Shooting Exercise III: Slope Dope

Scenario 1

You're out with your favorite shootin' iron and you've taken a break atop Mt. Shmuckatelli when you spot a (insert desired target type here) in the valley below. Now, you're kinda lost so you've got your map out...it says that Mt. Shmuckatelli is 950ft tall, and it looks like the horizontal distance to your target on the map is 758 yds.

1) what is the slope range to your taget?
2) what is the slope angle?
3) for your windage adjustments, which distance would you place on your scope...the slope range or the horizontal range...and why? (for the sake of argument, let's forget about hold-offs)

Scenario 2

You've come down to inspect the (insert desired target type here), when you spot another (insert desired target type here) up on the hill opposite Mt. Shmuckatelli. Your map gives the elevation of target's location as 408ft and, with the help of your inate trigonomic knowledge, you've SWAGed the shot angle at 17 degrees.

Find:
1) the horizontal range
2) the slope range

S.

 

A man has got to know his limitations.

Scenario 1: With my rig and my marksmanship, the trig and ballistics considerations are moot: at 758 yards, I still have a substantial stalk ahead of me.

Both scenarios need editing to clarify various distances. You give only the elevation of one end of the shot. Are we to take these as relative height changes?

 

There's a little gizmo that can help you with this problem, coincidentally called the slope doper.

 

OK...I'll make it clearer:

Scenario 1
...Shooter's ele: 950ft ASL, target ele: 0ft ASL


Scenario 2
...Shooter's ele: 0ft ASL, target ele: 408ft ASL


That's all the data you need, including what's given in the original post.

S.

 

Ahhh...I know the answer. I was good at these trick problems in high school:


The real Mt. Shmuckatelli is 926 feet tall, not 950!













Ha! Gotcha!

 

OK, I'll take a shot at it, even though my trig is a little rusty.

Scenario #1:

Slope range = 2464'
Slope angle = 67 degrees
Windage adjustment = I won't mess with my windage adjustment, as the problem gives no info. for win drift. I will set elevation adjustment that corresponds to horizontal range, as that is what affects drop.

Scenario #2:

Horizontal range = 1334.5'
Slope range = 1395.5'

Rounding will produce non-material differences in the answers.

 

Ooops, guess I'd better post an answer!

1: vertical=950ft, horizontal=2274ft, slope=2464ft (thank you Pythagoras.) I would label the slope angle as the depression of the line of sight below horizontal: -22.67 degrees or arcsin 950/2464. Dope the drop on the horizontal, dope the wind on the slope. A nit picker (or orals board) might want to consider the gravitational acceleration effects, though they probably won't figure in practical marksmanship.

2: vertical=450, angle=17 degrees, slope range=450/sin 17=1539ft, horizontal=1472 (thanks, again, Pythy.)

Rounded haphazardly and sig figs ignored.


Edit for transcription error(oops):

2: vertical=408 , angle=17, slope range=408/sin 17=1395.5feet, horizontal=1334.5feet

 

Bravo, guys...

Scenario #1
Given:
-Horz distance = 758yds
-Vertical distance = 950ft / 316 yds

Find
- Slope range to target

(a squared + b squared = c squared) ;
= square root of (750 squared + 316 squared)
= square root of 674420
= 821 yds / 2463 ft = slope range

- Slope angle

If (Horz disp = slope disp X cos slope angle), then
cos slope angle = 758yds / 821yds = .923264311815
so cos-1 = 22.592 degrees.

