Physical principles for the golf swing
Biology
Dave Tutelman
 February 28, 2023
Energy again
Let's
wrap up the section on "Biology" by exploring the intersection of
biology and physics. In fact, isn't
that intersection exactly what biomechanics is? Let's try a
problem that we should have enough information to solve, but will call
on our knowledge and also some resources that are available and we
should learn to use. Let me urge
you to follow this problem and solution; it teaches many of the things
you will have to know in order to to read about, and perhaps even do
your own, biomechanics.
Here is a problem that is simple to state and should not be all that
difficult to solve. Not conceptual difficulty, anyway. But it will take
a lot of effort to solve.
The important thing is to break down the problem into smaller problems
that we know how to solve. So this solution will go on for quite a bit,
but each
little calculation we do should be simple and familiar. There is a lot
in biomechanics modeling like that. In fact, you will encounter it in
lots of science and engineering, and other endeavors as well. It is
worth getting used to.
We are going to revisit the energy question. Remember a couple of
exercises we did earlier?
Review those previous examples to make sure you understand them. We
are about to take the next step.
The
problem:
Let's use that information, along with what we now know about
joints and the limitations of the human body, and estimate how
much muscle a human body needs in order to make an effective golf
swing. For this exercise, we will need all sorts of measurement
information about the human body. As a byproduct, we will learn about
resources we can use to look up the "bio" part of biomechanical
information.
Let's start with the specifics of the problem. We want to find the
amount of muscle needed for the following golf swing:
 The clubhead speed at impact is 100mph (44.7m/s).
 The club is a driver, with the relatively standard
specifications of 200g head weight, 55g shaft weight, 55g grip weight,
and 45in (1.14m) in length.
 The golfer is a 6ft tall male (1.83m) weighing 170 pounds
(77kg). We will assume his proportions are typical of the information
in the resource tables. His strength is typical of the average muscle
fiber strength we determined
earlier. .
When
we know the amount of muscle, we will try to relate it to how much of
the body must necessarily get involved in order to hit a decent drive.
Here are some resources available on the internet, which will give us
mechanical information about human body parts:
 Skeletal muscle accounts for 40% of the human body's mass.
 Here is a set of tables of data about body segments,
including:
 Segment weight as a percentage of total body weight.
 Segment length as a percentage of total body height.
 Location of segment center of mass. It is specified as a
distance from the proximal
end of the segment, that distance expressed as a percentage of total
body height.
The first thing we need to do is set out a strategy. Here is how I am
going to attack the problem.
 Find the
kinetic energy for the clubhead, the club, and some
fastermoving distal body
parts at the moment of impact.
That is how much energy the body will have to generate.
 But the body doesn't have to do it all by using muscle
during the downswing.
Remember that the club and the arms start the downswing much higher
than they finish. So we can get some of that kinetic energy from the potential
energy of the club and arms that had been generated by muscles
during the backswing. Subtract that from the required
kinetic energy, and we are left with what we need from the muscles in
the downswing.
 Convert the remaining kinetic energy requirement to how much
muscle tissue is required for the work.
 Now apply
limitations
to figure out how much more muscle we will need in real life. For
instance, every muscle used will have an antagonist muscle that, at
best, is dead weight but also required because it is part of the body.
Let's begin.

1. Find the kinetic energy
Unlike the
problem we solved
a few chapters ago, this
is no longer just the KE of the clubhead. We need to consider the
energy of the shaft and some moving body parts. To keep it simple,
we'll start with the fastmoving parts and try slower movers until they
add negligible kinetic energy to the total. We will start with the
arms, hands, and golf club. Then we will look at the body (thorax,
abdomen, and pelvis). If the body has nonnegligible kinetic energy, we
will include the legs.
In order to do that
computation, we can use the approach described by Dodig: approximate each
body part as an easily characterized geometric solid. (The diagram is
from the Dodig paper.)
We
will represent
the arms as cylinders, but the cylinders are so long compared to their
width that we can ignore
their width and just focus on their length; they are more rods than
cylinders. We are going to treat the hands as solid spheres 0.1m
diameter,
since they are curled up gripping the golf club. The thorax, abdomen,
and pelvis will be separate cylinders stacked atop one another, each
rotating at its own speed around a common axis, the centerline of the
cylinders.

