Endless belt and release physics

[12 piece bucket]

Donn

I posted this photo



You then responded as follows-: "The picture of Tiger, sure shows the Impact Circle Axis and the Left Shoulder Center of Impact Circle that V.J. presents in his book."

You are making a big mistake. In constructing my composite photo, I only used one image of Tiger Woods, and superimposed the yellow lines on that photo to show the general change in position of the left arm in the downswing time sequence. However, those yellow lines do not point at the same point, because the left shoulder is always changing position.

Here is the true reality.



I used a spline tool to trace the movement of Tiger's left shoulder during the mid/late downswing. You can see that his left shoulder continuously moves upwards and to the left as the downswing progresses - see the three white dots with interconnecting red lines. VJ's representation of Ben Hogan's left shoulder socket as being a single fixed point in space is an oversimplification, and it doesn't represent true reality. It is a gross over-simplification - equivalent to representing Ben Hogan's downswing clubshaft plane with a single plane line when the clubshaft is actually changing planes continuously as it moves down from the turned shoulder plane (at the end-backswing) to the hand plane (at impact).

Jeff.

Jeff . . . you make some good points here. I'd say that the point of the VJ illustration and what Homer would say is that the center of the shoulder turn needs to be far enough FORWARD to contol low point. That's all.

[Jeff]

12 piece bucket

I agree with you - the left shoulder must be well forward at impact to have a low point somewhere near the inside of the left heel.

Different golfers get the left shoulder to the "correct" impact location via different mechanisms. A S&T golfer, who has a small amount of left spinal tilt at the end-backswing (leftwards-centered swing style), will have to move the left shoulder far less to get the left shoulder to the 'correct" impact location than Tiger Woods, who has a rightwards-centered swing style.

By the way, I have altered the Tiger Woods photo by adding dotted yellow lines to demonstrate that the radius of the circle increases while the center of the circle moves leftwards/upwards/backwards.



Jeff.

[dkerby]

Hello again Jeff. I am quoting a post by EdZ:

If you imagine the path of your hands traveling in your
motion as a wheel (7-23), the 'widest possible' path of
the hands would be a circle around your center of
balance.

In a straight line delivery, your hands go from the 'edge
of the rim' at the top, directly in a straight line INSIDE
the circle toward your aiming point/both arms straight
and at both straight they are again 'touching the
wheel rim'. A 'feel' of 'narrow to wide" (top to both arms
full extended). Adelayed release/max trigger delay and
a more powerful (but tougher to time) motion.

In a circle delivery path, your hands effectively 'attempt'
to STAY on the rim all the way 'around the circle'. A very
'wide' feel in which you release fully and smoothly from
the top - very nearly throw away. A very consistent, but
much less powerful motion, much simple to judge distance.

You post on pulley wheels seems to focus on the circle or angled
line delivery path rather than the straight line delivery path.
I am looking at the pictures of Diane on pgs 203-205 of the yellow
book.

Some where, I think that I remember Yoda saying to set the
#3 pressure point at the top and take it straight to the ball.

Are we missing that their are three possible TYPES of paths
the hands can take toward impact.

Donn

[Jeff]

Donn

You wrote-: "In a straight line delivery, your hands go from the 'edge of the rim' at the top, directly in a straight line INSIDE the circle toward your aiming point/both arms straight and at both straight they are again 'touching the wheel rim'."

I can understand the idea of mentally "directing" the hands to go in a straight line direction towards the aiming point, but I have never seen a straight hand arc in a "real life" golf swing. All the hand arcs that I have seen are curved arcs/paths that are C-shaped or slightly J-shaped.

If you have access to a golf swing showing a straight line hand path, please let me know - I would like to analyse it using my swing analyser program.

I can see Diane mimicing a straight line hand path in the early/mid downswing by pulling her hands down that white path on page 203 of TGM, but I have never seen that phenomenon in a "real life" professional golfer's swing.

