94").

 

[Precision Underground]

I was curious about this a while back and quizzed the resident engineers here. I can’t remember the exact numbers but what I suspected was true. It takes a ridiculously small movement at the stock to make a big movement on target. This is why rear bag technique/design and recoil management are so important. Guys wonder why their zero wanders or their groups aren’t consistent. The answer is in the time between the beginning of the trigger pull and the bullet exiting the barrel. It’s a tiny fraction of a second but it’s the moment of truth. 1/2 of it is focus and not blacking out/anticipating/flinching when you pull the trigger. The other half is connecting the rifle to your body/the rear bag consistently/properly so you get a the same recoil direction every time.

 

[Tokay444]

It depends on the length of the gun from the butt to the pivot point.

 

[hlee]

This is so easy, you don’t need trig. Deflection at butt, multiplied by distance from front rest to target, divided by distance from front rest to butt, equals deflection at target.

 

[Tokay444]

Pythagorean theorem works too.

 

[DocRDS]

Don't say I didn't warn you.

The setup:

D = distance to the target
y = the movement across the paper
x = "roughly the distance to the target" (we're gonna get to that shortly)

Why in the hell did I draw a circle? By drawing the circle and setting up the problem, x and y are related to D via the trig functions.

y = D sin Angle
x = D cos angle

However, lets take a known Angle and distance. Your distance is 100 yards or 3600 inches (D). If I pick the angle to be 1 Minute of Angle, We actually know y, its the infamous 1 inch! (actually 1.047, but lets make the math easy).

Here's the fun part. x and D are essentially the same length. If D = 3600 then x = 3600 Cos (1/60)

Now while you kids were out partying, maths people learned that the cos of a small angle is 1. the actual value is 0.99999995. So x is exactly 3599.9998.

Hopefully we can agree that this is "approximately" 100 yards.

So to move the bullet 1 inch at 100 yards, you need to distrub the end of the barrel by 1MOA.

So the question becomes how far is the red distance when the fat black lines are your barrel. When you aim the gun, you are moving the barrel. We know the angular movement, so what is the physical movement.

So if you barrel length is 20 inches, the red part is just 20 sin (1/60) (in inches). That value happens to be .0003
So you need to disturb end the barrel 0.006 inches

Another 'trick" is that when the angle is small sin x = x, BUT the angle has to be in radians (note NOT MILLIRADIANS!)

But I'm not here to math flex--how do you go from degrees to radians? multiply by pi/180. But pi is roughly 3. 3/180 = 60

so sin x = x = 1/(60*60) first 60 is MOA 2nd 60 is from conversion to radians.

so x is 1/3600 or a circle. But we know our circle radius, so that arc length is the angle times the distance.

Distance is 3600 inches. sin x is roughly 1/3600

3600/3600 = 1

Shouldna skipped math class.

PS if you want to figure our how much you need to move the stock, it works the same way. Take your LOP (roughly) and form another triangle. if your LOP is 14" and you need 1MOA of movement:
14 * 0.0003 = 0.004 Inches. So a bump of 4 thou at the butstock moves your bullet 1 inch at 100 yards.

This works in all directions, left/right/up/down.

If you want 1/2 an inch. divide by 2.

Non trig way:
I want the movement to be 1 inch.
Keep the ratios the same

barrel/3600 = x

x = .0003 if the barrel is 20 inches.

 

[Enough Said]

I was curious about this a while back and quizzed the resident engineers here. I can’t remember the exact numbers but what I suspected was true. It takes a ridiculously small movement at the stock to make a big movement on target. This is why rear bag technique/design and recoil management are so important. Guys wonder why their zero wanders or their groups aren’t consistent. The answer is in the time between the beginning of the trigger pull and the bullet exiting the barrel. It’s a tiny fraction of a second but it’s the moment of truth. 1/2 of it is focus and not blacking out/anticipating/flinching when you pull the trigger. The other half is connecting the rifle to your body/the rear bag consistently/properly so you get a the same recoil direction every time.

