Endless belt and release physics
 

From my reading of the TGM book, I gather the impression that HK believed that the club released in the downswing because of a change in direction of movement of the hands from a straight line path to a curved path, and that the endless belt model explains the release phenomenon. Conceptually, the endless belt model theorizes that the club will travel at the same speed as the hands until the hands enter the pulley section of the endless belt model. Then, as the hands turn around the pulley, the club will accelerate relative to the hands and this represents the release phenomenon. However, this conceptual model is dependent on there being a straight line path movement of the hands in the early/mid downswing followed by a curved hand path later in the downswing, and "reality" doesn't seem to confirm this fact. If one looks at the hand path of professional golfers, their hand path is always curved and there is no straight line section. Here are two examples.

Bobby Jones strobe photo.



Tiger Woods photo



I used the slow motion Nike commercial swing video of Tiger Woods to produce this image. The red lines represent the clubshaft. The yellow lines represent the left arm (and I only plotted the left arm's location in the later downswing so as not to have too many lines cluttering the image). The green line is a hand-drawn line that plots the sequential movement of the hands over time. Point X is the point when release of the club is definitely apparent, although there is seemingly a small degree of release earlier.

Note that the hand path in both these images is near-circular, and there is no straight line path. Therefore, "reality" doesn't support HK's endless belt concept as explaining the release phenomenon.

Surely, a much better explanation is this mathematical model, which demonstrates that the club develops angular acceleration at all time points in the downswing when the pull on the grip is at an angle to the COG of the clubshaft.

http://nmgolfscience.tripod.com/release.htm

Jeff.

Tyreffic Pic!
 

You need a big gunner for this one! I command a cracked straw with ample spid wads, so I will bow low. BUT...Thanks for the Jones pic, it is the best example of the how concentric circles apply to the endless belt effect . I never saw it that clearly until now...THANKS!

If you permit me to opine on this endless field of debate, namely TGM. I have never considered a straight line delivery to be absolutely straight seeings that it J's out at the bottom, for that matter I know that my frozen bent and level right wrist is not perfectly level and probably bends and flattens a little extra during the course of my procedure (but as close as this human can get it) I also suggest to you concerning myself that if my tripod were used to take wedding pictures the bridesmaids would not make the shot! Unless of course Bucket it taking the picture! I do know that if I keep my hand path moving straighter relative to the circular delivery line of the club head then I can utilize the effect that conceptually is best explained by the endless belt effect. Kinda like a swingers utilizing that irrepressible fictitious fleeing force! The golf stroke is not the county fair ducky disappearing around the corner bit! I may add something cheeky here. If he of the Impregnable Quadrilateral and The One moves in a straighter line I dare say they could improve! :salut:

I am but a bantam weight on a forum of super heavy weights so I am probably not furthering this debate or enhancing this thread. But I reserve the right to speak my brand of twaddle! :laughing9

Here is some additional commentary that I think may have significant relevance to understanding the release phenomenon (as explained mathematically by nmgolfer in that link).

Consider this image of two waterskiers being pulled behind a motorboat.



Image 1 shows the two skiers (1 and 2) traveling at the same speed as they are pulled behind the motorboat. Presume that the motorboat travels at a constant speed, which means that the pull force on the connecting rope is constant, and that both waterskiers will therefore travel at the same speed (which is the same speed as the motorboat).

Image 2 shows what will happen if waterskier number 2 angles his skis outwards so that he travels along a curved path. He will start to travel faster than waterskier number 1 even though the pulling force exerted by the motorboat (which is traveling at a constant speed) along the rope is unchanged. Why does he travel faster? I believe it is due to the fact that the constant pulling force is angled relative to the direction of travel of the waterskier, and therefore the waterskier acquires angular acceleration. With the progression of time, waterskier number 2 will travel along a curved path, and travel through point A, pass point B and eventually end up at point C. During this time period, the waterskier number 2 will be travellng faster-and-faster, and eventually the waterskier will catch up to the boat (point C). Interestingly, the pull force exerted via the rope (as a result of the motorboat traveling at a constant speed) remains constant, but the waterskier eventually catches up to the boat. I think that this catch-up phenomenon is "equivalent" to the release phenomenon described by nmgolfer in his release mathematics explanation - that a golf club develops angular acceleration when the constant pulling force on the grip is at an angle to the COG and directional momentum of the club, and that the clubhead end of the club will eventually catch up to the grip end.

