Canting - the right answer

[Brown Dog]

JBM,

Thanks for the explanation, but I still can't make sense of what the pdf is showing …..I'm sure that if we were face to face sketching on a whiteboard it'd be immediately apparent! /ubbthreads/images/graemlins/smile.gif

Meanwhile, I notice my diagrams weren't clear to you either:

[ 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.


[/ QUOTE ]

....that's exactly what the pics were trying to show!! /ubbthreads/images/graemlins/smile.gif….... LOS fixed with the applied elevation rotating about it by 10deg

...I'm not sure that it's important whether we regard the rifle as having rotated 'under' the LOS
....or the line of departure having rotated above the LOS

... the endstate is the same:

-LOS fixed on Tgt.
-Elevation (intended to be applied vertically) now being applied at 10 degrees from the vertical.

Hopefully these sketches are clearer:

Correcting my earlier QE=TE+AS confusion:

We now have the new QE (quadrant elevation) …..8.8mils
The TE (tangent elevation) calculated and applied is unchanged (because the target range and bullet drop are unchanged)
Therefore we have, in effect, reduced our AS (angle of sight) by 0.28mils
Therefore we will miss low by the distance subtended by 0.28mils at 914m

(as well as miss right by the distance subtended laterally by the new bearing!)


I think I'll 'draw stumps' there! ...keen to receive feedback, but I'm done sketching!…..if nothing else, I've improved my PowerPoint skills! /ubbthreads/images/graemlins/smile.gif

Thanks to all for a good brain workout!

 

[JBM]

[ QUOTE ]
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

[/ QUOTE ]

I certainly wasn't taking it as you being stubborn. I think this is a good discussion -- it's making me think, that's certain.

I tend to think of things more mathematically. My online programs will give the canted drop and windage in two coordinate systems. I know it works and it agrees with both our formulas (yours in the LOS and mine in the target coordinate system).

I just haven't see a derivation of the formulas you posted. I think I would have to transform mine from the target coordinate system to the LOS coordinate system, but that's not so easy.

 

[JBM]

[ QUOTE ]

....that's exactly what the pics were trying to show!! /ubbthreads/images/graemlins/smile.gif….... LOS fixed with the applied elevation rotating about it by 10deg


[/ QUOTE ]

I used drawings just like this to derive my equations that I originally posted. Note that your drawing is in the Target 's coordinate system or at least the angles that I found are. I'm just trying to make the leap to the the LOS coordinate system for the way I shoot (using a multi-dot reticle) that's giving my problems.

JBM

 

[JBM]

[ QUOTE ]

I tend to think of things more mathematically. My online programs will give the canted drop and windage in two coordinate systems. I know it works and it agrees with both our formulas (yours in the LOS and mine in the target coordinate system).


[/ QUOTE ]

This was wrong, I think they are both in the target coordinate system, but we're working two different problems.

Would someone be so kind as to work two problems for me using your equations and post the results here. For the same bullet, muzzle velocity, sight height, etc calculate the canted windage and drop at 1000 yards for:

1) A rifle zeroed at 100 yards
2) A rifle zeroed at 1000 yards

Thanks!

JBM

 

[Gustavo]

JBM, taking the same conditions as in the example (first post)

100 yds ZERO

500(x) = 03.27"
500(y) = 47.30"

1000 yds zero

500(x) = 030.80"
500(y) = 108.84"

drop (500) = 65.8"

 

[JBM]

[ QUOTE ]
JBM, taking the same conditions as in the example (first post)

100 yds ZERO

500(x) = 03.27"
500(y) = 47.30"

1000 yds zero

500(x) = 030.80"
500(y) = 108.84"

drop (500) = 356.7"

[/ QUOTE ]

For a 100 yard zero, I have 3.500 moa of elevation.

Plugging this into my formulas for a 10 degree cant, I get a new elevation and azimuth of:

elevation = 3.446 moa
azimuth = 0.608 moa

For a 500 yard target (I'm assuming that's what 500(x) means):

0.608 moa at 500 yards = 3.183"

For a change of elevation of 3.500 - 3.446 = 0.054 moa, this gives a change in drop of

0.283" or a drop of 49.5 - 0.283 = 49.78"

[My 500 yard drop value is -49.5"]

My online calculator gives: -49.7" drop and 2.9" of windage, but the bullet starts
left of the line of sight by 0.3" so total deflection is 3.2". Which agrees.

This is the same 10 degree cant I used as my first example that had about a 6.1" deflection at 1000 yards (but I think some parameters were slightly different). But since the range is twice as far I would expect twice the deflection. You can see this from Brown Dog's nice PowerPoints. More elevation means more azimuth when canted. My first example had a zero range of 100 yards -- very small elevation.

So they agree at 500 yards. I don't know what your change in drop was, but I'm guessing it was close.

For 1000 yard zero, I get 34.61 moa of elevation. Plugging this and the 10 degree cant into my formulas, I get the following adjusted elevation and azimuth:

elevation = 34.084 moa
azimuth = 6.010 moa

At 500 yards, the windage value is comes out to 31.47". You have 30.8". My online calculator gives me 31.2" of windage, but again, it starts at -0.3" to the left (due to the cant), so total deflection is 31.5". My online calculator only prints to one significant figure, but it's certainly close enough. Your formula is off by about 3/4".

