Canting - the right answer

[Brown Dog]

Just spotted this part of JBM's eg: [ QUOTE ]
Elevation in this example is zero.

[/ QUOTE ] Apologies JBM! /ubbthreads/images/graemlins/blush.gif ...'300 down' it is! /ubbthreads/images/graemlins/blush.gif /ubbthreads/images/graemlins/smile.gif

 

[jhogema]

[ QUOTE ]
Brown dog
at 90 degrees
X = 300
and
y = - 300

[/ QUOTE ]
I agree that at 90 degr cant you get an equal x (left or right) and y (down) offset. Indeed, totally in line with the equiations Gustavo started this thread with.

Personally, I think math and ballistic calculations are just fine, but I do prefer tests to verify that I made no math or programming errors. What I can offer in this respect can be found at this Shooting sports ballictics page
Admitted, air rifle @ 10 m is not exactly long range hunting, but cant is still cant...

 

[Brown Dog]

Jhogema,

Yes, but (as I did) you have failed to notice that in JBM's 90 deg cant example he stated 'for simplicity elevation = zero'.

Under those conditions (albeit theoretical -when would you ever have a rifle sight set such that it gave no elevation to the barrel?) the miss will be x= 0 y = -300 regardless of the rifle's cant.
Only if the rifle has had it's 'up 300' elevation applied will it have the x = +300 y = -300 with a 90deg cant. /ubbthreads/images/graemlins/smile.gif

 

[Gustavo]

JBM, yes I'm using Army Metro (as in Sierra's model), however I cannot see this is the cause to find such differences in Drop values. Please could you verify if your dataset is the same as I posted?

 

[JBM]

It looks like my sight height may have been off (probably 1.5"). So here it is again:

For 100 yard zero, no cant, I get the following:

<font class="small">Code:</font><hr /><pre>
Range Drop Drop Windage Windage
(yds) (in) (moa) (in) (moa)
0 -2.0 *** 0.0 ***
100 -0.0 -0.0 0.0 0.0
200 -2.8 -1.3 0.0 0.0
300 -11.0 -3.5 0.0 0.0
400 -25.5 -6.1 0.0 0.0
500 -47.5 -9.1 0.0 0.0
600 -78.0 -12.4 0.0 0.0
700 -118.6 -16.2 0.0 0.0
800 -171.2 -20.4 0.0 0.0
900 -237.9 -25.2 0.0 0.0
1000 -321.3 -30.7 0.0 0.0
</pre><hr />

Elevation angle is 3.98 moa.

Plugging this into my formula with a 10" cant, I get:

Elevation = 3.920 moa
Azimuth = 0.691 moa

Change in Elevation = 3.98 - 3.92 = 0.06 moa or 0.31 inches lower at 500 yards.

Change in Azimuth = 0.0 - 0.691 = 0.691 moa or 3.62 inches at 500 yards.

My calculated trajectory for a 10 degree cant is:

<font class="small">Code:</font><hr /><pre>
Range Drop Drop Windage Windage
(yds) (in) (moa) (in) (moa)
0 -2.0 *** -0.3 ***
100 -0.0 -0.0 0.4 0.4
200 -2.8 -1.4 1.1 0.5
300 -11.1 -3.5 1.8 0.6
400 -25.8 -6.1 2.5 0.6
500 -47.7 -9.1 3.3 0.6
600 -78.3 -12.5 4.0 0.6
700 -119.0 -16.2 4.7 0.6
800 -171.7 -20.5 5.4 0.6
900 -238.4 -25.3 6.2 0.7
1000 -321.9 -30.7 6.9 0.7
</pre><hr />

This shows the calculated drop to be 47.7" and windage to be 3.3 - -0.3 = 3.6" which agrees well with the values calculated from the formula.

For a 1000 yard zero, I get the following trajectory:

<font class="small">Code:</font><hr /><pre>
Range Drop Drop Windage Windage
(yds) (in) (moa) (in) (moa)
0 -2.0 *** 0.0 ***
100 32.1 30.7 0.0 0.0
200 61.5 29.4 0.0 0.0
300 85.4 27.2 0.0 0.0
400 103.0 24.6 0.0 0.0
500 113.2 21.6 0.0 0.0
600 114.8 18.3 0.0 0.0
700 106.3 14.5 0.0 0.0
800 85.8 10.2 0.0 0.0
900 51.3 5.4 0.0 0.0
1000 -0.0 -0.0 0.0 0.0
</pre><hr />

With an elevation angle of 34.66 moa.