The question regarding wind is a little less concrete. As a general rule of thumb, at the longer ranges and / or extreme slope angles, the windage adjustments for the slope ranges are used. For example, you're an HRT DM, called out for a hostage situation on an airplane on a runway. Your .50 cal is positioned 1600m away from the aircraft, on a 22 degree down slope toward the aircraft. Going through the math, you get a horizontal range of 1400m and change. So you set your elevation for the 1400+m range, HOWEVER, the bullet still has to travel 1600m in the elements...rain, wind, heat, baro P, etc. All those elements will effect the bullet and it's trajectory for 1600m, 200 extra meters than the 1400+m horizontal range. For the shorter shots and / or smaller slope angles, the overall difference may not warrant any additional changes to windage settings. Say your slope range was 550m on a 15 degree slope. Go through the math and you'll find the horizontal distance is about 530m. Not enough to get the panties in a bunch over.


Scenario #2

Given:
- elevation of target above shooter: 408ft (136yds)
- angle of 17 degrees

Find
-Slope range to target
elevation above shooter / sin slope angle
= 136/sin 17
= 465yds / 1395ft

-Horizontal range
elevation above shooter / tan slope angle
= 136 / tan 17
= 444yds / 1332ft


I used the equations found in The Angle Fire Primer , but obviously there are a couple of way to find these ranges. One point remains constant: when taking a sloped shot, the shooter must use the horizontal range for his elevation adjustments. As you can see in the first scenario, the shooter already has the horizontal range. Acquiring the slope range of 821yds might have an impact on the choice of windage adjustments needed depending on the circumstances.

An interesting point was made by benewton in the Angle Fire post regarding other factors that can effect the shot's trajectory and must be accounted for in order to obtain accurate first shot hits. Temp and barometric pressure are far and away the two largest environmental and meteorlogical influences on any long range shot. These subjects have been covered elsewhere, but the correction factors for all the elements are based on the targets range. As you can see from the first scenario, there can be a large difference between the slope range and the horizontal range. Failure to correct for the slope angle (if it exists) is a quick and certain way to earn a miss. When faced with these conditions, the shooter must correct for slope before any other factors are considered.

Thanks for posting!
S.

p.s. you WILL see this material again!

 

The real answer is take the M1A with a high cap mag and walk the rounds to the target.

Did I win? Where do I claim my prize?

 

My Oehler Ballistic Explorer allows fiddling with atmospheric conditions and slopes. Borrowing from echo3mike's example, I grabbed the only .50BMG load in the ammo library, PMC's 660 grain FMJ at 3080fps. I ran the drop table at standard conditions: temp=59 degrees, elevation=0, humidity=79%, barometric pressure=29.53 inHg , slope=0; standard conditions with a 15 degree slope; and current conditions in Baghdad, 93 degrees, 113 ft elevation (Saddam Airport), 15% humidity, 25.43 inHg barometric pressure, with 0 slope and 15 degree slope. The drops calculated at 600 yards are:

Standard conditions
slope 0 degrees= 92.81 inches
slope 15 degrees=89.81 inches

Baghdad SDA
slope 0 degrees=86.48 inches
slope 15 degrees=83.67 inches

Deflection by a 10 mph wind from 3 o'clock are:

Standard conditions
slope 0 degrees=29.25 inches
slope 15 degrees=29.42 inches

Baghdad SDA
slope 0 degrees=22.74 inches
slope 15 degrees=22.89 inches

With a bad guy shielded by a hostage, those 6+ inch differences look pretty big, specially if you are the hostage.

 

Uhhhmmm... the .50 cal was intended more for hard target interdiction... shooting out the engines and keeping the plane on the ground vs. an anti-personell role. Having that rifle in that role at that range (1600m) is unrealistic, especially when there are better options available.

Interesting data, though. I think if you take those numbers out to 1700yds or so you'll see some pretty large deviations in the SAC numbers and the numbers at differing conditions.


S.

 

Yikes!

25.43 inches Hg is not even on the scale of my aneroid barometer! Yahoo lied to me, twice! More later when I can get the real barometric pressure in Baghdad.

