Calculating windage error due to rifle cant

[Tim K]

Was down at the PD match on Saturday and got to thinking about this, so I did the math.

BD = Bullet Drop
A = Angle rifle is canted from vertical
E = error, left or right

E = BD tan A

Example:
At 425 yards, my .308 drops 31" with a 100 yard zero. If my rifle is canted 5° from vertical, the windage error is:

E = 31" X tan 5°

E = 31" X 0.0875

E = 2.71"

Here's another example because 5° is a lot and 425 yards isn't.

Same rifle at 1,000 yards, bullet drop is 321". Assume cant of only 2°, which seems more realistic.

E = 321 tan 2°

E = 11.2"

[BigBear]

I should've paid attention in math... When did letters start taking the place of numbers?









Yes, I'm kidding. Sort of. hehehe.

[Tim K]

I should've paid attention in math... When did letters start taking the place of numbers?

Math always uses letters. Numbers are for arithmetic.

[BigBear]

Gee... see, I should've paid more attention in LIFE! HAHA.

So Cant is something I don't quit understand. If the bore of a rifle is cylindrical, and your optics are zeroed... with NO wind present - why does it matter if it's tilted one way or the other?

[Graves]

Morgan338LM just turned me on to this sweet little app:
http://www.knightarmco.com/bulletflight/

[Tim K]

Gee... see, I should've paid more attention in LIFE! HAHA.

So Cant is something I don't quit understand. If the bore of a rifle is cylindrical, and your optics are zeroed... with NO wind present - why does it matter if it's tilted one way or the other?

It's difficult to explain without waving my hands, but I'll try.

Since bullets travel in an arc, the barrel and the scope are not "pointing" to the same place. The barrel is actually pointing to some spot above where the scope is. Depending on the range to the target, the difference can be a lot or a little. In my 1,000 yard example, the difference was 321". So, when the bullet leaves my barrel, it moves up 27 feet and then comes back down that same amount to the target. If the rifle is canted to the right, that up and down motion will now have a component of right in it.

Thinking to the extremes sometimes helps illustrate things. Consider what would happen if you held the rifle at 90° to the right. All the compensation for bullet drop has now been turned sideways and the error would be huge (roughly 27' X 2).

If that doesn't help, and I wouldn't be surprised if it didn't, draw a ballistic arc on a piece of paper and cut it out so you have a thin strip about 1/8" wide. Hold it vertically to see what uncanted looks like, then tilt it over 45°. When you tilt it and look down from the top, you'll notice the bullet would have to move first to the right, then back left to follow the arc to the target. Clearly impossible. Adjust your paper arc so it's canted 45° and also a straight line when looking down from the top and you'll see precisely why cant can be a big deal.

[esaabye]

Good info Tim, I wondered how much that mattered when I shoot highpower. Not a great amount but measurable and will cost points.

I know I tend to torque my rifle under heavy sling and I have to keep after it and verify cant or I will stitch left to right across the target during a slowfire string (left handed shooter).

[Not_A_Llama]

Was down at the PD match on Saturday and got to thinking about this, so I did the math.

BD = Bullet Drop
A = Angle rifle is canted from vertical
E = error, left or right

E = BD tan A

Example:
At 425 yards, my .308 drops 31" with a 100 yard zero. If my rifle is canted 5° from vertical, the windage error is:

E = 31" X tan 5°

E = 31" X 0.0875

E = 2.71"

Here's another example because 5° is a lot and 425 yards isn't.

Same rifle at 1,000 yards, bullet drop is 321". Assume cant of only 2°, which seems more realistic.

E = 321 tan 2°

E = 11.2"

With your way of viewing things, I think you're actually looking for the Sine, not the tangent - consider the behavior of tangent at 45 degrees.

There's actually more calculations that go in, and there's a reference point problem - it's a cool problem, I'm writing more up. My basic contention of there being a problem is that the way you're modeling it basically looks like 1) sight in 2) fire 3) rotate 4) fire 5) measure difference, while the more "accurate" behavior looks like 1) sight in 2) fire 3) rotate 4) sight in whilst canted 5) fire 6) measure.

[Tim K]

With your way of viewing things, I think you're actually looking for the Sine, not the tangent - consider the behavior of tangent at 45 degrees.

There's actually more calculations that go in, and there's a reference point problem - it's a cool problem, I'm writing more up. My basic contention of there being a problem is that the way you're modeling it basically looks like 1) sight in 2) fire 3) rotate 4) fire 5) measure difference, while the more "accurate" behavior looks like 1) sight in 2) fire 3) rotate 4) sight in whilst canted 5) fire 6) measure.

Run the numbers, sine and tangent are essentially equal at small angles. Doesn't matter which one you use. It's true there's more to it. For instance, it's more properly calculated using the max ordinate, not the drop based on a 100 yard zero. I get it. I just don't care. The intended point of my post is not to calculate it with utmost precision but to illustrate that avoiding a canted rifle is important for precision distance shooting.

[Not_A_Llama]

It's not meant as a slight. I just thought it was odd you reached for the kinda-off function that's just as hard to calculate.