Plugging these into my formulas for a 10 degree cant, I get:

Elevation = 34.133 moa
Azimuth = 6.019 moa

Change in elevation is 34.66 - 34.133 = 0.527 moa or 2.76" at 500 yards.

Change in azimuth is 6.019 moa (started out at zero) or 31.52" at 500 yards.

My calculated trajectory with a 10 degree cant and 1000 yard zero is:

<font class="small">Code:</font><hr /><pre>
Range Drop Drop Windage Windage
(yds) (in) (moa) (in) (moa)
0 -2.0 *** -0.3 ***
100 31.6 30.2 6.0 5.7
200 60.4 28.9 12.3 5.9
300 83.8 26.7 18.6 5.9
400 100.8 24.1 24.9 5.9
500 110.5 21.1 31.2 6.0
600 111.5 17.8 37.5 6.0
700 102.5 14.0 43.8 6.0
800 81.5 9.7 50.1 6.0
900 46.3 4.9 56.4 6.0
1000 -5.5 -0.5 62.7 6.0
</pre><hr />

The difference in drop from the trajectories is 2.7" and for windage 31.2 - (-0.3) = 31.5".

Both agree very closely with the values calculated from my formula.

Gustavo -- what are the values you are getting? I see the final drop and windage, but what are the changes? What did you get for a change in drop and windage using your formula? That's what we ought to be comparing. That way we can eliminate the differences in the trajectory calculations.

 

[TiroFijo]

JBM, not Gustavo, but:

the "drop" number in your example is not "real drop" (vertical distance between the bullet and its line of departure) in my nomenclature, but the "bullet path" ((height of the bullet's point of impact in relation to sight line), and it is related to it by the following formula:

BP = -Drop(R) - R/R0*(SH + Drop0)- SH

SH = sight height
R = range
R0 = zero range
Drop0 = drop at zero
BP = bullet path

Remember the formula we use: H(R) (height of bore line in relation to sight line, as a function of range R) is the same formala as above, less the -Drop(R) part:

H(R) = R/R0*(SH + Drop0)- SH

X(R)=H(R)*sin ß
Y(R)=H(R)*cos ß - Drop(R)

I don't know the value of the "true drop" at any range in your example, but the results seem to correlate very, very well. If you can tell me the "true drop" at 100 yds I'll be able to check with more precision.

 

[JBM]

You're correct, I don't usually differentiate between the two drops (I know they are different), I just don't really think about the line of departure that much. It's not something I use.

The point I was trying to get at is post the corrections given by your formula, not the drop plus the correction. If post the corrections, we can compare the formulas without having to worry about the differences in trajectory calculation tools, atmospheric models, etc.

Sorry -- didn't answer your question:

True drop would just be range*elevation angle + sight height + the drop listed above:

100 yards * 3.920 moa - 1.5 inches + 47.8 inches = 4.1" - 1.5" + 47.8" = 50.4" (Does this look correct?)

This is for the 100 yard zero, 10 degree cant. I got the 47.8 from the 47.5 inches of drop added to the 0.31" of extra drop due to the cant. Is this what you want?

JBM

 

[Gustavo]

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

 

[TiroFijo]

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

Sorry, JBM, we are latin!

Drop at 100 yds is just the drop with the barrel parallel to the ground, with sight height = zero.

 

[JBM]

[ 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

 

[Jimm]

Yall having fun now ? .....................I hope ! /ubbthreads/images/graemlins/laugh.gif /ubbthreads/images/graemlins/cool.gif /ubbthreads/images/graemlins/crazy.gif

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 /ubbthreads/images/graemlins/laugh.gif)

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 /ubbthreads/images/graemlins/grin.gif But remember , my head hurts , my eyes are bugging out .........................

Jim B.

 

[TiroFijo]

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 /ubbthreads/images/graemlins/laugh.gif

 

[Brown Dog]

I've had a Eureka moment.

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 /ubbthreads/images/graemlins/smile.gif .
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.
…..QE is an irrelevance in cant calculations /ubbthreads/images/graemlins/blush.gif /ubbthreads/images/graemlins/smile.gif.

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 /ubbthreads/images/graemlins/smile.gif ]

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.

I think you all owe me a beer! /ubbthreads/images/graemlins/smile.gif /ubbthreads/images/graemlins/smile.gif

 

[edge]

Now my head hurts /ubbthreads/images/graemlins/grin.gif

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

edge.