Lets
start by quantifying the velocities, both angular and linear. In the
diagram, we have simplified the swing by limiting it to the arms and
club, and considering the arms and the club each as a single rigid
body. Actually, this is fairly reasonable, since for this exercise we
are only interested in what is happening at impact (remember, right now
we are looking for the kinetic energy at
impact).
The quantities shown in the diagram are:
 v
= the linear velocity of the wrist, which is also the butt of the club.
 V
= the linear velocity of the clubhead due to club
rotation.
 w
= the angular velocity of the arms assembly, in radians per second.
 W
= the angular velocity of the club in the swing plane.
I know from lots of computer simulations that the total linear velocity
of the clubhead is due in part to the linear motion of the hands and
the rest due to the angular velocity of the club itself. In all the
simulations I have done of decent swings, the breakdown is within a
percent or two of 20% hand speed and 80% club rotation. (If you want to
experiment with a computer simulation yourself, download
Max Dupilka's program SwingPerfect.) For purposes of this exercise,
we really don't want to bother with notsodecent swings, because we
are trying to find how much muscle we need to
swing the club at 100mph. A notsodecent swing will require more than
that minimum muscle; that is why is it not a good swing in the first
place.
In correspondence with other biomechanists, the 8020 split is not far
off; if anything, it may be even more lopsided. We will use 8020 for
this problem.
The statement of the problem says the clubhead speed is 44.7m/s. If
that is an 8020 split between V and v, then V=35.8m/s and v=8.9m/s
Now let's find the angular velocities. For that, we will need some
lengths, because what we know so far is linear velocity. Linear and
agular velocity are related by a radius, which is a length. Let's refer
to the body
segment lengths, and reduce what we need to a table.

Assembly

Linear
velocity

Radius

Angular
Velocity

Arms

v=8.9m/s 
Upper
arm = 17.2% of height
Forearm = 15.7% of
height
Total = 17.2+15.7 = 32.9%
Golfer's height = 1.83m
Radius = 32.9%*1.83m = 0.602m

w
= 8.9/.602 = 14.8rad/s

Club

V=35.8m/s 
1.14m

W
= 35.8/1.14 = 31.4rad/s

In order to make the table calculations easier to follow, we will
colorcode where we got the information from.
 Numbers that come directly from the problem statement
in green.
 Numbers that come from our reference on body
segment data in red.
 Numbers that come from a previous
calculation that we did in blue.
We
will keep that colorcode convention for the rest of the calculations
as well.
So now we know the motions of the two major assemblies. Let's use that
to get the motions of each moving part. That will give us what we need
to compute the kinetic energy of each part, which we will do in a
separate table.
We will start by recording or, if necessary, computing the dimensions
of each moving part. The dimensions we will need are:
 Radius from the pivot to the center of gravity* (CG) of
each part.
 Mass of each part.
 Moment of inertia of each part about its CG, generally
referred to the principal moment of inertia and designated I_{o}.
* Note that center of
gravity and center of mass
are the same thing. CoM is more recent, and I see it more often in
modern biomechanics papers. CG is the term I used in engineering school
around 1960, and I still see often in engineering tracts. No matter;
you can use them interchangeably, and I will. For this problem I will
use CG, because it is what our reference for body segment dimensions
uses. (* For
the nitpickers, yes there
is a subtle difference, but it does not apply for anything happening on
the surface of the Earth whose dimensions are relatively small. In the
case of a golfer, the largest object we are concerned with, the
difference between CG and CoM is less than one part in 10^{12}.)
The two
assemblies are the arms and the club. Let's deal with each.
First, the arms assembly; we will do the calculations for one arm and
double it. In the diagram, the crossed blue centerlines are the pivot
point at the shoulders, and the bottom of the assembly is the blue
arrow, which has a linear velocity of v=8.9m/s.
(Remember, the blue linear velocity number means we know this from a
previous
calculation. From now on, I will apply the colorcoding convention
without mentioning it.) Let's compute the properties for the
upper arm (aqua) and the forearm (tan). We will not compute anything
(yet) for the hands; they rotate at the speed of the club assembly, so
they are not part of the arms assembly, which rotates more slowly.
Upper
arm
 The radius is 1.83m
(the golfer's height) times 17.2%
(the segment length) times 43.6%
(the segment length to the CG location). The last two are in two
tables in our body segments reference. I won't mention this
again; you'll have to know it by the red color code. That result is 0.137m.
 The mass is 77kg
(the golfer's mass) times 3.25%
(the segment mass) = 2.5kg.
 The principal moment of inertia I_{o} is
calculated as if each arm segment is a rod of
the same length. So we will use the formula for moment of inertia of a rod
rotated around its CG. We will ignore the diameter of the arm; I have
done some quick calculations and concluded that our error in doing so
is less than 1%.
I_{o}
= mL^{2}/12 = 2.5kg*(1.83m*17.2%)^{2}/12 = 0.0206kgm^{2}
Forearm
 The radius (r in the
diagram) is the distance from the pivot to the CG. For the forearm, it
is the entire length of the upper arm (d_{1}),
plus the distance from the elbow to the CG of the forearm (d_{2}).
d_{1}
turns out to be 1.83m (the
golfer's height) times 17.2%
(the upper arm segment length). That result is 0.314m.
d_{2} is 1.83m
(the golfer's height) times 15.7%
(the segment length) times 43%
(the segment length to the CG locationi). That result is 0.123m.
r = d_{1}
+ d_{2} = 0.314+0.123 = 0.437m
 The mass is 77kg
(the golfer's mass) times 1.87%
(the segment mass) = 1.44kg.
 The principal moment of inertia I_{o} is
calculated as if each arm segment is a rod of
the same length. So we will use the formula for moment of inertia of a
rod rotated around its CG.
I_{o}
= mL^{2}/12 = 1.44kg*(1.83m*15.7%)^{2}/12 = 0.01kgm^{2}
Since there are two arms involved, we will have to double the energy
when we come to that step.