If you find my waterskier analogy reasonable, then you may also consider another fact. Waterskier number 2 travels faster-and-faster along the curved path as he progressively moves through point A and point B to point C. Also, the closer the waterskier gets to the boat (and the faster he travels), the less the drag load he imposes on the connecting rope. So the question becomes, does COAM apply to this situation? Does the motorboat have to slow down (given an unchanged level of engine thrust) if waterskier number 2 progressively speeds up along the curved path as described?

Jeff.

Left Shoulder Center of of Impact Circle
 

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.

Some analogies don't hold water
 

Assume your water skier has already reached point C. In doing so, he could have arrived there at a constant rate of acceleration, or accelerated to say point B and then stayed at a constant rate. He will never catch up to the boat on muscle power alone unless he had a head start from the other side, or he is very strong. In addition, as he nears the boat he will actually slow down until he reaches his original speed, yet the pull on the rope will be significantly greater than the pull on skier 1. You have to have the arms of a power lifter to stay at point C all day due to the resistance of the water against the bottom of the angled board you are standing on.

When he makes his turn to go back towards point A it must be a sharp turn while keeping the rope taut or else he will temporarily sink into the water. Once started, he won't have nearly the effort of pull as he did catching up to the boat, rather a slight turning of the ski will whip him down to point A and momentum will hurl him over the wake, likely catching his fin on Skier #1 rope and thereby plunging both into the water.

A better analogy might be the water-skier preparing to jump the wake or a competition slalom skier which I once was, but due to the nature of skiing which is essentially a flat board resisting the force of water, I don't think there are parallels to be found.

The Golfing Machine model of the endless belt involves an intentional straight line thrust of the hands to their aiming point, not a literal, ungolf-like straight line path of the hands to the bottom of the swing arc. The intent is another practical application which is difficult to measure mathematically. The actual straight line of the hands can sometimes be seen for a few inches at start down, but it depends on the procedure.

Longitudinal Acceleration
 

In the picture of Tiger, counting from the right, the 6th
red line for the clubshaft seems to be length wise for
for Longitudinal Acceleration using a rope handle
technique. I would think that the blue line from red
line 6 to redline 7 would be straight rather that curved. Your curved line appears to suggest radial acceleration. Red line 6 is length wise down
to approx hip level. Would this be the point where
pulley starts?

Do you buy Homers concept of Longitudinal Acceleration
vs radial acceleration?

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.

Donn - you wrote-: "Do you buy Homers concept of Longitudinal Acceleration vs radial acceleration?"

I will deal with a swinger's swing - where HK implies that the hands supply a constant pulling force along the length of the club longitudinally (along its longitudinal axis). It is true that as the left hand moves in space it is pulling the grip end of the club in a particular direction, and that the directional force can be perceived to be pulling the grip end of the club along its longitudinal axis. However, the true reality is that the club develops angular acceleration from the very start of the downswing (because the hands are constantly changing direction) and the hand pull cannot therefore remain along the longitudinal axis of the clubshaft throughout the entire downwsing (because the club has inertia and the COG of the club doesn't move instantaneously in the same linear direction with respect to the pulling force). As the downswing evolves, and as the club progressively releases, then the hand pull direction is increasingly at an increasing angle to the longitudinal axis of the shaft.

You also wrote-: "I would think that the blue line from red line 6 to redline 7 would be straight rather that curved. Your curved line appears to suggest radial acceleration. Red line 6 is length wise down to approx hip level. Would this be the point where
pulley starts?"

I don't think that the hand movement arc is ever straight in a golf swing, and that it is always near-circularly curved. I actually think that there is no "pulley" in the hand curve of "real life" golfers (like Tiger Woods and Bobby Jones) because the hand arc curve is essentially C-shaped, with no abrupt changes in direction as would occur with a J-shaped curve.

Jeff.

Bagger - thanks for commenting.

I agree that although the rope pull is theoretically constant (because the boat is traveling at a constant speed) that the actual pulling force exerted on waterskier number 2 must become greater when the skier angles the skis to the side and moves along a curved path (instead of a straight line path) - because of the resistance of the water. Do you have the know-how to estimate/calculate the magnitude of this greater pulling force at point A, point B and point C - taking into account the facts that the i) waterskier is progresively moving faster as he moves along that curved path, and ii) the rope is an increasing angle relative to the directional movement of the skier?

