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# Canting - the right answer

#50
05-22-2006, 03:47 PM
 Silver Member Join Date: Jun 2007 Location: Buenos Aires, ARGENTINA Posts: 131
Re: Canting - the right answer

JBM,

TiroFijo is right on the money regarding your values of "Drop" and "True Drop". I too differentitate both values.

So, my data corresponds to Drop not Path (in other words NOT taking into account Sight Line (Drop) and in Path considering it)

Using the same dataset as in the previous example

PATH(500) = -47.0" with a 100yds ZERO
PATH(500) = -111.5" with a 1000yds ZERO

of course DROP(500) = -65.8" with both ZEROS
__________________
regards, Gus

http://www.patagoniaballistics.com
#51
05-22-2006, 05:39 PM
 Bronze Member Join Date: Dec 2002 Location: Asuncion, Paraguay Posts: 39
Re: Canting - the right answer

JBM, Gustavo, we're in a crossfire [img]/ubbthreads/images/graemlins/smile.gif[/img]

Sorry, JBM, we are latin!

Drop at 100 yds is just the drop with the barrel parallel to the ground, with sight height = zero.
#52
05-22-2006, 08:16 PM
 Silver Member Join Date: Jan 2004 Location: New Mexico Posts: 113
Re: Canting - the right answer

[ QUOTE ]
JBM,

TiroFijo is right on the money regarding your values of "Drop" and "True Drop". I too differentitate both values.

So, my data corresponds to Drop not Path (in other words NOT taking into account Sight Line (Drop) and in Path considering it)

Using the same dataset as in the previous example

PATH(500) = -47.0" with a 100yds ZERO
PATH(500) = -111.5" with a 1000yds ZERO

of course DROP(500) = -65.8" with both ZEROS

[/ QUOTE ]

Of course this whole conversation started about a formula for the canted path (your term) and windage. I STILL haven't seen a worked example. I've posted two worked examples including the data I start with (the trajectories). All you post are the final answers. That makes it hard to see what is going on.

JBM
__________________
JBM Small Arms Ballistics -- http://www.eskimo.com/~jbm
#53
05-22-2006, 08:37 PM
 Platinum Member Join Date: Jun 2007 Location: west of Little Rock ,Ark. Posts: 1,271
Re: Canting - the right answer

I am making fun but I do appreciate the info that is tied up in all those posts and now am wanting to se if youse guys can talk dummy, err , I mean Jimba speak .

All my life I have made my living doing physical things ( except that little while I worked for Unc Sam [img]/ubbthreads/images/graemlins/laugh.gif[/img])

Point is that if a vertical line is out of whack , it looks out of whack and I make adjustments . Of course I cant get it perfect ! but I can get it so close as to be indistinguishable . I am sure that this is not only pertinent to me !

So ..............what cant angle is necessary for it to be detrimental to 1000 yd shooting ? I know you have provided the info here for me ( uuuhh uuuhhh) to figure that out [img]/ubbthreads/images/graemlins/grin.gif[/img] But remember , my head hurts , my eyes are bugging out .........................

Jim B.
#54
05-22-2006, 08:56 PM
 Bronze Member Join Date: Dec 2002 Location: Asuncion, Paraguay Posts: 39
Re: Canting - the right answer

JBM, could you please tell me this number to "translate" into my system: drop at 100 yds, with the barrel parallel to the ground (no elevation dialed in), with sight height = zero; or how much elevation you need for a 100 yds zero, stating the SH you use.
TIA

jimm, why do we spent time in stuff like this? I do ask myself the same question sometimes [img]/ubbthreads/images/graemlins/laugh.gif[/img]
#55
05-23-2006, 05:07 AM
 Gold Member Join Date: Sep 2003 Location: Blighty Posts: 637
Re: Canting - the right answer

Standby…

I’ve been having trouble with the idea being mooted here that at 500yds, the effect of a 10 deg cant for a rifle zeroed at 1000yds is roughly 10 times bigger than for a rifle zeroed at 100yds ………to use JBMs figures:

At 500yds
100yd zero 10 deg cant: x = 3.3” y = -0.2”
1000yd zero 10 deg cant: x = 31.25” y = -2.76”

This has been, for me, been entirely counter-intuitive [img]/ubbthreads/images/graemlins/smile.gif[/img] .
The effect should be the same in both examples……

..I’m embarrassed to say, it was only this morning whilst walking the dog, that the reason for the error in these calculations popped into my head:

Bottom Line Upfront: JBM et al; the erroneously low ‘100yd zero’ values you are producing are caused by the fact that you are basing your calculations on the fired QE (Quadrant Elevation)………..you should be basing them on the TE (Tangent Elevation) at the target range.

Here’s why cant effect calculations ‘factoring in zero distance’ are erroneous (and I hope you will see, counter intuitive):

(Again using JBM’s figures)

Current reasoning based on fired QE:

Effect at 500yds:
100yd zero (Fired QE = 3.98 MOA)10 deg cant: x = 3.3” y = -0.2”
1000yd zero (Fired QE = 34.66 MOA)10 deg cant: x = 31.25” y = -2.76”

So, virtually a factor 10 difference due to the 100yd zero? ….No!

Consider this:

[this would have been easier to illustrate if the original data had been for cant effects at 1000yds rather than 500yds…..but I don’t want to spend time trying to recreate JBM’s data [img]/ubbthreads/images/graemlins/smile.gif[/img] ]

Had the 500yd target been set up in a 47.5” dip in the ground (ie a small gully 4ft lower than the firing point) a rifle fired at the '100yd zero' QE of 3.98 MOA would have resulted in a target round.
ie the rifle would now be zeroed at 500yds

So,
QE = 3.98 MOA produces a 100yd zero when AS (angle of sight) is zero
And
QE = 3.98 MOA produces a 500yd zero when the target is 4ft below horizontal

I hope you can now see that it is entirely counter-intuitive to assume that when we call QE 3.98 the ‘100yd zero’ it will produce a smaller cant error than the same QE when we call it ‘500yd zero (target 4ft below horizontal)’.

What has changed? ……..The Angle of Sight (AS) has changed; it has changed 'unwittingly' and in a sense that could almost be considered 'virtual' (ie by firing at 500yds with a 100yd zero, you have introduced a virtual change to the AS). The TE (tangent elevation – that, in this case, describes the elevation required to account for bullet drop at 500yds) is identical between the 2 examples.

Remember QE = TE + AS

You must run your calculations on the fired TE at the target range.

If you base your calculations on the fired QE you are, in effect, failing to take into account a changed AS

[ie the QE applied for a ‘100yd Zero’ would also be the QE applied for a ‘1000yd zero with target 26ft below horizontal’. The proportion of that QE that is TE is the same in both cases (the bullets drop [NB not ‘path’] at 1000yds is the same regardless of the assumed zero distance) what has changed is the proportion of the QE that is AS. …..when we treat the ‘100yd zero’ as a ‘1000yd zero with target 26ft below horizontal’, in effect, a negative value AS has been applied. Calculation based simply on QE do not take account of this ‘virtual’ AS change.]

In an earlier post I was on the money when I stated:

[ QUOTE ]
it will be impossible [I'd now amend this to " you mustn't forget"] 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'.

[/ QUOTE ]

But I failed to make the mental connection that TE is the critical value here.

#56
05-23-2006, 06:22 AM
 Platinum Member Join Date: Nov 2005 Posts: 1,088
Re: Canting - the right answer

Can we just take the scope off of this rifle, cuz its broken anyway, and shoot the rifle with and without a cant?

edge.

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