Now
let's do the same thing for the assembly containing the club. It also
contains the hands, because the hands are holding the grip of
the club and rotating with the angular velocity of the club. Here is an
image for the components of that assembly. Again, the crossed
centerlines represent the pivot. When we go to compute velocities of
each segment to get kinetic energy, we will have to remember that the
pivot is already moving with linear velocity v, so we have
add v to
the velocity of each segment in this assembly.
Let's get the radius, the mass, and the moment of inertia of the hands,
the shaft, and the clubhead.
Hands
Because the hands are wrapped around the butt of the club, they are
more like spheres than the rectangular prisms that Dodig used in his body model. We will model them
as solid spheres, each of diameter 10cm (0.1m, or 4").
 The radius, the distance of the CG from the pivot, is the
radius of the sphere itself. That is half the diameter, which we have
chosen as 0.1m. So the radius is 0.05m.
 The mass is 77kg
times 0.65% = 0.5kg.
Since there are two hands in this assembly, that's a total mass of 1kg.
 The moment of inertia is calculated as that of a solid
sphere, using the formula
I_{o} = 2/5*mr^{2} = 0.4*0.5kg*(0.05m)^{2} = 0.0005kgm^{2}.
Since there are two hands in this assembly, that's a total inertia of 0.001kgm^{2}
Even with two of them, that is miniscule. We'll carry it along this
time, but if it turns out to add negligible energy, we can probably
remember to ignore it in the future.
Notice that we have not included the grip of the club here. That is
because it is going to be so small as to get lost in the wash. It
exists roughly where the hands do (of obvious necessity), with a mass
only about 1/20 of the hands' mass. We could just lump it into the
hands, where it will give about a 5% error in the hands' contribution.
If the hands' contribution to the total energy is substantial, we can
go back and include the grip explicitly. (Hint  that won't be
necessary.)
Shaft
As with the arm segments, we are going to model this as a rod. Since
the lengthtodiameter ratio is much larger than an arm, it is closer
to an ideal rod. We will assume that the CG is midway along the rod.
Remember that the hand is holding the butt end of it, so the shaft
extends all the way to the wrist, which is the pivot for the assembly.
 The radius is half the shaft length of 1.14m, or 0.57m.
 The mass is 55g,
or 0.055kg.
 The moment of inertia is that of a rod.
I_{o}
= mL^{2}/12 = 0.055kg*(0.57)^{2}/12 = 0.018kgm^{2}
Clubhead
 The radius (the distance to the pivot) is just about
exactly the shaft length, or 1.14m.
 The mass is 200g,
or 0.2kg.
 The moment of inertia is more interesting. There is a Rule
of Golf that limits the clubhead moment of inertia around the vertical
axis at 6000gcm^{2}, or 0.0006kgm^{2}. When you think
about the design criteria for a driver head and what the head looks
like, you come to the conclusion that the MOI around the vertical axis
is going to be the largest. We are looking at an axis in a different
direction, probably with a value of 0.0004 or 0.0005. Let's go with 0.0005kgm^{2}. If it turns out this contributes
significant energy to the system, we can revisit it. (But it won't.)
Now we can go and compute the actual kinetic energies contributed by
each segment. The formulas we will use are the ones we have already
worked with for linear
and rotational
energy:
 Linear energy = 1/2 * mass * linear velocity squared
 Angular energy = 1/2 * moment of inertia * angular velocity
squared.
Since most of the linear velocities of the segments come from rotating
a rod around a
pivot, let's explicitly state how we get the linear velocity in the
table:
 Linear velocity = (radius * anglular velocity) + velocity
of the pivot
We'll capture it all in a table for linear energy and another table for
angular energy. No need for our color code here; everything
is the result of our calculations above.
Linear Energy
Segment