You also seem to imply that the pulling force felt by the skier at point C would be at its greatest. I would imagine that it is at its least at point C - because of two factors - i) The waterskier is now moving much faster than the boat, and ii) the angle of the rope is nearly at right angles to the direction of travel of the skier. I would think that the pulling force felt by the waterskier would be at its greatest in the earliest part of the curved path (prior to point A).

Jeff.

Jeff,
Some brief comments are bolded above in relation to your post. Also, by the link that the author has for basic physics I would presume that the author doesn't "believe" or isn't capable of explaining the concept centrifugal force- another reason to question the degree of intellectual acumen of the author.

Jeff,

I can't help you with the math. Depends on the horsepower of the boat, the effective boat draft, how hard the skier pulls the boat sideways, prop pitch and size, water surface conditions, ski design, rope length, etc. The greater the angle from the center of the boat to point C, the greater the angle of the ski "blade" otherwise the skier is decelerating.

Trust me, not much application to the golfers release but if you want to do the math, there are plenty of resources to put the puzzle pieces together. Not my cup-o-tea, I was only concerned about rounding the buoy's without dropping an elbow or rope in the water. That takes time off the score and ruins rhythm.

Some are Performers and some are Engineers. Rarely the two meet in a single individual, but we need each other collectively. I appreciate what you are attempting here Jeff.

About that left shoulder movement. Can you quote VJ in context so there is no misunderstanding?

Thanks,

Bagger - you wrote-: "The greater the angle from the center of the boat to point C, the greater the angle of the ski "blade" otherwise the skier is decelerating."

I cannot understand this point. If the curve passing through all the points A,B and C is a defined curve of a certain shape, then the angle of the ski blade must be constant at all times during the skier's passage along that curved path - irrespective of the angle of the skier to the boat. The angle of the rope changes, but the skiers blade angle must surely remain constant if he maintains a curved path of constant curvacture.

What do you want me to clear up about the left shoulder movement? I only made the point that the idea that the left shoulder remains "fixed" in space as the center of a circle with a fixed radius is not what happens in a "real life" golf swing in the late downswing - because i) the left shoulder socket is moving upwards, leftwards and backwards in the late downswing and ii) the distance between the left shoulder socket and the clubhead constantly changes as the clubshaft becomes progessively more in-line with the left arm in the late downswing.

Jeff.

I've always wished I could combine my passions but I could never reconcile skiing and golfing.

I'm not interested in discussing the skier analogy, but I am interested in your comment regarding VJ's view of Hogan's left shoulder.

"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 -"

Just want to make sure that VJ's representation is rooted in reality. I haven't read his book so I'm trusting that this statement is fair and contextually accurate.

Bagger

Here is a copy of VJ's Impact Circle photo from his book.



Jeff.

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" contolled by hand path and #3 angle.

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.

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.

Straight Line delivery path vs Circle Delivery Path
 

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

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.

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

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.

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.
http://www.youtube.com/watch?v=-JVXk...eature=related

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.

Bagger

You would have to supply a solid physics/mathematical explanation for your assertion that a skier cannot accelerate between point B and point C if he keeps his skis constantly angled for me to consider your assertion seriously. Those U-tube waterskier competitors straightened their skis after accelerating sufficiently, so that they could have a straight-line directional launch from that launching platform.

There is a big difference between a waterskier between point B and point C (and Tiger Woods' club between point B and point C in his swing). Because of the presence of waterdrag and an increasing angle between the rope and the the skier's curved path, the skier will slow down as he passes point C and approaches the boat. By contrast, in a golf swing, air-resistance doesn't impede the release phenomenon in a golf swing, and there is no reason why the club cannot continue to accelerate between point B and point C. In fact, the club does and the clubhead passes the hands after impact.

The waterskier example obviously has limited analogy to a golf swing because the left hand undergoes a 90 degree rotation during the release swivel, as you pointed out, and that is an obvious confounding variable.

My only point of using the waterskier analogy is to give an analogous, easy-to-understand, example of the principle that underlies the release phenomenon - that it is due to angular acceleration developing because the hand pull on the grip is at an angle to the club's COG-momentum and that i) variations in hand speed and ii) hand arc curvature at all time-points during the downswing will cause variations in the degree of angular acceleration during the downswing, and therefore release variations - between different golfers (as nm golfer's mathematical explanation predicts).

