During the 1971 fall, I biomechanically analyzed my fastball, breaking ball and screwball. I borrowed Temple University’s Red Lakes Lo-Cam 500 frames per second camera to use with Michigan State University’s Agricultural Engineering Department’s 400 frames per second camera and the recently purchased MSU’s Physical Education Department’s Red Lakes Lo-Cam 500 frames per second camera. To coordinate film speed with true time intervals, I used the PE Department's Research Laboratory’s recently constructed sophisticated one-thousandth second three display timer system. I placed known measures in the three cameras' views.

Side view and rear view cameras anchored twenty-five feet from a spot three feet in front of the pitcher's plate. The third camera suspended fifteen feet above that location.

The cameras held four hundred feet of sixteen millimeter film. One foot of film contains forty frames. Therefore, four hundred feet of film contains sixteen thousand frames. At 500 frames per second, 400 feet of film permitted thirty-two seconds of action. We needed three seconds per pitch. Therefore, we filmed three attempts at three different pitches.

We started the cameras when I started the down motion of my windup and stopped the cameras when the baseballs contacted the catcher’s glove. With three camera operators at the ready, I threw my fastball, breaking ball and screwball. A coordinator signaled when to simultaneously start the cameras.

With the developed film in a 16 millimeter stop action analyzer projector, I focused critical film frames on graph paper with the known measure at the same scale for the three views. The high speed timers synchronized the three views’ critical film frames.

The drew a circle around the baseball at the moment I released the pitch and wrote an (R) in the circle. I reversed the film five frames and circled the baseball again. Because pitchers rapidly accelerate pitches from forearm leverages through baseball releases, I repeated this process until I came to the moment where I first started to raise my forearm from horizontal. From that point on, I reversed the film ten frames and circled the baseballs until the baseball disappeared into my glove.

From baseball’s appearance out of the glove to release required 0.444 seconds. From baseball’s appearance to baseball’s first movement towards home plate required 0.240 seconds. I define baseball's first movement towards home plate as **leverage**. From leverage to release required 0.204 seconds.

For this discussion, I include only my analysis of my fastball. To biomechanically analyze my fastball, I had to develop displacement, velocity and acceleration graphs for my force application that accelerated my fastball towards home plate. Therefore, I started my measurements at leverage.

I labeled the baseball at leverage, Baseball #0. To precisely coordinate the three camera views, I synchronized the films with the high speed timer. The overhead view clearly defined Baseball #0. Baseball's #1 through #6 occurred between leverage and the start of forearm acceleration. Baseball's #7, #8, #9 and R occurred between the start of forearm acceleration and release. The side view film permitted me to circle the last ball (LB) before the baseball left the view. The overhead view film verified that the baseball traveled parallel to the camera. I used the displacement and time between (R) and (LB) to calculate release velocity. I used the rear view film to determine the time from release to catcher's glove. I had earlier measured the distance from the front of the pitcher's rubber to the catcher's glove. Therefore, I could calculate the baseball's velocity as it contacted the catcher's glove.

The side view film provided towards home plate and vertical displacements, but not side-to-side. The rear view provided side-to-side and vertical displacements, but not towards home plate. The overhead view provided towards home plate and side-to-side displacements, but not vertical. Therefore, side and overhead views verified towards home plate displacements, side and rear views verified vertical displacements and rear and overhead views verified side-to-side displacements. I measured each of the intervals of each of three views ten separate times and averaged the totals for each interval. From these displacement measures, I determined the two sides of the right triangle of each view. For example, from baseball #0 to #1 in the side view film, I had the towards home plate displacement and the vertical displacement. From the two side measures, I calculated the hypotenuse which was the true displacement. I repeated this process for each interval for each view. To determine the true displacement for each interval, I averaged the hypotenuse measures for each interval from the three views.

The Fastball Displacement Table contains the pitch displacements and their time intervals.