My online calculator gives me 113.5 inches of drop at 500 yards (remember it's zeroed at 1000 yards).

For a change of elevation change of 34.61 - 34.084 = 0.526 moa, this gives a change in drop of

2.75" or a drop of 113.5 - 2.75 = 110.75". My online calculator gives 110.7". You have 108.84", but I don't know what the uncorrected drop is so the difference may vary well be atmospheric models, etc.

The reason that I compare to my online calculator is that it does the velocity vector transformation by cant and los angles. It doesn't use an approximation or my formulas I derived above -- it's a different method and independent check.

So I still think my formula is right. I don't know if yours is an approximation or not because I haven't seen a derivation (I'm not really sure it would matter that much, we agree pretty well). I do know that Rinker's book says that for longer ranges, you'll see more lateral deflection than multiplying by the sin() will give you, which tells me that it is an approximation.

JBM

 

[Gustavo]

JBM,

Sorry, I made an error in the drop value!

The post was edited and corrected.

Just to make sure we were using the same data :

MV: 2900fps
BC: 0.490
LOS: 2.0"
Temp: 59°F
Press: 29.53 inchHg
Humidity: 78%

then two Zero Ranges, one at 100yds and the other at 1000yds

Please, could you check your data?

regards, Gustavo

 

[JBM]

[I posted this earlier but don't see it so I'm reposting.]

I didn't use your 500 yard drop value. The equations don't need it. They only depend on the zeroed elevation and the cant angle. I think my answers are OK. What would help would be for you to post the intermediate numbers and how you arrived at those numbers. If you can find a difference in drop (projection?), we can compare that and remove most of the effects of using two different trajectory models. Or you could recalculate using my online program so that we have a common trajectory to start with.

As for the differences in drop, I see you have 78% for the humidity which tells me that you are using the Army Standard Metro for the atmospheric model. My programs don't use it -- they use the ICAO model. Even the Army hasn't used the Army Standard Metro for years. It does have a smaller density which would lead to slightly better downrange performance in tables. I've always suspected this might be one reason industry is reluctant to switch. There is probably some differences in the way the trajectories are calculated too. What did you use to find the drop? What method does the program use?

JBM

 

[Jimm]

My tummy hurts now , my eyes are starting to bulge out and they hurt too . I'm beginning to think that a trip to the dentist would be a fun outing .

I started to cook some food for my family , then after a long time ( 30 or 40 minutes ) I realized that I was still starting to cook food for my family .

/ubbthreads/images/graemlins/laugh.gifJim B.

 

[Brown Dog]

Jimm: /ubbthreads/images/graemlins/grin.gif /ubbthreads/images/graemlins/grin.gif

Anyone without tummy ache:
The test targets at this link may be of interest: http://www.riflescopelevel.com/cant_tests.html

 

[edge]

[ QUOTE ]
SNIP
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.

SNIP

[/ QUOTE ]

All of the BULLET drop does become windage( in the scope view), BUT your "aiming error/vertical compensation" has now become Windage "from the shooters view"!

IMO, Brown Dog is pretty close ( 50+ inches )!

IMO, you WOULD miss right by 300 inches! Your bore is pointed 300 inches ABOVE the scope centerline( under normal conditions), and if you rotate 90 degrees right, then the bore is now 300 inches to the right of the scope! ( assuming a 1000 yard target).

edge.

 

[JBM]

[ QUOTE ]
[ QUOTE ]
SNIP
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.

SNIP

[/ QUOTE ]

All of the BULLET drop does become windage( in the scope view), BUT your "aiming error/vertical compensation" has now become Windage "from the shooters view"!

IMO, Brown Dog is pretty close ( 50+ inches )!

IMO, you WOULD miss right by 300 inches! Your bore is pointed 300 inches ABOVE the scope centerline( under normal conditions), and if you rotate 90 degrees right, then the bore is now 300 inches to the right of the scope! ( assuming a 1000 yard target).

edge.

[/ QUOTE ]

This was an example to show the different coordinate systems. Elevation in this example is zero. So as I said it's a 300 miss at the target, down, not in windage.

 

[Brown Dog]

[ QUOTE ]
it's a 300 miss at the target, down, not in windage.

[/ QUOTE ] Surely it would be a 300 El and 300 Az miss?! /ubbthreads/images/graemlins/smile.gif


.....sorry, quiet Friday afternoon!

 

[Buffalobob]

Brown dog

Correct

If one assumes the gun is zeroed at the center of the coordinate system ( y is vertical and X is horizontal) and one neglects second and third order terms and the range is the range of the scope zero.

At zero degrees

X = 0
and
Y= 0

at 90 degrees

X = 300

and

y = - 300


At upside down

X = 0

and
Y = -600

at 270 degrees

X = - 300

and

Y = - 300


This assume the equation of a circle with a center at X = 0 and Y = -r

Equation is

X^2 + (y - r)^2 = r^2

where X = rsin a

and Y = r cosa


I have to go run some errands while it is daylight but I might have a minor misrememberance in the y term. It should be the same as what Gustavo put up once you get terms and assumptions the same.

Don't worry about Jimm, I have filled his head with dreams of a new gun that will shoot faster than light and distances that reach infinity and the brass never needs triming and the barrel never wears out. His stomach will be allright once Roy sends him some bread pudding.