Nesbit or Kwon or MacKenzie?

Dave Tutelman  --  Dec 6, 2018


This appendix covers some of the more technical points in the analysis on the second page. If you're not into math, you don't need to read this (and probably won't get much from it). I go into some of the equations here, but there is no attempt to be completely rigorous. This would never be accepted as a degree thesis by a graduate school. That doesn't mean it's wrong, just not a complete analysis. The point here is to give some idea of the math and the engineering knowledge that led me to my assertions on the second page.

When I was in engineering school around 1960, there wasn't as much specialization as today. My degree was in Electrical Engineering, but I had to take courses in Mechanical and Civil Engineering as well. I learned that there were direct analogues of electrical systems in both ME and CE. By "direct analogues", I mean that the mathematical equations are the same, and you can even identify the quantities that represent one another in the different disciplines. On several occasions, I used those analogues to analyze something about mechanics or materials that surprised my professors; ME or CE students would not have gotten to those particular analysis tools before senior year or even graduate school. (In fairness, there were relatively few components that electrical engineers had to learn about, compared with mechanical engineers. And circuits using those few components required frequency and transient analysis. So EEs had both the opportunity and the need to learn the techniques early in their education.)

So I do have engineering experience and intuition in the technical details here. That said, let's look a little more deeply into analysis of the shaft bend during the downswing.

The basic bending equation

The most elementary, fundamental formula for flexural bending goes back over 250 years. About 1750, Euler and Bernoulli (names ranking just below Newton in pantheon of physics and math) proposed an equation for the flexural bending of a beam. It is still the basic equation at the beginning of the chapter on bending in any engineering text.


  =   M


  • R is the radius of curvature of the beam at some point x along the beam. (See the diagram to the right.) By the way, a golf shaft is a tubular beam of varying diameter. It is definitely governed by the Euler-Bernoulli equation.
  • M is the bending moment at point x along the beam. Most of the art of using Euler-Bernoulli is figuring out M(x) at each point on the beam.
  • E is the elastic modulus of the material at point x. (Always positive.)
  • I is the area moment of inertia of the cross-section of the beam at point x. (Always positive.)
  • You can't measure E nor I easily for a golf shaft. But it is possible to measure the shaft's stiffness non-destructively, which is E times I. Fortunately, all the bending equations have the product EI rather than either one separately. Almost all engineering situations dealing with shaft bending deal with EI as a single quantity that varies with x. (An exception to this rule is the design itself of the shaft. The shaft designer has to choose materials, orient them for strength where he wants it, and deal with diameters and wall thicknesses. That engineer deals with E and I separately to accomplish the EI(x) that will govern the shaft's performance.)
  • The inverse of the radius of curvature is the curvature itself. So the left side of the equation is the curvature of the shaft, the sharpness of the bend, at point x.
In engineering texts, where you are assumed to know calculus and the mathematical meaning of curvature, the Euler-Bernoulli equation is usually written as:


  =  – M


In calculus, the second derivative of a nearly horizontal curve is its curvature. Expressing it this way enables its solution as a second order differential equation. Also -- important to this specific debate -- it explicitly controls the sign (the direction) of both moment and curvature.

Before we leave this simple equation, let's do some simple algebra and solve for M.
M  =  - EI d2y

Let's call this the "moment form" of the Euler-Bernoulli equation. We will be using the equation later, so it is convenient to have a name for it.

Bending needn't be any more complicated than this equation. If you know the curvature at point x, then you know the bending moment at point x. You certainly know the direction of the bending moment, which is what this debate is about.

So what do we know about the direction of curvature near the handle? In the 1990s, TrueTemper built an instrument called ShaftLab. It was data capture of shaft bend during the swing, and the bend was measured directly by strain gauges attached to the shaft just below the grip of the club. Maybe not precisely at the hands, but quite close -- definitely close enough.

And what did ShaftLab show? At impact, the shaft is always bent forward where the strain gauges are. The Euler-Bernoulli equation says that means the bending moment (torque) applied to the shaft by the hands is negative at impact.

But this is static. Steady state. We want to know about a shaft while it is being swung.

Yes, there are some differences, some additions. But you ought to note that the Euler-Bernoulli equation still rules, though there may be some dynamic forces as well as static ones producing the bending moment. The main examples are:
  • If an adjacent segment of the shaft (at z-Δz) is deflected, that will produce a bending tug on the adjacent segment at z.
  • The bending of the shaft produces friction. The friction force is in a direction opposite to the motion producing it, and is proportional to the rate of bending; faster bending means more friction.
Let's look at the equations of motion for dynamic bending.