Another added point - I have a PingMan-type driver-testing machine at the golf facility where I practice, and I have studied that machine's release action. It has a passive hinge joint that can rotate >90 degrees to simulate the release swivel. Interestingly, it always rotates perfectly during the release even though the hinge joint is totally passive - I presume that it has something to do with the COG of the clubhead causing the clubhead to automatically rotate to a square alignment at impact. I also have noticed that the central arm's swingarc, and therefore hinge joint's (between the central arm and clubshaft) swingarc is circular, and that the central arm travels at a constant speed. There is no endless belt pulley analogy that is applicable to that machine and yet it releases the clubshaft perfectly/naturally - the only "release phenomenon" explanation that presently makes sense to me is nm golfer's mathematical explanation.

Jeff.

The skiier will slow down when he can no longer maintain his ski angle to the boat and since there is upwards of 800lbs of rope tension while accelerating as you near point C, that becomes difficult for anyone to maintain.

As an exercise, think of the ski angle to the rope as fixed during initial acceleration prior to point A at 22.5 degrees (direction of intended travel). When the rope angle to the boat has reached 22.5 degrees to the boat, the skier must then have a 45 degree relationship of the rope to the ski in order to continue accelerating on his original linear 22.5 degree direction of travel. Otherwise the effective angle of his ski to the boat is reduced towards parallel with the boat, with accompanying deceleration towards the boats constant speed. So as the skiers ski travels on an arc with an original ski angle at 22.5 to the boat, the angle of the ski must increase for every degree of increase in the angle of the rope to the boat, otherwise acceleration stops.

So yes, at the end of the day, its all about drag and skier strength. You simply cannot maintain a constant ski angle in relation to the boat in order to accelerate to point C.

I'm happy to agree with you on that.
I tried to dig up some of the physics on skiing but didn't have any success. I hope my simple geometry make sense.

I think we all agree that the hands travel in an arc around a rotating left shoulder joint. Also that clubs are designed to rotate to "square" with normal centrifugal release motion. I think you may be struggling with the concept of aiming the hands with a linear intent to their aiming point, similar to the endless belt diagram.

Everything in a golf stroke is seen as circular but like the skier analogy, what is not seen is linear intent.
But it is the lines that are our guide-lines to good golf.

Bagger

Thank you for your explanation of the effect of water drag load on the limits of a waterskier's physical ability to maintain a curved path. That makes the waterskier analogy less useful. Hopefully, other forum members will imagine the waterskier situation as an "imaginary situation" that exists in the absence of any water drag load, so that they can basically try to get a visual picture of nm golfer's mathematical explanation. I wonder if there is a better visual analogy that can better convey the "essential idea" behind his mathematical explanation, which seems to be very sound.

Jeff.

I would have said Thank You to Baggerfor taking the time to explain instead pissing on him. Maybe on your other forum this is protocol.

No Harm
 

Its cool 6B. Given Jeff's distinguished Medical background and attention to detail, I can appreciate where he is coming from.

I took his response as a compliment and kind of enjoyed re-experiencing my Skiing time anyway.

6BMike

You wrote-: "would have said Thank You to Baggerfor taking the time to explain instead pissing on him."

Wow! I am flabbergasted at this comment. I did thank Bagger. I wasn't insulting him. I acknowledged his greater knowledge re: the effect of water drag load on a waterskier's skis, and I therefore concluded that I would need to find another visual analogy to better illuminate nmgolfer's mathematical explanation.

Jeff.

that reminds me in the arm stroke,

there are paw, pick , and a very interesting one Pause...

alot of interesting points..

I find you being flabbergasted hold to imagine. But nothing personal ol chap. It's just you seem less than affable in many replies. Not many rocket scientists on this site but a whole lot of good golfers that know how the sweet spots moves on plane to achieve impact.

First of all, Thank you Jeff for asking some very interesting questions regarding TGM - in this thread as well as in others.

And even though the endless belt analogy has some merits, I certainly agree that concepts that are close to the real thing are better than those who are farther away.

But angular acceleration isn't what this is about. Angular acceleration is often regarded as one of the most differentiating factors between swinging and hitting, and is possible one of the most misunderstood concepts in golf.