Table 30.1: Fastball Displacement-------------------------------------------------------------------------- | Baseball # | Cumulative Displacement(feet) | Cumulative Time(second) | -------------------------------------------------------------------------- | 0 | 0.00 | 0.000 | | 1 | 0.46 | 0.024 | | 2 | 1.13 | 0.048 | | 3 | 1.88 | 0.072 | | 4 | 2.39 | 0.096 | -------------------------------------------------------------------------- | 5 | 2.87 | 0.120 | | 6 | 3.68 | 0.144 | | 7 | 4.99 | 0.168 | | 8 | 5.96 | 0.180 | | 9 | 7.26 | 0.192 | | R | 8.79 | 0.204 | -------------------------------------------------------------------------- | LB | 10.91 | 0.220 | --------------------------------------------------------------------------

The definitional average velocity formula calculates segment average velocities.

where:

1. v stands for average segment velocity

2. s

3. s

4. t

5. t

1. v = s

2. v = 0.46 feet - 0.00 feet / 0.024 seconds - 0.000 seconds

3. v = 0.46 feet / 0.024 seconds

4. v = 19.17 feet / second

1. v = s

2. v = 1.13 feet - 0.46 feet / 0.048 seconds - 0.024 seconds

3. v = 0.67 feet / 0.024 seconds

4. v = 27.92 feet / second

1. v = s

2. v = 1.88 feet - 1.13 feet / 0.072 seconds - 0.048 seconds

3. v = 0.75 feet / 0.024 seconds

4. v = 31.25 feet / second

1. v = s

2. v = 2.39 feet - 1.88 feet / 0.096 seconds - 0.072 seconds

3. v = 0.51 feet / 0.024 seconds

4. v = 21.25 feet / second

1. v = s

2. v = 2.87 feet - 2.39 feet / 0.120 seconds - 0.096 seconds

3. v = 0.48 feet / 0.024 seconds

4. v = 20.00 feet / second

1. v = s

2. v = 3.68 feet - 2.87 feet / 0.144 seconds - 0.120 seconds

3. v = 0.81 feet / 0.024 seconds

4. v = 33.75 feet / second

1. v = s

2. v = 4.99 feet - 3.68 feet / 0.168 seconds - 0.144 seconds

3. v = 1.31 feet / 0.024 seconds

4. v = 54.58 feet / second

1. v = s

2. v = 5.96 feet - 4.99 feet / 0.180 seconds - 0.168 seconds

3. v = 0.97 feet / 0.012 seconds

4. v = 80.83 feet / second

1. v = s

2. v = 7.26 feet - 5.96 feet / 0.192 seconds - 0.180 seconds

3. v = 1.30 feet / 0.012 seconds

4. v = 108.33 feet / second

1. v = s

2. v = 8.79 feet - 7.26 feet / 0.204 seconds - 0.192 seconds

3. v = 1.53 feet / 0.012 seconds

4. v = 127.50 feet / second

Because average velocities occur at segment mid-points, locate the average velocities halfway between the time segments.

Table 30.2: Fastball Velocity------------------------------------------------------------------------ | Baseball # |Average Velocity(feet/second)|Cummulative Time(seconds)| ------------------------------------------------------------------------ | 1/2 | 19.17 | 0.012 | | 1 1/2 | 27.92 | 0.036 | | 2 1/2 | 31.25 | 0.060 | | 3 1/2 | 21.25 | 0.084 | |----------------------------------------------------------------------| | 4 1/2 | 20.00 | 0.108 | | 5 1/2 | 33.75 | 0.132 | | 6 1/2 | 54.58 | 0.156 | | 7 1/2 | 80.83 | 0.174 | | 8 1/2 | 108.33 | 0.186 | |----------------------------------------------------------------------| | 9 1/2 | 127.50 | 0.198 | ------------------------------------------------------------------------

