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All About Spines - Appendix

Dave Tutelman

Appendix A - Calculation of torque threshold

Here we calculate two things:

The amount of torque the golfer exerts on the grip in order to square the clubface.
The amount of torque due to the bend of a misaligned shaft.

The goal is to find the size of spine that causes #2 to be equal to 5% of #1. We will do this for a driver, because drivers seem to be the clubs most sensitive to spine effect.

The physics is done in MKS (meter-kilogram-second) units, to keep them consistent. We will convert from the more familiar (to Americans, at least) inches and pounds.

(1) Torque exerted by golfer to square the clubface

To find the torque, we will need the moment of inertia of the clubhead about the shaft, which we will call I. We will start by finding Io, the moment of inertia about the center of the clubhead. Let's approximate the clubhead as a thin-walled spheroid 4.4" wide, 4.4" front-to-back, and 2.5" high. (The height does not figure into the formula, so it doesn't matter what we choose for it.) The clubhead is 200g, as most driver heads are.

Io
= (2/3)mr2

where:

m = mass = 200g = 0.2Kg
r = radius = 2.2" = 0.056m

Io
= .00042 kilogram meters squared

Sanity check: The USGA has recently proposed a limit on clubhead MOI of .00059 kilogram meters squared. Given the proposed limit, meant to curtail a recent trend to higher MOI, we should expect a head a few years old to be in the range of .0004 to .0005. That is where our estimate lies, so we pass the sanity check.

The shaft is situated .056m (2.2") from the center, so

I
= Io + md2 = .00042 + 0.2*.0562
= .001 kilogram meters squared

(Actually, it's .00105, but let's keep the numbers round.)

How much torque does it take to close the clubface? The clubface closes 90º, which is π/2 radians, in the last 100msec before impact. The time actually varies with the golfer; 100msec is a good swing with late release. As Wishon notes repeatedly in his shaft-fitting writings, there is not much shaft bend in the vicinity of impact when the release is early, so spine should be much less of a problem there. Therefore, let's focus on the late release swing.

Back to physics 101. (I have been tutoring my son in freshman college physics, so this stuff is still fresh.)

T
= Ia

where:

T = torque in newton meters
I = the moment of inertia we computed above
a = the angular acceleration of the shaft, in radians per second squared

So we need a good estimate of the angular acceleration. Let's assume it's approximately the same over the release; it isn't, but the estimate is pretty good. According to my son's physics text, with a constant acceleration:

Angle
= 1/2 (angular acceleration)*(time)2

π/2
= 1/2 a (0.1)2

Solving for angular acceleration:

a
= 314 radians per second squared

We plug this back into the equation for torque to get

T
= Ia = .001 * 314 = 0.314
newton meters

If you prefer to think in non-metric terms, that's 2.8 inch pounds or just under a quarter of a foot-pound.

(2) Torque due to bend of misaligned shaft

It is easier to compute the torque for a few values of spine than to come up with a closed-form expression and solve for the torque. So that is how I did it. (But see below) The spine that gave 5% of the torque applied by the golfer was 3.6cpm. Let's just show the calculation for 3.6cpm.

A few numbers we will need later:

We are using a 230cpm shaft. 3.6cpm is 1.6% of 230cpm. Since stiffness varies with the square of frequency, this is actually 3.2% when measured as a percentage of stiffness.
I measured the shaft using an inverted deflection board (specifically an NF-4). I determined that a deflection of 1 inch creates a load of 1.8 kilograms. We will need the spring constant K of the shaft, which is 1.8 kilograms per inch, or (converting to the proper units) 693 newtons per meter.
From ShaftLab data, the shaft bend in the vicinity of impact is less than 1.5" for almost all golfers. In order to maximize the spine effect, we'll use 1.5", or 0.038 meters.

Let's use potential energy to figure out the torque on a bent shaft. The potential energy in a deflected shaft is given by:

PE
= (1/2) Kx2

where:

K = spring constant. We computed this above to be 693 newtons per meter. But it's 3.2% different for the NBP and the spine; the spine would have a K of 715 newtons per meter, 3.2% higher.
x = shaft deflection. We decided this would be 0.038 meters.

We will find the difference in potential energy between the shaft bent 1.5" at its spine and at its NBP. Then we will find the torque needed to create this amount of energy as the shaft turns 90º from NBP to spine. The PE difference is:

PE
= [(1/2)(715)(.038)2] -
[(1/2)(.693)(.038)2] = .0158
newton meter

So we need to find the torque that does .0158 newton meters of work when rotated through π/2 radians. If the torque were constant, it would be simple.

Work = T a or .0158 = (π/2) T

which solves to T = 0.01 newton meter.

But the torque varies as you move between an NBP (zero torque) and a spine (also zero torque -- an unstable equilibrium to be sure, but still an equilibrium). In between, the torque varies much like a sine wave, being at its maximum halfway between NBP and spine. So let's indulge in a little calculus, and solve it as a sine wave.

Work
= ∫ t(θ) dθ

where

t(θ) = T sin 2θ
T is the maximum torque during the rotation. In other words, it's the torque when the shaft is aligned as badly as it can be.
The formula is sin 2θ instead of sin θ because the sine wave goes through 180º while the shaft rotates through 90º.

Integrating, we get:

Work = ∫₀^π/2 T sin 2θ dθ = T/2 [ cos 2θ ]₀^π/2 = T/2 [2] = T

But we know that work = .0158 newton meter, so we can solve for T. In fact, the solution is trivially easy, since Work=T. So the torque in the worst possible alignment is 0.0158 newton meter.

Looking back at #1 above, the torque applied by the golfer was 0.314. Doing the math:

0.0158 / 0.314 = 5%

QED

(3) Closed form for torque of bent, spiny shaft

With 20-20 hindsight, the closed-form expression isn't that difficult. I repeated the steps using symbolic instead of numerical representations, and it solved to:

=

T

Kx2

where:

c = The spine magnitude in CPM
F = The base frequency of the shaft (in the NBP plane) in CPM
T = The torque at the worst alignment position, the maximum torque in newton meters
K = The spring constant of the shaft in deflection, in newtons per meter. (Yes, this is necessarily related to F, but we don't want to get into that now. It would require introducing tip weights and beam length, an unnecessary complication for a spine problem.)
x = The shaft deflection, in meters

Last modified -- 3/2/2008