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# Canting - the right answer

#1
05-15-2006, 10:19 AM
 Silver Member Join Date: Jun 2007 Location: Buenos Aires, ARGENTINA Posts: 131

I've just finished the "research" on the subject of canting. First draw is that the accepted models are plain wrong, a suspicion I had since my first post on this matter.

After looking at some papers, especially an excellent one written by Jeroen Hogema, a dutch gentleman, I started to see some discrepancies among the usual published stuff (in this and other forums) regarding the correct solution to calculate the effects of canting.

One thing that especially alerted me was the “fact” that vertical deflection was almost regarded as minimal and not to worry about…weird to say the least.

Also, I contacted Rubén Nasser ("Tiro Fijo") from Paraguay and a poster here.

So between Jeroen, Rubén and myself after a very enjoyable exchange of emails, the correct solution showed up.

In short, the correct solution MUST account for LOS and Zero Range, and thus VERTICAL DEFLECTION is an ISSUE.

So the right formula to account for cant is :

X(R)=H(R)*sin(cant)
Y(R)=H(R)*cos(cant)-DROP(R)

where H(R) is height of bore line with respect to sight line (as a function of range R)

I really don't want to make a tedious thread with many formulas, but if someone is interested, just let me know and I'll post them.

Example :

MV: 2900 fps
BC: 0.490
LOS : 2.0 inches
Zero : 200 yards
Cant : 10 degrees

Dist / H( R ) / X / Y
yards / inch / inch / inch
0 / -2.00 / -0.35 / -1.97
100 / 3.55 / 0.62 / 1.30
200 / 9.10 / 1.58 / -0.14
300 / 14.65 / 2.54 / -6.97
400 / 20.20 / 3.51 / -20.21
500 / 25.75 / 4.47 / -40.64
__________________
regards, Gus

http://www.patagoniaballistics.com
#2
05-15-2006, 02:57 PM
 Gold Member Join Date: Sep 2003 Location: Blighty Posts: 637
Re: Canting - the right answer

[ QUOTE ]
I really don't want to make a tedious thread with many formulas, but if someone is interested, just let me know and I'll post them.

[/ QUOTE ]

.......at a quick glance your Y values are not very different to what my ballistic program gives for an uncanted rifle [ie your Y values look to be showing a roughly normal bullet path relative to LOS]:

For example, using your rifle/bullet data, at 400yds I get a bullet path relative to LOS of -21" for an uncanted rifle;

.....the Y value given in your table for a 10deg cant at that range is -20.21

......given that we're probably using different ballistic programs, I'm not sure that the 0.8" discrepancy is quite a EUREKA! moment [img]/ubbthreads/images/graemlins/smile.gif[/img] !

....I would have said that 0.8" is a very small vertical change at 400yds for a massive 10 degrees of cant [img]/ubbthreads/images/graemlins/confused.gif[/img] [img]/ubbthreads/images/graemlins/smile.gif[/img]

....it's bedtime at this location -no doubt I've missed something [img]/ubbthreads/images/graemlins/smile.gif[/img]
#3
05-15-2006, 05:38 PM
 Platinum Member Join Date: Jan 2003 Location: The rifle range, or archery range or behind the computer in Alaska Posts: 3,830
Re: Canting - the right answer

+1 on post everything!!
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Long range shooting is a process that ends with a result. Once you start to focus on the result (how bad your last shot was, how big the group is going to be, what your buck will score, what your match score is, what place you are in...) then you loose the capacity to focus on the process.
#4
05-15-2006, 09:29 PM
 Silver Member Join Date: Jan 2004 Location: New Mexico Posts: 113
Re: Canting - the right answer

Like so many things in shooting, I don't think it's that easy. My ballistics programs solve this problem by rotating the muzzle velocity vector around the y axis (x axis is to the shooter's left, y axis is downrange and z axis is straight up).

The canting formulas posted here are first and foremost an approximation. Probably a pretty good one, but an approximation. What you have to do is find the new elevation and azimuth angles after the rotation about the y axis by the cant angle. I've done the trig and (I think) the new angles are:

elevation = acos(rd)
azimuth = acos(cos(e)/rd)

where rd = sqrt( (sin(e)*sin(c))^2 + cos(e)^2 )

where acos is the Arccos, a is the elevation before canting and c is the cant angle.

I think most formulas out there do not do the rotation, but use a simple cos/sin approximation to the new velocity vector. This will certainly introduce more error.

