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 (meterkilogramsecond) 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 thinwalled spheroid 4.4" wide, 4.4" fronttoback, 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)mr^{2}
where:
 m = mass = 200g = 0.2Kg
 r = radius = 2.2" = 0.056m
So
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 + md^{2} = .00042 + 0.2*.056^{2}
= .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 shaftfitting 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}
or
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 nonmetric terms, that's 2.8 inch pounds or
just under a quarter of a footpound.
(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 closedform 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 NF4). 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) Kx^{2}
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
2020 hindsight, the closedform expression isn't that difficult. I
repeated the steps using symbolic instead of numerical representations,
and it solved to:
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