Angular acceleration does not increase the speed of whatever is being rotated. Angular acceleration doesn't do anything but change the direction while conserving energy. It is only longitudinal force that can increase the club spead. Or shoud I say: "Geometrically Orienteted Linear Force" - G.O.L.F

In a pure swing, the swing center is shifting. It is always moving a little ahead of the rotation center. Rotation of left shoulder joint is key here. I personally believe that the role, position and movement of the left shoulder joint is underestimated in TGM as far as swinginging is conserned.

This shifting of center is creating a longitudinal force component in addition to the centripetal force component of the "pulling string". The pulling string is used to pull & rotate at the same time. The longitudinal force component is what increases speed. The centripetal force component only takes care of the circular part and doesn't produce speed.

Think of David Leadbetters illustration when he spins a little something attached to a tiny rope. He has to move his hand in a circle to pick up speed.

Only longitudinal force produces work. And work is required to increase the speed. It is the longitudinal force of the swing that increases the clubhead speed. Or should I say: Geometrically Oriented Linear Force: G.O.L.F.

And Newton has explained why there is no other way.

It is possible to do a pure swing and a pure hit from the same top position, and have the same club alignment at impact and basically the same speed too. These facts are strong indications that the two stroking methods basically containes the same amounts of angular acceleration (the part that caused the rotation) and longitudinal acceleration (the part that increases the speed). One of them would miss the ball othervise.

I don't think the water skiing analogy works without water drag. It is the water drag that enables the skier to go in another direction than the boat, and this is a sort of overtaking action. As long as the skier is coming up from behind he will have larger speed than the boat - even relative to the course the boat is going. But the skier will have largest speed when he goes in the direction that is most different to the boat - when he comes back from a far left, is straight behind the boat and aiming far right. The true analogy to golf here is, I believe an overtaking action. A transfer of energy from boat to skier. A skilled and heavy skier will significantly slow down the boat. Some of the energy will be wasted to water shuffling, but some of it will accelerate him to the right.

If the boat went in circles, and managed to keep with up the skier, he could be straight behind the boat aiming far right and picking up speed all day long.

Looking at the Bobby Jones sequence, the swing seems to have a very steady linear force as the club is acellerating gradually.

Looking at Tiger's swing, something interesting seems to be happening around 9 o'clock in the downswing. Is the clubhead slowing down for a brief moment? Is he manipulating the geometry in order to increase the G.O.L.F from that point and through the ball?

Bernt - thanks for posting your opinion. I am always interested in reading other people's opinions, because I may learn something "new" that will change my mind about an "issue". Unfortunately, I first have to understand another person's opinion before I can change my mind, and I cannot understand many points that you seem to be expressing.

To start, you wrote-: "Angular acceleration does not increase the speed of whatever is being rotated. Angular acceleration doesn't do anything but change the direction while conserving energy. It is only longitudinal force that can increase the club spead. Or shoud I say: "Geometrically Orienteted Linear Force" - G.O.L.F."

I am not sure what you mean by longitudinal force. Consider nm golfer's mathematical explanation. He stated that the hands are always changing direction and speed at every moment of the downswing. Do you regard that as a longitudinal force, a G.O.L.F force? Secondly, he stated that if the hands are pulling the grip end of the club in the same direction that the hands are moving, and at the same speed as the hands are moving, that the clubhead end of the club would be angularly accelerated at every fractional time-point of the downswing, and that the cumulative effect of many thousands of time-points of angular acceleration inputs would cause the clubhead end of the club to progressively speed up. In other words, he is seemingly implying that angular acceleration doesn't only change the direction of the clubhead's movements, it also causes the clubhead end of the club to speed-up. Do you disagree? Secondly, nm golfer's mathematical explanation doesn't state anything about "conservation of energy" because there is not a "fixed" amount of energy in his hand/clubshaft system. The hands can constantly receive additional energy throughout the downswing from a variety of power sources (eg. release of power accumulator #4).

You also wrote-: "In a pure swing, the swing center is shifting. It is always moving a little ahead of the rotation center."

Could you please define "swing center" and "rotation center". I cannot develop a mental picture of your developing argument. You further wrote-: "This shifting of center is creating a longitudinal force component in addition to the centripetal force component of the "pulling string". The pulling string is used to pull & rotate at the same time. The longitudinal force component is what increases speed."