Angular
velocity
(rad/s)

Pivot
velocity
(m/s)

Radius
(m)

Linear
velocity
(m/s)

Mass
(kg)

Linear
energy
(J)

Upper
arm

14.8

0

0.137

2.03

2.5*2

10.3

Forearms

0.437

6.47

1.44*2

60.2

Hands

31.4

8.9

0.05

10.47

1

54.8

Shaft

0.57

26.8

0.055

19.7

Clubhead 
1.14

44.7

0.2

200

Total






345

Angular Energy
Segment

Angular
velocity
(rad/s)

Moment
of inertia
(kgm^{2})

Angular
Energy
(J)

Upper
arms

14.8

0.0206*2

4.5

Forearms

0.01*2

2.19

Hands

31.4

0.001

4.9

Shaft

0.018

8.9

Clubhead 
0.0005

0.25

Total



20.7

One more thing we ought to check out before we claim we know the
kinetic energy. The body (the thorax, abdomen, and pelvis) is rotating
only, not much linear velocity.
But it is a lot of mass, so let's just check and see if its rotational
energy is worth worrying about. How can we find out the body's
rotational speed? Here is our first look at the kinematic sequence of a good golf
swing, adapted from an article
by Phil Cheetham. We will see a lot more of this important
biomechanical graph later; for now, all we are interested in is the
value at impact.
The vertical axis is angular velocity. (It is in degrees per second,
but is easily converted to radians per second, and they agree fairly
well with the numbers we have been using so far.) We will need to deal
with a few outages that we will have to approximate:
 The "body" is the thorax and the hips in this graph. But
they are rotating at different speeds.
 Our measurements of body segments include the thorax,
abdomen, and pelvis, which are not onetoone with the parts for which
we know the rotation speeds.
Here is what we will do to resolve it. (If this were a serious
measurement, we would have to do more. But since we are looking for an
approximate, ballpark answer, we can get away with approximations.)
 We will consider the body to be made up of three cylinders,
the thorax, abdomen, and pelvis. The thorax and abdomen will each be
cylinders with a 6" (.15m) radius, and it does not matter what the
pelvis size is because it is not rotating and therefore will contribute
zero no matter what the radius is. The formula for the moment of inertia is ˝mr^{2}.
 The thorax will rotate at about 250deg/sec at impact, as in
the graph. That is 4.4rad/sec. The mass is 77kg*20.1% or 15.5kg.
 The abdomen will rotate at half the speed of the
thorax, or 2.2rad/sec. The mass is 77kg*13.1% or 10.1kg.
 The hips will be considered stationary; they have run out
of speed by the time impact occurs. Therefore, they have no kinetic
energy at impact.
Now let's get busy with the table of calculations.
Angular Energy
Segment

Angular
velocity
(rad/s)

Mass
(kg)

Moment
of inertia
(kgm^{2})

Angular
Energy
(J)

Thorax

4.4

15.5

0.174

1.7

Abdomen

2.2

10.1

0.114

0.3

Pelvis

0





0

Total




2

So the bottom line is that there isn't a lot of kinetic energy in the
body at impact. There is plenty of mass, but the velocities are pretty
small at impact.
The total kinetic energy of the arms, hands, and golf club is
345+20.7+2=
368Joules . A few points worth
noting before we move on:
 More than half the total is the linear kinetic energy of
the clubhead. Why? It is moving by far the fastest, and energy comes
from velocity squared.
 Almost none of the energy, less than a tenth of a percent,
comes from clubhead's angular energy. That is due to the small moment
of inertia of the clubhead, even though it is rotating at 31.4rad/sec,
which is about 1800deg/sec.
 In fact, less than 10% of the energy is in angular energy
at all (22.7J out of 368J is only about 6%).
 When working through any reallife problem  or a
mathematical model that purports to represent real life  it is
incumbent on us to review our assumptions and see their impact on the
results. It is a check on both the reality and the usefulness of what
we learn from our model. So let's look at that for a bit.
 The linear kinetic energy of the clubhead is 55% of all
the energy. That's more than half the energy, even at only a quarter of
a percent of the mass. We didn't need much of an assumption for that;
the linear velocity was given, and we [should!] all know that almost
every driver head is just about 200g.
 When dealing with the properties of the arms, we got good
estimates (or at least consistent estimates from a reputable source) of
the mass and the position of the CG; therefore, we can be pretty sure
our linear energy estimates are valid. But we made several
approximations in finding the moment of inertia; we assumed a uniform
cylinder, and then simplified it further to a rod. But, looking at our
results, the linear energy of the arms is more than ten times their
angular energy. So even if our assumptions about moment of inertia were
off, they only affect a small portion of our answer  less than 2% of
the answer, in fact.
 We took some major liberties with the shape of the hands
gripping the club. But the linear energy hardly depends at all on the
shape, just the location and mass  and we have a pretty good grip on
that (pun intended). The shape assumptions would show up in the angular
energy, and for the hands that is less than a tenth of the linear
energy and less than 2% of the total energy.
 We assumed the grip is negligible in the kinetic energy
calculation. Certainly the angular energy would be negligible.
Even the linear energy would be less than 3J. How do I know? The grip
is where the hands are, so the linear and angular velocities will be
the same as the hands. But the grip's mass (and probably its moment of
inertia as well) is only about 1/20 of that of the hands.
 As we already noted, the rotation of the body contributes
less than 1% to the kinetic energy, so any gross assumptions we made
about it don't really matter much.
So we can probably trust the energy estimate to
be good enough for educational purposes. Note that if we were doing a
serious biomechanics experiment to see the effect of changing sometime,
we might need to be more precise.
Now that we know the kinetic energy of the highenergy pieces of the
swing, we next look at where we can find a supply for that energy.

