Stability of the swing plane
Dave Tutelman
 September 19, 2016
I had an email discussion with Todd
Dugan about the implications
of stability of the swing plane. Todd is a golf instructor and
I am an engineer, so the first thing we had to do was to reconcile what
"stability" actually means; it has a very specific meaning in
mechanics. Ultimately, Todd wanted to make a statement about the
putting stroke, but I found it very instructive to start with the full
swing. My own takeaway from the discussion is:
 Dynamic forces do a good job of stabilizing the swing plane
for a fullswing drive.
 Gravity dominates the forces affecting the putting swing
plane; dynamic stability has relatively little to do with it.
Here is how I came to that conclusion.
Stability and the
swing plane
The first step in solving a problem is a precise statement of what the
problem is. In this case, both "stability" and "swing plane" need to be
defined carefully. The former has a scientific definition that is not
exactly the same as the colloquial use, even though it is consistent
with it. The latter is controversial among instructors and
biomechanists; but we are going to limit the discussion to a part of
the swing that everybody agrees is pretty much planar.
What is stability?
Engineers and physicicsts talk about a concept called "equilibrium". It
refers to an object which, when placed in a particular position, will
stay in that position. There are three recognized kinds of equilibrium:
stable, unstable, and neutral. Wikipedia has an excellent article explaining the
difference, and I'm going to borrow a few diagrams from it to present
the definition in my own words. Each diagram shows a ball resting on a
surface of some shape.
Stable
equilibrium:
If you move the ball left or right, gravity provides a force that wants
to return the ball to its original position. 
Unstable
equilibrium:
If you move the ball left or right, gravity provides a force that wants
to push the ball even further from its original position. The original
equilibrium is a delicate balance, not stable at all. 
Neutral
equilibrium:
If you move the ball left or right, it will stay where you put it.
Gravity does not have a preference for any position on this flat
surface. 
We will use the term "stability" in this spirit. A swing plane is
considered stable if any perturbation of the club from the plane causes
a force that wants to return the club onplane. In fact, that happens
 and to greater or lesser degrees with different kinds of swings.
What is
the swing plane?
I'm
going to borrow another diagram to show
what I'm using as the definition of the swing plane. There have been
many studies that show the downswing not to be on a single plane.
And if you've ever watched Peter Kostis draw lines on the SwingVision
image on TV, I'm sure you believe that. It's way more complicated than
a plane, and probably more complicated than anybody needs to think about.
But there is a part of the swing that stays remarkably close to a
plane. It is the yellow sector of the circle in the diagram, from
shafthorizontal on the downswing, at least to the ball, and probably
to shafthorizontal on the followthrough. A significant portion of the
biomechanics world has dubbed this the "functional swing plane", and there
are good physical reasons it is planar. In fact, the reason is stable
equilibrium, as we will see. 
Stability of the
fullswing plane
The
clubhead moves in a plane, the functional swing plane, for the 90°
arcs just before and just after impact. Let's look at the dynamics of
that part of the swing, to see whether the swing has equilibrium there,
and whether that equilibrium is stable, unstable, or neutral.
The diagram shows two forces that the clubhead exerts on the shaft in
the vicinity
of impact:
 The inertial force in reaction to the pull that the
shaft exerts to curve the path of the clubhead. It is an "equal and
opposite reaction" (in Newton's terms) to the centripetal force, and is
commonly referred to as "centrifugal force". Without getting into a
fight about nomenclature or whether the force is "real", it can be
usefully quantified. The
clubhead is moving on a curved path that is at least approximately
planar, and the force must thus act in that plane. Its direction is
away from the center of curvature (to be equal and opposite to the pull
of the shaft), and its magnitude is:
where:
m
is the mass of the clubhead.
v
is the velocity of the clubhead.
R
is the radius of curvature of the clubhead's path.
 Gravity acts in a vertical direction, straight down.
Its magnitude is
F_{g} =
mg
where:
m
is the mass of the clubhead.
g
is
the acceleration of gravity, a constant. It is 32 feet per second per
second in English units and 9.75 meters per second per second in MKS
units.
These are the forces that tug on the shaft and on the hands that will
either keep the club inplane or move it outofplane. The
stability of the swing plane can be measured by comparing the inplane
forces to the outofplane forces.
If the forces that keep the swing inplane dominate, then we can
say the plane is stable. That is the basis of this
investigation!