One can try and pull the PP#3 point in a straight line direction straight down to the aiming point, but the hand arc/path will still be curved because while the hands are moving down in the direction of the aiming point, the body is pivoting and the arms are moving around the body at the same time, and the end-result is a curved hand path.


Jeff.

[dkerby]

Jeff, I don't have a picture of a straight line delivery path.
If one comes across I will let you know.
Maybe their is a curved arc or J-shaped path. The procedure
is called an Arc of Approach. Homer does say in 10-23-0
that the paths are taken by the hands - Not the club head
Your descriptions refer to the hands - correct. In 10-23-A,
STRAIGHT LINE, Homer says "This pattern holds the Hands
on a Delivery Path that is a straight line leading from the
Top-of-the-line hand position directly at and through the
Aiming Point (2-J-3) when there is no Plane Shifts (10-7)."
Homer mentions in 2-K that Only the Circumference is NOT a
straight line.

Even with a slight curve of the Arc, I do not believe that this
would be enough to invalidate the concept of the Endless Belt
Effect. The pictures of Diane swinging do not seem unusual to me.

Yoda put up some excellet pictures of of the Endless Belt Effect.
I do not know that he would want to get into the dissussion.

All the best, Donn

[Jeff]

12 piece bucket

You wrote-: "Jeff . . . keep in mind that Homer was simply using the Endless Belt to illustrate a CONCEPT. I don't think Homer ever thought that hands move on anything but an Arc. It is the player's intent to move the hands in a straight line . .. but as you say in reality that don't happen. I'm sure that Homer was certainly aware of that fact.
The Endless Belt is a nice way to think about things particularly with different "pulley diameters" controlled by hand path and #3 angle."

I agree with you that the endless belt is a nice way to think about it, but there are alternative ways of thinking about the release phenomenon that may be more compatible with reality.

If you look at the hand arc of many professional golfers swinging a driver, you will see that the hand arc is C-shaped with a broad curve of near-constant radius (near-circular or elliptical). There is no J-shaped hand arc that would fit in with the endless belt analogy. So, the question becomes - how best to explain the release phenomenon in those golfers who have C-shaped hand arcs (like Bobby Jones, Tiger Woods and the Pingman machine).



I think that the best mathematic explanation was provided by nmgolfer in that link. Even David Tutelman has recently accepted his mathematical explanation as a better alternative than thinking about centrifugal forces (which is a controversial concept). The fundamental mathematical principle underlying the release phenomenon is that a golf club will develop angular acceleration when the pull force on the grip is at an angle to the COG of the club, and that the club progressively builds up angular momentum because of the cumulative effect of these angular acceleration forces at every moment of the downswing.

Although Bagger doesn't think that the waterskier analogy is useful in trying to understand the release phenomenon, I think that he is not thinking of the waterskier analogy in the correct manner. He was talking of water resistance, when one should ignore water resistance and consider the situation from the following perspective.



Image 1 - there is a constant pull force from the connecting rope because the motorboat is traveling at a constant speed. Therefore, the two waterskiers will travel at the same speed as the boat because the rope pull force is linearly in line with the angle of the skiers skis. Image 2 - now, if waterskier number 2 angles his skis to the side so that they would intentionally carve a curved path of constant radius, thereby creating a C-shaped path, he starts to accelerate relative to waterskier number 1 and the boat. Why? It is secondary to the fact that the pull force from the rope is now at an angle to the direction of his skis. He subsequently experiences angular acceleration. The amount is very small at first, but imagine that there are 1,000 time-points between the start of his curved path and point A. At every one of those 1,000 time-points, he develops an additional amount of angular acceleration because the constant pull force is at angle to his direction of travel. The effect is compounded and his velocity increases. Now imagine another 1,000 time-points between point A and point B. At every one of these 1,000 time-points, he continues to accrue even more angular acceleration because the pull force from the rope is at an increasing angle to his direction of travel. Therefore, it is not difficult to understand how waterskier will be travelling much faster at point B than point A - even though the boat and waterskier number 1 are travelling at a constant speed. This process of accumulating additional angular acceleration in very small incremental amounts continues to point C when the waterskier is traveling at maximum speed and catches up the boat. In other words, his speed increases constantly due to the cumulative effect of additional amounts of angular acceleration at every moment of his C-shaped curved path (of constant radius).