Graph 30.1: Fastball Velocity130 . . . . . . . . . . . 120 . . . . . . . . . . . 110 . . . . . . . . . . . 100 . . . . . . . . . . . 90 . . . . . . . . . . . 80 . . . . . . . . . . . 70 . . . . . . . . . . . 60 . . . . . . . . . . . 50 . . . . . . . . . . . 40 . . . . . . . . . . . 30 . . . . . . . . . . . 20 . . . . . . . . . . . 10 . . . . . . . . . . . 0 .___.______.______.______.______.______.______.______.______.______. 0.012 0.036 0.060 0.084 0.108 0.132 0.156 0.180 0.204 Time(seconds)

When pitchers uniformly accelerate pitches, velocity tables maintain uniform upward slopes. Clearly, until time 0.108, the slope failed to uniformly move upward. Even thereafter, while the slope moves upwardly, it starts upward slowly, then rapidly and then slowly again.

The average velocity for the tenth segment (Baseball #9 to R) is 127.50 feet / second. However, 127.50 feet / second is not the release velocity. Pitchers cannot increase pitch velocities after they release them. Therefore, the last segment’s final time always occurs at release. Consequently, calculating precise release velocities is not possible. However, because the side view camera captured Baseball R (Release) and Baseball LB (Last Ball) and the initial and final segment times, calculating the fastball’s initial velocity is possible.

1. v = s

2. v = 10.91 feet - 8.79 feet / 0.220 seconds - 0.204 seconds

3. v = 2.12 feet / 0.016 seconds

4. v = 132.50 feet / second

Because fastballs cannot go faster than the pitcher’s fingertips at release, I achieved a release velocity of at least 132.50 feet/second.

The definitional average acceleration formula calculates the segment average accelerations.

where:

1. a stands for average segment acceleration

2. v

3. v

4. t

5. t

1. a = v

2. a = 27.92 ft/sec - 19.17 ft/sec / 0.036 seconds - 0.012 seconds

3. a = 8.75 feet / second / 0.024 seconds

4. a = 364.58 feet / second

1. a = v

2. a = 31.25 ft/sec - 27.92 ft/sec / 0.060 seconds - 0.036 seconds

3. a = 3.33 feet / second / 0.024 seconds

4. a = 138.75 feet / second

1. a = v

2. a = 21.25 ft/sec - 31.25 ft/sec / 0.084 seconds - 0.060 seconds

3. a = -10.00 feet / second / 0.024 seconds

4. a = -416.67 feet / second

1. a = v

2. a = 20.00 ft/sec - 21.25 ft/sec / 0.108 seconds - 0.084 seconds

3. a = -1.25 feet / second / 0.024 seconds

4. a = -52.08 feet / second

1. a = v

2. a = 33.75 ft/sec - 20.00 ft/sec / 0.132 seconds - 0.108 seconds

3. a = 13.75 feet / second / 0.024 seconds

4. a = 572.92 feet / second

1. a = v

2. a = 54.58 ft/sec - 33.75 ft/sec / 0.156 seconds - 0.132 seconds

3. a = 20.83 feet / 0.024 seconds

4. a = 867.92 feet / second

1. a = v

2. a = 80.83 ft/sec - 54.58 ft/sec / 0.174 seconds - 0.156 seconds

3. a = 26.25 feet / second / 0.018 seconds

4. a = 1458.33 feet / second

1. a = v

2. a = 108.33 ft/sec - 80.83 ft/sec / 0.186 seconds - 0.174 seconds

3. a = 27.50 feet / second / 0.012 seconds

4. a = 2291.67 feet / second

1. a = v

2. a = 127.50 ft/sec - 108.33 ft/sec / 0.198 seconds - 0.186 seconds

3. a = 19.17 feet / second / 0.012 seconds

4. a = 1597.50 feet / second

Because average accelerations occur at segment mid-points, locate the average accelerations halfway between the time segments.