You seem to be implying that there is longitudinal force component that causes the increase in clubhead speed. What is this longitudinal force component and where is it operant in the PingMan machine's swing? If you are implying that the left shoulder socket is moving left-laterally in space, while the left arm rotates from the fulcrum point of the left shoulder socket point, and you are implying that the constant movement of the left shoulder socket in space represents the longitudinal force, how can the hands "know" what percentage of their force of forward movement comes from the movement of the left shoulder socket in space versus the rotational movement of the left arm?

Could you please explain what you see in Tiger's swing at 9 o'clock?

Thanks.

Jeff.

Jeff,
Reading the posts here is like having a Homer Kelley nightmare for me, in other words- I see the same lack of clarification here that Kelley has in his book in regards to linear speed and angular speed and the lack of clarificaiton or definitions. Since the post essentially is about angular acceleartion I think it would be important to define angular acceleration and also show me in the Bobby Jones swing above- where this exists.

Thanks,
Mike O.

Mike O

I don't have a precise knowledge of physics, so my use of the term 'angular acceleration' may be incorrect. However, if you look at the hand arc and clubhead arc in the Bobby Jones photo, and look to see how much the hands move per unit time relative to the clubhead movement per unit time, the clubhead is moving a greater distance than the hands - and the clubhead movement per unit time is increasing progressively as the downswing evolves (even though hand speed remains relatively constant). Therefore, I perceive that the clubhead is developing angular acceleration and that this represents the release phenomenon.

Jeff.

Jeff,
Sorry for the bluntness in advance but you're four pages into your own thread dealing with angular acceleration among other things in search of the "truth" and you state "I don't have a precise knowledge of physics, so my use of the term 'angular acceleration' may be incorrect." That's problem number one! Iin finding the "truth" you need to precisely define what you are talking about. With the internet- it's not that difficult to look up the definition. Here is Wikipedia's definition- "Angular acceleration is the rate of change of angular velocity over time. In SI units, it is measured in radians per second squared (rad/s2), and is usually denoted by the Greek letter alpha."

While the clubhead in the Bobby Jones photo is acquiring Angular acceleration (you can measure the radians or degrees between the end of each shaft near the clubhead and see that from one to another the distance is increasing)- however that's only part of the release factors. As it's the linear speed of the clubhead that is the real intent of the release and the clubhead can be picking up clubhead linear speed with an extension of the swing radius and at the same time have no increase in the angular acceleration of the clubhead.

Here is the definition of angular velocity from Wikipedia: "In physics, the angular velocity is a vector quantity (more precisely, a pseudovector) which specifies the angular speed, and axis about which an object is rotating. The SI unit of angular velocity is radians per second, although it may be measured in other units such as degrees per second, degrees per hour, etc."

As you indirectly stated by comparing the differences between the hands and the clubhead movements- it's important to identify which part of the club that you are talking about and what type of speed/velocity/acceleration that you are talking about when having these discussions. For example in regards to the Bobby Jones photo the end of the clubshaft and the clubhead are picking up angular acceleration but the grip end of the shaft is not picking up angular acceleration. However, it's the clubhead speed not the angular velocity or angular acceleration of the clubhead that I think is the essential issue and those are not directly related.

The way you've answered the post above in the quote- it appears that you're implying that the relationship of the hand motion versus the clubhead motion has something to do with the angular acceleration formula of the clubhead - which is doesn't in the strictest sense.

You say "and the clubhead movement per unit time is increasing progressively as the downswing evolves" - here again you can't tell if you are talking angular velocity of the clubhead or the linear speed of the clubhead or angular acceleration of the clubhead? Three completely different things- 1) for angular velocity ignoring COAM- extending the radius has no effect on it - as opposed to 2) the clubhead linear speed which is greatly influenced by an extension of the radius. For example you could have a situation where you have angular deceleration yet with an extension of the radius an increase in the linear speed of the clubhead.

Those terms....

I was mixing angular acceleration with radial acceleration in my last post.

