Brain teaser for the day...

[Erik Cortina]

Since no explanation required, I'll simply say, it will go right and low.

 

[savagedasher]

If a gun, say a slow shooting heavy bullet 308, is dead nuts on the X at 1000 yards, where will the point of impact be if you cant the rifle 45 degrees to the right and shoot?

No explanations for now, just your answer.

I would make this a poll but I don't know how to poll dance.

With out the barrel being perfectly straight. It would just be a guess. We all know they never have made a perfectly straight barrel. Larry

This is assuming a perfect barrel. No wind.

I 'm going to NOUN out of this Is my answer Larry

 

[fayettefatts]

I agree with high left.

 

[Erik Cortina]

Well, are you going to say where the bullet went?

 

[sailhertoo]

I have'nt tried it. Id just guess low left.

There's a bigger spred of answers here than my 600 yard group!

Please tell us why Eric.

eta- why right and low. Not why my 600 group is everywhere.

 

[savagedasher]

Till you explain what Cant means .Their is no answer or even close. Example are you just rotating the barrel 45 degrees. or are you moving the gun or barrel 45degrees .or are you moving every thing at a 45 degrees low right. Larry

 

[Erik Cortina]

I have'nt tried it. Id just guess low left.

There's a bigger spred of answers here than my 600 yard group!

Please tell us why Eric.

eta- why right and low. Not why my 600 group is everywhere.

Think of the line of sight as the axis. You have to point the barrel up to get to 1,000 yards. So the barrel is pointing up at 12 o'clock in the front and touching the line of sight in the back, if you spin the barrel one rotation it will make a circle around the axis (line of sight). If you cant it to the right, shot will go right and low.

 

[dkhunt14]

If you keep the crosshair on the x and rotate the rifle the barrel will move up and left.

 

[Erik Cortina]

If you keep the crosshair on the x and rotate the rifle the barrel will move up and left.

Barrel can't move up nor left. Read my description above and visualize it.

 

[338 Mollett]

You guys are gonna put me out in the field if you don't solve this.

 

[Immike]

You guys are gonna put me out in the field if you don't solve this.

I wouldn't let too many people watch you do this without explaining first that you are conducting an experiment, lol. Or just do it during one of those deer hunting sight in days and fit right in.

 

[Ggmac]

It would impact at X same as before . You did not mention a scope or sights above the bore , so canting the bore still puts you dead on .
If a point above the bore , sights scope etc , the higher above the bore the more it would move your poi if canted r it would be low and right . But you did not mention sights or scope , so it would still be at X.

 

[KenO]

I said left and high. I shoot sling, if I am on target with the front sight, and cant to the right, the barrel is going to rotate counerclockwise, with the target at the base of the axis.

My thinking in a real shooting scenero. Unless this is a trick question and the sights have nothing to do with it, it has to be up and left.

If the barrel is laying in a rest, and your just rotating it, then impact wont change, unless your counting the bend in the barrel.

 

[69chevelle]

gstaylorg is correct. with 30 moa for elev bullet would impact approx.7.67ft. low and 18.51ft. right. with 36moa for elev bullet would impact approx.9.2ft. low and 22.21ft. right if my calculations are correct. think about it this way if you had 30moa elev your bullet drop is approx.26.18ft. if you canted your rifle to the right 90 degrees(laying flat on its side)you would have 30moa right windage with no elev.this means bullet impact would be 26.18ft. right and 26.18ft.low.

 

[savagedasher]

Simplest scenario: no windage on rifle, only elevation. The barrel is perfectly parallel with the line of sight through the scope in the horizontal plane (ie. perfect right/left alignment with regard to the line of sight). Barrel is pointing upward with regard to the scope line of sight in the vertical plane (up/down) so that the bullet will cross the line of sight at some distance out from the muzzle on its upward trajectory toward the target. Now rotate the rifle 45 degrees right, keeping the line of sight through the scope crosshair centered on target. The boreline of the rifle (which is what matters) will rotate to the right and down as Erik stated, not up and left. As goes the bore, so goes the bullet.

Imagine a very long cleaning rod sticking out of the muzzle. What matters is where the rod is pointing, and therefore where the shot will end up at some distance past where the bullet crosses the line of sight, not where the muzzle itself may seem to end up in relation to the scope line of sight. During a complete rotation of the barrel about the line of sight scope axis, the cleaning rod (and muzzle) will describe a cone because the barrel is pointed up with regard to the line of sight at the start of rotation. Canting the rifle to the right will move the impact right and down by pretty close to the values I posted. This is not speculation, it's simple trigonometry.

When a rifle is perfectly level and sighted in on a target center at some distance with no windage adjustment on the scope, the amount/effect of dialed elevation is 100% vertical. However, when the scope is rotated right (or left), elevation is converted into windage because the rifle barrel was aimed up to begin with. The question is by how much? If the rifle was turned 90 degrees right (completely on its side), all of the elevation adjustment would be gone, completely converted into right windage. So at 0 degrees rotation, we have full elevation ( 100%, or 1.0), and at 90 degrees rotation, we have none (0%, or 0.0). The trigonometric functions that describe either 1.0 or 0.0 at either 0 degrees or 90 degrees are cosine and sine. That is why they can be used to approximate the effect of canting the rifle on elevation and windage.

+ me Larry

 

[DCRYDER]

How's about this? An ordinary setup with gun and scope level, the bullet path should follow the vertical crosshair from just below center to way above center and back to center when it hits the target at 1k. Now look at an old analog clock and picture a scope centered in it. Now picture the upper part of the vertical crosshair pointing between the 1 and 2. Knowing the bullet follows this line up the crosshair but gravity ain't bringing it back down how can it possibly impact to the left? Same goes for iron sights, no difference.

 

[DCRYDER]

If its canted around the exact centerline of the bore, it should hit dead nuts where the first shot impacted. Canted around point of aim, probably high left or right depending on right or left handed shooter.

Point of aim. Cross hairs are on the X.

Here's the mention of a sighting system.

I still don't agree with "scenario three".

 

[Nomad47]

Assuming a scope and sighted in for 1000 yards, low and right is the correct answer. And some of the posters are forgetting that gravity begins acting on the bullet the instant it leaves the barrel.

 

[mr45man]

So far: :
low R- 15
low L -5
High L -7
right - 1
berm - 1
tree - 1
no change? -1
John H.

 

[Nomad47]

Looks like 15 got it right.

I'm so certain of my answer that I'm willing to bet a new pickup truck (which I need) on it.