Table 30.3: Fastball Acceleration--------------------------------------------------------------------- | Baseball # |Average Acceleration(ft/sec^{2})|Cumulative Time(second) | --------------------------------------------------------------------- | 1 | 364.58 | 0.024 | | 2 | 138.75 | 0.048 | | 3 | -416.67 | 0.072 | | 4 | -52.08 | 0.096 | |------------------------------------------------------------------ | | 5 | 572.92 | 0.120 | | 6 | 867.92 | 0.144 | | 7 | 1458.33 | 0.165 | | 8 | 2291.67 | 0.180 | | 9 | 1597.50 | 0.192 | ---------------------------------------------------------------------

The average acceleration for the ninth segment (Baseball #8 1/2 to #9 1/2) is 1597.50 feet / second^{2}. However, 1597.50 feet / second^{2} is not the release acceleration. Pitchers cannot accelerate pitches after they release them. While calculating pitchers’ accelerations at pitch release would be valuable information, mathematically, only calculating acceleration at the start of the last segment is possible.

Sir Isaac Newton’s Law of Acceleration provides the definitional force formula with which to calculate the segment average forces.

where:

1. F stands for average segment force

2. m stands for baseball’s mass

3. a stands for average segment acceleration

Baseball’s mass remains constant for all segments. Baseball’s mass equals baseball’s weight divided by earth’s gravitational acceleration. While earth’s gravitational acceleration varies with altitude, the accepted average earth gravitational acceleration is 32 feet / second^{2}. Therefore, baseball’s mass equals 5.25 ounces / 16 ounces / 32 feet / second^{2} or 0.328125 / 32 feet / second^{2} or 0.0102539 feet / second^{2}.

1. F = (0.0102539)(a

2. F = (0.0102539)(364.58)

3. F = 3.738 pounds

1. F = (0.0102539)(a

2. F = (0.0102539)(138.75)

3. F = 1.391 pounds

1. F = (0.0102539)(a

2. F = (0.0102539)(-416.67)

3. F = -4.272 pounds

1. F = (0.0102539)(a

2. F = (0.0102539)(-52.08)

3. F = -0.534 pounds

1. F = (0.0102539)(a

2. F = (0.0102539)(572.92)

3. F = 5.875 pounds

1. F = (0.0102539)(a

2. F = (0.0102539)(867.92)

3. F = 8.900 pounds

1. F = (0.0102539)(a

2. F = (0.0102539)(1458.33)

3. F = 14.954 pounds

1. F = (0.0102539)(a

2. F = (0.0102539)(2291.67)

3. F = 23.499 pounds

Table 30.4: Fastball Force------------------------------------------------------------- | Baseball # |Average Force(pounds)|Cummulative Time(second)| ------------------------------------------------------------- | 1 1/2 | 3.738 | 0.024 | | 2 1/2 | 1.391 | 0.048 | | 3 1/2 | -4.272 | 0.072 | |-----------------------------------------------------------| | 4 1/2 | -0.534 | 0.096 | | 5 1/2 | 5.875 | 0.120 | | 6 1/2 | 8.900 | 0.144 | | 7 1/2 | 14.954 | 0.165 | | 8 1/2 | 23.499 | 0.180 | -------------------------------------------------------------

Positive forces occur when pitchers increase towards home plate forces to their pitches. Negative forces occur when pitchers decrease towards home plate forces to their pitches. Uniform accelerations require uniform force applications. When force applications vary, pitchers unnecessarily stress their pitching arms. However, acceleration graphs better serve to illustrate force application mistakes.

The Fastball Acceleration Table provides the segment average accelerations with which to construct the Fastball Acceleration Graph. The vertical line (ordinate) plots acceleration magnitudes. The horizontal line (abscissa) plots time. Segment average accelerations range from 2291.67 feet / second^{2} down to -416.67 feet / second^{2}. Therefore, the acceleration graph has 220 feet / second^{2} increments with a zero line and ranges from 2310 feet / second^{2} down to -440 feet / second^{2}.

