So here I am on the side of a steep slope, it is 900 yds. down to target level and the target is 900 yds. in front of me. The bullet I am useing goes subsonic at 1,050 yrds. my conclusion is that the target is 222yds. out of range for me. Whats your conclusion?

So would that be a 45 degree downhill shot? A little confused. what did the range finder and ADI say?

Accepting that a target is out of range when it is farther than your bullet's ability to remain supersonic, then yes, you are correct. A2 + B2 = C2 A = Opposite B = Adjacent C = Hypotenuse In this case: 9002 + 9002 = C2 810,000 + 810,000 = 1,620,000 √1,620,000 = ~1,272

Your math is correct, assuming a 45 degree angle. I think in reality, if you are hunting game, the maximun range is usually far less than the subsonic speed would indicate. Most hunting bullets aren't designed for proper expansion and performance at that low of a velocity.........Just passing on a thought.

As Trebark said and demonstrated with his drawing, you are correct. One thing I think gets missed when we are talking long range shots at steep angles is: THE SHOT IS STILL THE FULL UNCORRECTED DISTANCE. The angle only effects the distance gravity has a downward pull on the bullet. So, when the rangefinder says 1200 yards, and the angle corrected dial up is 950......... just remember it is still a 1200 yard shot. Wind, and other environmental factors have to be calculated at 1200, not 950 in that example.

Thanks guys. My buddy was having a hard time understanding that and needed to read it from someone other than myself. I picked the numbers for simplicity. Why 900 horizontal yards is possible and a calculated 900 + yd hold is different.

It should also be said that this calculation is true regardless of wether the shot is up-hill or down-hill. Also, looks like some of the formatting was lost from my first post. A2 = A(squared) or in this case: 900x900 + 900x900 = C(squared) 810,000 + 810,000 = 1,620,000 square root of...1,620,000 = ~1,272

The pythagorean theorem (A-squared + B-squared = C-squared) applies to right-triangles. So it actually, requires a 90 degree angle (the angle formed by A and B). The other two angles (formed by A/C and B/C) can be different than 45s.

That's right, and I knew that. Guess I shoud've worded it differently. I was assuming 45 degree angle just because he stated 900 below and 900 away. One of the sides would have to be a different length for the angle to be something other than 45 if it were a true right-triangle. Good catch.

Fun subject, sort of the meat and potatoes of long range shooting. It's even more fun or frustrating useing bullets that never went to school, they only receive 2 seconds of flight training. I'm pretty sure my buddy will progress a lot faster if I can talk him into going to a LR shooting class that has a formal structure.

Get him to take a class or....set up a 12" steel plate at 600 yards and when he hears the first 'ping' he will be hooked!

Hello Treebark. Shooting gongs and my favorite "tic tac toe" with cheap bottles of cola ( we remove our trash as do all good shooters) is a very fun off season hobby. It's addictive and thought worthy. Getting people to that competitive point is rewarding. I have an extremely difficult time in getting people past that point. I have an extremely hard time teaching the supportive and necessary fluff that distinguishes a shooter from the target wise. If I were a better instructor I could turn people into something better than marksmen or sharp shooters. I am very proud to help people to this point but I am totally incompetent to teach beyond this level. The school of hard knocks has taught me well' but after point X they should go to a school named something other than " hard knocks." However my opinions on this are based on my own inabilities and not the ability of others.