Physical principles -- biology 4

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?

We found the energy required to accelerate a clubhead to 100mph. Then we calculated how much of that can come from the potential energy of the clubhead cue to gravity.
We figured out the work that a given mass of muscle can do, based on measured force for a given percentage contraction of the muscle.

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 faster-moving 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 fast-moving 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 non-negligible 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 not-so-decent swings, because we are trying to find how much muscle we need to swing the club at 100mph. A not-so-decent 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 80-20 split is not far off; if anything, it may be even more lopsided. We will use 80-20 for this problem.

The statement of the problem says the clubhead speed is 44.7m/s. If that is an 80-20 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 color-code 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 color-code 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 10¹².)

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 color-coding 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 = mL²/12 = 2.5kg*(1.83m*17.2%)²/12 = 0.0206kg-m²

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 (d₁), plus the distance from the elbow to the CG of the forearm (d₂).
d₁ turns out to be 1.83m (the golfer's height) times 17.2% (the upper arm segment length). That result is 0.314m.
d₂ 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 = d₁ + d₂ = 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 = mL²/12 = 1.44kg*(1.83m*15.7%)²/12 = 0.01kg-m²

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*mr² = 0.4*0.5kg*(0.05m)² = 0.0005kg-m².
Since there are two hands in this assembly, that's a total inertia of 0.001kg-m²
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 length-to-diameter 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 = mL²/12 = 0.055kg*(0.57)²/12 = 0.018kg-m²

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 6000g-cm², or 0.0006kg-m². 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 0.0005kg-m². 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	14.8	0	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 (kg-m²)	Angular Energy (J)
Upper arms	14.8	0.0206*2	4.5
Forearms	14.8	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 one-to-one 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, ball-park 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 ½mr².
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 (kg-m²)	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 real-life 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 high-energy 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/sec²).
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 non-obvious 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 368-56.6 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 368-43.4 = 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 stretch-shorten 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.
Torque-velocity 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 technology-based 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/cm² 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 well-conditioned 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 stretch-shorten cycle. Let's do that now. (Spoiler: it will help, but nowhere near enough to make up the difference by itself.)

Stretch-shorten cycle

We mentioned the stretch-shorten cycle (SSC) earlier. It is purported to be able to give a boost to the muscles, if there is a counter-move (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

Last modified -- Apr 5, 2023

Physical principles for the golf swing

Biology