Graph 30.1: Fastball Acceleration2200 . . . . . . . . . . 1980 . . . . . . . . . . 1760 . . . . . . . . . . 1540 . . . . . . . . . . 1320 . . . . . . . . . . 1100 . . . . . . . . . . 880 . . . . . . . . . . 660 . . . . . . . . . . 440 . . . . . . . . . . 220 . . . . . . . . . . 0 .------------------------------------------------------------------ -220 . . . . . . . . . . -440 ._____.______.______.______.______.______.______.______.______.____ 0.024 0.048 0.072 0.096 0.120 0.144 0.168 0.192 0.216 Time(seconds)

Uniformly accelerated fastballs have horizontal lines across acceleration graphs. Clearly, this fastball did not uniformly accelerate. To understand why not, simultaneously examine the fastball acceleration graph and the side and overhead view schematics.

Because what happens before the pitch’s first towards home plate movement influences its movements thereafter, first examine the fastball’s movement from the glove to Baseball #0 (Transition).

The side view shows the fastball significantly moving downward and slightly backward. The overhead view shows the fastball dramatically moving medially.

Baseball #1 accelerated at 364.58 ft/sec^{2}. Uniform accelerations of 364.58 ft/sec achieve 81.67 feet/second (55.67 miles per hour) release velocities.

Table 30.5: Velocities for 364.58 ft/sec------------------------------------------------------------------------ | Baseball # |Average Velocity(feet/second)|Cummulative Time(seconds)| ------------------------------------------------------------------------ | 1/2 | 8.75 | 0.012 | | 1 1/2 | 17.50 | 0.036 | | 2 1/2 | 26.25 | 0.060 | | 3 1/2 | 35.00 | 0.084 | |----------------------------------------------------------------------| | 4 1/2 | 43.75 | 0.108 | | 5 1/2 | 52.50 | 0.132 | | 6 1/2 | 61.25 | 0.156 | | 7 1/2 | 67.81 | 0.174 | | 8 1/2 | 72.19 | 0.186 | | 9 1/2 | 76.56 | 0.198 | |----------------------------------------------------------------------| | R - LB | 81.67 | 0.212 | ------------------------------------------------------------------------^{2}Uniform Accelerations

Achieving 140.00 feet/second (95.43 miles per hour) release velocities require uniformly accelerations of 625 feet/second^{2}.

Table 30.6: Velocities for 625.00 ft/sec------------------------------------------------------------------------ | Baseball # |Average Velocity(feet/second)|Cummulative Time(seconds)| ------------------------------------------------------------------------ | 1/2 | 15.00 | 0.012 | | 1 1/2 | 30.00 | 0.036 | | 2 1/2 | 45.00 | 0.060 | | 3 1/2 | 60.00 | 0.084 | |----------------------------------------------------------------------| | 4 1/2 | 75.00 | 0.108 | | 5 1/2 | 90.00 | 0.132 | | 6 1/2 | 105.00 | 0.156 | | 7 1/2 | 116.25 | 0.174 | | 8 1/2 | 123.75 | 0.186 | | 9 1/2 | 131.25 | 0.198 | |----------------------------------------------------------------------| | R - LB | 140.00 | 0.212 | ------------------------------------------------------------------------^{2}Uniform Accelerations

Therefore, instead of 364.58 feet/second^{2}, uniform accelerations of 625 feet/second^{2} achieve desired 140.00 feet/second (95.43 miles per hour) released baseball initial velocities. Further, 625 feet/second^{2} uniform accelerations require uniform force applications of only 6.409 pounds.

Between Baseballs #0 and 1, the fastball significantly moved upward and slightly towards home plate and lateral. The fastball should significantly move towards home plate and slightly upward. However, because the fastball is much too far medial, it has to move lateral.

Baseball #2 accelerated at 138.75 ft/sec^{2}. Therefore, between Baseballs #1 and #2, the acceleration reduced 225.83 ft/sec^{2}. Uniform acceleration requires uniform acceleration, not reduced accelerations.

