why do shots drop more when shooting at a small incline at extreme range?

My ballistic calculators predict more drop at a small incline when shooting at extreme ranges, like 2000 yards. Without the calculators, i would have held for the horizontal distance, or multiplied the required hold-over by the cosine of the line of sight. I obviously missed a lesson somewhere. WHY DO THESE SHOTS HIT LOW?

Re: why do shots drop more when shooting at a small incline at extreme range?

Its not as easy as reducing a 2000 yard slope shot to say a 1900 yard flat shot by simply multiplying the slope distance by the cosine angle, because the projectile still has to travel 2000 yards and suffer the due effects of flying through the air that far. ie loss of velocity etc.

Re: why do shots drop more when shooting at a small incline at extreme range?

Well, when shooting at range of, let's say, 1100 yards, a small incline of 5 degrees means the shot will hit slightly high. It might be small enough of a difference to be" lost in the noise", but it does hit higher than a flat shot would. Now move to a target 2000 yards away. Why would the same 5 degree incline cause it to hit low?

Re: why do shots drop more when shooting at a small incline at extreme range?

5 degrees is a very small correction. It should not make much of a difference to your slope to horizontal distance. Not enough angle to see the effect that I described.

You need a much greater angle to get the difference between the slope distance and horizontal distance to get this effect.

I ran the numbers with one of my loads and with this very small angle, the elevation required is still less by 0.2MOA. So maybe you made a mistake in your input? However if you input 1993 yards and shoot it as 0 angle you need less by another 0.3MOA so it depends on how you reduce the distance.

Re: why do shots drop more when shooting at a small incline at extreme range?

This is hard to describe and I am not doing a great job of describing what is in my head, so I will give some examples using my load.

2000 yards flat.......... 76.6 MOA required.
2000 yards down 5 degrees 76.4 MOA required. (So if you used 76.6 MOA without taking the angle into account, you would hit high)

2000 yards x cosine 5 degrees = 1992.4 yards.
1992.4 yards flat requires 76.0 MOA. (So if you used 76.0 MOA over a 2000 yard slope distance you would hit low).

So by simply using the cosine angle and reducing it to 1992.4 yards will give you less elevation than a true 2000 yard shot down a 5 degree slope by some 0.4 MOA.

As I said before, it has to do with how you calculate the shot and the total distance that the projectile travels through the air in real life.

Re: why do shots drop more when shooting at a small incline at extreme range?

I realized this when I found I could have a 140 moa range if I removed some material from my scope base. Now, this is on a 223, so it probably wouldn't have any use, but I was curious. 140 will take my load to 2000 yards at this altitude. So I was checking to see how probable a hit would by estimating different variables like speed deviation and angle error, among others. Well speed deviation means huge errors with a 223, but I found more hold over was required with each degree of incline up to 7 degrees, at which point the amount of holdover would decline as i assumed it would. It is only .5 moa, but at 7 degrees, I would have been turning my turret the wrong way. Look it up. G7 bc=.208 v=2890 alt=4300 temp=59 bar=78. I confirmed the prediction on two ballitic programs, ballistic fte and kac bulletflight. There is some science behind these prediction, and I wish I understood this.