JBM, I believe the basic formulas I posted are widely used, perhaps with a different twist or additional refinements. When Gustavo, our dutch friend Jeroen (the author of the article in the link I posted) and I were trying to find out more about canting, I wrote to Bill McDonald of Sierra, and he told me: "I believe your equations are precisely correct. Drop is the radius of the circle as measured from the extended bore line."

Of course I don't want this or any discussion to turn into a "magister dixit" kind of argument, I'm just sharing what we have found and the formula we came out to correct at any range, cant angle, and zero range.

The values you posted for horizontal deflection seem too small, by a factor of ten or so, compared to everything I've ever read on this topic.

In all cases the important thing is to know the angle of the LOS to the bore line, and its projection at any given distance.

For example, if we take a 175 SMK @ 2680 fps, std. cond., SH = 1.75", the Sierra Infinity program predicts a drop of 541.13" at 1000 yds

To zero at 1000 yds, we have to compensate for SH + drop = .75" + 541.13" = 542.88"

this is: 542.88"/10.47 = 51.85 MOA = 0.864º

the angle between LOS and bore line is 0.864º

the two lines intersect at: 1.75" / sin 0.864º = 116.05" = 3.224 yds

With any cant angle to the right, the POI is going to move to the LEFT, and UP, from the muzzle up to about 3 yds; and from that point on is going to move RIGHT and DOWN.

this angle subtends at 1000 yds: (1000 - 3.224) * sin 0.864º = 15.030 yds = 541.13" (this is the drop at this distance)

so the POI movement, in relation to cant angle, is:

[ QUOTE ]
JBM, I believe the basic formulas I posted are widely used, perhaps with a different twist or additional refinements. When Gustavo, our dutch friend Jeroen (the author of the article in the link I posted) and I were trying to find out more about canting, I wrote to Bill McDonald of Sierra, and he told me: "I believe your equations are precisely correct. Drop is the radius of the circle as measured from the extended bore line."

[/ QUOTE ]

I'm not arguing that. I just don't understand the equations and I'm a little skeptical that the final equation is so simple. That's all.

[ QUOTE ]

The values you posted for horizontal deflection seem too small, by a factor of ten or so, compared to everything I've ever read on this topic.

[/ QUOTE ]

Read my previous post -- different cooridinate systems -- RELATIVE TO THE TARGET, they will be less. The PDF I posted shows why. When I was interested in cant errors, I wanted to know how much I would miss by when I canted accidently. I don't intentionally cant and then correct in my scope for it. What I found, was the relative to the target (which is what matters to me) the shift of the point of impact was much less than the typical formulas give you which were relative to the line of sight. In other words, if your cant formula gives me 57" of windage error in the line of sight coordinate system, that doesn't mean that I'm going to miss the TARGET by 57". It's actually less at the target. Much less. That was my point, for a an accidental cant of less than a degree, I don't worry about it.

[ QUOTE ]

In all cases the important thing is to know the angle of the LOS to the bore line, and its projection at any given distance.

[/ QUOTE ]

Which is exactly what my original equations give, again, relative to the target. They provided the changes in elevation and azimuth for a rotation around the line of sight. I've shown them to be almost exact, even for large angles.

[ QUOTE ]

For example, if we take a 175 SMK @ 2680 fps, std. cond., SH = 1.75", the Sierra Infinity program predicts a drop of 541.13" at 1000 yds

[/ QUOTE ]

You might want to check that drop number, that's about 100" off of what they predict on their website, for a .308 175gn MK. My online program gets about 430" drop at 1000 yards which is close to what they say on their online tables. Am I using the right bullet?

[ QUOTE ]

To zero at 1000 yds, we have to compensate for SH + drop = .75" + 541.13" = 542.88"

this is: 542.88"/10.47 = 51.85 MOA = 0.864º

the angle between LOS and bore line is 0.864º

the two lines intersect at: 1.75" / sin 0.864º = 116.05" = 3.224 yds

With any cant angle to the right, the POI is going to move to the LEFT, and UP, from the muzzle up to about 3 yds; and from that point on is going to move RIGHT and DOWN.

[/ QUOTE ]

I agree.

[ QUOTE ]

this angle subtends at 1000 yds: (1000 - 3.224) * sin 0.864º = 15.030 yds = 541.13" (this is the drop at this distance)

so the POI movement, in relation to cant angle, is:

horizontal projection: X = drop*sin ß

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

[/ QUOTE ]

Again, I'm not disagreeing with you, I was just skeptical. I need to work out the equations myself to see it.

