Quote:
Originally posted by NIump50
Quote:
Originally posted by bob jenkins
Quote:
Originally posted by NIump50
If the ball because of speed and backspin flattens out a 1/2 degree from it's downward trajectory isn't that for all practical purposes a rise?
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And, we're back to where we started -- defining the term.
As I (and I think others) are using the term, I mean that if F1 releases the ball such that it's instantaneous /initial vector is parallel to the ground, the ball will never be farther from the ground than it is at it's initial point. In addition, the ball's distance above the ground at any position will be (a) less than at any point prior to that position and (b) greater than at any point after that position.
I also think that everyone agrees that if two balls are thorwn at the same initial angle relative to the ground that (all other things being constant) (a) the faster ball will "fall less" over the same distance and (b) the ball with the greater backspin rate will fall less.
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How much drop does a 95 mph fastball have in 53'
I contend the initial vector is at the eventual target 53 feet away. I don't see any arc in a 90 + mph fastball.
Can it be figured mathematically?
Sat Mar 11, 2006, 11:01pm
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