[ QUOTE ]
at 90 degrees
X = 300
y = - 300
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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...
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. [img]/ubbthreads/images/graemlins/smile.gif[/img]
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.
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
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.
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: