Was going to post something on this a while back, but the method with JBM calculator seemed too convoluted. The other day, cayugad posted a link to this calculator: http://www.handloads.com/calc/index.html


What's great about this one is that it has a total drop calculation. Total drop is the drop from the bore-line. The bore line is determined by where you sights are set and it is line the bullet would travel if there were no gravity. When the muzzle velocity is slower, then the drop is greater, because it takes longer to get to a given range.

Below is load I'm considering for Elk using 410 BB HP. GOAL, be able to keep all my bullets in a 8 " circle at 150 yards.

Let's assume my sighting precision is 2.0 MOAwith open sights.

Let's assume that bullet instability is 1.5 MOA (this is rifle/powder/bullet combination's natural spread with zero velocity distribution) It would be like a machine rest spread where all the muzzle velocities were identical. Bullet instability MOA is a function of range so one can't just keep protracting it out forever.

Lets add them together, 2.0 + 1.5 = 3.5 MOA. So at 150 yards I might expect to keep all my shots in a (3.5 x 1.5) 5.25" inch circle, PROVIDED, my velocity distribution is zero (all my MV's identical)

Of course they won't all be. What is clear, I must keep the velocity component of the spread to less than 2.75" or I won't meet the goal of keeping all the shots in a 8" circle. From the tables below, if I chrono my MV's between 1300 fps and 1375 fps, then the total drop will range from 29.53 and 32.54, a difference of 3.01"or about 2 MOA at 150 yards.The vertical distribution component is range dependent,meaning that the further one goes out, the greater the MOA of the distribution.

Adding the 3.01 inches to the the 5.25" (8.26"), I find I haven't met my goal. In order to achieve it, I must improve my sighting precision or reduce the velocity spread, which are the only two factors I am able tochange. Assuming all the above assumptions true, in the absence of wind, I would find my distribution to form an elliptical shape which is 5.25" wide and 8.26" tall. From a machine rest the shape would be 2.25" wide and 5.26 in" tall.

For those who remember UC's post of a target using the BS conicals. It had a very narrow width and was somewhat tall. The very narrow width was an indication of very good sighting precision and bullet stability. The somewhat tall portion of the pattern was an indication that the velocity distributions were somewhat broad.

Happy Hunting, Phil