V
Vibe
Guest
The new relative wind has a vector component of the negative of the remaining forward velocity, plus the 10mph at right angles to it. Now the bullet is being accelerated in both of those directions - negative to the direction of travel. If 0° is X then the crosswind is along the Y axis.,,,,,with our discussion.
Let's assume the straight-line over-the-ground course from the muzzle to the target at 1000 yards is due north or better referred to as heading '0' degrees. Our bullet starts out on a heading of '0' degrees, its ground track is also '0' degrees as there is no crosswind in the first 500 yards. If the bullet is launched at 3000 fps it feels a relative wind right on the nose of 3000 fps. Of course, it is constantly decellerating after leaving the muzzle.
Reaching 500 yards, our bullet enters a direct crosswind from left to right at 10 mph. The crosswind results in a bit of drag on the left side of the bullet and it turns half a degree left, into the crosswind, to streamline itself with the new relative wind.
OK. But only because it will not drift in a vacuum.The bullet is now flying perfectly streamlined thru the airmass on a heading of 359.5 but its nose is pointing half a degree left of course relative to the ground. If not for the fact that it is constantly slowing down, the bullet would not drift at all!
But neither would it change heading or pointing direction.
"Heading" actually will change slightly as the direction of travel and the direction of the crosswind become less and less perpendicular...but that's a bit of nit picking in this example. It really only becomes important in Als example of the upward "wind" seen during falling. This is because, while under the influence of the crosswind, the ground track path will be curved, not straight - and the longer it is in the crosswind, the more "off course" the bullet will seem to be.As velocity continues to decrease, on a heading of 359.5, the bullet begins to drift downwind over the ground. Heading does not change but ground track does, and that's what we are concerned with; ground track.
the problem is that while the bullet WAS in the crosswind - it was accelerated in the direction the wind was going, and therefore still retains all of the velocity transferred to it in the Y axis direction. Once the crosswind is no longer accelerating it, this becomes a constant Y direction velocity in the direction of the wind...until an opposite crosswind cancels it out...Of course this Y axis component is also now degrading and a rate exactly proportional to the rate that the X axis velocity is degrading.Perhaps the reason so many have trouble accepting my explanation is because they think the bullet turns much farther than it actually does. Of course, I realize we often exagerate things to make a point, but Alinwa's drawing is a bit misleading in that it depicts the wind correction angle as approximately 30 degrees! No one would believe a bullet turns 30 degrees back to its original heading upon exiting the crosswind at 601 yards and of course, it doesn't. It was only crabbed (relative to the ground) half a degree (heading 359.5) to begin with so when it exits the crosswind at 601 yards, the bullet turns half a degree to the right, streamlines itself with the new relative wind and is now on its original heading of '0' degrees. From 601 yards to the target winds are calm, there is no crosswind component so there can be no drift.