Mathmatician needed ....

L

Larry Willis

Guest
If your scope is canted one degree to the right, and your rifle is sighted dead on at 100 yards, how far to the right will your rifle shoot at 600 yards after adding 16 MOA vertical?

- Innovative
 
Mathmatician I am not...

.......but IF you impact 1" to the right at 100 yards you should also impact 1" to the right at 600 yards, IF you were shooting in a tunnel; this is IF the raise is in fact vertical and not angled :confused: JMO
 
Jan said:
.......but IF you impact 1" to the right at 100 yards you should also impact 1" to the right at 600 yards, IF you were shooting in a tunnel; this is IF the raise is in fact vertical and not angled :confused: JMO
Click to expand...

huhhh??

First of all Larry's scope IS canted..... that's the question at hand.

And secondly, how cannot 1"@100 be 6"@600 using your example?

:confused:



butinnyway....

Larry, mathematician is to me as frog is to Farrah Fawcett but......IF:

-your scope is 1.5" above your bore....and centered...

Wow, this is really tough :eek: how do you know if it's indeed above the bore? You must first of all establish this relationship..... and even then the problem isn't simply geometric, you'll have to account for velocity and BC to establish drop from boreline....

I'll be interested to see can anyone parse this into simple terms. IMO the modeling, the reduction, will be the toughie.

fun

:)

al
 
Is the rifle shooting to POA at 100 with no cant, and then adjusted and canted? If the rifle was held vertical, would the adjustment put the POI on the POA at 600? It may be that we are so used to projecting angular dispersion to longer ranges that we are misapplying it to this situation. It may be that starting at a point where the LOS and trajectory intersect when the rifle is vertical, that the deviation of the POI from POA will be the same at a longer range, because we are dealing with a degree of rotation in a plane that is more or less perpendicular to the bore. In this case, I would trust a test over a visualization.

Al, you have a range at your house. Run a test and let us know what happens.
 
Well......

............a few years back I was at a shoot and noticed my X-hair starting to rotate to the right. As the day progressed it finally ended up ( vertical X-hair) at about 4:00. My point of aim stayed the same thoughout the shoot, at 100 and 200 even though I did click up for 200. May as well rotate the scope to the correct position IMO. Canted, because of the rings, not lapped ?
Yeah, what Boyd said..:)

PS......we're talking 1° vs. MOA
 
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The math isn't overly difficult (you can do it on an ordinary calculator easily), but I don't have the formula in front of me. If anyone has a copy of Harold Vaughn's book, it's covered in there fairly briefly.
 
Larry Willis said:
If your scope is canted one degree to the right, and your rifle is sighted dead on at 100 yards, how far to the right will your rifle shoot at 600 yards after adding 16 MOA vertical?

- Innovative
Click to expand...

Aproximately 1.74"

Greg
 
Charles,

I rounded a little on an input and a factor and came up with 1.745..." but since that was actually for an arc segment and I didn't figure what the true length was I dropped the 0.005...". As is necessary for me this was an arithmetic solution, no calculus or even algebra. And I could be wrong! I hope somebody that knows proofs this. I'm curious myself. I think Larry knows the answer and wanted to see how many would take the bait.

Greg
 
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Greg .....

Your answer sounds close to what I'm seeing with a "measured" one degree angle on the crosshairs. How did you come up with that answer?

- Innovative
 
Maybe we can learn something by looking at the most extreme case of cant that is possible.

To sight in at any distance, if the longitudinal axis of the scope is horizontal and in line with the POA, the axis of the bore must be canted upward to varying degrees depending on the distance. (For purposes of this discussion let us visualize an external adjustment scope. We actually have the same situation with an internal adjustment scope, we just can't observe the exact orientation of the erector tube.) If we then rotate the assembly on the axis of the scope 90 degrees, what was formerly an upward cant of the barrel, relative to the line of sight becomes entirely horizontal...does it not? and the vertical angle goes to zero. This would have both the CL of the bore and the CL of the scope horizontal, so the bullet would drop from that line accordingly. Viewed from above there would be an angular divergence equal to the upward cant needed to compensate for the drop at the specified distance. Since the amount of upward cant relative to a horizontal LOS increases with the specified distance, the lateral displacement of the bullet's impact would likewise increase, minus the scope height. Yes? no? Maybe?
 
Larry Willis said:
Greg .....