 

Not to be the snotty new kid, but could those of you who would use the horizontal distance please explain why the Sniper Training manual FM 23-10 slope dope table says:

RANGE IS GIVEN IN SLOPE RANGE (METERS), NOT MAP DISTANCE
Table 3-3, Bullet Rise at given angle and range in minutes

And as I recall, gravity equals 32 feet per second squared. The number of seconds the bullet takes to get there is proportional to the ACTUAL distance it flies, which is the slope distance, not the horizontal distance.

 

If you were to shoot straight up, how far would the bullet move perpendicular to its original path as a result of gravity? It wouldn't. A straight up shot is just an extreme angle shot. The principle is the same.

In answer to your question, I don't know. If the manual says that bullet drop will be a function of slope distance rather than horizontal distance, then it is wrong.



Scott

 

Come to think of it, you could create a table & method using either horizontal or slant distance. I just don't know which would be easier to comprehend & use. probably depend on the individual.

All the other stuff I can dig up on the subject is very ambiguous as to whether the author is using horizontal or slant distance.

Seems to me it is easier to determine the slope distance and I would prefer to use it in any calculations instead of having to figure the horizontal distance. the table in the FM is easy to comprehend; just subtract that amount from your horizontal dope.

Now a SWAT guy can just use his laser ranging inclinometer to get the exact data, and radio it to another guy in front of a computer, who plugs it into the ballistic program.

Don't understand the point re shooting straight up.

 

intheblack...

I think you may want to check your facts. There is a fairly decent body of data that strongly supports the use of the horizontal range over the slope range when taking a sloped shot. If the slope range to target was accurate and produced hits, there wouldn't be a need for the data in the first place. I hate to bust your chops in a public forum, but you're wrong on a number of counts.

If I range a target at 800yds, 0 slope angle, my load of a 175gr SMK at 2710fps will have a drop of ~197" at SAC. If the angle were increased to 10 degrees, the horizontal range would come to 788yds and the bullet drop is ~180"... 17" might be considered a minor difference, but it's over 2 MOA off. At a 35 degree slope, the horizontal range comes out to 655yds, with a drop of 111.5". An 85.5 yd difference in range estimation at these distances will cause a miss. When the distance isn't as extreme, the difference isn't as noticable: at 300yds sloped range, a 10 slope angle gives a horizontal range of 295yds, and a 35 degree angle gives a horizontal range of 246yds. Not quite as much of a difference.



You also misread the table in FM 23-10:

The values in the left hand column are the sloped range to target. You'll multiply this range by the values found in the table to get your horizontal range, and subsequently use that value for your scope adjustments. It's fairly accurate for a number of conditions, however the system using

(scope adjustment for slope range in MOA) X cos slope angle = actual scope adjustment

is more accurate over more conditions.

S.

 

You said:

"And as I recall, gravity equals 32 feet per second squared. The number of seconds the bullet takes to get there is proportional to the ACTUAL distance it flies, which is the slope distance, not the horizontal distance."

You referenced gravity being related to actual distance. In the context of the discussion, this would be important because of the impact gravity would have on the path of the bullet. In a shot fired straight up, gravity has no impact on the PATH of the bullet. In a shot fired horizontally, it impacts it at the rate of 32" per second squared. As the angle of the shot goes from horizontal to vertical, the impact of gravity on the path of the bullet decreases to zero.

 

I don't mind being shown wrong, I just hate being confused.

Echo3mike, I still don't get your interpretation of the table; were you taught to use this particular table? Let's run these numbers your way:

slant range = 500
incline angle = 30 degrees
This is a 3-4-5 triangle; trigonometry the horizontal range is 400 and the height is 300.

your interp says to go along the 500 row to the 30 column, and find the number 2.3. Multiply 2.3 x 500 = 1150, which cannot be the horizontal range because the horizontal range is 400.

Unless you mean that the sight elevation equal to a "horizontal zero" at 1150 is also the correct setting for a 30 degree target at 500 ???

In which case you still have the intermediate step of figuring out the change from your actual zero to a zero at 1150 ???

I would be interested in understanding the method used to create the table.

