Canting - the right answer

[Gustavo]

BrownDog and Meichele,

Please see the post from TiroFijo. There you will find the complete package.

God! this post is becoming more and more interesting!

 

[Gustavo]

JBM, please could you run your software with the data I posted and publish the output (first post at the top) ?

I'm pretty much interested in seeing the values your run produce. regards!

 

[Brown Dog]

JBM, Gustavo, Tiro et al,

Firstly, I need to get out more! /ubbthreads/images/graemlins/smile.gif but this is interesting, and JBM's very low (in my opinion! /ubbthreads/images/graemlins/smile.gif) horizontal displacement value is causing me dissonance!

JBM, I still can't see how you're getting such small (0.5MOA) lateral deviation.

I've sketched things out for myself to check my reasoning (and having looked at Tiro's link; I think I'm singing off the same song sheet as him! -Tiro correct me if I'm wrong!)

If the whole thing is examined in terms of angular subtension; the horizontal displacement for 10 deg cant at 1000yds must be far greater than 0.5MOA……I still think you're off by a factor of 10 or so!

[I worked this eg out for myself, metres and mils are easiest for me. Standby for a pleasing mix of metric and imperial /ubbthreads/images/graemlins/smile.gif :
My eg is 3000fps, BC 0.5, scope height 1.5in, zero range 100m. For simplicity, the boreline and the angle of departure are considered to be the same]

……well, I did say they were intended to make the issue clearer for me!

…as regards needing to factor in 'range'; to my mind this is done automatically if you think in terms of angular subtension…......in zeroing (at 0 cant) at a particular range I 'shoot out' range (drop and subtension) issues unwittingly.

In my example, the 10 deg cant effect is the same as if I had changed my scope setting by Az 3.28mils and El 0.28mils.

I can't see why it's more complicated than that, the cant simply has the same effect as a bearing and elevation change. Drop remains unchanged. The miss simply equates to the magnitude of the bearing and elevation change /ubbthreads/images/graemlins/smile.gif /ubbthreads/images/graemlins/confused.gif /ubbthreads/images/graemlins/smile.gif


Right, please now all tell me why I'm wrong! /ubbthreads/images/graemlins/smile.gif /ubbthreads/images/graemlins/smile.gif



PS...before Gustavo's original post on the 5th of this month, I have to admit that my 'interogation' of cant had never gone deeper than remembering the mnemonic "ELXEL" for work ....top prize to anyone who recognises it!

 

[JBM]

I found why I was off when I tried to verify my results again this morning -- I had the wrong coordinate system checked. I think that is also why we are have some disagreement as to what the right answer is.

If you note in my first post, I said I checked the box "Drop and Windage Relative to Target" in my online calculation. This provides drop and windage values in a target reference frame (e.g. relative to the target). My formula calculates the angles in that reference frame so the results of the trajectory are in the same reference frame. If you uncheck this box, you get much larger windage values. I had to think about this for a while and drew a picture so that I could see it. (You can see the picture at cant.pdf). Both are correct, but in different reference frames.

You can play with these online calculators at my site, JBM. Just click on "Calculations" at the top and then Trajectories. You can calculate in either reference frame and see the differences in drop/windage due to different cant angles and zero ranges.

If you look at the picture, there are two sets of axes. One is rotated by 10 degrees, this is the LOS, or canted reference frame. The other is the target reference frame. Also labeled on the target is the impact point. Look at the difference in windage in the two reference frames and you'll see why we got different answers. Also as the drop increases, the difference is more.

I'm still not convinced that the simple sin/cos formulas are correct, but they are in a different coordinate system. I still think mine is correct for the uncanted coordinate system. It's just that it tells you where on the target that the bullet is going to hit. If you are intentionally canting the firearm, my formulas will tell you where it will hit RELATIVE TO THE TARGET. The other case is that if you want to correct for a cant -- In this case you can correct relative to the target or line of sight (LOS), but since you are looking through the scope/sights, the LOS coordinate system is probably easier to do the correction in. In which case, my formulas don't do a lot of good.

Keep in mind that when you zero at a far range you have a much bigger elevation angle. If you look at my formulas, you'll see end up with much larger elevation and azimuth angle changes. This leads me to believe that to minimize the effects of cant, you zero close to minimize the elevation angle (angle between the line of sight and the bore axis). Granted this isn't always convenient, but would certainly minimize the errors due to canting.

Let me think about them some more, I should be able to come up with formulas in both coordinate systems.

JBM

 

[JBM]

[ QUOTE ]
JBM, I still can't see how you're getting such small (0.5MOA) lateral deviation.

[/ QUOTE ]

See post below -- it's the different reference frame.