Between Baseballs #1 and #2, the fastball significantly moved towards home base and upward and slightly lateral. However, because the fastball is so far medial, it has to move lateral.

Baseball #3 decelerated at 416.67 ft/sec^{2}. Not only did the fastball not uniformly accelerate, it decelerated. This deceleration indicates that I had not begun upper arm acceleration.

Between Baseballs #2 and #3, the fastball significantly moved towards home plate and upward, but not lateral. However, the fastball remains too far medial.

Baseball #4 decelerated at 52.08 ft/sec2. At least the deceleration magnitude decreased. Perhaps, I have again started to accelerate the fastball.

Between Baseballs #3 and #4, the fastball significantly moved towards home plate and slightly upward. With the slight upward movement, perhaps the fastball has reached its rectilinear drive line. However, because the fastball remains too far medial, it has significant lateral movement to accomplish.

Baseball #5 accelerated at 572.92 ft/sec^{2}. I have entered upper arm acceleration. I doubt I can continue this uniform acceleration. Like teenage drivers ruining cars with race starts, panic stops and race starts, rapid fastball accelerations, decelerations and fast accelerations ruin the pitching arm.

Between Baseballs #4 and #5, the fastball moved considerably towards home plate, downward and lateral. The fastball should move only towards home base. However, we have a new problem. The fastball moved downward. Now, the fastball has to stop its downward movement before it moves back upward to the rectilinear drive line. Overcoming this downward movement further unnecessarily strains the shoulder’s rotator cuff muscles.

Baseball #6 accelerated at 867.92 ft/sec^{2}. The acceleration has finally increased greater than the required 625 ft/sec^{2}. However, to achieve 140 ft/sec release velocities in the shortened force application time remaining, I would have to uniformly accelerate at much higher levels.

Between Baseballs #5 and #6, the fastball significantly moved lateral, considerably downward and slightly towards home plate. The significant lateral and considerable downward movements waste towards home plate force.

Baseball #7 accelerated at 1458.33 ft/sec2. The acceleration has significantly increased, but it has to become uniform.

Between Baseballs #6 and #7, the fastball significantly moved lateral, considerably towards home plate and slightly upward. The significant lateral movement wastes towards home plate force. At least, the fastball has stopped its downward movement.

Baseball #8 accelerated at 2291.667 ft/sec^{2}. The acceleration continues to significantly increase. Unfortunately, incremental accelerations severely stress the pitching arm. (Because the film advanced only five frames from Baseball #7 through release, the diagrams do not depict the rapid acceleration increase.)

Why did this acceleration jump occur? The side view diagram shows the distal humeral end rapidly moving forward. The overhead view diagram shows that the fastball towards home plate than lateral. Centripetal acceleration caused the extreme lateral movement between Baseballs #5 and #8. The lateral movement did not accelerate the fastball towards home plate.

Between Baseballs #7 and #8, the fastball significantly moved towards home plate, lateral and upward. The forearm is just starting to accelerate the baseball. While the forearm accelerates the fastball towards home base, the elbow muscles fight the lateral centripetal force to straighten the pitching arm.

Baseball #9 accelerated at 1597.50 ft/sec^{2}. The acceleration reduced 694.17 ft/sec^{2} from Baseballs #8.

Between Baseballs #8 and #9, the fastball significantly moved towards home plate and slightly lateral and upward.

Between Baseballs #9 and R, the fastball significantly moved towards home plate and slightly medial.

Between Baseballs R and LB, the fastball significantly moved towards home plate and slightly medial. Released baseballs move as the release forces require. Therefore, baseballs become ballistic projectiles. The only forces operating on ballistic projectiles are gravity and the interaction between surface friction and its fluid medium. In pitching, the fluid medium is atmosphere (air molecules).

Scrutiny of the Fastball Acceleration Graph simultaneously with the side view and overhead view schematics identified many mistakes. These mistakes propose several pitching motion corrections.