I believe that the same phenomenon occurs in the Pingman machine and professional golfers like Tiger Woods.

Consider the situation of Tiger Woods.



Note that at point A, Tiger still has a 90 degree angle between the left arm and clubshaft and the club has not released, while it is released by impact (point C). Both the hand arc and clubhead arc is C-shaped between point A and point C. So, how does the release phenomenon occur? It occurs because at every moment between point A and point C, the pull force on the grip end of the club is at an angle to the COG of the club, and the club therefore develops angular acceleration. Between point A and point B, the cumulative effect of the angular acceleration increments is very small so the degree of release is very small by point B, but between point B and point C the release happens faster-and-faster because of the cumulative effect of incremental amounts of additional angular acceleration at every moment in that C-shaped curved hand/clubhead path.

There are many logical/intellectual benefits to nmgolfer's mathematical explanation.

1) It doesn't invoke the idea of centrifugal forces, which some people believe is an abstract concept.

2) It explains why the clubshaft reaches maximum speed at impact, while the endless belt concept incorrectly predicts maximum clubhead speed at the time of "going through the acute J-shaped curve bend". HK has to provide an additional explanation for the fact that clubhead speed increases all the way to impact, and he writes about factors that must maintain hand speed all the way to impact.

3) It is not depend on any COAM-belief. HK invoked COAM in his TGM book and apparently implied that the hands have to slow down as the clubhead speeds up, unless the golfer did "something" in addition to ensure that the hands maintain a constant speed. In nmgolfer's mathematical explanation, there is no COAM-effect, and the hand speed can easily remain constant while the club constantly develops angular acceleration at every fractional moment of the release phase of the swing.

Now, if any forum member can demonstrate a flaw in nmgomfer's mathematical explanation, I would like to learn of that flaw, so that I can more correctly understand the physics of the release phenomenon.

Jeff.

[Bagger Lance]

Although Bagger doesn't think that the waterskier analogy is useful in trying to understand the release phenomenon, I think that he is not thinking of the waterskier analogy in the correct manner. He was talking of water resistance, when one should ignore water resistance and consider the situation from the following perspective.
.

Jeff,

You've put significant thought and effort into this so lets explore the boat analogy one more time. I've just invented some magic dragless skis so we that we can continue. Likewise, I can put on ice skates and you can pull me across a frozen lake on your snowmobile in order to reduce the drag effect of plowing water in your skiing example.

First a couple of questions:
When the skier reaches point C what speed is he traveling?
If faster than the boat, does he pass the boat?
If he passes the boat, how far does he go before ceasing to accelerate?
Check out these cats who are looking for the maximum acceleration point.

To save time, I'll answer the above.

The skier will travel at the same rate of speed as the boat when he reaches point C. In other words, somewhere around point B he loses the ability to accelerate and momentum will carry him to point C as he decelerates to the speed of the boat. If he had a head start like the ski jumpers, he might be able to pass the boat from momentum alone. Somewhere between point A and point B (a 22.5 degree angle), he will not be able to accelerate much further no matter how much he angles the ski. In other words, if you are going the same rate as the boat while at a 22.5 degree angle to it, you will be hard pressed to gain any speed to get to point C no matter how hard you lean into it.

In the golf swing, we are talking about wristcock faciliating angular momentum by use of the flail or left wrist. Why doesn't the clubhead overtake the hands (boat) or decelerate when nearing their inline condition in a good swing?

Because the golfer transfers all of that wristcock angular momentum into his turning left wrist which is #3 accumulator roll power. There is no skiing analogy for the transfer of wrist cock momentum into roll power.