To
this end, I have made a spreadsheet to compute the inplane and
outofplane components of the inertial and gravitational forces. You
can download
it
yourself and play with the numbers if you're not happy with my choice
of numbers. The inputs to the spreadsheet are what we have already seen
 m,
v,
and R
 plus the angle of elevation of the swing plane. In the vicinity of
impact, this corresponds to the proper dynamic lie of the golf club.
The
diagram illustrates two ways of thinking about this stability figure of
merit. They are mathematically identical, just different ways you can
view it.
 The ratio of the inplane force to outofplane
force. (Those are the dotted red force vectors.) The larger this
number, the more stable the swing plane. A large number means that the
forces that want to restore the clubhead onplane dominate the forces
that want to move it offplane.
 The angle between the resultant
force and the swing plane. (The resultant is the solid red force
vector.) That shows just how far outofplane is the total force on the
clubhead. If the angle is small, then the forces are close to the plane
and the clubhead will stay more or less onplane.
Why do I say
they are mathematically identical? The ratio of inplane to
outofplane force (figure of merit #1) is the cotangent of the angle
between the resultant and the swing plane (figure of merit #2). So from
either one you can easily get the other.
Notice that the diagram
includes both the inertial force (green) and the gravity force (blue).
Most of that is inplane, but at least part of the gravity force is
outofplane. 
I
want to belabor the point, so that it is completely clear why stability
is governed by the relationship between inplane and outofplane
forces. Look at the picture of Rory McIlroy, one of the best drivers of
the golf ball, near impact. I have added the forces at the two ends of
the shaft: Rory pulling in on the shaft with his hands, and the
clubhead pulling out on the shaft.
If the two forces are not in
the same line (that line being the shaft axis), then they are parallel.
What does that mean? They will form a moment  a torque  equal to
the magnitude of the forces times the distance between them. If they
are not along the same axis, the resulting torque will tend to turn the
club so that the forces do
line up.
If you look again at the defintion of stable equilibrium, that is
exactly what it is about. If the forces are parallel to the plane but
not in the plane, they will turn
the club to move it back into the plane.
Now
that's the inplane force; it stabilizes the swing plane. But suppose
the force has an outofplane component. That will tend to accelerate
the clubhead away from the swing plane. It does not have any tendency
to provide stable equilibrium to the swing plane, unlike the inplane
force. If the outofplane force is tiny, if it is dominated by the
inplane force, then the overall effect is a stable swing plane. But,
to the extent that the outofplane force is at least a significant
fraction of the inplane force, either the club will go out of plane or
the hands will have to provide some [beta] torque to keep the club on
plane to make good impact with the ball.

Finally,
let's get to some numbers. I used the formulas (which are implemented
in the spreadsheet) to compute the inertial force F_{i},
the gravity force F_{g},
and how they align with the swing plane. I used a driver swing, with
the 200gram clubhead moving at 100mph at impact. I placed the center
of rotation of the clubhead's path a few inches above the hands, and
the functional swing plane at 50° from horizontal. What I found was:
Inertial force 
= 71 pounds 
Gravity force 
= 0.4 pounds 
That
certainly suggests that inertia dominates the forces the
clubhead
exerts on the shaft. So we should expect this to be a very stable swing
plane. And it is. When we resolve inplane and outofplane forces, we
find:
Inplane force 
= 71 pounds 
Outofplane force 
= 0.3 pounds 
Intoout ratio 
= 252 
Resultant angle offplane 
= 0.23° 
So
the inplane forces are more than 250 times as large as the
outofplane forces, and the resultant force is aligned within less
than a quarter degree of the swing plane. Conclusion: a
very stable swing plane!
One more point worth making: These numbers are a snapshot of the swing
plane at impact.
That is where the velocity is highest and the clubhead path is likely
the most curved. (Not necessarily the most curved part, but you have to
do something special to flatten out the curve near impact.) So this is
almost certainly the most stable instant of the swing. If we averaged
the stability over the entire functional plane, it would be somewhat
less stable  though still dominated by the inertial inplane forces. 
Stability of the
putting swing plane
Now let's look at the putting swing. Does the same
thing hold true? Is the putting plane kept stable by inertial forces?
Todd's
motivation for the question is based on whether the "straight back,
straight through" putting stroke is physically sound. If inertia
effectively stabilizes the putting swing plane, then the most stable
swing would be an "around" swing, not a straight line. Todd's geometric
reasoning is perfect; see the diagram for this discussion.
 If the stroke is stabilized by inertia, then the
hands have to move in the same plane as the putter head.
 A putter lie is typically in the 70°75° range 
tilted, not vertical. So that plane must be tilted.
 The putter head (and hands) swing on arcs in the
plane. Since the plane is tilted, the horizontal movement of the putter
head (the projection of putter head movement onto the ground) cannot be
a straight line. It is intosquaretoin.
So the only question that remains
is the inertial stability of the putting plane. If inertial forces do a
good job of stabilizing the plane, that is a fairly strong argument
against a putting stroke that is straight back, straight through. It
can still be a rocking motion in a single plane. But the plane
is tilted and not vertical, so the horizontal component of the path is
not
straight.