2. Find and subtract
potential energy
The first place we will look for an energy supply is the potential
energy due to gravity. We did a simpler version of this problem in the earlier chapter on "Energy".
In that exercise, we considered only the clubhead's potential and
kinetic energy. Now we are enlarging that to the whole body. But we
have found that the kinetic energy has little contribution except for
the arms, hands, and club. This should be even more so for potential
energy, because it depends on a difference in altitude between a body
part at impact and that same body part at the top of the backswing. So
let's stick with that same collection of segments.
Here
is the picture we are working with, the relevant segments at transition
and again at impact. The potential energy lost  really, converted to
kinetic energy  is the vertical difference between its center of
gravity at transition and at impact, times the mass of that segment. We
have already computed the mass
and the position of the CG of each of the segments in the previous section; we will use those values
here.
For each segment, we will find the difference in height between the
segment's CG at transition and at impact. The diagram is simplified. In
our work, we will consider the segments:
 Upper arm.
 Forearm
 Hand
 Club shaft
 Club head

This
diagram
gives a little more detail on how we will do the computation.
 The boundary between transition and impact is at the red
measurement "Shoulder Height".
It is the combined length of the arms and the club. We compute that by
adding together the length of the upper arm, the lower arm, and the
shaft. (The shaft passes through the clubhead and the hands, so we
don't have to include their length.)
ShoulderHeight = 1.83m*(17.2%+15.7%)+1.14m = 1.74m
 At transition, everything is at an angle.
 We assume that the wrist cock is 90°. That is fairly
typical for a good driver swing.
 We assume that the turn of the lead arm is not quite
180°, but rather 10° short of that. Therefore:
 The vertical projection of the arms is multiplied by cos(10°), or 0.98
 The vertical projection of the club shaft and head is
multiplied sin(10°)
or 0.17.
 The club at transition begins in the middle of the hands
at "Grip Height".
This is ShoulderHeight plus the vertical extension of the arms above
the shoulders (at an angle of 10° from vertical, of course), plus half
the diameter of the hands.
GripHeight = ShoulderHeight+(1.83m*(17.2%+15.7%)+0.05m)*cos(10°) =
ShoulderHeight+0.64m
We
will compute everything we need for potential energy from these
basics, in the following table.
 The potential energy is the vertical difference times the
weight (force) of the segment. Force, if you recall, is mass times the
acceleration of gravity (9.8m/sec^{2}).
 I assume by now you know how to use the
body segment tables, so I won't go through the details to get those
numbers.
 For other nonobvious calculations, there are notes below
the table.
Segment

Length

Radius,
shoulders
or grip
to
CG

Vertical
distance
below
shoulders
(impact) ^{3}

Vertical
distance
above
shoulders
(transition)

Vertical
difference ^{6}

Mass

Potential
energy

Upper
Arm

0.315m

0.137m

0.137m

0.135m
^{4}

0.272m

5.0kg

13.3J

Forearm

0.287m

0.437m

0.437m

0.430m
^{4}

0.867m

2.88kg

24.5J

Hands

0.1m

0.652m

0.652m

0.642m
^{4}

1.294m

1.0kg

12.7J

Shaft

1.14m

1.172m
^{1}

1.172

0.74m
^{5}

1.912m

0.055kg

0.98J

Clubhead



1.742m
^{2}

1.742

0.84m
^{5}

2.58m

0.2kg

5.1J

Total







56.6J

Notes:
 Add lengths of upper arm and forearm to half the length of
the shaft; that gets you to the CG of the shaft.
 Add lengths of upper arm and forearm to the length of the
shaft; that gets you to the CG of the clubhead.
 Same as "radius" in previous column.
 Radius times cos(10°).
 GripHeight minus Shoulder Height (which is 0.64m), then add
the distance from hands to CG times sin(10°).
 Add the vertical distance below shoulders to the vertical
difference above shoulders.
So the potential energy due to gravity is 56.6 joules, which is about
15% of the total swing kinetic energy of 368 joules. That means we have
to account for 36856.6 or 311 joules.

