Okay, Here's what I know-
Caliber=375 H&H
Bullet=300gr North Fork Softpoint
Velocity=2550fps
Bullet Drop=17 inches
Rifle=CZ 550 American

Here's what I don't know-
Distance
Ballistic Coefficient

I was in RSA in April and shot a Kudu that was by far the longest shot I've ever taken, we did not have a range finder with us but we did know this criteria and I thought it would be simple to calculate how far the shot was. I was wrong! All the tables I've found only go by ballistic coefficient and North Fork doesn't have that(at least that I could find). Any help figuring this out would be greatly appreciated. Thanks Joe

If you can't find the BC for your bullet use one of approximately the same shape. Your estimate of drop will have as much or more impact on the answer. If your rifle happens to shoot a 1" group at 100yds then your are talking 3-4 inch at best. So 17 could be at the extreme so the drop might be 14-20".

All you need is the velocity and the drop; the distance can be calculated by physics and the calculus. I was a dumbass in school, but I'm sure one of the physicists or engineers on this site can figure it out. This problem is not unlike an artilleryman figuring out how to hit an enemy emplacement on a hillside at an unknown distance or in the same vein, how to hit a target that's well below the field piece. A body accelerates as it falls through space at an ever-increasing rate until it achieves terminal velocity. At near-earth altitudes, that rate of drop is 32 feet per second per second, meaning the body will drop 32 feet in one second during the first second and an additional 64 feet during the second second. By the end of two seconds, the body has fallen 96 feet. Aerodynamic drag will eventually cause the object to stop accelerating (the BC is a measurement of the drag) and thus achieve its terminal velocity.

I have an email for the answer in to my dad, who is a civil engineer and likes to calculate these kind of things.

I was a bit wrong-- we also need the angle of elevation above zero degrees (horizontal). The formula is something like distance = arctan of the angle of elevation x velocity (with aerodynamic drag in this instance to be considered negligable). The above is from my memory, which ain't too good no more.

Assuming a few things, like a Woodleigh SP and you Northfork have a similar BC.
100yd zero, 17" low at about 300yd
200yd zero, 17" low at about 345yd

Thanks everybody, I appreciate your input. I was wondering about substituting the BE from a different bullet but I wasn't too sure if that would work. Thanks again. Joe

My dad figured it out for sea level on the Earth's equator and for at the North Pole: about 248 yards, plus/minus a yard or two. I can send you the email of the solution if you PM me with your email.

I would say your dad is pretty close is you just calculate drop from the muzzle. Question would be is the 17" from the muzzle or below point of aim? I believe the angle the muzzle is pointed upward will affect the answer.

This is a two second calculation for a ballistic program, if you can tell us your POI (sight in height) at 100 yards. (ie. so Tailgunner doesn't have to guess)

If you give your normal group size at 100 yds, one can also estimate the probably range of distances from which the shot was taken. (per ramrod340's initial response).

May I ask how you arrived at the 17 inches of drop?

Okay, sorry for the delay. We have my rifle sighted in for 1&1/2 inch high at 100 yrds, The rifle groups 1 inch or less if I do my part and don't jerk.
The 17 inch measurement came from measuring where the bullet hit to where I was aiming when the shot was taken. I was aiming at the top of the "mane" above the shoulder and the bullet hit 17 inches low from there which we measured when the animal was there.
Saeed, you've hunted with Pierre', he was doing something with his thumb which he uses to "guesstimate" distances, did he tell you how he comes up with the distances using his thumb?
Thanks again. Joe

Joe
The thumb thing is simple "ratio and proportian" (remember similar triangles from highschool geometry?) 24" (brisket height) at X yards = Y amount of thumb at arms length.
Same as using part of your reticle to estamate ranges.
Knowing your sight offset I can re-calculate the range, again with some of the same assumptions about BC, elevation and atmospherics.
Edit Estamated range 315-325yd (Barnes solid BC .305, Woodleigh RN BC .340).

Tailgunner, Thanks for the yardage and the thumb thing. Let me try this out, if I go to the range and see how much target (in inches) is covered by my thumb at 100,200,etc and I know roughly how much it covers, I just transfer to size of animal?
So if my thumb (or nail) covers say 10 inches at 100yds and a deers shoulder is 15-20 inches I can guess how far it is? I know these numbers aren't correct but I'm just trying to grasp this concept. I've never tried it before and it seems like a simple idea.
Thanks Joe

Joe
That's the concept.
To continue your example, if your thumb nail is the same height as a 10" target at 100yd, than a 15" deer would be 150yd out and a 20" deer would be 200yd out when there the same size as your nail.
To solve it mathmaticly, you need the distance from your eye to your thumb (at full arms length) and the height of the portion of your thumb being used (knuckle to tip, nail, width, nail "moon", etc), to work out the angle created at your eye (height divided by distance = tangent of the angle). To find the distance to a 20" target, divide 20 by the tangent and divide that answer by 36 (to convert to yards). Example 1.1" thumb, 19" arm = a Tan of .0579, 20" / .0579 = a distance of 345" or 9.5yd
Obviously the smaller the referance object (part of thumb) and the larger the target being covered by that portion is (crat vs moose) is the further out the object is.
I find it easest to work it out on paper, and just remember some of the basic distances IE knuckle to tip, width, nail (what ever your numbers work out to).