Let's dive in and look at the numbers. A typical
putter has a 320gram head, though there are plenty with heavier heads
around. (Actually, while clubhead mass will affect the forces, it will
not change the ratios nor angles of those forces; so the choice of head
weight is not going to make a difference.) The typical putter lie angle
is 72°, so we will use that for the angle of the functional swing
plane. And let's assume the putter stroke is a pendulum from the
shoulders, so the radius of curvature is about 60 inches.
The most interesting input is the clubhead speed. Also the most
critical, since inertial force depends on velocity squared. I happen
to have previously
computed the ball speed for a fairly typical putt, a 10foot
putt on a green that stimps at 10. That came out to 5mph. So the
clubhead speed is 5mph divided by the smash factor. When you plug
putter head numbers (mass=320g, COR=.77) into the formula for smash
factor, you get 1.55 for the factor and 3.23mph for the
clubhead speed.
Here is what the spreadsheet tells us:
Inertial force 
= 0.1 pounds 
Gravity force 
= 0.7 pounds 
Hmmm! This time, the inertial force is considerably smaller that the
force due to gravity. That is not a strong argument for stable
equilibrium. Let's resolve inplane and outofplane forces, as a check
on that:
Inplane force 
= 0.77 pounds 
Outofplane force 
= 0.22 pounds 
Intoout ratio 
= 3.5 
Resultant angle offplane 
= 16° 
The inplane forces are bigger than outofplane. But the ratio
is less than 4:1  hardly a domination. And the resultant force is 16°
out of the swing plane; that suggests the plane is not going to be
stable.
Wait! How does this even happen? The inertial force
was much lower than
gravitational force. How did we ever get more inplane force than
outofplane? The diagram of forces tells the story.
Yes, the inertial force (green) is considerably smaller than the
gravitational force (blue). But the tilt of the swing plane is only 18°
from vertical. (That's 90° vertical minus 72° lie angle.) So most of the
gravitational force is in the same direction as the shaft axis 
nominally inplane. That is how we got a larger inplane force. But it does not suggest a stable swing plane. In
fact, the stable plane is with the club hanging nearly vertical.
You can see this when you look at the angles. The resultant force is
16° offaxis. That means there's a lot of force tending to kick the
club out of plane. Most telling is the fact that the plane itself it
tilted 18°; if the resultant force is 16° from the plane, it is only 2°
from straight up and down.
Bottom line: the putter wants to hang vertically. Inertial forces are
only helping with 2° out of an 18° deficit. That is not a
swing plane with stable equilibrium  not even close.

Comparisons and
conclusions
Here is the data  the raw numbers  from the two cases I've run on
the spreadsheet.
Description 
Mass
(grams) 
Speed
(mph) 
Radius
(inches) 
Lie
Angle (deg) 
Dynamic
Force
(lb) 
Gravity
Force
(lb) 
Inplane
Force
(lb) 
Out
of plane Force
(lb) 
Force
ratio 
Resultant
angle from plane
(deg) 
Driver
100mph 
200 
100 
50 
50° 
70.7 
0.439 
71.071 
0.282 
252.1 
0.227° 
10ft
putt, stimp 10 
320 
3.23 
60 
72° 
0.098 
0.702 
0.766 
0.217 
3.531 
15.81° 
It is clear from these numbers that the driver swing plane enjoys
stable equilibrium, while the putting stroke has a lot less stability.
A good look says that the putting stroke has almost no stability; there
is seven times the tendency to hang straight down as to stay onplane.
What does this say about the viability of a straight
back straight
through putting stroke, compared to a tiltedplane putting stroke?
Let's return to the picture of the swinging putter head. But now
we'll look at two possibilities: a curved path and straight back
straight through.
Curved
path
We've
already see this one. There is a swing plane that contains the shaft,
and obviously both the hands and the putter head. The projection of the
putter head's path on the ground is curved  in to square to in.

Straight
back, straight through
This
stroke has the putter head moving in one plane (the blue vertical
plane) and the hands in another (the pink vertical plane). This can be
accomplished with a shoulder rock; this time the rocking is in a
vertical plane not a tilted one. The result is a putter head path that
is straight back and straight through.

This says that there is no argument based on stable equilibrium to
favor one style over another. One may in fact be better; I don't know.
But stability of the swing plane cannot give us the answer.
Slightly generalizing this conclusion... During our discussion, Todd
tried out the principle, "The
conditions resulting from an inplane swing establishes it as most
proper; the golfer should always reproduce this condition,
even in lowpower swings where he may have to provide the restoring
force himself."
The words "most proper" and "should" may be a hint that this is a value
judgement, not an absolute. The principle may be true. Or it
may be false. Or it may be true for some golfers and not others. Or it
may even have some truth and some falsehood for every golfer.
There may even be a way to determine whether it is true, false, or
something else.
Anatomy? Maybe.
Motor learning? Maybe.
Biomechanical experiments? Maybe.
But the physics of
stable equilibrium simply doesn't help us with that question.
Last
modified  Sept 20, 2016
