Andy, (or should i say Redmist?)
I have been following the very same thread and our good friend Pete Lincoln has solved the mystery.
You will be able to read about it yourself, but for the benefit of the guys here, it seems a important bit of imformation was missing from the equation.
The 282yd target was in fact at a higher elevation then the 100yd target, when shot from the same bench.
Below is Pete's explanation with all the maths,the guy in question is also called Pete.
Pete, you need some instruction in the art of rifle shooting mate. not to mention a reminder of some of the math you learned at school.
When a rifle is zeroed in on a level or nearly level range and then is fired either uphill or downhill, the rifle will always shoot high. This effect is well known among shooters, particularly hunters and Military and Law to work out the cosine
In a right-angled triangle, the size of any angle is related to the ratio of the lengths of any two sides by the trigonometric functions. The basic functions are sine, cosine, and tangent. These functions are based on the similarity of triangles that have a right angle and one other angle in common.
If we imagine the 3 sides of a right angled triangle as bieng, AB, BC, AC.
A bieng the point at where the rifle is, B bieng the target, and C bieng the point directly below or above (depending on uphill of down hill shooting) the target at the same elevation as A. we need to work out the angle at A and thuis determin the cosine of this angle,. we do this by using the following:
The cosine of angle A is the ratio of the lengths AC to AB.
Am i making sense so far? ( Squaddies get taught this with regards to sniper rifle employment and laying artillery)
Imagine you the shooter being at your bench A and the target is sitting on a hill at B. you use a laser range finder from your pos A and determin that the target is 282 yards away. You use your Mildot Master, a protracter or theodolite and determin that the angle from level up to the target is a 35 degrees slant. By using a calculator (Line of sight distance x cosine of firing angle) you can thus determin the distance AC, which is the distance for which you would need to adjust your trajectory to make a direct hit on target at B.
hence we have 282y ( AB) x . 0.81915204428899 (the cosine of the 35 degree angle)
which gives us 231 yards.
hence.. what i am getting at, ???
If : the target at 282 yards lazer measured distance from you up the hill was at an angle of 35 degrees elevation, then your claimed trajectory of 1 ich low doesn't reflect the actualt level trajectory of 282 yards but an actual drop value for the distance of 231 yards,
once again for the slow..
given a zero of 1"high at 100yards on a target level at the same elevation or altitude with the rifle, you shoot at a target up a 35 degree slope and see a drop of only 1 inch below point of aim ( 2inch below 100y point of impact)
however as the target is up a 35 deg slope, the actual drop value refelcts a point in the trajectory for a distance from muzzle of 231 yards
which, when we put the elevation oif target above launch site into the trajectory calculations, suddenly we are much nearer the computer generated figures of
1 inch high at 100y. smack on at 200 yards and -4inch at 280 yards.
i would bet that if you measure the angle of elevation to your target at 282m from your bench, work out the cosine of that angle in order to get you the actual distance value for which you would adjust, compute that distance into the computer program, you would come out with your claimed only 1 inch drop.
angle to target, be it up or down hill is an important part of any trajectory mathematics.
There is an excellent tool to help with this problem, manufactured by a friend of mine, Ward Brian, and the name of that gadget is the Angle Cosine Indicator
I'm sure you have seen it.
Ive used it often,
for instance, i shot a goat last summer in Mallorca (infact i shot 9)
tough old billy at a laser measured 400m.
down on the side of the opposite mountain, i happened to be about half way up the side of the highest mountain in Mallorca and the goat heading up hill out of the valley bottom, we first saw him at about 800m distance and we closed the gap until we could get no nearer due to a 300m sheer drop down a cliff.
I worked out the cosine of the angle to the goat, goat was at a 40 degree angle below at 400m, cosine of 40 deg angle was(and still is) 0.7660. concurrent with the markings on my Angle Cosine Indicator
, when i held the reticle on the goat i saw the number on the ACI or 76.
hence i made a quick mental calculation of 400m x .76 = 304.
i knew that i had top adjust my 4-16x50PMII elevation turret for a distance of 304meters.
this i did, up 12 clicks ( i normaly use 13 clicks but due to the altitude and heat i knew from experience i wouldn't need the full value) i also added 3 clicks right windage to compensate for the approx 5mph wind.
i put 2 rounds through the boiler room and one through the kneck with in the space of about 8 seconds and now have a very nice goat head hanging on my wall.
now if i hadn't allowed for the angle, i would have adjusted the full 23 clicks needed for 400m, minus 2 for the altitude and temp etc, and would have over shot my aim point the goat by about 12 inches. i might have been lucky and just clipped his spine, but a it wouldn't have been what i'd call surgical precision.
what i'm getting at is that the elevation of your target above level, may be the bug in the system when it comes to matching your tested point of impact to the computer date provided by Nowler and Redmist.
you may all be correct. but your claim of 1inch low may be a product of the fact that the target is elevated at whatever degree and thus would refelect a drop value for a distance nearer than the actual target distance.
I hope i made sence.
Pete LEnforcement Snipers or Marksmen,
you need to use the following formula
Line of sight distance x cosine of firing angle