Equations of motion

The homogeneous equation

The model of the shaft is a second-order differential equation. It is solved as a homogeneous equation (that is, the free-body response with no external forces or torques applied) and a set of boundary conditions representing those external influences. Let's start with the homogeneous equation, equation (2) in the paper.

The first thing to notice is that the equation tells us the three dimensions of the analysis:
  • The t dimension, of course, is time.
  • The z dimension is the axial distance, the measurement along the shaft's axis.
  • The y dimension is the deflection in the swing plane. This is a 2D analysis, where the plane of interest is the swing plane. (The "alpha" direction.) There was no attempt to model cross-plane deflection (called "droop" or "beta deflection") nor rotation around the shaft axis ("gamma"). Of course y changes with time. It also changes with z; the deflection is different as you move along the shaft.
The next thing to look at is what the three terms of the equation represent.
  • In the first (blue) term, mz is the mass per unit length, and the second partial derivative is the acceleration of the deflection. So it is F=ma for each increment of shaft length. What it says is the propagation of deflection along the shaft is determined by the mass per unit length. If the mass of the shaft itself were zero, the term would not be there and there would be no wave propagating along the shaft. Put another way, the mass of the shaft is the only thing preventing the entire shaft to bend like an ideal spring. The shape would be similar at all times, with only the amplitude being different.
  • In the second (yellow) term, c is a damping or friction factor, and the partial derivative is the velocity of the deflection. This accounts for damping of the shaft's vibrations, specifically the damping due to the shaft's internal friction. The damping term is acting as a dashpot (if you're mechanically inclined) or a resistor (if you're electrically inclined) where it sits in the equation.
  • The third (red) term looks a little like the Euler-Bernoulli equation. Look at the part inside the square brackets. It is what we earlier named the "moment form" of the Euler-Bernoulli equation. Note that EI is correctly identified as a function of z; that is, EI varies along the length of the shaft. The butt of any golf shaft will be stiffer than the tip, and that variation is mathematically expressed as EI(z). The variation of EI shown in Figure 7 of the Nesbit-McGinnis paper is very reasonable for a shaft, rather middle-of-the-road. So that can't be the difference between Nesbit's model and well-known shaft behavior.
The terms of the homogeneous differential equation say mathematically that the wave motion along the shaft and the damping of the shaft have to balance out the moment necessary to create a bend in the shaft.

Initial and boundary conditions

  • The homogeneous equation is the behavior of the system after it has been tweaked or plucked, as it settles back to steady state.
  • The initial conditions specify exactly the state at some point in time (t=0 in this analysis, and in most analyses).
  • The boundary conditions are where the external drivers (forces and torques, in a mechanical problem like this) are added to the system.
Let's look at how Nesbit and McGinnis set up the initial and boundary conditions, based on this view of the shaft.

The z-axis runs from zero at the hands to L at the tip. All the external forces and torques are applied by the clubhead to the tip of the shaft. Here are the equations for the boundary conditions.

Equations (3) and (4) are the boundary conditions at z=0 for the entire time of the simulation. The equations constitute the enforcement of the cantilever assumption. They say that at z=0, where the shaft is under the hands, the position (y) is zero and the slope (y') is zero -- meaning the shaft comes out of the hands with no slope, perfectly flat. These equations completely determines the influence of the hands on the shaft, and setting them to zero assures that the hands will behave as a clamp. There will be no damping at the hands at all. From our damping experiment, we know this is wrong.

If we were going to make the hands a damping grip instead of a clamped cantilever, equation (4) would be the place to do it. Instead of setting the slope to zero, the slope should be on a stiff, lossy, angular spring. I haven't worked the problem in detail, and I haven't set up a problem like this in over 50 years, but my instinct tells me equation (4) should be something like:
A δy(0,t)

 +  B δ2y(0,t)

δz δt
 =  0
In the first term, A is a spring coefficient for the stiffness of the very stiff pivot (instead of a perfectly stiff cantilever). The second term has B as a damping coefficient which causes more energy loss the faster the slope changes.

Getting back to Nesbit, equations (5) and (6) are the boundary conditions at z=L for the entire time of the simulation. The equations are based on the forces and moments on the tip of the shaft. In the diagram above, those are Fy, Fz, and Mx. The functions Fy(t), Fz(t), and Mx(t) are prescribed by earlier simulations that determined the forces and moments applied to the tip of the shaft.

