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

[ajhardle]

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?

 

[Topshot]

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.

That is why it hits lower.

 

[ajhardle]

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?

 

[Topshot]

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.

 

[Topshot]

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.

 

[ajhardle]

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.

 

[royinidaho]

Do a search on this topic. Years ago Shawn Carlock, I think it was, lead a discussion. Buffalobob chimed in.

The result as I recall was that multiplying the range by the cosine of the angle is an incorrect calculation.

I worked it out on a spread sheet, back then, it was spot on. Can't recall the proper placement of the cosine in the equation.

I rely on "Shooter" at this point. Haven't missed much since switching to the software.

 

[ajhardle]

It has do with the actual distance a bullet travels through the air. Of course. But now I am more confused because if I enter a negative angle, hold over declines. Following the arc of a bullets path, a negative would lengthen the actual distance travelled correct? Now my holdover went from 141.5 to 138 with -7degrees. Why?????

 

[ajhardle]

I assume this is gravity helping the downhill shot, but I would like a more detailed description from some one who knows.

 

[Augustus]

Guys, when shooting at angles the line of sight is a longer distance than the actual horizontal distances. For our purposes gravity only effects the projectile for the shorter horizontal distance; therefore, if you enter the line of sight in your software it will give a larger value for the holdover causing the Rd to go high.

You really don't need to worry about inclination angles unless the angle is fairly steep or you are shooting at extreme range. I did the trig on a 5 degree inclination at a mile and there were only a couple of yds difference in the LOS vs the horizontal distance.

 

[Augustus]

Oh, it matters not a whit if the inclination is uphill or downhill. If you enter your LOS distance on a steep slope up or down your Rd will strike high. It's all about the distance gravity acts on the projectile not the LOS.

If you want to get very deep into the subject there is another part to the puzzle but it seems to be negligible and we have other things far more important to worry about.

 

[ajhardle]

......there are much more important things, and at that range, I think finding a proper drag curve is number one. So.... let' say I'm shooting my 10 shot groups and I enter 5.9 degrees when it was negative 5.9. My calculator says flat is 2964", 5.9 is 2918, and -5.9 is 2978 @ 2000 yards. No joke here. A five foot difference between uphill and downhill. Now using the wrong angle, my drag curve would be far enough off to be confusing. The scientist generating the formulas seem to know about something we don't.

 

[Augustus]

You sir, are more than a mite confused. There is no way in hell there is a 5 ft difference in your solution when shooting uphill vs downhill at 2000 yds with a 6 degree slope. If the LOS is 2000 yds the horizontal distance would be around 1989 yds. That is 11 yds difference, if you use LOS distance for your input your POI will be 11 inches high regardless if you are shooting uphill or downhill. This info is for my 375 Chey-tac. I have a couple of locations at 1770 yds and 2000 yds, the slopes are just a shade over 6 degrees. I have fired many rds from both ends of the ranges, trust me, there is not any difference in the solutions.

 

[ajhardle]

Augustus, i am confused. You are a more experinced shooter, and i assume more skilled shooter, than I, and your knowledge is more valuable than my findings that may never affect my shooting. That said, I am just relaying info. It is there, you are missing two key points.
First is that am talking about the difference between uphill and downhill. You may have understood that, but make sure to use the negative sign in the calculations.
Second point is EXTREME distances. The 375 cheytac is amazing and makes extreme much less extreme compared to varmint cartrige.
So I implore you to try two things: Calculate your 375 at 3000 yards at both uphill and downhill angles of 6degrees, and also try my 223 load with g7 bc of .208 at 2890 fps.
I know how ridiculous 2000 with a 223 sounds, I am just using it as an example to show that at extreme distances things vary between uphill\downhill shots.
(Ballistic FTE predicts 70" difference between 6 and -6degrees for rm 375 rbtat @3400fps at a range of 3000 yards. I thought that was interesting.)