Hold on
a
second! We have been computing this as if the downswing is completely
vertical. It isn't! The swing plane
is on an angle to the vertical.
Check out the picture; it shows a swing plane of 50°, which is 40° from
vertical. This is fairly typical of a driver swing. So each of the
vertical distances and vertical differences are only 0.77 times the
numbers we calculated. (0.77 is the sine of 40°, or the cosine of 50°.)
That means the potential energy is only 0.77 of what we calculated.
Here's a sanity check that the swing plane cannot be vertical. The
radius we computed from shoulders to clubhead is 1.742m. But if we take
the body height of 1.83m and subtract the head and neck we get from our body
segment reference, we get 1.63m  which is considerably less than
the 1.74m radius. So if the swing plane were vertical, the shoulders
would have to be higher than they actually are. Conclusion, the swing
plane must be slanted.
How slanted does it have to be? Well a vertical 1.63m actual shoulder
height compared with a 1.74m required shoulder height would give a
ratio of 0.936, which is the sine of 69°. That is not as low as the 50°
angle we know is typical for a driver swing, but there are other
factors as well. For instance, The golfer is not standing tall at
impact (unless it was a swing with "early extension", a swing fault).
If the golfer has a slanted back (true of a good posture at impace),
the shoulders would be even lower, and the swing plane flatter than
69°. Add up all these little things, and we get the picture we see at
the right.
The bottom line is that the potential energy is only 0.77 of what we
calculated when we assumed a vertical plane, or 43.4 joules. That is 12% of the
total kinetic energy we computed before, 368 joules. So we still need
to
find 36843.4 =
325joules to power the swing.

3. Compute muscle mass for the
rest of the energy
The remaining 325 joules of energy in producing the motion of club and
body segments we assume to come from the muscles. At the end of this
section, we will take a quick look at the possibility of the
stretchshorten cycle playing a part. That may come from muscles,
fascia, or tendons. For purposes of this tutorial, it doesn't really
matter which kind of tissue produces the SSC, just whether it can
happen at all and how much power it can produce. But first, let's just
look at normal muscular contraction and see how much muscle it would
take to produce 325 joules by simple contraction.
Earlier,
we generated this graph estimating the work each kilogram of muscle
fiber can do with a single contraction. (Built into this solution is
the assumption that, during the
downswing, no individual muscle is going to contract, relax, and
contract again. If a muscle is involved in the downswing, it only
contracts once during the complex of motions that is the dowswing.)
The skeletal muscles of the typical human male can generate 9.5N of force per
square centimeter of muscle cross section. Any skeletal muscle will
lose that force density the more it contracts, down to zero force at
about 60% contraction. Beyond that it can't do work, which means it
can't contribute to energy. From that information, we derived this
graph, which flattens out at 60% contraction to a work capability of
about 30 joules per kilogram of muscle.
So let's do some very simple arithmetic.
 We need to find 325 joules of work to power our golf swing.
 A kilogram of muscle can do 30 joules of work.
 So a golf swing needs 325/30 = 10.8kg of muscle. (That's 28.3
pounds of muscle.)
To
see how much of the body's muscle this uses, we can't just
assume
that all 77kg of the body is muscle. In fact, about 40% of the body is skeletal muscle. (BTW, both the
fraction and strength deteriorates with age, so let's assume that our
subject is a young, strong golfer. This fact explains a lot about why
we lose driving distance with age.) Therefore, only 40% of our
subject's 77kg is muscle, or 30.8kg.
So we are using about a third of
the body's muscle mass to make a golf swing.
Well, not quite. We have to account for a bunch of limitations of our
body's power system, which we have discussed
earlier.