1. Maintain controlled, uniform transition velocities.

2. Maintain 150^{o} bent elbows throughout transitions and upper arm accelerations.

3. Start transitions with humeral longitudinal axes pointing downwardly.

4. Minimize transition distances from gloves to first towards home plate movements.

5. Start baseballs’ towards home plate movements when humeral longitudinal axes achieve horizontal.

6. Move baseballs slightly elliptically downward, lateral and upward to first towards home plate movements.

7. Drive shoulders and upper arms behind baseballs throughout the upper arm accelerations.

8. Drive with forearms and hands behind baseballs throughout upper arm and forearm accelerations.

9. Rotate upper torsos and shoulders to point acromial process lines towards home plates.

10. Uniformly accelerate baseballs through releases.

Unfortunately, in early 1972, I did not understand all of these mistakes. Learning these mistakes required almost twenty years. However, I did make some immediate adjustments. I minimized the medial transition movement. I rotated my torso and shoulders more. I drove my forearm inside my elbow. I uniformly accelerated. I made my forearm a blur through release.

While anecdotal information deserves severe skepticism, comparing my 1971 statistics (before this high speed film study) with my 1972 statistics (after this high speed film study) may have some value. However, please do not read too much into this comparison.

1. In 1971, I pitched 111 innings in 66 games. In 1972, I pitched 116 innings in 65 games. Therefore, I pitched about the same amount.

2. In 1971, I walked 50 batters, gave up 100 hits and struck out 85. In 1972, I walked 47 batters, gave up 82 hits and struck out 95. Therefore, with five additional innings, in 1972, I walked 3 fewer batters, gave up 18 fewer hits and struck out 10 more batters. While walking 3 fewer batters is insignificant, giving up 18 fewer hits and striking out 10 more batters had significant impact.

3. In 1971, I gave up 53 earned runs for an Earned Run Average (ERA) of 4.30. In 1972, I gave up 23 earned runs for a 1.78 ERA. While 18 fewer hits and 10 more strikeouts might account for some fewer earned runs, other factors must explain most of the 30 fewer earned runs.

4. Relief pitchers cannot control wins. Nevertheless, with teammate and manager help, I received credit for 5 wins in 1971 and 14 wins in 1972.

The average velocity formula calculated the fastball’s initial velocity at 132.5 ft/sec. The side view schematic shows that I released the fastball 6.05 feet in front of the pitcher’s rubber. The catcher caught the fastball precisely three feet behind home plate. Therefore, the fastball traveled 57.45 feet (63.5 feet - 6.05 feet). The rear view camera measured pitch release to catcher’s glove time at 0.467 seconds.

1. v stands for fastball total average velocity

2. s

3. t

1. v = s

2. v = 57.45 / 0.467

3. v = 123.02 ft/sec (83.86 mph)

Rearranging the mathematical average velocity formula provides the final velocity (v_{}f) formula.

1. Mathematical Average Velocity Formula v = v_{i}+ v_{f}/ 2 2. Multiple both sides by (2) (2)(v) = (2)(v_{i}+ v_{f}) / 2 3. Because 2/2 = 1 2(v) = v_{i}+ v_{f}4. Add (-v_{i}) to both sides 2(v) - v_{i}= v_{i}- v_{i}+ v_{f}5. Because (-v_{i}+ v_{f}) = 0 2(v) - v_{i}= v_{f}6. Change sides v_{f}= 2(v) - v_{i}

1. v = 123.02 ft/sec

2. v

1. v

2. v

3. v

4. v

The fastball decelerated from an initial velocity of 132.5 ft/sec (90.32 mph) in the 2.12 feet and 0.016 seconds after release to a catcher’s glove final velocity of 113.54 ft/sec (77.40 mph). Therefore, the fastball decelerated 18.96 ft/sec (12.92 mph) in its flight to the catcher’s glove.