Equation (5) is a balance of forces at the tip of the shaft, those forces in the y direction (across the shaft in the alpha plane). Fy obviously fits this description, but the other term? We recognize the other term as the derivative (slope) of the moment (it's the moment form of Euler-Bernoulli) taken along the z axis. Engineers with experience in beam analysis instantly recognize the derivative of the bending moment as the shear force (internal force) across the beam. So the equation says that the imposed y force at the tip has to balance the internal shear force at the tip.

Equation (6) is a balance of moments, again at the tip of the shaft. The third term is obviously a moment; it's the externally imposed moment at the tip. The first term we should by now recognize as the moment form of Euler-Bernoulli, so it is a moment. That leaves the second term to be interpreted as a moment. Let's look at that. Fz(t) is the component of force F(t) along the z axis. With a straight shaft, that would be pure axial force on the shaft, trying to stretch it rather than bend it. But we are dealing with a deflected shaft; the derivative in the second term is the slope of the shaft at the tip. So Fz(t) resolved itself into an axial component (trying to stretch the shaft) and a radial component (trying to bend back towards straight). If we multiply Fz(t) by the slope, we get the radial (bending) component. Well, there's this little confusion between the tangent and the sine; but for small enough angles they are close to the same (and the tip is bent maybe 2-3 degrees, which is small enough).

For the record, I just gave a hand-waving justification, not a rigorous one. I'm unhappy with it myself. I just described a term that is a force, not a moment. I am unable to find a moment arm, a distance, to go with the bending force I described. Having drawn a few diagrams, I think I would be happier if the term were:
- Fz(t) y(L,t)
That would at least be a bending moment in a direction to reduce the bend, and would also be consistent with the diagrams I drew.

Rather than boundary conditions, equations (7) and (8) are initial conditions. They apply at t=0, and they apply for all z -- that is at every point on the shaft.

Equation (7) says the shaft is straight at the beginning of the downswing, which is just plain silly. During the transition (the start of the downswing), the shaft is bent anywhere from 40% to 100% of the maximum it will bend during the downswing. Where it lies in this range depends on the way the golfer loads the shaft, but I have never heard of a swing with anything vaguely resembling a straight shaft at transition. Try it yourself. Find any good golfer's driver swing on YouTube, face-on slow-motion. Freeze it at the exact moment of transition and lay a straight edge along the shaft as it exits the hands. You'll find the tip substantially in lag due to bend.

Equation (8) says the shaft is not going into any bend at the moment of transition. Not only is it not bent, but not moving towards bending; there is no derivative of y with respect to time at any point along its length. This also sounds wrong, but isn't as obviously wrong as equation (7).

In any event, I'm not making much of a fuss about this error. I doubt that it has much to do with the question on the table here, which is the direction of shaft bend and hand couple at impact. I'm sure it warps other results somewhere, but I'm not going to spend time on it here.

An electrical engineer's view

As I mentioned above, undergraduate electrical engineers learned certain analysis techniques earlier and perhaps more intensively than other engineering disciplines. Most of those techniques were to deal with transient behavior of systems, often reactions to step or impulse function inputs or an alternating input at some frequency or combination of frequencies. My experience with this sort of analysis on analogues of mechanical systems gives me a little insight into the elastic shaft that is the subject of this investigation.

Let's look at two analogues, and see what we can learn from them.

Tuned circuit

Here's an example of a perfect analogue between mechanical and electrical systems. It's not a random example by any means; it relates directly back to oscillation and damping, like we saw in the damping experiment.

The two simple systems above, one mechanical and the other electrical, are direct analogs of one another.
  • In the mechanical system, the mass will bounce back and forth if the dashpot doesn't introduce too much damping.
  • In the electrical system, the voltage will oscillate up and down if the resistor doesn't introduce too much damping.

Looking more closely, here are quantities that are direct analogues in the two systems:

Mechanical system Electrical system Role in the system
Position of mass Charge on capacitor Main analysis variable x
(or, for electrical, Q)
Force Voltage The "push" in the system
Velocity of mass Current First derivative of x or Q
Acceleration of mass Derivative of current Second derivative of x or Q
Spring Capacitor Proportional to x,
F = Kx
v = Q/C
Dashpot Resistor Damping,
energy dissipation,
proportional to x',
F = cx'
v = RQ'
Mass Inductor Proportional to x",
F = mx"
v = LQ"

Not only that, the equation of motion/charge is exactly the same:

  +  2cω dx

  +  ω2x  =  0
This is a second-order differential equation where:
  • t = time.
  • x = position or voltage, depending on which system we analyze.
  • c = damping factor. We'll get back to this in a minute.
  • ω = the resonant frequency of the system in radians per second.
If you tug on either system and let go, there will be a transient response, which may well be oscillation Let's look at the language an electrical engineer would use to describe the different types of oscillation.