TMs and FMs are not much good at explaining what the reader doesn't already know already, but the table legend says (re-arranging for better grammar):

"(table cell shows) bullet rise in minutes, at given angle (column) and range (row)"

Anyone want to run it both ways with a ballistic program? The table is probably for the 173 grain M118 special ball ammo, since the book includes the M24 rifle.

 

Just had a look at the army "Sniper Training & Deployment" TC 23-14 from 1989, and it has NOTHING on inclined fire.

What the heck are the current FM/TMs for sniper training in the various services?

Haven't done a web search yet, but are there any field usable formulas & tables online?

I read the exteriorballistics.com article on inclined fire; still don't have complete understandings of the step-by-step for the 3 methods discussed.

 

Whoa...you're doing too much work. I thought it was a different table: sorry. All the table is saying is how far over the target you'll be (in MOA) if you don't correct for slope. It's not giving ANY info on how to correct the adjustments. The numbers I'm getting for a .308 175gr SMK at 2600fps are pretty close to these amounts using Infinity.

FMs and TMs are notorious for glossing over some subjects. In the 4 or 5 week course at Ft. Benning there's information that's more important to teach than a bit of arcane ballistic trivia that can't really be explored fully in the backwoods of Georgia, and most military / LEO snipers I've spoken with say that basically they're just taught to "plug and chug". It's more mission critical to know how and when to use the formula than it is to know why the formula works.

You can find beaucoup FMs and TMs at Bigger Hammer...at least the ones that are declassified. FYI, the U.S.M.C. S/S manual doesn't have all that much on the subject either. That may change with the newest revision but don't expect much.

There are some methods that are more field expeidient thant the equation I posted above but they're less accurate. The most commonly found table of angle / cosines can be found at Marine Scout Sniper's formulas page. So simple even a Marine can use it...find your slope range, multiply that range by the corresponding value and you'll have your horizontal range. In theory you can apply these values to the elevation (in MOA) for the slope range to acquire the elevation (in MOA) for the corrected range elevation.

 

Well I've spent Sunday cogitating on this; turns out my initial gut perspective that the distance is important means that I was thinking in the direction of the Improved Rifleman's Rule, where you apply the Cosine correction to the drop, not the distance. I worked out an example to myself and put it into the discussion at

-More On Range Estimation And Angle Fire-
in the Rifle Country forum of The High Road.
http://www.thehighroad.org/showthread.php?threadid=29445&goto=newpost

I knew about Biggerhammer, but I'll check out the other URL.

More theoretically, suppose you zero your rifle at 500 yards, but actually the target was, say, 25 yards away from horizontal. I wonder when that sort of additional error in the mix becomes significant.

And how the heck do you notice slope on long shots in the first place? At 1000 yards, if the target is actually 87 yards above horizontal, that's a 5% angle. I doubt you could see it visually, and scopes & spotting scopes don't have inclinometers built-in.

5 degrees at 1000 comes to 1003.8 yards slant range. That's something less than 6" of extra drop (my shareware won't give an exact enough interval above 500 yards).

And on short steep shots, I wonder what amount of error in the slope estimate becomes significant? Does an error of 5 degrees on a roof-to-ground shot screw the pooch?

Speaking of which, the only ballistic program I can find for the Mac is the shareware Ballistik Pro (by a German guy).

 

InTheBlack,

A 30, 60, 90 degree triangle is not a 3-4-5 triangle. The ratio of the sides is 1-2-3^1/2 (one to two to the sq. rt. of three).

You can check this by multiplying 500 by the cosine of 30. Equals 433 which is the horizontal distance, not 400. Elevation is 250, not 300.




Scott

 

Oops, didn't mean for it to necesarily be 30 degrees; I was using 300-400-500 ranges for something else and made a triangle out of it. Should have done the math.

Anyone have other tables? Its good to have different ways of solving a problem; you can pick the most convenient or the one you can remember easiest.