Question

with the 12" diameter regulation softball (not sure on actual weight...possibly 6 oz)...going from home plate to 2nd base in the time of 1.46 seconds...the speed is__________....then convert down the size and weight to a baseball and to the dimensions of the baseball length...not just the time/distance to get the speed of the throw....but how fast and at what clocked time would it be if it were a regulation baseball thrown the same way on the dimensions of a baseball field's 2nd base?
In Softball, from home plate to 2nd base the distance is 84 feet 10 1/4 inches. In Baseball, the same distance is 127 feet, 3 and 3/8 inches.
Can someone please convert this from Softball to what it would be equivalent to in Baseball?

Answer 1

Let's calculate everything in meters per second, and then we can convert it to any desirable units.

Given:

1 - Baseball: distance from Home plate to 2nd base is $D_B=38.741\text{ m}.$

Drag coefficient for the baseball is $c_B=0.4.$

Diameter of the baseball is $d_B=74\text{ mm}.$

2 - Softball: distance from Home plate to 2nd base in softball is $D_S=25.704\text{ m}.$

Time is $t_S=1.46\text{ s}.$

Diameter is $d_S=97\text{ mm}.$

Drag coefficient is $c_S=0.3$

Calculation:

A softball speed is thus

$$v_S=D_S/t_S=25.704/1.46=16.92\text{ m/s}.$$

The drag force for a ball (in the following expression for the softball, notice subscript $S$) is proportional to its cross-sectional area, speed and its drag coefficient:

$$F_{drag.S}=kv_S^2c_SA_S=kv_S^2c_S[\pi d_S^2/4].$$

If we assume that from the beginning (strike) the ball goes with constant speed, it means that the force with which it was thrown equals the drag force.

If we threw a baseball along the same distance with the same force (with the same drag force) and the weather is the same as for the softball, its speed would be higher since it is more heavy and its cross-sectional area is smaller. Moreover, we can express this speed in terms of the softball's speed:

$$F_{drag.B}=kv_B^2c_BA_B=kv_B^2c_B[\pi d_B^2/4],$$

$$v_B=\sqrt{\frac{F_{drag}}{kc_B[\pi d_B^2/4]}}=\sqrt{\frac{kv_S^2c_S[\pi d_S^2/4]}{kc_B[\pi d_B^2/4]}}=v_S\frac{d_S}{d_B}\sqrt{\frac{c_S}{c_B}},$$

$$v_B=16.92\cdot\frac{97}{74}\sqrt{\frac{03}{0.4}}=19.21\text{ m/s}.$$

Of course this approximate result is not 100% true since we ignored Magnus force (when the ball rotates) and force of gravity, but it scientifically shows significant difference in speeds which affects strategies and players' behavior during the game.

Since we know the speed of the baseball and the softball, you can calculate time necessary for the baseball (or softball) to cross the softball (or baseball) field by dividing the corresponding distance by the speed.