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# Canting - the right answer

#8
05-16-2006, 06:56 AM
 Silver Member Join Date: Jan 2004 Location: New Mexico Posts: 113
Re: Canting - the right answer

rd is defined above. "e" is a type and should be "a" -- it's the elevation.

Keep in mind that these formulas don't take into account the sight height and they assume that the initial azimuth is zero. It's considerably more complicated if it's not.
__________________
JBM Small Arms Ballistics -- http://www.eskimo.com/~jbm
#9
05-16-2006, 07:02 AM
 Gold Member Join Date: Sep 2003 Location: Blighty Posts: 637
Re: Canting - the right answer

[ QUOTE ]
This windage angle would give a little over 6 inches of deflection at 1000 yards which is what we see above (5.8 - -0.3 = 6.1 inches).

[/ QUOTE ]

...only 6" of horizontal deflection at 1000yds for a 10deg cant seems an extraordinarily low value.

Working this out the 'other' way ( http://longrangehunting.com/ubbthreads/s...p;page=1#109139 ) (taking bullet path -294 as meaning bullet drop is approx -328") gives a horizontal deflection of 57" (ie approx 10 times your value!) and a vertical change of -5"

...I haven't tried to rework your calculations. Have you only worked on 1 deg rather than 10? ....or quoted the vertical value as the windage value (5" being almost the same as 6") .......or are these differing calculation approaches giving values that are different by a factor of 10?! [img]/ubbthreads/images/graemlins/smile.gif[/img]

...if 10 deg cant only gives a 6" deflection at 1000 yds; it's not really worth worrying about a more realistic degree or 2!
#10
05-16-2006, 07:03 AM
 Silver Member Join Date: Jan 2004 Location: New Mexico Posts: 113
Re: Canting - the right answer

[ QUOTE ]
The basic formulas we are using are:

horizontal projection: X = drop*sin ß

vertical projection: Y = drop*(1 - cos ß)

ß = cant angle

[/ QUOTE ]

Do you have a derivation for these?

[ QUOTE ]

I think these formulas should be pretty accurate for the small angles we are discussing, normal cant in LR shooting should be 6º or less.

[/ QUOTE ]

If I have time, I'll see what I can do with my formulas and a small angle approximation.

[ QUOTE ]

The sight height has no effect when you zero the scope at any distance, since you are basically converging the LOS and bore line at that range, and then compensating for drop (see images A, B and C in this article: http://www.tirofusil.com/canting01.php )

[/ QUOTE ]

The effects of sight height are an offset to the zero in both the elevation and windage. It's very small and probably be neglected though. I didn't because I wanted everyone to see the accuracy when everything is taken into a ccount.

[ QUOTE ]

When you cant the rifle you do it rotating on the LOS, so drop is the "diameter of the circle". This is normally done in target or long range shooting.

[/ QUOTE ]

I'm not sure what you mean by "drop is the diameter of the circle".

[ 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 ]

I'd just like to see the derivation for the formulas. I'll check the link you posted.
__________________
JBM Small Arms Ballistics -- http://www.eskimo.com/~jbm
#11
05-16-2006, 07:11 AM
 Gold Member Join Date: Sep 2003 Location: Blighty Posts: 637
Re: Canting - the right answer

JBM,

We posted almost simultaneously, could you have a look at my question?
#12
05-16-2006, 07:14 AM
 Platinum Member Join Date: Dec 2005 Posts: 2,483
Re: Canting - the right answer

[ QUOTE ]
The basic formulas we are using are:
horizontal projection: X = drop*sin ß
vertical projection: Y = drop*(1 - cos ß)
ß = cant angle

I think these formulas should be pretty accurate for the small angles we are discussing, normal cant in LR shooting should be 6º or less.

[/ QUOTE ]I 'cant' disagree with the above. An excellent review indeed! That's exactly what I've used for over 40 years. And the sight height above bore wasn't considered in my use 'cause it was based on the sights zeroed for the range used.

It's a bit fun to intentionally cant a rifle when using aperture sights to make about a half or full minute change in windage when shooting long range (or 100 yards with a rimfire). With a spirit level on the front sight and knowing how much cant angle there is when the bubble's a given amount off center saves coming out of position to make a sight change. In a crosswind situtation, I could cant the rifle a bit into the wind when I feel it pick up; away from the wind when it lets off. Keeping shots centered is a piece of cake doing this.
#13
05-16-2006, 07:42 AM
 Gold Member Join Date: Sep 2003 Location: Blighty Posts: 637
Re: Canting - the right answer

So Bart, on JBM's eg figures, are you getting 57" horizontal or 6" ?

I'm starting to wonder if I'm missing something here [img]/ubbthreads/images/graemlins/confused.gif[/img] [img]/ubbthreads/images/graemlins/confused.gif[/img] [img]/ubbthreads/images/graemlins/smile.gif[/img] ....Gustavo appears to have been dancing on a pinhead with regard to the calculation you posted on the first 'cant' thread (ie using the same method but thinking it's different)
....but I can't understand why my calc is factor 10 different to JBM's.

Grateful if you (or someone!) would 'idiot check' my calc!
#14
05-16-2006, 08:07 AM
 Silver Member Join Date: Jan 2004 Location: New Mexico Posts: 113
Re: Canting - the right answer

[ QUOTE ]

...only 6" of horizontal deflection at 1000yds for a 10deg cant seems an extraordinarily low value.

[/ QUOTE ]

Granted, I haven't run out to the range to check since last night (I wish!), but it makes sense. The 10 degree rotation adds about 0.5 moa of windage. That's a little over 0.5" for every hundred yards.

It agrees with my online calculator which does a velocity vector rotation, not an approximation.

[ QUOTE ]

Working this out the 'other' way (taking bullet path -294 as meaning bullet drop is approx -328") gives a horizontal deflection of 57" (ie approx 10 times your value!) and a vertical change of -5"

[/ QUOTE ]

That does seem excessive to me. In my example, the elevation angle is 3.35 moa -- not much, the elevation angle after canting is going to be somewhat less (but greater than zero) so at most (with a 90 degree cant), you're only going to get about 3.35 moa change or about 50" at 1000 yards. With a 10 degree cant the change in azimuth is very small, about 0.05 moa.

Think about it this way, if the elevation angle (angle between the line of sight and the bore) was zero, and the sight height was zero you wouldn't get any change. Yet the other forumulas appear to me to show a change in drop and windage. My formula does not. Would somebody run that case with the other formulas? I probably just don't understand them.

[ QUOTE ]

...I haven't tried to rework your calculations. Have you only worked on 1 deg rather than 10? ....or quoted the vertical value as the windage value (5" being almost the same as 6") .......or are these differing calculation approaches giving values that are different by a factor of 10?! [img]/ubbthreads/images/graemlins/smile.gif[/img]

[/ QUOTE ]

Pretty sure. That's why I verified it with my online programs using a different method.

[ QUOTE ]

...if 10 deg cant only gives a 6" deflection at 1000 yds; it's not really worth worrying about a more realistic degree or 2!

[/ QUOTE ]

I guess it depends on what you're trying to hit!

buffalobob asked about a 180 degree cant. Yes my formula will work before I simplified it for angles less than 90. I removed a term of y/|y| (y is the y component of the velocity unit vector) which would include a minus sign for large angles. It would then give an elevation angle of -3.35 moa and azimuth of 0.0 moa which is what you would expect.

If I have some time, I'll include this on my website with some graphics.
__________________
JBM Small Arms Ballistics -- http://www.eskimo.com/~jbm

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