[12 piece bucket]

I initially thought that a four barrel hitting stroke occurred according to the following sequence 4:1:2:3, and that a four barrel hitting action occurred when a right shoulder thrust downplane released PA#4 slightly before PA#1 became actively operational, while a triple barrel hitting action was only 1:2/3 without any preceding PA#4 release via a right shoulder thrust action.

However, I recently read this archived Yoda post.

"The Swinger's active Right Shoulder accelerates the Left Arm, and Centrifugal Force and its Throw-Out Action then powers the Clubhead. The Right Arm remains passive -- except for its Right Triceps Extensor Action and the #3 Pressure Point sensing and directing the Clubhead Lag -- and thus the Stroke remains Three-Barrel. In contrast, the Hitter's active Right Shoulder serves as the 'launching pad' for the Right Arm and its Muscular Drive-out Action that powers the Clubhead. Since the Right Arm then drives the entire Primary Lever Assembly (including the Left Arm), the Full-Power Hitting Stroke can properly be classified as Four-Barrel."

That Yoda post suggests that the full power hitting stroke can properly be classified as four barrel because the right arm drives the entire primary lever assembly, which includes the left arm. OK. Then what represents a triple barrel hitting action - ? where the right arm doesn't drive the left arm?

Jeff.

Jeff . .. . In hitting the right arm ALWAYS drives the left arm . . . . even if it's just single barrel. Ted describes 4-Barrell hitting as "it's easier to push a cannon ball that is already rolling than one that is at rest" The pivot takes up the slack in then the right arm drives everything out. If the right arm doesn't drive . . . it' ain't hitting.