JBM
Well-Known Member
[ QUOTE ]
I've had a Eureka moment.
Standby…
I've been having trouble with the idea being mooted here that at 500yds, the effect of a 10 deg cant for a rifle zeroed at 1000yds is roughly 10 times bigger than for a rifle zeroed at 100yds ………to use JBMs figures:
At 500yds
100yd zero 10 deg cant: x = 3.3" y = -0.2"
1000yd zero 10 deg cant: x = 31.25" y = -2.76"
This has been, for me, been entirely counter-intuitive /ubbthreads/images/graemlins/smile.gif .
The effect should be the same in both examples……
[/ QUOTE ]
It's entirely intuitive to me and I think it would be to you (maybe not -- many people think of the same problem many different ways...) if you understand about what is happening when you cant. A certain part of the elevation angle is transformed under rotation to a new elevation angle and new azimuth angle. The more elevation angle you have, the more azimuth angle you end up with after rotation. That's why the zero range matters.
Look at what Gustavo posted:
[ QUOTE ]
100 yds ZERO
500(x) = 03.27"
500(y) = 47.30"
1000 yds zero
500(x) = 030.80"
500(y) = 108.84"
drop (500) = 65.8"
[/ QUOTE ]
For different zeroes he shows different cant -- by about a factor of 10. As I've said before, this agrees closely with my answers. So we have three methods (my derivation, the formulas Gustavo posted and my online calculator) saying almost the exact same thing.
[ QUOTE ]
..I'm embarrassed to say, it was only this morning whilst walking the dog, that the reason for the error in these calculations popped into my head:
Bottom Line Upfront: JBM et al; the erroneously low '100yd zero' values you are producing are caused by the fact that you are basing your calculations on the fired QE (Quadrant Elevation)………..you should be basing them on the TE (Tangent Elevation) at the target range.
…..QE is an irrelevance in cant calculations /ubbthreads/images/graemlins/blush.gif /ubbthreads/images/graemlins/smile.gif.
[/ QUOTE ]
They are not erroneous. The error caused by canting is entirely dependent on the angle between the bore and the line of sight (within the limits of the flat fire approximation which is very good for everything were talking about here). If it's zero you get no error due to canting. That was the whole point of showing the difference between the two zero ranges.
[ QUOTE ]
Here's why cant effect calculations 'factoring in zero distance' are erroneous (and I hope you will see, counter intuitive):
[/ QUOTE ]
Zero distance only enters into it because that effects what I call the elevation angle -- the angle between the barrel bore and the line of sight.
[ QUOTE ]
Had the 500yd target been set up in a 47.5" dip in the ground (ie a small gully 4ft lower than the firing point) a rifle fired at the '100yd zero' QE of 3.98 MOA would have resulted in a target round.
ie the rifle would now be zeroed at 500yds
[/ QUOTE ]
No, the rifle is still zeroed at 100 yards, you just aimed 47.5" high at 500 yards.
[ QUOTE ]
So,
QE = 3.98 MOA produces a 100yd zero when AS (angle of sight) is zero
And
QE = 3.98 MOA produces a 500yd zero when the target is 4ft below horizontal
[/ QUOTE ]
That's not zeroed. All you're saying is that you'll shoot low when shooting past the zero range. I agree.
[ QUOTE ]
I hope you can now see that it is entirely counter-intuitive to assume that when we call QE 3.98 the '100yd zero' it will produce a smaller cant error than the same QE when we call it '500yd zero (target 4ft below horizontal)'.
[/ QUOTE ]
Canting error is produced by a rotation about the line of sight, the bore of the scope. The angle that matters is the angle between the line of sight and the bore because it's the cause of the error.
Take a look at the formulas Gustavo posted:
[ QUOTE ]
The formulas Gustavo posted take this into account:
X(R)=H(R)*sin ß
Y(R)=H(R)*cos ß - Drop(R)
where H(R) is the height of bore line in relation to sight
line, as a function of range R:
H(R) = R/R0*(SH + Drop0)- SH
SH = sight height
R = range
R0 = zero range
Drop0 = drop at zero
[/ QUOTE ]
Notice the factor R/R0*(SH + Drop0) - SH. This can also be written R*Elev - SH, where Elev is the elevation (angle between the bore and the line of sight), because (SH + Drop0)/R0 is the amount you raise the barrel to zero at range R0 (again to within the flat fire approximation)
Now, X(R) = H(R)*sin(B) = R*Elev*sin(B) - SH*sin(B)
and, Y(R) = H(R)*cos(B) - Drop(R) = R*Elev*cos(B) - SH*cos(B) - Drop(R)
or if you substitute
e' = Elev*cos(b)
a' = Elev*sin(b)
You have
X(R) = R*e' - SH*sin(B)
Y(R) = R*a' - SH*cos(B) - Drop(R)
I think you could, with good agreement, use my formulas that I originally posted and find e' and a', then plug these into Gustavo's posted formulas and you should get about the same answers. Again, I don't know what Gustavo, et. al. are using for Drop(R) so there may be some difference there too.
The only problem I have with these equations -- the only problem I've EVER had with these equations -- is what is essentially a new elevation and azimuth of Elev*cos(B) and Elev*sin(B). It's not entirely correct. The actual angles under a rotation are what I've shown in my first post. These are very close, but they aren't entirely correct. I ONLY brought it up because I have ballistics book(s) saying that it's an approximation and that it may not work at longer ranges and this is after all LongRangeHunting.Com. If this were ShortRangeHunting.Com I wouldn't have ever posted.