Here's a pretty good description of radial and linear acceleration / tangential acceleration. Tangential acceleration is similar to linear acceleration in the sense that it works in the same direction that the subject body is moving. The only difference is that tangential acceleration changes direction all the time. This is by the way illustrated by the notation "AT" in the figure below, indicating that linear acceleration is also tangential acceleration. This was writting for alpine skiing, but it is valid for the golf swing too:



Comment from me:

This could be a golf swing - but for a split second only. Unfortunately, the direction is similar to a left handed golf swing. But my comments are right handed.

The radial acceleration will always be 90degree to where the club head is heading and the tangential force will always be in the same direction as the club head is moving. The "rope" - using a rope handling technique will always be pointing in the same direction as the sum of the two vectors Ar and At. By applying torque to the swing (something that is heavily used in a hit, and is possibly also unavoidable in parts of a "pure" swing), the "rope" will be a virtual rope, and it could easily be anchored far outside the left shoulder.

As long as there is some lag in the swing, and as long as the swing is picking up speed - the rope will be offset to origo. And this offset is lag to the primary lever. Lag doesn't have to be limited to the relation between arm and club either. It can also be related to the relation between the whole primary lever and origo.

1) Early in the downswing the ratio of tangential vs total force is relatively big. When there's a huge angle between left arm and the clubshaft, a lot of tangential force is applied to the club head. And a lot of the resistance, or weight that we feel is because the club is picking up speed.

2) Later in the downswing the ratio changes towards more radial force (force that doesn't increase speed). A lot of the resistance we feel now is from this radial force.

3) Within this frame of reference, clubhead lag is to a very large degree the ratio of tangential vs total force. With a 100% pure rope handling technique, the shaft orientation vs origo (at any time) is probably a decent indicator of this ratio.

4) Clubhead lag seems to be referring to the secondary lever most of the time. But the offset between Origo and the center of the pulling action (which could be left shoulder joint) represents lag to the primary lever assembly. You can still have lag pressure in the swing even though the clubface has caught up with your hands. Not that I recommend it though.


Original explanation continued:

As a few other terms I used in the post, I could certainly have picked a better one. I should have used the term tangential force, ref my other post above.

I agree that the hands are always changing direction and speed. In addition, the hands transfer radial and tangential forces.
I was confucing "angular acceleration" with radial acceleration. Sorry about that.

I basically agree with the above, but I think it makes sense to decompose angular acceleration. Angular acceleration is created by radial acceleration and tangential acceleration. But it is only the tangential component that increases the swing speed.

There are basically two ways of generating tangential acceleration.

1) By applying torque (as in trying to bend the shaft)
2) By partly pulling the club in a direction that has a tangential component as well as the radial component.

"Swing center" was another one of my ill-defined terms. Unfortunately, I don't have a good expression available at the moment. But what I mean is that the total force applied to the club (tangential and radial force) is not pointing at the rotation center. An offset is required to generate a portion of tangential force. This offset could be the location of left shoulder vs. head/neck which is likely to be the rotation center.

But may I restate that radial force only changes direction and conserves the velocity energy that has been transmitted to the clubhead so far in the swing.

The pingman machine has its mechanical left arm attached to a mechanical shoulder. This shoulder is rotating arount the center of rotation, so that an offset is created.

It is moving in an arch. How close this arch is to a circle probably varies from golfer to golfer. But the point is that the pulling must stay "ahead" of the swing center to increase speed. There i said it again. I meant center of rotation. "Swing center" is probably an equally correct phrase. But since I've already abused the term "swing center" i will try to let that rest at least in this thread.

But regardles of swing technique, somewhere in the body, torque must be applied. In a pure swing, torque is applied to rotate the left shoulder. And this torque enables the left shoulder to generate radial and tangential acceleration up on the club by simply pulling the arm.

I don't understand this question.
[quote]

Could you please explain what you see in Tiger's swing at 9 o'clock?
QUOTE]

When the clubhead is around 10 o'clock, it is loosing speed beween two frames. If the illustration is correct. 10 o'clock as seen from our side, that is. I'm wondering whether there's some downstroke loading going on.

But it could as well be just an inaccuracy in the illustration.

Edit: Unfortunately I am not able to present it as anything else but a link.



I do not like the impact circle consept, because it makes it seem like the rotation centre of the golf swing is at the left shoulder. Which it woould be if the sholder was kept still.

I've added some "momentum" to the drawing.

The centre of rotation is IMO close to the golfer's nose or neck.