Your answer sounds close to what I'm seeing with a "measured" one degree angle on the crosshairs. How did you come up with that answer?

- Innovative
Click to expand...

Larry,

16 moa at 600 yds is almost exactly 100". A 1 degree segment is 1/360 of the circumference of a circle. We know the radius of that circle is 16 moa at 600 yds or 100" so 2rPi equals 614.32..." divided by 360 equals approx 1.745". The true length of the 1 degree arc segment is a little longer than the distance between the end points so I rounded down to the second decimal.

I have a scope that appears to have more than 1 degree disagreement between the axis of the retical and the plane of the elevation mechanism. It screws with your observation of what you think must be spin drift or coriolis effect. But we get sighter shots so what the hey anywho.

Thanks for the teaser.

Greg
 
Greg Culpepper said:
Charles,

I rounded a little on an input and a factor and came up with 1.745..." but since that was actually for an arc segment and I didn't figure what the true length was I dropped the 0.005...". As is necessary for me this was an arithmetic solution, no calculus or even algebra. And I could be wrong! I hope somebody that knows proofs this. I'm curious myself. I think Larry knows the answer and wanted to see how many would take the bait.

Greg
Click to expand...

So in your opinion the answer is the same whether you're shooting a 6.5X.284 or a Sharps?

al
 
Hi Al,


Neither a 6.5x284 nor a Sharps will display 16 moa at 600 yds. But if they both did the solution would be the same for both. Your right though, angle of departure determines the offset if both are zeroed for the same distance. But the question was posed for a specific elevation to reach zero at 600 yds.

Greg
 
Greg .....

I was thinking about the effect of canting your rifle or having your crosshairs slanted and not knowing it. Tactical rifles don't have the option to fire sighter rounds, and shooting without a level can add almost 2 inches of error to your shot at 600 yards. (Actually, I thought it was slightly more than that.)

The question came up after designing a tool for leveling a scope within half a degree. (It takes about one minute) I can also shoot with this tool attached. Thanks for the calculations. I knew that someone out there would figure it out pretty quickly.

- Innovative
 
Larry,

The problem is that induced windage error with elevation adjustment comes from the mechanical plane of the elevation adjustment mechanism not being vertical (not being aligned with gravity or not plumb). The angle of the crosshair doesn't matter at all except that it will cause the shooter to cant the scope's mechanism in an attempt to level the crosshairs. The rifle cant doesn't matter either if it doesn't vary. The azimuth of bullet departure remains constant as long as the cant is constant. You could shoot the rifle at a 45 or even a 90 degree cant and the bullet would travel true for windage with elevation adjustment as long as the axis of the scope's elevation adjustment is plumb. John Whidden shoots with lots of cant in his rifle but the mechanical axis of his scope or sights are plumb when he is in position. Same deal for Dave Tubb and lots of other non-rest type shooters.


I guess the easiest way to say this is that the angle between the sights and rifle doesn't matter. What matters is that elevation adjustment moves through a plumb plane just like the bullet must. Gravity always desribes up and down.

Greg
 
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Greg .....

I've seen the extreme amount of cant used by David Tubb. I'm sure it works for him, but when I see crooked crosshairs it drives me crazy. I agree that when your crosshairs are canted and you raise the elevation, you're only adding "extra" horizontal when the crosshairs are not level while taking the shot.

However, it's pretty easy to have your scope off by one degree and not notice it. After shooting with a level, I've been surprised to see how often my scope was not level. It's easy to align your crosshairs parallel to the target frame, but in tactical shooting (or long range hunting) there is rarely a perfectly level (or vertical) reference. Just another thing to think about ....

- Innovative
 
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Charles E said:
Um, Greg, how many decimal points would you want to go to to drop the "approximately"?
Click to expand...

Charles,

The answer for 16 moa, 600 yds and 1 degree cant is 1.754507308" without much rounding, but I had to use the sine function on a calculator to get there. About two 1/8 min clicks. But as long as we're talking about how many angels on the head of a.........

Greg
 
If your sight is on at 100 yards even though it is canted 1º it will still be on a vertical line at 600 yards.

Think about it for a while.

Concho Bill
 
Bill,

If the vertical adjustment has a horizontal component because of cant, the farther you sight in, the greater the horizontal offset becomes. With cant you can only avoid that error at one distance. The greater the distance and the greater the cant the greater the error.

Do what Boyd suggested and exaggerate cant and trajectory to help your modeling

Greg
 
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