PRESS…
We press triggers.
;-)

 

[Jack Master]

@gamewarden

This calculation is dependent on the distance from the rear bag to the bipod. - This is for 1" at 100 Yards. Divide this in half for 1/2"

Here is the math

Sin[(1/3600) (Tan-1)] x length from rear bag to bipod = Movement
= 0.000277778 x distance

Playing with this a little more - its simply the distance from rear bag to bipod divided by 3600. Example 26/3600=0.00722 (same as table)

 

[Tokay444]

 

[Enough Said]

The "Answer" to this problem is a FIRM, ERECT Rear Bag.
:-]].

https://www.snipershide.com/shootin...ing-the-answer-the-compromise-grande.7128412/

--Taylor

 

[Precision Underground]

The "Answer" to this problem is a FIRM, ERECT Rear Bag.
:-]].

https://www.snipershide.com/shootin...ing-the-answer-the-compromise-grande.7128412/

--Taylor

 

[gamewarden]

@gamewarden

This calculation is dependent on the distance from the rear bag to the bipod. - This is for 1" at 100 Yards. Divide this in half for 1/2"

View attachment 8199806

Here is the math
View attachment 8199813

Sin[(1/3600) (Tan-1)] x length from rear bag to bipod = Movement
= 0.000277778 x distance

View attachment 8199830

Playing with this a little more - its simply the distance from rear bag to bipod divided by 3600. Example 26/3600=0.00722 (same as table)

Much appreciated.

 

[acudaowner]

you asked ...

 

[Maggot]

Don't say I didn't warn you.

The setup:
View attachment 8196409

D = distance to the target
y = the movement across the paper
x = "roughly the distance to the target" (we're gonna get to that shortly)

Why in the hell did I draw a circle? By drawing the circle and setting up the problem, x and y are related to D via the trig functions.

y = D sin Angle
x = D cos angle

However, lets take a known Angle and distance. Your distance is 100 yards or 3600 inches (D). If I pick the angle to be 1 Minute of Angle, We actually know y, its the infamous 1 inch! (actually 1.047, but lets make the math easy).

Here's the fun part. x and D are essentially the same length. If D = 3600 then x = 3600 Cos (1/60)

Now while you kids were out partying, maths people learned that the cos of a small angle is 1. the actual value is 0.99999995. So x is exactly 3599.9998.

Hopefully we can agree that this is "approximately" 100 yards.

So to move the bullet 1 inch at 100 yards, you need to distrub the end of the barrel by 1MOA.

View attachment 8196415

So the question becomes how far is the red distance when the fat black lines are your barrel. When you aim the gun, you are moving the barrel. We know the angular movement, so what is the physical movement.

So if you barrel length is 20 inches, the red part is just 20 sin (1/60) (in inches). That value happens to be .0003
So you need to disturb end the barrel 0.006 inches

Another 'trick" is that when the angle is small sin x = x, BUT the angle has to be in radians (note NOT MILLIRADIANS!)

But I'm not here to math flex--how do you go from degrees to radians? multiply by pi/180. But pi is roughly 3. 3/180 = 60

so sin x = x = 1/(60*60) first 60 is MOA 2nd 60 is from conversion to radians.

so x is 1/3600 or a circle. But we know our circle radius, so that arc length is the angle times the distance.

Distance is 3600 inches. sin x is roughly 1/3600

3600/3600 = 1

Shouldna skipped math class.

PS if you want to figure our how much you need to move the stock, it works the same way. Take your LOP (roughly) and form another triangle. if your LOP is 14" and you need 1MOA of movement:
14 * 0.0003 = 0.004 Inches. So a bump of 4 thou at the butstock moves your bullet 1 inch at 100 yards.

This works in all directions, left/right/up/down.

If you want 1/2 an inch. divide by 2.

Non trig way:
I want the movement to be 1 inch.
Keep the ratios the same
View attachment 8196425

barrel/3600 = x

x = .0003 if the barrel is 20 inches.

Being numerically dyslexic I hate people who are good at math. You qualify.