WildRose
Well-Known Member
If you are shooting downhill into an upsloping wind the effect is the same as if you are shooting into a headwind on flat ground.Before I attempt another explanation, let me throw out 2 disclaimers. #1. I am not a teacher and therefore do not have the giftings of bringing to light the mysteries of shooting. #2. I am endowed with very limited communication skills. With these 2 disclaimers I will cinch up my belt and attempt another round of explanations.
From what Songdogger told us, he shot out to 500 yards to verify point of aim versus point of impact. That being said, it is not his shooting ability off of lead sleds, bipods, etc. You won't miss a 24" target at 450 yards when you have already been practicing shooting at 500 yards.
The observed problem is that Songdogger shot 8 times and all of them were high. The only thing in the thread that I did not see was what his aiming point was in relation to where the bullets were impacting.
In the old days, rangefinders ranged a target and returned your data in a line of sight format. From this reading, it was entered into a ballistic program along with the incline ange to arrive at a shooting solution.
With Songdoggers balistic program already in use through his binoculars, his shooting solution already compensated for the incline angle, wind, barometric pressure and all the other things that go into a solution for a first round hit.
With fear and trepidation, let me enter into a realm that is difficult to understand and more difficult to explain. For you mathmaticians out there, please bear with me. If I were to draw a triangle on a piece of paper and at the top of the apex I drew a horizontal line that was parallel with the base of the triangle and then measured the distance between the 2 horizontal lines with a piece of string it is going to be shorter than the slope of the triangle. If I were looking through ballistically corrected binoculars, the shooting solution of 450 yards is shorter than the actual distance to the apex. This is important only from the stand point that the bullet is actually traveling further than the 450 yards shown in the shooting solution, but it also means that the environment has a longer affect on the bullet than what one might expect.
On flat ground, a head wind generally means that the wind is traveling parallel with the ground. In the case of shooting down hill, we are not dealing with a headwind in the sense that it is moving in a horizontal position. In this particular scenario, the wind is traveling upslope generating a lifting affect.
Because of the unique form of a bullet, when it is fired from a rifle, the projectile travels in a parabolic curve. Ugh...this is difficult. The rifle is shot from 13" above ground level with the full force of 10 mph wind on the underside of the bullet, generating an unknown lift. This was proven by missing 7 of the eight shots that were too high to impact a 24" target. I suspect that the shot that actually connected with the target was shot with the crosshairs of the scope below the belly line of the animal trying to compensate for the lifting force of the wind on the bullet while in flight.
So, how do you factor in a wind force that is lifting your bullet upwards as it speeds to the target? I don't have any ballistic program that will give me a solution. So I have devised a method, although it is not anywhere perfect, to try to compensate for the uplift on a bullet that is shot downhill with with an unknown uplifting force.
This seems to work, but the logic is full of holes. My ballistic program is designed to compensate for horizontal drifting because of wind. Because of the bullet trajectory while traveling down hill I have settled on using a full value wind deflection and deduct that from my shooting solution in MOA increments. In the case of Songdoggers experience, his ballistic program indicates 4.5 MOA for 450 yards.
In both cases the wind is traveling perpendicular to the ground.
An upslope wind only causes a rise in your round when you are shooting across it, say from one ridge to another.