Kinesiologists calculate in feet per second (ft/sec). To find the conversion factor with which to convert feet per second to miles per hour, divide 5280 feet in a mile by 3600 seconds in an hour for a conversion factor of 1.467. To convert feet per second to miles per hour, divide by 1.467. For example, 132 ft/sec divided by 1.467 gives 89.98 mph. To convert miles per hour to feet per second, multiply by 1.467. For example, 90 mph multiplied by 1.467 gives 132.03 ft/sec.

Professional baseball scouts sit behind home plates, aim their radar guns at pitchers and, after every pitch, record what their radar guns read. What radar guns read determine which pitchers receive professional baseball offers and how much.

Many companies manufacture radar guns. However, companies use two time intervals between radar beams. Some radar guns pulse radar beams every 0.2 seconds. Other radar guns pulse radar beams every 0.05 seconds. Radar gun manufacturers claim to provide release and home plate velocities.

The fastball study pitch decelerated from an 132.5 ft/sec (90.32 mph) initial velocity in the 2.12 feet after release to an 113.54 ft/sec (77.40 mph) final velocity at the catcher’s glove. From release to catcher’s glove required 0.467 seconds. Without air molecule density decelerating the fastball, the 132.5 ft/sec fastball would cover the 57.45 feet to the catchers’ glove in 0.434 seconds. Therefore, air molecule density decelerated the fastball 0.033 seconds. Air molecules uniformly decelerate baseball pitches from releases to catchers’ gloves.

If radar guns pulse beams every 0.2 seconds, then almost one-half of 132.5 ft/sec fastballs have passed. The velocity reading is an average for that time period. The fastball study pitch decelerated 12.92 mph from release to catcher’s glove. In the first 0.2 seconds, the fastball study pitch decelerated approximately 6.46 mph. Therefore, at mid-interval, the fastball study pitch decelerated approximately 3.23 mph. In the final 0.2 seconds, the fastball study pitch also decelerated approximately 6.46 mph. Therefore, at mid-interval, the fastball study pitch decelerated approximately 3.23 mph.

If radar guns pulse beams every 0.05 seconds, then almost one-ninth of 132.5 ft/sec fastballs have passed. The fastball study pitch decelerated 12.92 mph from release to catcher’s glove. In the first 0.05 seconds, the fastball study pitch decelerated approximately 1.436 mph. Therefore, at mid-interval, the fastball study pitch decelerated 0.718 mph. In the final 0.05 seconds, the fastball study pitch also decelerated approximately 1.436 mph. Therefore, at mid-interval, the fastball study pitch decelerated 0.718 mph.

In conclusion, no radar gun provides true release velocities or true home plate velocities. However, the 0.05 second beam interval radar guns provide closer readings than the 0.2 second beam interval radar guns. However, neither radar guns operated as effectively as the 0.016 initial velocity time interval for the fastball study pitch and the catcher’s glove final velocity calculation.

Professional baseball scouts do not precisely align their radar guns’ beams in pitch flight lines. Therefore, radar beams angle to pitches. Angled radar beams create the Doppler Effect. The Doppler Effect alters the displacements with which radar guns calculate velocities. Therefore, angled radar guns provide incorrect readings.

Professional baseball scouts hide inabilities to recognize pitching talent behind radar gun readings. Quality pitchers require more than high velocity fastballs. Nevertheless, even with the preceding limitations, 0.05 second pulse beam radar guns are useful teaching instruments. In young pitchers, their proprioceptive awarenesses frequently fail to correctly assess their force application techniques that generate their best release velocities. Best release velocities frequently occur when pitchers proprioceptively experience the least pitching arm stress. Further, while pitchers maintain maximum pitching arm velocities, pitches’ effectiveness improves at specific velocity differences from maximum fastball velocities. Pitchers should throw breaking pitches 14.67 ft/sec (10 mph) less than maximum fastball velocities. Pitchers should throw change-up pitches 29.34 ft/sec (20 mph) less than maximum fastball velocities. Therefore, as a teaching/learning instrument, radar guns have value.