[ QUOTE ]

I've sketched things out for myself to check my reasoning (and having looked at Tiro's link; I think I'm singing off the same song sheet as him! -Tiro correct me if I'm wrong!)


[/ QUOTE ]

I'm not quite sure what you're sketching. Your first couple of sketches make it seem like you're using a triangle with a 10 degree angle to calculate the windage. This isn't really physical. Think about what happens when we cant a firearm. You rotate about the centerline of the scope/sights. For a clockwise cant, this causes the the barrel to swing to the right -- higher azimuth angle and lower elevation angle. I use two pens, one parallel to the ground, the other under it, at an angle upward to simulate the barrel. The amount of added azimuth is small, no where near 10 degrees -- hence the rotation formulas I originally derived.

The issue and I think major point of contention here is that my differences in elevation and azimuth are relative to the target -- NOT the line of sight. There is a major difference here -- take a look at the pdf I posted and you'll see what I mean.

Like I said, I'll have to think about it in the LOS coordinate frame and see what kind of formulas I can come up with.

JBM

 

[Brown Dog]

JBM,
I've been puzzling over this; I have to say, my brain has just gone "tilt".

I understand where you're going with trying to calculate the changes to Az and El caused by rotating the elevated bore around the LOS. ...I've just tried doing this several times and have come up with junk each time (and can't track down my error -hence my brain frazzle!)

(I'm assuming the statement QE =TE + AS [Quadrant Elevation = Tangent Elevation + Angle of Sight] is in use on your side of the pond):

[apologies, but here I hit full mumbo jumbo! /ubbthreads/images/graemlins/blush.gif /ubbthreads/images/graemlins/smile.gif]

It strikes me that even once the Az change is identified; it will be impossible to calculate what portion of the newly calculated QE is AS and how much is TE. ...and thus you'll have to run an iterative 'reduction' process to work out what portion of the QE is TE, and what portion is AS. As with artillery reduction routines, this would involve plotting and replotting the fall of shot until the shot to shot change to the postulated AS reaches an acceptably small value between 'shots'.

....but where to start? I suspect the 'simple' method in my pics is 'close enough for government work' but when I try to model it at the gun end I, like you, am getting wildly small values.

(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)


Right, I'm off to wrap my head in a cold towel /ubbthreads/images/graemlins/smile.gif

 

[Buffalobob]

Gustavo and TiroFijo


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

And

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


are using left hand and right hand reference frames

This will introduce a small error in your calculation (or more appropriately in your ability to communicate smoothly with each other) at a latter date if you refine your equations to include other interactions.

Using your convention, I would suggest that you modify your equations to add a function which we can call F of theta { F(theta)}. This function will account for such things as right hand twist barrels and vertical gravity velocity vectors. One of the underlying assumptions/simplifications you have used is a horizontal line of sight.

A 30MOA angle between the line of sight and line of departure of the bullet will result in a gravity induced velocity difference at a 1000yds of about 0.3 fps when the gun is in the zero cant position and in the 180 degree cant position. This introduces an error of about 0.1 inches in your answer.

The modified equation would look something like this


Y(R)=H(R)*cos(cant)-DROP(R) + F(theta)

Where F(theta) = {k1(0.5 gt^2) + k2(spin)}sin(theta)

K1(0.5gt^2) can be readily calculated from the JBM's ballistic calculator or it can be simply had by approximation and parameterization.
K2(spin) must be determined experimentally for each gun and bullet, however, because the effect itself is small a constant approximation may be adequate. It all goes back to my original question on your original post. What is it that you really want. Accuracy of the calculation only should be tailored to your needs.

Here is a quote from Kregg Slack and you will notice he has five degrees (not minutes) of angle between scope and barrel.


"this here is the unlucky dog that i shot at 3125yds with the 338 lapua. 300 gr seirra bullet time of flight was a little over 8-sec it took 325 min of elvation to get there at 2750 fps at 5000 ft elvation. the bases on this rifle were solids made to shoot only at that range.but i had 60 min or so in the scope for fine tuneing. fine cross hair 25x50x50mm scope from dick thomas."



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.

 

[TiroFijo]

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:

horizontal projection: X = drop*sin ß

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

 

[JBM]

[ 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.

 

[TiroFijo]

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

 

[JBM]

[ QUOTE ]
JBM,
Sorry about the numbers in my example, it should read meters instead of yards /ubbthreads/images/graemlins/smile.gif
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.

 

[Brown Dog]

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 /ubbthreads/images/graemlins/confused.gif /ubbthreads/images/graemlins/smile.gif

...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. /ubbthreads/images/graemlins/smile.gif

 

[Gustavo]

[ 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

 

[JBM]

[ QUOTE ]

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


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


[/ 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).

I hope this helps.

JBM