Should be about 0.020".   Edit, assuming a 30-32" barrel.

Not .2"?  Making it easy for me... Based upon a 36" sight radius or using a scope, regardless whether the movement is at the rear or front sight, I believe will result in a movement of .1 inch per 100 yards.

CajunRebel wrote on Apr 16^th, 2016 at 8:54pm:


.200 is the correct answer.  This question can be set up as a simple proportion.  Assuming a 36" sight radius, .001/36" = X/7200". Cross multiplying we get 36X= 7.2.  7.2/36 = .200".  Using the same method it is easy to find how much sight adjustment is needed to move the point of impact a desired distance at a given range.  Fortunately, this is one of the few things I remember from math classes.

JS

If you are lucky the wind will blow it back on target  I am never lucky!

Jack

Tricky! 

More specific information defining "as the bullet leaves the bore" and in what fashion the movement takes place is required to have any hope of accurately answering this fiendishly diabolical question.  Theoretically, if the bullet is at all still being controlled by the barrel as the shift occurs, then the point of impact will also shift.  If the bullet has in fact exited by as little as a few tenths (.0001") precisely as the muzzle shift occurs, no impact shift should occur.  However, even then some impact shift is possible due solely to that minute crown movement redirecting gas flow for a split second against the projectile base to a path slightly different in relationship to the direction of projectile travel.  Even then, the degree of shift from such crown redirection of gasses will vary depending on the mass of the projectile, its size, shape, velocity, and the residual gas volume and pressure within the bore at the time of projectile exit.  Lightweight, high velocity, little projectiles are usually more affected by crown variations than big old lumbering buffalo slayers.  The amount of shift would be nearly impossible to accurately predict with only the information provided and way beyond my skills to calculate even if more information were given.  My geezer brain starts to short out after just so much of this treachery. 
 
However, theoretically, assuming the projectile is still under control of the barrel as movement occurs, if the movement of the muzzle shift is perfectly parallel in relationship to sight alignment, the impact will shift an equal amount regardless of distance.  BUT, if the .001" shift is angular in relationship to the sight alignment, while the projectile is still under control of the bore, then the amount of impact shift will depend on the MOA that .001" muzzle movement represents.

Oh, that is a deviously asked question!

Fortunately, I do not have to concern myself with such intricacies.  My groups these days look a lot like a robin that ate too many mulberries left a deposit.  I’m happy if the holes even just sort of look like they might be friends on the target.

You hard holders will have to worry about it!

Hayface

If the barrel moves at the literal instant the bullet moves .001" it will only be off by .001" at any distance.  The only way it will be more is if you move the muzzle before the bullet gets to the exact muzzle then it will be a trig problem by using the distance of the bullet to the muzzle when the muzzle was moved.
Draw a line on a piece of paper and shift that line up .001" and then draw that line again and at the end it will be only .001".  They will both run parallel to each other no matter what as long as the bullet leaves exactly at the muzzle when barrel is shifted.
Sight line has nothing to do with it since the bullet is a point in time as it exits the muzzle.

How about this?

If John's question is re-written to state that the muzzle moves .001" at the instant that the bullet leaves the muzzle, then, mathematically the deviation from POA approaches infinity.

Consider the .001" movement as the "y" component.  As the "x" component (the bullet's position with respect to the muzzle) approaches zero (exits the muzzle), the angle between the x and y components approaches 90 degrees.   

JackHughs