The curved arrow around the golfer's head is momentum created by the rotating body and applied to the left shoulder. This momentum results in a pulling force on the left arm and further on the left club. This force is illustrated with the red arrow above the golfer's shoulder.

Since the sholder is offset to the centre of rotation, the pulling force from the shoulder can be decomposed in a component that is purely radial and another component that is purely tangential. That would be the yellow arrows by the clubhead.

The small arrow at the golfer's right hand is some of the force from extensior action. This could also be described as torque applied to the left arm, which is required to deal with clubshaft vs left arm lag.

Mike

You don't have to apologize for being blunt. I encourage all forum members to attack my arguments rigorously without restraint, because I believe so strongly in the Popperian falsification principle. My ideas/opinions are only valid to the extent that they cannot be falsified, and I don't know if they can be falsified if people don't vigorously challenge my opinions.

When I stated that my knowledge of physics-terms could be inexact, then it doesn't mean that I am knowingly using those "terms" imprecisely. My understanding of the term "angular acceleration" is totally compatible with the Wikipedia definition. However, I believe that angular acceleration doesn't always refer to a point-object moving around a "fixed" point in a circular motion, but it can also apply to a linear structure (eg. clubshaft) moving around a fulcrum hinge point (eg. hands) which is itself in motion. In nm golfer's mathematical explanation, he is referring to the clubshaft when he talks about angular acceleration, and not solely the clubhead. When I stated that one can see the clubhead progressively moving a greater distance per unit time during the release phase, and that this represents angular acceleration, I really meant that the clubhead is a "marker" for the rate of angular acceleration of the clubshaft, which means that the grip end of the club must be experiencing the same degree of angular acceleration as the clubhead end (because they are both simply point-locations on the same clubshaft).

I therefore don't understand your comment-: "For example in regards to the Bobby Jones photo the end of the clubshaft and the clubhead are picking up angular acceleration but the grip end of the shaft is not picking up angular acceleration."

You also made the following statement-: ""As it's the linear speed of the clubhead that is the real intent of the release and the clubhead can be picking up clubhead linear speed with an extension of the swing radius and at the same time have no increase in the angular acceleration of the clubhead."

I cannot understand your opinion - from the perspective of nm golfer's mathematical explanation of the release phenomenon. You seem to be implying that the clubhead (which is merely a point-location on the clubshaft) can pick up linear speed due to an extension of the swing radius without any angular acceleration of the clubshaft. When you refer to "swing radius" I presume that you are talking about the distance between the left shoulder socket and the clubhead. If my understanding is correct, then I think that you may have your logical argument back-to-front. The swing radius can only increase (in the presence of a left arm of constant length) if the 90 degree angle between the left arm and clubshaft changes toward a 180 degree situation (left arm and clubshaft become more in-line). That happens as a result of the release phenomenon - which is due to the clubshaft acquiring angular acceleration during the release phase. If one accepts my reasoning, then this other statement of yours doesn't make sense - "For example you could have a situation where you have angular deceleration yet with an extension of the radius an increase in the linear speed of the clubhead." How can the swing radius be increasing if angular acceleration of the clubshaft is decreasing.

It is true that we are mainly interested in the linear speed of the clubhead at impact, but I think that the linear speed of the clubhead is derived to a large extent from the fact that the clubshaft is angularly accelerated during the release phenomenon.

Jeff.

Left Shoulder
 

Jeff & BerntR, Seems to me that you guys are sticking with
the premiss that the left shoulder is offset to the center of
rotation. This is true, in most cases, with the shift and turn
concept. With the Hogan move, that V.J. Present, the left
shoulder is already even or ahead of the ball at the top of
the swing so that Hogan could have a left shoulder impact
circle center. On pg. 83 of "The Final Missing Piece" V.J.
says: "Review sequences of Hogan's sing in figure 43. From
this vantage point, the white spot above his right hip pocket
actually moves backsward or away from the target at the
start of his downswing. If the hips had moved forward to
start the downswing, this white spot would also move
forward or at least stay in the same position'. The pictures also
indicate that the left was at the ball at the top of the swing.

With shift and turn in the downsing, I can see many of your
points. But using V.J. and Hogans pictures from V.J.'s book to illustrate
what happens in a shift and turn method can certainly be confusing.
V.J. pointed out that his book not a method. It is an explanation of one Man's method.