4. Apply limitations
Let's consider how the efficiency of the golf swing is affected by each
of the limitations we considered:
 Load  we have
already considered this. It is how we came up with the graph of work vs
contraction.
 Antagonist
muscles  this turns out to be a big factor, as we will see
below.
 Range
of motion  this will limit the amount a given muscle can
contract. For any limit less than 60% (where the energy vs contraction
graph flattens out), we will need more muscle to get a given amount of
energy.
 Torquevelocity curve
 it is impossible to take this into account without knowing a lot of
detail of exactly which joints have exactly what torques applied and
how fast those joints are moving. So we are going to ignore this
effect. But know right now that there is at least one joint (the wrist)
that is rotating so fast the last 40msec of the downswing that the
torque provided by the muscles is actually negative. So our answer to
the original question is optimistic; we just aren't sure how optimistic.
Antagonist muscles
We have already noted that each skeletal
muscle has an antagonist counterpart on the opposite side of the joint.
If the agonist muscle (the one we are using to power the golf swing) is
of a certain size, and thus able to do a certain amount of work, the
antagonist muscle should be of roughly the same size. So at best we
will need
twice the muscle we just calculated: once to perform the action and
then the same amount again sitting relaxed on the other side of the
joint.
I said "at best" because the antagonist muscle may not be perfectly
relaxed. In our previous look at antagonistic muscle pairs, we noted
that there is often some tension in the antagonist muscle. We might be
able to train it out for a specific motion, like a golf swing, but I'm
not sure of that, nor how much training of what kind it will take.
Nevertheless, let's ignore the antagonist's tension for now. We have no
way to quantify it, nor do we know pragmatically how to minimize it, so
let's just keep our answer on the optimistic side.
That means that antagonistic pairs will merely double the amount of
muscle needed for a golf swing, to 21.6kg.
We're up to 2/3 of the body's skeletal muscle.

Range of motion
It isn't
immediately obvious how range of motion can have an effect on our
calculations but it, combined with the scale of the joint and muscle,
can indeed. I adapted this diagram from Figure 6 in our reference on the strength of muscles. The
black is the original diagram from the paper, and I have added things
in red.
The paper was interested in how much force the quadriceps muscle
(shaded in red) can produce. The quad muscle extends the knee. The knee
joint is the fulcrum of a lever. The quad has a lever arm R_{o}
identified in the paper as having a typical length of 0.037m. By
measuring the force F_{s} at
the ankle and multiplying by the lever arm ratio R_{s}/R_{o},
they found the force F_{o}
exerted by the quad muscle.
That was their goal. We have a different goal in mind.
We want to see how much the quad is able to contract before the knee is
fully extended and cannot move any more  the range of motion limit.
The quadriceps runs the length of the thigh, which our body segment
reference tells us is 23.2% of the height. For our 1.83m golfer, that
would be 0.425m; we will call it L, the length of
the thigh. If the muscle at the knee in the position shown wraps around
the pivot at a radius of R_{o},
then the length of that wrap is πR_{o}/2,
which is 0.058m. Summarizing what we have so far:
 The length of the thigh is 0.425m
 The extra length of quadriceps wrapped around the joint is
0.058m.
Now we know that during the golf swing the knee is never flexed more
than 90°. We also know that it is limited to a straight line because of
range of motion; any more than that could cause a hyperextension
injury. So we can reason that:
 The maximum length of the quadriceps is 0.425+0.058=0.483m
 When the quads are contracted to full extension of the
knee, the length is down to 0.425m, the length of the thigh.
 So the contracted muscle is 0.425/0.483=0.88, meaning 88%
of the relaxed length. Therefore, the muscle has contracted by 12% (that's 100%88%).
The significance of this is that the quads' contraction during the golf
swing is limited by the geometry of the muscle and joint, including the
range of motion. While the muscle tissue itself may be capable of 60%
contraction, it is limited in this configuration to a contraction of
just 12%. Going back to our work vs contraction graph, we are no longer
getting 30 joules of work per kilogram of muscle. If we are limited to
just 12% of the muscle's contraction, then we can only get about 10
joules per kilogram.
Now not all muscles are limited by range of motion to only 12%
contraction. But those driving the knee and elbow joints are. If we
were to guess that an average active muscle in the golf swing can have
20% contraction  averaged over all the muscle mass involved  then
we can expect about 16 joules of work
from every kg of muscle.
(Note: that is pure guesswork. We know that the elbow and knee are much
less efficient, and we are hoping that enough other important muscles
 lats, obliques, glutes, etc  are less constrained, sufficient to
get the overall average up to 20%. I do not know enough anatomy to
actually calculate that.)
Sanity check
Let's review what we have found out and see if it makes sense. Let me
stress that I don't see enough attention to sanity checking in classes
that I have taught at the college and graduate level. I hope it is not
becoming a lost art, because it is absolutely vital in any science or
technologybased society. 'Nuff said, I hope.
 The body/arms/club have a kinetic energy of about 368
joules at impact. That means we have to find some way to exert 368J of work on the clubhead to get
it to speed.
 We can get 43.4J from the potential energy of the arms and
club dropping from their position at transition to their position at
impact. We have to get the remaining 325J
from muscle contraction.
 Assuming we can get 16J of work on average per kg of muscle
(limited by range of motion), we will need 325/16=20.3kg of active muscle.
 Not all muscle is active muscle. Due to antagonistic muscle
pairs, about half of all muscle is inactive for any given motion (like
a golf swing). So we actually need more like 40.6kg of muscle to have 20.3kg of
active muscle.
 Only about 40% of the body is skeletal muscle. We have a
77kg golfer, so we have 30.8kg
of muscle in the body.
Folks, we have a deficit here.
It probably isn't a true deficit; we
know that there are 77kg golfers who can swing the clubhead well over
100mph. So remember that our information is pretty approximate. We made
a bunch of assumptions any time we needed information we couldn't
easily find. I tried to make the assumptions on the optimistic side,
but I was not optimistic enough. Let's look at places where we might
have been too pessimistic:
 We assumed an average male golfer. Maybe the male golfers
who can swing the club at 100mph are not average at all. Our reference on muscle strength said that there was a
relatively wide variation in that number of 9.5N/cm^{2}
of muscle force. But the study found a range of 7.1 to 12.6. That's a
pretty wide range. At the upper end of the range, the muscle work
needed for 100mph is available.
 The assertion that a body is 40% muscle is also based on an
average. A wellconditioned athlete will have a greater percentage of
muscle mass, if only because training puts more muscle on the same
frame. Maybe you have to be well conditioned to achieve 100mph of
clubhead speed.
 In a number of places, I imposed an assumption in the
absence of actual information. For instance, our average of 20%
contraction of each active muscle might be low. My intuition says it is
not, perhaps it is even high, but my intuition about anatomy is not
well educated. I could have been off by enough to make the difference.
 We assumed that the potential energy, which reduces the
amount of energy needed to be supplied by muscle, is based on a 50°
swing plane. 50 degrees is typical, but there is certainly variation
even in tour players' swing planes. Maybe a more vertical swing plane
is required for high clubhead speed. Or perhaps it is a very upright
backswing and a significant shallowing, a "drop into the slot", is
helpful.
 We did not consider the stretchshorten cycle. Let's do
that now. (Spoiler: it will help, but nowhere near enough to make up
the difference by itself.)