For both the mechanical and the electrical oscillatory system, there is a common description of the behavior. The graph above shows the major named types of behavior:
  • Undamped, or no damping - If the damping factor c is zero, then the oscillation continues forever.
  • Underdamped - If the damping factor c is less than 1, the response crosses zero at least once. For c considerably less than 1 (say, less than 0.2 or so), the response oscillates at an ever-decreasing amplitude. "Underdamped" means it is damped, but still manages to oscillate somewhat. The red curve has a damping factor of 0.02, so it gets a lot of oscillation in before it dies out for all practical purposes.
  • Overdamped - If the damping factor is more than 1, the response never crosses zero. It really isn't oscillating, just dying out. The blue curve has a damping factor of 4.
  • Critically damped - If the damping factor is exactly 1, that is the fastest the system can "come to rest" for practical purposes. It is also the lowest damping value where the response never crosses zero. The green curve has a damping factor of 1.
Now that we have a language, let's see if we can apply it to the damping experiment we did when we plucked a club with different ways to secure the handle. We choose a damping factor that makes the curve look like the response we observed when doing the experiment.

  • Clamped - This is seriously underdamped; the damping factor is .005 and it loses only about 2.5% of its amplitude each cycle. This matches the damping I measured in stop-frame on the video.
  • Braced - A damping factor of .2 leaves us about 3 cycles before the amplitude is too small to notice.
  • Hands - The clubhead passes zero, but then just limps back to zero. Not critically damped, but certainly not highly underdamped. This graph uses a damping factor of .5, or half of critical damping.
In any event, the damping for "clamped" is more than an order of magnitude lower than either "braced" or "hands". (The ratios are 40 and 100, respectively.) That says that only 1% to 2.5% of all system damping is in the shaft. But the equations Nesbit uses for the simulation only include damping in the shaft.

Transmission line

The tuned circuit is doesn't completely explain the behavior of the shaft, though it gives a strong hint. But it doesn't talk about the flex wave traveling along the shaft. It only addresses how the shaft mounting or grip deals with signals coming up the shaft. So let's look at traveling and standing waves.

A good (admittedly not perfect) analogue of the shaft is an electrical transmission line. This includes the mighty power lines running across the countryside on tall towers, as well as the lowly coaxial or even twin-lead cables that carry a TV signal around your house. Let's look at the latter.

The figure shows an equivalent circuit for a transmission line. Yes, it also has resistance; it is lossy. But that is mostly a factor in power transmission. TV signals within a home don't suffer much loss, so let's just look at the inductors (L) and capacitors (C) in the equivalent circuit. Why restrict ourselves to lossless lines? Because our damping test showed that the shaft itself (the transmission line) is not where the losses in the system are happening.

The equivalent circuit models the transmission line as a large number of small LC segments. The more and the smaller the segments, the more accurate the approximation. The actual, perfectly accurate, model of the transmission line has infinitesimal segments (approaching zero), and is better described by a continuous differential equation than the discrete circuit in the diagram.

The L and C obviously vary from cross-country power transmission to in-home coax cable, and even coaxial is different from twin-lead. Let's look at some interesting properties of a transmission line:
  • The ratio of L to C is constant for a particular design of transmission line, no matter how large or small the approximating segments. If you halve the length of a segment, you halve both L and C, so the ratio is the same. That ratio is a characteristic of the line. In particular, the square root of the ratio [ sqrt(L/C) ] is called the "characteristic impedance" of the cable, and has an interesting real-world implication we will explore.
  • If I send a signal down a transmission line, it reaches the end and is reflected back. Or not. The reflection depends on the load.
    • If the load at the end of the transmission line is an open circuit (infinite ohms), then it is reflected back with no loss of energy.
    • If the load is a short circuit (zero ohms), then it is also reflected back completely, but with a polarity reversal, + to -.
    • If the load is in between those extremes, part of the energy will be absorbed in the load, and the rest will be reflected back with or without reversal; low-ohm loads will reverse the polarity.
    • If you're still with me, it seems obvious that there is some size of the load that will completely absorb the signal, with no wave reflected back. That would be the load impedance which is barely too small for positive reflection and barely too large for negative reflection. Remember the "characteristic impedance" from the first bullet point? That is exactly the size of load to absorb all the energy and reflect nothing back.
  • Depending on how much of the wave is reflected at each end, the transients running up and down the line may reinforce or cancel one another at different places along the transmission line and form a "standing wave" on the line. Standing waves mean that the signal transfer becomes very sensitive to both the frequency and the exact length of the line.
  • But... with a load at or near the characteristic impedance (also called a "matched load"), any signal cannot survive much longer than one trip down the line. A matched load gives a much smoother frequency response. If the signal is a transient pulse of some sort, rather than a periodic wave, the signal is received undistorted by the load and never appears again (because there is no reflection to contend with).
A lot of the properties of a transmission line are shared by our golf shaft.

