Baseball is a game of speed, despite not having a game clock, play clock, or any other sort of time constraints. The difference between being called out and safe can be determined by a fraction of a second. That call could be the difference between winning and losing a game. If you spend twenty minutes at any high school, collegiate, semi-pro, or professional baseball team’s facility, you will see coaches with stop watches timing players running, throwing, and transferring the ball. Coaches and players are constantly working on improving their times in order to execute defensive plays more efficiently, which will result in more put outs. Arguably the most significant and thrilling put out is a play at the plate. The offensive team has an opportunity to
…show more content…
s(t¬2) = 10w (1- e^((-t2)/10) ) = x e^((-t2)/10) = 1 - x/10w t2 = -10 ln ( (10w-x)/10w )
This is the equation for the second part of the relayed throw, from the cutoff to the catcher.
The Relayed Throw Equation
Now that both time equations for the relayed throw are formed, an equation for the entire time of the throw can be created. Once the throw reaches the cutoff, he must transfer the ball from his glove to his throwing hand, get a good four-seam grip on the ball, pivot, and throw. To account for this transfer, a half of a second is added to the equation to calculate the speed of the relayed throw. tw(x) = 1/2 + t1 + t2 tw(x) = 1/2 + [-10 ln ( (810+x)/1100 )] + [-10 ln ( (10w-x)/10w )] tw(x) = 1/2 -10 [ ln ( (810+x)/1100 ) + ln ( (10w-x)/10w ) ]
In order to find the minimum time with this equation, the derivative must be found. dt/dx = -10 [ 1/(810+x) - 1/(10w-x) ]
By solving for x, based on the denominators of the derivative, the minimum distance from the cutoff to home can be determined after plugging in the shortstops initial velocity.
922 + x = 10w – x
922 + 2x = 10w
2x = 10w – 922 x = 5w –
s(t¬2) = 10w (1- e^((-t2)/10) ) = x e^((-t2)/10) = 1 - x/10w t2 = -10 ln ( (10w-x)/10w )
This is the equation for the second part of the relayed throw, from the cutoff to the catcher.
The Relayed Throw Equation
Now that both time equations for the relayed throw are formed, an equation for the entire time of the throw can be created. Once the throw reaches the cutoff, he must transfer the ball from his glove to his throwing hand, get a good four-seam grip on the ball, pivot, and throw. To account for this transfer, a half of a second is added to the equation to calculate the speed of the relayed throw. tw(x) = 1/2 + t1 + t2 tw(x) = 1/2 + [-10 ln ( (810+x)/1100 )] + [-10 ln ( (10w-x)/10w )] tw(x) = 1/2 -10 [ ln ( (810+x)/1100 ) + ln ( (10w-x)/10w ) ]
In order to find the minimum time with this equation, the derivative must be found. dt/dx = -10 [ 1/(810+x) - 1/(10w-x) ]
By solving for x, based on the denominators of the derivative, the minimum distance from the cutoff to home can be determined after plugging in the shortstops initial velocity.
922 + x = 10w – x
922 + 2x = 10w
2x = 10w – 922 x = 5w –