JBM
I've had a Eureka moment.
Standby…
I've been having trouble with the idea being mooted here that at 500yds, the effect of a 10 deg cant for a rifle zeroed at 1000yds is roughly 10 times bigger than for a rifle zeroed at 100yds ………to use JBMs figures:
At 500yds
100yd zero 10 deg cant: x = 3.3" y = -0.2"
1000yd zero 10 deg cant: x = 31.25" y = -2.76"
This has been, for me, been entirely counter-intuitive /ubbthreads/images/graemlins/smile.gif .
The effect should be the same in both examples……
[/ QUOTE ]
It's entirely intuitive to me and I think it would be to you (maybe not -- many people think of the same problem many different ways...) if you understand about what is happening when you cant. A certain part of the elevation angle is transformed under rotation to a new elevation angle and new azimuth angle. The more elevation angle you have, the more azimuth angle you end up with after rotation. That's why the zero range matters.
Look at what Gustavo posted:
[ QUOTE ]
100 yds ZERO
500(x) = 03.27"
500(y) = 47.30"
1000 yds zero
500(x) = 030.80"
500(y) = 108.84"
drop (500) = 65.8"
[/ QUOTE ]
For different zeroes he shows different cant -- by about a factor of 10. As I've said before, this agrees closely with my answers. So we have three methods (my derivation, the formulas Gustavo posted and my online calculator) saying almost the exact same thing.
[ QUOTE ]
..I'm embarrassed to say, it was only this morning whilst walking the dog, that the reason for the error in these calculations popped into my head:
Bottom Line Upfront: JBM et al; the erroneously low '100yd zero' values you are producing are caused by the fact that you are basing your calculations on the fired QE (Quadrant Elevation)………..you should be basing them on the TE (Tangent Elevation) at the target range.
…..QE is an irrelevance in cant calculations /ubbthreads/images/graemlins/blush.gif /ubbthreads/images/graemlins/smile.gif.
[/ QUOTE ]
They are not erroneous. The error caused by canting is entirely dependent on the angle between the bore and the line of sight (within the limits of the flat fire approximation which is very good for everything were talking about here). If it's zero you get no error due to canting. That was the whole point of showing the difference between the two zero ranges.
[ QUOTE ]
Here's why cant effect calculations 'factoring in zero distance' are erroneous (and I hope you will see, counter intuitive):
[/ QUOTE ]
Zero distance only enters into it because that effects what I call the elevation angle -- the angle between the barrel bore and the line of sight.
[ QUOTE ]
Had the 500yd target been set up in a 47.5" dip in the ground (ie a small gully 4ft lower than the firing point) a rifle fired at the '100yd zero' QE of 3.98 MOA would have resulted in a target round.
ie the rifle would now be zeroed at 500yds
[/ QUOTE ]
No, the rifle is still zeroed at 100 yards, you just aimed 47.5" high at 500 yards.
[ QUOTE ]
So,
QE = 3.98 MOA produces a 100yd zero when AS (angle of sight) is zero
And
QE = 3.98 MOA produces a 500yd zero when the target is 4ft below horizontal
[/ QUOTE ]
That's not zeroed. All you're saying is that you'll shoot low when shooting past the zero range. I agree.
[ QUOTE ]
I hope you can now see that it is entirely counter-intuitive to assume that when we call QE 3.98 the '100yd zero' it will produce a smaller cant error than the same QE when we call it '500yd zero (target 4ft below horizontal)'.
[/ QUOTE ]
Canting error is produced by a rotation about the line of sight, the bore of the scope. The angle that matters is the angle between the line of sight and the bore because it's the cause of the error.
Take a look at the formulas Gustavo posted:
[ QUOTE ]
The formulas Gustavo posted take this into account:
X(R)=H(R)*sin ß
Y(R)=H(R)*cos ß - Drop(R)
where H(R) is the height of bore line in relation to sight
line, as a function of range R:
H(R) = R/R0*(SH + Drop0)- SH
SH = sight height
R = range
R0 = zero range
Drop0 = drop at zero
[/ QUOTE ]
Notice the factor R/R0*(SH + Drop0) - SH. This can also be written R*Elev - SH, where Elev is the elevation (angle between the bore and the line of sight), because (SH + Drop0)/R0 is the amount you raise the barrel to zero at range R0 (again to within the flat fire approximation)
Now, X(R) = H(R)*sin(B) = R*Elev*sin(B) - SH*sin(B)
and, Y(R) = H(R)*cos(B) - Drop(R) = R*Elev*cos(B) - SH*cos(B) - Drop(R)
or if you substitute
e' = Elev*cos(b)
a' = Elev*sin(b)
You have
X(R) = R*e' - SH*sin(B)
Y(R) = R*a' - SH*cos(B) - Drop(R)
I think you could, with good agreement, use my formulas that I originally posted and find e' and a', then plug these into Gustavo's posted formulas and you should get about the same answers. Again, I don't know what Gustavo, et. al. are using for Drop(R) so there may be some difference there too.
The only problem I have with these equations -- the only problem I've EVER had with these equations -- is what is essentially a new elevation and azimuth of Elev*cos(B) and Elev*sin(B). It's not entirely correct. The actual angles under a rotation are what I've shown in my first post. These are very close, but they aren't entirely correct. I ONLY brought it up because I have ballistics book(s) saying that it's an approximation and that it may not work at longer ranges and this is after all LongRangeHunting.Com. If this were ShortRangeHunting.Com I wouldn't have ever posted.
JBM
