Robbo81303: If someone could explain why simple physics doesn't apply in this circumstance, I'd love to know why. Thank you.
Okay... I'll try to explain the physics to the best of my understanding.
Robbo81303: It's only the 'better' balls I have issues with. Yes, the higher arc will get knocked down more but that hardly explains the shorter distances under heavy tailwinds.
The apex (or higher arc, as you put it) is going to depend largely on what spin you put on the ball. You will get maximum apex with full backspin, regardless of whether the wind is a headwind or tailwind. With a headwind, the apex will be higher due to "ballooning effect". With a tailwind, the apex will be lower, so it will appear that the ball will be "knocked down" more - and, consequently, will carry a shorter distance.
Robbo81303: Even knocked down, the better ball will still have a much higher arc than a b-xd and the winds are (or, at least, should be) more forceful at higher altitudes.
The better ball will have more spin than the B-XD. That's the reason for its higher arc.
I believe that the winds may have more of an impact at higher altitudes (apexes) on certain holes and certain courses. On an open course (like St. Andrews or RSG), there's probably not much (if any) difference. On a course with lots of trees (like Olympic or Congressional), where the trees can act as a wind screen, you may see more wind effect on your ball when it's above the tree line and less when it's below. But I don't know if WGT has programmed it to be that sophisticated.
Robbo81303: Moreover, the better ball is significantly lighter in weight than a b-xd, which should mean the wind would carry the better ball farther. And yet, no.
All balls in WGT are the same weight. No difference between a B-XD and a better ball.
That's my understanding of the physics, somewhat simplified.