The shortcut posted was Jeff Hoffman's modification of the of the long hand formula, sort of a modified British Method
It still requires a constant that has to be calculated it's just a smaller number,
Hoffman’s Formula As initially developed, his formula is intended for .308 Match Cartridges, either the 168- gr. BTHP round, or its cousin, the 175-gr. round. His formula has three basic steps:
1. Make an accurate range estimate and preliminary wind speed calculation as if it were a 10 MPH full-value wind
2. Adjust this calculation for actual wind speed
3. Adjust again for actual wind value His formula’s First Step requires estimating the range and then using this figure as RANGE (in hundreds of yards) –1 (a constant) = Required Correction in Minutes of Angle.
For the sake of the formula, assume this correction is for a full-value, 10 MPH wind. This means, if your target is 700 yards away, you would calculate: 7 (hundreds of yards) – 1 (Jeff’s constant) = 6 Minutes of Angle Compensation. I emphasize: This first step does NOT consider the actual wind velocity or direction. It is only the First Step of a three-step formula. His formula’s Second Step: Adjust the Minutes of Angle Compensation for the actual wind velocity, expressed as a percentage of a 10 mph wind. Continuing with our first example, with a 700-yard target, let’s now consider that the actual wind speed is 5 mph. Already we found that 6 Minutes of Angle [7 – 1 = 6 MOA] was yielded, so we now multiply that 6 by 0.5 because the A increment riflescope. Or, you would hold the equivalent of 2 MOA into the wind – at 700 yards, that means holding 14 inches into the wind.
Restating Our Example To keep things clear, let’s follow the example again, from beginning to end. Your target is 700 yards away. So expressing it in hundreds of yards, we have 7 – 1 (Jeff’s constant) = 6 Minutes of Angle. In the second step, that 6 MOA is multiplied by the actual wind velocity, expressed as a percentage of a 10 mph wind. Thus, with a 5 MPH wind, we take that 6 MOA and multiply it by 0.5 = 3 MOA of correction. And finally, in the third step, you apply this 3 MOA to the angle of the wind on your bullet’s flight. At 45 degrees, that means ¾ value, so multiple the 3 MOA x 0.7 = 2.1 MOA, rounded off to 2 MOA. You apply this final result to your scope so you can hit dead-on in this wind. One point Jeff emphasizes is that this formula works just fine for the .308 Winchester 175-gr. Match load all the way to 1000 yards, but he recommends not employing it for the 168-gr. load after 600 yards. As well, he recommends rounding down your range estimate a bit because his formula slightly overstates the wind at longer distances – thus, consider a 735-yard target as 700 yards. And when you find yourself in especially hot weather or high altitude, he found that his Step One constant is better as -2 rather then -1.
Applying the Formula to Other Cartridges The Hoffman Formula is caliber-specific for .308 Winchester Match loads, but Jeff experimented with other popular sniping rounds, too. For the .223 cal. 77-gr. Match load, use no constant – in Step One: 700 yards yields 7 MOA. Calculate it the regular way in Steps Two and Three. In firing tests he’s found that this is accurate only to 600 yards because beyond that the wind increasingly drifts the lightweight, 77-grain bullet. What about the .300 Winchester Magnum? The 190-grain Match projectile is 30 percent less affected by wind than a .308 bullet. In other words, you need only 70 percent of the indicated wind compensation. The easiest way to calculate this is to follow all three steps exactly as explained above, then, as an additional step, multiply the final result by 0.7 (70 percent).
Had our earlier example involved a .300 Winchester Magnum, 190-grain Match load, we’d take the final result, 2 MOA, and multiply that by 0.7 which yields 1.4 MOA, rounded off as 1.5 MOA. Thus, dial six clicks on a ¼ MOA increment scope, or hold 1.5 MOA into the wind – at 700 yards that would be 10-1/2 inches. Similarly, Hoffman found that .338 Lapua Magnum projectiles were 40 percent less affected by wind than a .308 bullet. Therefore, calculate exactly as if the .338 were a .308 round through all three steps, and then multiply the final result by 0.6 (sixty percent) for the windage adjustment. I congratulate Jeff Hoffman for his contribution to long-range shooting, both through
The other shortcut is the straight up British method, which has already been talked about.
It's not new or unknown, or special, nor is it the same level of adaptability to other calibers. There is still a bit of fudge to it, just read the text.
You all ruined the other thread with a ton of BS it was not removed because it contained an answer I was not looking for, Hell here and the other I showed where it came from, nobody else knew the origin of it.
