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Originally Posted by Bainer
-the line from the pitcher's hand to the catcher's glove is not straight- it is arced, though accepted as 'normal'.
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Are you saying there is an arc in a 98mph fastball?
From it's trajectory coming out of the hand, how far does a straight 98 mph fastball drop or 'arc'?
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Originally Posted by Bainer
-Now, a reduction of that arc is simply a reduction- not a rise. For a rise to occur, the ball would have to achieve a position which falls between it's initial release point and lowest point- ie, drop and come back up- or RISE.
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I disagree!
I do not contend that once the ball begins to drop (from gravitational force) it will then reverse course and then rise.
My definition of a rising fastball is a ball that ends up higher at the plate than the original trajectory of the ball out of the hand of the pitcher.
Therefore, the ball can still be on a downward angle from pitchers hand to intended target and the rise comes from physical forces, causing lift among other things, to lessen the descent angle.
The downward angle and descent angle has nothing to do with gravity. Everything is based on original trajectory of ball coming out of pitchers hand.
Since the pitcher releases the ball above the batters shoulders most all pitches start out at a downward angle. A rising fastball also starts out at a downward angle and maintains a downward angle even after it "rises". Because of the rise, the angle of descent becomes less than the original trajectory.
The explanation given for why a ball 'appears' to rise is that it doesn't drop as much as expected and therefore gives the illusion of rising.
Let's define drop.
If drop is merely the difference in height between release point and where it crosses the plate then I am in total agreement. Because there is a less than expected drop in a rising fastball relative to it's original trajectory.
If drop is defined as how much gravity has pulled the ball downward below it's original trajectory, then I disagree. A straight 98 mph fastball does not deviate from it's original trajectory in 56', therefore there is 0 drop. If there is 0 drop there cannot be a reduction in drop, unless of course there is a negative reduction, which would then be a rise.