If you have a Black Hills .308 cartridge with a 175-grain bullet leaving your muzzle at, say, 2351 feet per second, and you THINK your gun is still zeroed at 400 yards from last Boomershoot but would like to verify this, and you aim at a 100-yard target without changing the scope settings, what is the impact point (height above point of aim) of the bullet at 100 yards that would indicate a 400-yard zero? A 600-yard zero?
(My local range only has 25-yard and 100-yard targets.)
David, try Nikon’s “Spot On” web based calculator
http://www.nikonhunting.com/spoton/
The Firearms Blog posted about it awhile back. I have a Nikon scope and it worked pretty good for me, but I’ll bet even if you don’t have a Nikon scope you can probably play with it with your load and a similar scope setting and get the correct holdovers. Worth a try anyway.
Hornady also has a ballistic calculator.
You really need to know the BC of the particular bullet though. And I didn’t find it on the Black Hills website.
A quick fiddle with it suggests 11.04 inches high at 100yds for a 400yd zero… sound about right? I have a 300 yd zero and I’m 5.5 inches high at 200yds. We’ll find out next weekend.
Chris is right, the Nikon calculator only lists SMKs in 175 grain. Does Black Hills use Sierra bullets?
David, the calculator wants to give that cartridge 2600 fps muzzle velocity and when you do it returns 8.69 inches instead of 11+. You’d better play with it and make sure… hate to make your spotter crazy with a bad zero 😉
That seems pretty slow, but using Brian Litz’s (experimentally calculated) G7 BC of .243, assuming the 175 is a Sierra MK, and using your velocity, I get 11.21″ high at 100 yards with a 400 yard zero.
A Palma or F/TR rifle could expect to get a bit over 2800 fps with that bullet in a .308–but with a long (30″ or so) barrel. 2300 seems *really* slow.
Thanks, guys! I’m shooting from a 16″ pistol barrel, so that’s why it’s slow. 🙂
I have the BC of that bullet written down somewhere with the Boomershoot stuff. I’ll see if I can dig it up. I think it is Sierra MatchKing, but with the moly coating.
You know Joe wrote Modern Ballistics, right? 🙂
BC is needed for a proper answer. All objects fall at the same rate … but a high BC shape sheds velocity faster.
Black Hills uses Sierra Matchkings. BC: 0.496.
Assuming your scope is 1.5″ above the boreline, a 400 yard zero at Boomershoot with conditions of 55 F, 3000′ above sea level, then testing with conditions of 65 F at your range at 200′ above sea level, then expect 10.9″ high at 100 yards.
Assuming 600 yard zero with the same conditions as above then expect 20.1″ high at 100 yards.
Modern Ballistics is your friend.
Kristopher, your first sentence is correct as far as it goes. The other statements are wrong.
Thank you, Joe!
…and the rest of us proles just drool at this display of exterior ballistics know-how…/grin/
BTW, THIS prole would suggest centering all the scope adjustments and starting all over again at the range…a year is a long time to have the pistol without proof of zero.
Yes, Joe … I got the coefficient-drag relationship backwards.
But, if you really want to nit-pick: are you claiming that ( low velocity drag not withstanding ) not everything accelerates towards the ground at 9.8 meters per second per second?
The problem is the theoretical and the practical versus the actual. Find the bullet/load that is the most accurate for your barrel by verifying the tightest groups. Now you know what works for that barrel, at that altitude, humidity and temerature. Other variables such as degree of fowling of the barrel, variabiltity of bullet weights and variability in the brass also come into play, but I tend to live with them. Now that you have the load and rifle all “tuned up”, travel somewhere with a 5-7,000 foot elevation change, 40-50 degree temperature differance, and no humidity. Apples and oranges.
The acceleration experienced by an object in an atmosphere depends on the drag. Terminal velocity, for example, limits the velocity of an object dropped from a great height. It turns out that when an object is traveling well in excess of terminal velocity in the horizontal direction starts dropping it experiences a vertical drag force greater than the vertical velocity component along would indicate. Hence two bullets, one dropped and the other fired precisely horizontal from a gun on a perfectly horizontal plane will not strike the ground at the same time. The one dropped will strike first.
Look at the drop and time to target for any ballistics table. Compute the drop based on the time to target using the expression D = 1/2 x A x T^2. You will find they do not match.
I know. It’s not what everyone believes. But it matches the facts. I am surprised as much as anyone when all the number crunching matches reality.