# Canting - the right answer

Discussion in 'Rifles, Bullets, Barrels & Ballistics' started by Gustavo, May 15, 2006.

1. ### GustavoWriters Guild

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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

2. ### Brown DogWriters Guild

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[ 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 /ubbthreads/images/graemlins/smile.gif !

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

....it's bedtime at this location -no doubt I've missed something /ubbthreads/images/graemlins/smile.gif

3. ### Michael EicheleWell-Known Member

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+1 on post everything!!

4. ### JBMWell-Known Member

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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

5. ### Jon AWell-Known Member

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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. /ubbthreads/images/graemlins/laugh.gif

6. ### BuffalobobWriters Guild

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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.

7. ### TiroFijoActive Member

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Dec 23, 2002
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

8. ### JBMWell-Known Member

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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.

9. ### Brown DogWriters Guild

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[ 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?! /ubbthreads/images/graemlins/smile.gif

...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. ### JBMWell-Known Member

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[ 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.

11. ### Brown DogWriters Guild

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JBM,

We posted almost simultaneously, could you have a look at my question?

12. ### Bart BWell-Known Member

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Dec 25, 2005
[ 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. ### Brown DogWriters Guild

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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 /ubbthreads/images/graemlins/confused.gif /ubbthreads/images/graemlins/confused.gif /ubbthreads/images/graemlins/smile.gif ....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. ### JBMWell-Known Member

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113
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[ 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?! /ubbthreads/images/graemlins/smile.gif

[/ 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.