So I calculated a trajectory with no cant, no wind, and a bullet with a G1 BC of 0.5 at 3000 ft/s:

<font class="small">Code:</font><hr /><pre>
Range Drop Drop Windage Windage
(yds) (in) (moa) (in) (moa)
0 -1.5 *** 0.0 ***
100 -0.0 -0.0 0.0 0.0
200 -2.9 -1.4 0.0 0.0
300 -10.8 -3.5 0.0 0.0
400 -24.6 -5.9 0.0 0.0
500 -45.0 -8.6 0.0 0.0
600 -73.3 -11.7 0.0 0.0
700 -110.6 -15.1 0.0 0.0
800 -158.7 -18.9 0.0 0.0
900 -219.2 -23.3 0.0 0.0
1000 -294.6 -28.1 0.0 0.0
</pre><hr />

The calculated elevation for the uncanted case is 3.35 moa.

For a 10 degree cant, you have the following trajectory:

<font class="small">Code:</font><hr /><pre>
Range Drop Drop Windage Windage
(yds) (in) (moa) (in) (moa)
0 -1.5 *** -0.3 ***
100 -0.0 -0.0 0.3 0.3
200 -3.0 -1.4 1.0 0.5
300 -11.0 -3.5 1.6 0.5
400 -24.8 -5.9 2.2 0.5
500 -45.3 -8.6 2.8 0.5
600 -73.6 -11.7 3.4 0.5
700 -111.0 -15.1 4.0 0.5
800 -159.1 -19.0 4.6 0.6
900 -219.7 -23.3 5.2 0.6
1000 -295.1 -28.2 5.8 0.6
</pre><hr />

The drop is slightly more because the elevation angle is slightly less. Note that the Windage is negative at the muzzle becauase the bullet starts below the line of sight in the unrotated (uncanted) coordinate system.

After canting, the bullet starts slightly to the left (negative x). Also the option "Drops Relative to Target" is checked so that the bullet drop and windage values are not relative to the rotated (canted) coordinate system.

If I plug in an elevation angle of 3.35 moa and an azimuth angle of 0.0 into my formulas above, I get a canted elevation of 3.299 moa and a canted azimuth of 0.582 moa. Plugging these values into my online calculator, I get the following trajectory:

<font class="small">Code:</font><hr /><pre>
Range Drop Drop Windage Windage
(yds) (in) (moa) (in) (moa)
0 -1.5 *** -0.3 ***
100 -0.1 -0.1 0.3 0.3
200 -3.0 -1.4 0.9 0.4
300 -11.0 -3.5 1.5 0.5
400 -24.8 -5.9 2.1 0.5
500 -45.3 -8.7 2.7 0.5
600 -73.6 -11.7 3.4 0.5
700 -111.0 -15.1 4.0 0.5
800 -159.1 -19.0 4.6 0.5
900 -219.7 -23.3 5.2 0.6
1000 -295.2 -28.2 5.8 0.6
</pre><hr />

I also had to set the sight offset to -0.3 inches to the left to simulate the canted case.

As you can see they agree almost exactly.

So to use the formulas, calculate a new elevation and azimuth and adjust the uncanted trajectory by the difference between the uncanted elevation and the elevation obtained from the formula. Do the same for the windage.

For example, the uncanted elevation was 3.35 moa and the calculated elevation when canted is 3.299. The different is about 0.05 moa less than the uncanted case or about 0.5 inches at 1000 yards which is about what we see above. Due to significant figures, you won't see it in the Drop (moa) column.

For windage, the uncanted windage is 0.0 moa, and the canted windage is 0.582 moa. 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).

JBM
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#5
05-16-2006, 12:08 AM
 Platinum Member Join Date: Dec 2001 Location: Mukilteo, WA Posts: 1,092
Re: Canting - the right answer

Great info guys. JBM, could you re-do the above at a more realistic cant, like 5 or 3 degrees or something? I don't think I could shoot a rifle canted at 10 degrees unless somebody had just hit me in the head with a hammer. [img]/ubbthreads/images/graemlins/laugh.gif[/img]
#6
05-16-2006, 05:29 AM
 Platinum Member Join Date: Jun 2001 Location: Potomac River Posts: 5,089
Re: Canting - the right answer

JBM

"rd"

And

"e"

A clue to the <font color="red"> correctness </font> of the calculation is when the cant angle is 180 degrees. In such a case the scope will be on the bottom and the barrel will be on the top.
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The Smokin Fur Rifle Club
#7
05-16-2006, 05:55 AM
 Bronze Member Join Date: Dec 2002 Location: Asuncion, Paraguay Posts: 39
Re: Canting - the right answer

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.

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 )

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.

But when you have a hunting rifle you don't normally change the scope's settings, so you may take a shot at 400 m even if your zero is 200 m using holdovers. In this case the angle between LOS and bore line corresponds to the 200 m zero and the effect of canting would be smaller than if the rifle was zeroed at 500. The sight height does have an effect in this case.

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

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