An Unbelievable Throw? - Page 10

NIump50 · #**136** (**permalink**) Sat Mar 11, 2006, 11:01pm

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?

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.

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?

SAump · #**137** (**permalink**) Sun Mar 12, 2006, 12:42am

Final PROOF; GRAVITY is not my HERO

100 mph = 146.6666666667 ft/sec

A 100 mph fastball reaches the plate 57 feet away in 0.3886 seconds.
A 100 mph fastball reaches the plate 60 feet away in 0.4091 seconds.
Let's not argue about the release point and compromise that the distance is covered in .4000 seconds.

A pitcher releases the pitch just over 7 feet towards the strike zone at 3 feet, a four foot drop.
This path can be represented by a simple linear equation, y = - 10 X + 7.
AT X = 0, Y = 7 and at X = 0.4000, Y = 3

Now consider the parbolic flightpath of a baseball with NEUTRAL/ZERO lift falling through gravitaional influences alone.
y = - 16 X squared + 7
AT X = 0, Y = 7 and at X = 0.4000, Y = 4.44

Did you read/see that? The difference in height after 0.4000 seconds is 1.44 feet lower than one following the natural laws of gravitation.

If the pitcher wants to release a 100 mph pitch into the strike zone, he must supply a downward FORCE to account for an additional 1.44 foot drop. He simply does this with a snap of the wrist in a DOWNWARD release angle.

And still you believe a 100 mph baseball cannot RISE.

ONCE AGAIN, I have provided you with physical, mathematical, psychological and SEXUAL PROOF. Yes SEXUAL, because after every math and physics proofs I have posted, I really love knowing which of YOU are full of HOT AIR.

PEACE

GarthB · #**138** (**permalink**) Sun Mar 12, 2006, 02:47am

I believe that the stealing R3 is out on the batter's interference since B1 struck out prior to interfering.

SAump · #**139** (**permalink**) Sun Mar 12, 2006, 02:46pm

From J/R, quoted in 2005 BRD, "when determining ordinary effort, wind is a factor; sun in the fielder's eyes is not a factor." BLAH, BLAH BLAH ....
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Clearly, you recognize the importance of WIND on a batted ball. But you (so often) neglect AIR resistance on a 100-mph ^rising^ fastball. Have you let an emotional condition leaning towards gravity CLOUD your better judgement?

Wouldn't you agree, "when determining a RISING fastball, AIR and VELOCITY are factors?

SanDiegoSteve · #**140** (**permalink**) Sun Mar 12, 2006, 03:35pm

It also has a lot to do with the loft of the driver and the subsequent launch angle vector.