The Beauty of 12-5

[Trig]

How many Accumulators should one use for 12-5-1?

Good question. Homer explicity says to zero out accumulator #3. So the answer is less than 4!

If I think about it, there is no wrist cock either, so that eliminates #2.

I think the answer is 2. You should use accumulator #1 and #4 while executing 12-5-1.

[bray]

Good question. Homer explicity says to zero out accumulator #3. So the answer is less than 4!

If I think about it, there is no wrist cock either, so that eliminates #2.

I think the answer is 2. You should use accumulator #1 and #4 while executing 12-5-1.

Trig,

I was trying to figure out how Acc. #4 was involved in a stage one stroke. Then I referred to 6-B-4-0 and read this; "Left Arm Power in any form or amount can still be considered #4 Accumulator Action."

So, yes Accumulator #4 is involved in the two barrel stage 1 motion.

I learned something today, now back to answering the pro shop phone.

Soting through the Duffer's Bible.

B-Ray

[tongzilla]

Trig,

I was trying to figure out how Acc. #4 was involved in a stage one stroke. Then I referred to 6-B-4-0 and read this; "Left Arm Power in any form or amount can still be considered #4 Accumulator Action."

So, yes Accumulator #4 is involved in the two barrel stage 1 motion.

Hmm...so for 12-5-1, when Homer lists Acc #4, was he referring to independent Left Arm Power, or Body Power? The latter will clearly make the zero pivot requirement harder to achieve. But it also has mechanical advantage (which is pretty irrelevant for 12-5-1).

[Yoda]

Hmm...so for 12-5-1, when Homer lists Acc #4, was he referring to independent Left Arm Power, or Body Power?

We're talking about a Stroke that is two feet back and two feet through. Body Power? I don't think so!

[tongzilla]

We're talking about a Stroke that is two feet back and two feet through. Body Power? I don't think so!

I didn't mean the full fireworks e.g. spinning the flywheel and Right Shoulder Drive to Load PP#4, etc...

I meant using just a little Pivot to give that Left Arm a little mini blast off, as opposed to using your left arm muscles to move to left arm (ala Ben Doyle...that's a discussion for another day )

[Yoda]

I didn't mean the full fireworks e.g. spinning the flywheel and Right Shoulder Drive to Load PP#4, etc...

I meant using just a little Pivot to give that Left Arm a little mini blast off, as opposed to using your left arm muscles to move to left arm (ala Ben Doyle...that's a discussion for another day )

The Right Shoulder Turn Thrust drives the Left Arm in a Pivot Stroke. And the Shoulders are a Dual Agent (2-H, 7-13), i.e., they are part of both the Power Package and the Pivot. So, Shoulder Motion (and Action) does not violate the requirements for Zero Pivot.

However...

In addition to Zero Pivot, the Basic Motion of 12-5-1 specifies that the Shoulder Turn also be Zero. Remember, this is a tiny shot. As I've previously stated, everybody wants to make this motion bigger than it is. Per 6-B-4-0 and 10-3-D, the Left Arm motion can be independent of the Pivot.

Point of information: In the as-yet-unpublished 7th edition, Homer Kelley changed the Zero Pivot requirement of 12-5-2 to Minimal Pivot. In addition to the already permitted Downstroke Shoulder Turn, this expansion satisfies the need for the Swinger's essentially inert Left Arm to pick up a little Body Momentum Transfer from a 'thrown' Right Shoulder via the Hip Action of 7-15. In addition, it also satisfies more than a little 'psychological need.'

Bottom line: I'm all for the Swinger's 'mini-blast-off' in 12-5-2 (Acquired Motion) and the major-blastoff of 12-5-3 (Total Motion), but...

In the Basic Motion of 12-5-1...

Nada.

The Left Arm is on its own.

Which, of course, is why Hitting -- Right Elbow Drive -- is so attractive in the Short Strokes.

[annikan skywalker]

Oh what the Hey.....You've gone through Stage One as in depth as never before...you've enticed us with Stage 2.....Let's finish the deal here...I dare ya....

[Yoda]

I was trying to figure out how Acc. #4 was involved in a stage one stroke. Then I referred to 6-B-4-0 and read this; "Left Arm Power in any form or amount can still be considered #4 Accumulator Action."

So, yes Accumulator #4 is involved in the two barrel stage 1 motion.

The Basic Motion of 12-5-1 is a One Accumulator Stroke (Single Barrel, 10-4-A). And that one Accumulator is an Arm Accumulator, either the Left (10-4-A-4) or the Right (10-4-A-1). If the player chooses to use "Left Arm Power in any form or amount," then the Thrust is Centrifugal (6-C-0-4) and the Stroke is a Swing. If the player chooses to use the Right Arm to drive the Primary Lever (Left Arm and Club), then the Thrust is Muscular (6-C-0-1) and the Stroke is a Hit.

Both Accumulators are referenced in 12-5-1 because both Arms are in Motion -- one Active and the other Passive. And that Motion is identical, no matter which Arm actually drives the Stroke.