- Dec 28, 2023
- 53
- 18
Great clip of bullet spin pitch.
Enjoy!
Enjoy!
Your description of IR reminded my of my attempt to show IR requiring flexion. Didn't turn out as well as I'd hoped.I know this is going to turn off a lot of people but it has to be done....
The upper arm internally rotating along its long axis (humeral rotation) with a bent elbow will move the ball through an arc with a radius equal to the distance of the ball from the humeral long axis (line perpendicular to the humeral axis). The speed of the ball along the arc transcribed by the end end of this line (ball) would be equal to the humeral rotation rate (radians per second; 360° = 6.28 radians) x the distance the ball is from the center line of the humeral arc to the ball..
The linear speed of the ball is equal to how fast the upper arm creates internal rotation times the distance that the ball is away from the center line of the upper arm rotation.
In previous post I did a hypothetical calculation:
Humeral internal rotation rate w = 9000° per second = 157 radians per second.
Distance from ball to humeral rotational axis r = 12" = 1 foot.
Ball speed = w x r
Rotational speed of ball along arc transcribed by internal rotation = (157 radians per second x 12 inches (1 foot) = 157 feet/sec = 107 MPH.
Great clip of bullet spin pitch.
Enjoy!