We'll revisit the model of the shaft we used to set up the boundary conditions for the differential equations. Now let's compare it to the transmission line. The forces and torques are applied at the clubhead (tip) end of the shaft, and the hands receive those impulses as the wave represented by a torque travels the length of the shaft to reach the hands. So, in transmission line terminology, the clubhead is the source and the hands are the load. Let's look point by point and compare the shaft with a transmission line.
  • I don't know of anything about a vibrating beam (or golf shaft) comparable to a characteristic impedance. Perhaps there is; if not, perhaps there should be. We will see why below.
  • If the shaft is sufficiently lossless, any signal or wave traverses it at some characteristic speed until it reaches the end and is reflected back. Or not. The reflection depends on the load.
    • If the load at the end of the shaft is a rigid clamp, then it is reflected back with no loss of energy.
    • If the load is a frictionless hinge, then it is also reflected back completely, but with a polarity reversal, + to -.
    • If the load is in between those extremes -- a lossy hinge -- part of the energy will be absorbed in the load, and the rest will be reflected back with or without reversal; low-friction loads will reverse the polarity.
    • If you're still with me, it seems obvious that there is some load that will completely absorb the signal, with no wave reflected back. That would be the hinge loss which is barely too small for positive reflection and barely too large for negative reflection. Let's call that the "matched load".
  • Depending on how much of the wave is reflected at each end, the transients running up and down the line may reinforce or cancel one another at different places along the transmission line. If instead of transients, we have a sinusoidally varying excitation, it can form a "standing wave" on the shaft. Standing waves mean that the signal transfer becomes very sensitive to both the frequency and the exact length of the shaft.
  • But... with a load at or near the matched load, any signal cannot survive much longer than one trip up the shaft. A matched load gives a clean transient response. The transient signal reaches the load intact and never shows up again.
The lesson here, guessed from knowledge of how a transmission line works, is that even if you have no damping in the shaft itself, you can still have very strong, even perfect, damping if the load at the grip end is close to the matched load.

What's new?

Mike Duffey asked me for a summary of what is new in this December 2018 version, compared with the October version. Here it is.

In the October version, I had a confusing (and probably confused) argument why higher mode vibrations (harmonics) were not likely responsible for significant changes of the shape of the shaft, as Nesbit and previously Brylawski had asserted. In order to clean up my muddy reasoning, I went into Nesbit and McGinnis' paper on shaft bend. As a result of that foray, here is my [tighter, I hope] reasoning.

Nesbit and McGinnis report a reverse-bend shape at impact. It is a lag curvature near the hands that devolves to a lead curvature for the bottom 40%-60% of the shaft. If true, that would also explain the positive couple, by simple application of Euler-Bernoulli bending. But is it true?

Nesbit and McGinnis make several shaky assumptions when they set up the differential equation for shaft bend. Three of those assumptions explain the different shaft shape and thus the torque. These assumptions are:
  1. All damping of the shaft's vibration is in the shaft itself; none is in the handle and hands. (The paper pays lip service to damping in the hands, but the equations don't cover it.) Experiment suggests less than 3% of the damping is in the shaft itself, and more than 97% is in the hands.
  2. At impact, the only force or torque bending the shaft is the eccentric loading due to centrifugal force at the CG of the clubhead. That assumption is refuted by both empirical data (from ShaftLab) and analytical modeling (a Sasho MacKenzie paper on shaft bend).
  3. The shaft is modeled as a cantilever beam, admittedly for ease of computation. (Nesbit asserts the shaft/hand interaction is not sufficiently understood to do otherwise.) This assumption is obviously questionable, but I don't have a precise refutation beyond its clearly incorrect implication for damping.
Fixing assumption #1 provides the basis for ignoring higher mode behavior, especially when you factor in the speed of propagation of a flex wave.. Any higher-mode wave has no chance to stick around long enough to affect shaft shape in an important way.

Assumptions #2 and #3 can together explain the reverse bend at impact even statically. We don't need flex waves, either stranding or traveling, to create the reported shaft shape. Since #2 is false and #3 is questionable at best, that negates the whole idea of a reverse bend in the shaft at impact, a shape also refuted by ShaftLab measurements.

Last modified -- Dec 21, 2018