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
Linear Mode