Stretchshorten cycle
We mentioned the stretchshorten cycle
(SSC) earlier. It is purported to be able to give a boost to the
muscles, if there is a countermove (a "stretch") immediately before
the actual power move (where the muscles contract, or "shorten"). I
provisionally concluded that this might result in as much as a 10%
improvement in power in a golf swing, based on a 10% improvement in
rowing. It might well be less, and I
doubt it will be more. But let's go with 10% for this exercise.
Suppose we can actually achieve that 10% gain. We need 325J from the
muscles. If we get that with 100% normal muscle activity plus 10% from
the SSC, that means we only need 325/110%=295joules from normal muscle
activity. Repeating our calculations for the amount of muscle we would
need:
 295 joules divided by 16 joules per kg (allowing for range
of motion) gives 18.4kg of active muscle needed.
 Doubling this number to 36.8kg
accounts for the antagonist muscles, which also have to be there even
if we aren't using them in this motion.
Remember that our golfer's body only has 30.8kg of skeletal muscle. So
we still aren't there yet. But SSC has reduced the discrepancy by about
40%.

Conclusions
The solution we have found says that an effective golf swing cannot be
just an arm swing. It has to recruit all the muscles in the body in
order to get the most clubhead speed you are capable of. Getting
stronger  bigger muscles  helps, as long as you don't become
musclebound. (You can't tell from this computation, but building up
the fasttwitch muscles is important as well. That is what "speed
training" is about.)
But there is another message as well. It involves how we examine the
golf swing, rather than about the golf swing itself.
Solving this problem was a lot of work. It was not deep mathematics;
almost all the math was something we learned (or should have learned)
in high school. Knowing what math to apply where is the
challenge. And that is probably true of much of what you will want to
know about the golf swing. You will spend more time finding the
references for your input and cranking through the calculations than
you had anticipated, but the math and science isn't all that deep. Let
me urge you
to also spend the time to double check and sanity check your
calculations. Your results may have been worth obtaining even if they
are
based on questionable assumptions, provided you make clear what your
assumptions are and how reliable they are. But your results are not
worth much if the arithmetic is wrong, or the principles are applied
incorrectly.
Last
modified  Apr 5, 2023