JBM,
Sorry about the numbers in my example, it should read meters instead of yards [img]/ubbthreads/images/graemlins/smile.gif[/img]
BTW, this "modified JBM" http://www.mega.nu:8080/traj.html
that uses several G1 BC's like the Infinity has almost identical results.

[ QUOTE ]
JBM,
Sorry about the numbers in my example, it should read meters instead of yards [img]/ubbthreads/images/graemlins/smile.gif[/img]
BTW, this "modified JBM" http://www.mega.nu:8080/traj.html
that uses several G1 BC's like the Infinity has almost identical results.

[/ QUOTE ]

Yeah, but it uses the old drag functions. It gives good results, but I don't know what he has modified so I use my new one. It has a bullet library -- just select the bullet and it will use the Sierra published BCs (all of them) to calculate a trjectory.

earlier I said [ QUOTE ]
interestingly, on your linked program, setting zero distance as 1000, cant 10 deg and ticking 'el corrn for zero range' 'az corrn for zero range' and 'drop and windage rel to tgt' .....the windage value it gives is 57" !! (same as my 'clockwork' calculation on your data earlier

[/ QUOTE ]

but you say: [ QUOTE ]
if your cant formula gives me 57" of windage error in the line of sight coordinate system, that doesn't mean that I'm going to miss the TARGET by 57". It's actually less at the target.

[/ QUOTE ]

...I'm failing to understand the differences here.... your own program gives 57" if everything is set to 1000 [img]/ubbthreads/images/graemlins/confused.gif[/img] [img]/ubbthreads/images/graemlins/smile.gif[/img]

...I suspect the answer will lie in the different coord sytems you refer to....which I'm not understanding at all [I'm sorry to say, I couldn't fathom what the pdf was showing]! ....grateful if you'd explain them. [img]/ubbthreads/images/graemlins/smile.gif[/img]

[ QUOTE ]
Gustavo and TiroFijo
Now then, if you are really bored and have time on your hands you can further eliminate certain other simplifying assumptions and the calculation will be exponentially more difficult. On of your assumptions is that at zero cant the scope/ line of sight is centered over the bore/ line of departure. There are many applications where the scope is offset to one side of the bore. I have such a gun.

Just some thoughts to keep the neurons firing.

[/ QUOTE ]

Yes, a good suggestion to keep the brain on the move.

However, I do not much appreciate the idea of having to deal with models, that while correct in terms of the detailed data they incorporate, tend to increase complexity without any real gain.

So far, I still believe that the math under the development here presented with Jeroen and TiroFijo is sound and correct above all.

However, the nice part here, is to have so many posters on board in order to discuss issues further.

Tiro, was in my opinion, very clear and concise on the explanation and the math reasoning behind the math.

Please DO NOT UNDERSTAND this as being stubborn! just that I still believe in the result obtained and the contrast with some field results and other software, like PCB and QuickTarget

...I'm failing to understand the differences here.... your own program gives 57" if everything is set to 1000 [img]/ubbthreads/images/graemlins/confused.gif[/img] [img]/ubbthreads/images/graemlins/smile.gif[/img]

[/ QUOTE ]

Check the "Drop and Windage Relative to Target", zero at 100 yards and see what you get.

[ QUOTE ]

...I suspect the answer will lie in the different coord sytems you refer to....which I'm not understanding at all [I'm sorry to say, I couldn't fathom what the pdf was showing]! ....grateful if you'd explain them. [img]/ubbthreads/images/graemlins/smile.gif[/img]

[/ QUOTE ]

Think about it this way, if I tilt my rifle to the right 90 degrees, ALL the drop becomes windage in my scope. As an example, if I my elevation angle is zero (just to make it simple), and I have a 90 degree cant, my 300" drop (made up number) becomes 300 inches of windage in my scope. It's still 300" of drop at the target -- I didn't miss the target by 300" to the right.

Take a look at my PDF. The tilted cross represent the scope crosshairs, the untilted ones represent an uncanted scope (the same as the target coordinate system). Notice that the windage of the impact point is greater than the windage in the untilted coordinates. The drop is almost the same. The more drop you have the greater the discrepancy. If you don't have any drop the differences between the two coordinate systems is small (maybe even negligible).