Physical principles for the golf swing

Energy and Momentum

Dave Tutelman  --  October 3, 2022


At first, I was not going to have this chapter in the tutorial. I changed my mind for a bunch of reasons:
  1. One of the goals of the tutorial is to allow you to read important articles in the field, so you can keep up with the science, not just carry away what I thought I knew when I wrote it. Some of the papers that should be most interesting to instructors and golfers are intimately entwined in the subject of energy or momentum.
  2. Everybody involved in golf uses the terms "energy", "momentum", "friction", and probably others introduced here. It would be nice if you had a precise understanding of what the words mean, not just a vague, colloquial idea.
  3. If you wanted to know what gravity or shaft bend might contribute to clubhead speed, this is the first place to look.
So let's get started!



Among the most famous laws of physics are the "conservation laws", commonly stated as:
  • Conservation of energy = Energy can neither be created nor destroyed; it is a constant.
  • Conservation of momentum = In a closed system, total momentum is a constant.
Let's put some precision and understanding into these statements. Let's understand what energy and momentum are, and how to think about them in the context of golf. That context keeps things relatively simple. We can stick to simple Newtonian physics, and not worry about relativity, quantum physics, or even forms of energy beyond mechanical. (We will talk a bit about other forms of energy, but not in detail and not quantitatively.)

Energy

Work

Well, then, within the context of golf physics, what is energy anyway? Energy is the ability to do work! Not helpful? We need to know what "work" is? You're absolutely right -- and physics has a precise and quantifiable definition of work.

Work is a distance that a body moves times the force in the direction the body is moving.

Let's look at some examples of work and non-work under this definition.
  • If I push against a wall, I am not doing work. I may be exerting a force and expending a lot of effort, but nothing moves so it is not work under the definition.
  • If I push against a heavy door and it opens, I am doing work. The force I exert (the push) times the distance the door moves while I push with that force is the work I did... or, put another way, the energy I expended.
  • If I push against that same door but it is locked, am I doing work? If you said "yes", go back to the beginning of this section and start over.
  • Here is a tug of war between the green team and the red team. Let's look at it in terms of work.
    1. The teams are well matched. They both exert a force of exactly F. Nothing moves, they just pull. No work is being done.
    2. The red team puts forth a little more effort. The forces are still both F within a fraction of a percent, but the red force is slightly larger. The resulting acceleration may be small, but the red team incrementally pulls the green team over the line. The red team did work (expended energy) equal to F times the distance the rope moved. The green team had work done to them equal to the same amount, F times the distance the rope moved.
Wait! "Work done to them"? What does that mean? It means the green team was receiving energyfrom the red team. In this case, the energy received was not useful, but there are cases where the work done to an object is recoverable energy. Sisyphus spent eternity learning this. He did work pushing the rock up the hill. As he neared the top of the hill, the rock would get away from him and roll back to the bottom.

But a mass moving at speed is a form of energy we will see later, called "kinetic energy". The Sisyphean myth can be described in terms of physics and energy. Sisyphus did work on the rock, pushing it up the hill. The fact that it was near the top of the hill meant that it had absorbed energy from Sisyphus, and now had a lot of "potential energy", something else we'll see later. With Sisyphus no longer pushing it upward, it rolled downward, picking up speed and converting that potential energy to kinetic energy.

So, under the right circumstances, work done to an object may be recoverable or partly recoverable.

When work is not recoverable, it is converted to some other form of energy. For example, work done against the force of friction is converted into heat, another form of energy. That is why you can start a fire by rubbing two wooden sticks together. At the end of this section, we'll look very briefly at non-mechanical forms of energy. But first, let's look at the most important kinds of mechanical energy, potential energy and kinetic energy.

Potential energy

When you do work on an object and that work is recoverable but has not yet been recovered, the object has acquired potential energy. (Let me be a bit more specific -- some might say anal -- about that definition. We're talking about mechanical work and mechanical recovery. It is certainly what we mean in the golf context, and most seem to use the term that way in general.) Let's look at two important examples, gravity and a spring. Each example teaches us something we can use later in golf physics.

Gravity

We have a ball of weight W sitting on the lower shelf in the picture. We pick it up and put it on the upper shelf, a height h higher. How much potential energy do we add to the ball?

The potential energy is equal to the amount of work we do against gravity to lift the ball to the upper shelf. Let's see...
  • We are lifting the ball, which will take an upward force of W to oppose the weight.
  • We apply a force of W to the ball through a distance equal to the height h.
  • Therefore, the work done is W*h.
So the added potential energy is equal to Wh.

But we glossed over the fact that the path of the ball is not just "straight up and stop"; the ball has to move the the shelf as well. It turns out that it doesn't matter. Let's look at the problem again, with a specific path, to see why we can ignore the path.

Here is the same example, but with a path made of four segments. We'll call the height of each segment y. Let's look at each segment to see how its work is Wy.
  1. This part of the path is obviously Wy, because we are lifting straight up a distance y with a force of W.
  2. Now the path is off vertical by an angle of a. That means the path is longer; it is y times 1/cos a. So wouldn't the work be greater? Remember, work is the distance times the force along the path. W is a vertical force, and the path is not vertical. So let's get the components of W that are parallel and perpendicular to path B. Turns out that the component in the direction of the path is smaller than W; it is W multiplied by cos a. So for path B, the work is still W times y, the height of path B, not the length of path B. That's because the length is divided by the same amount that the force component is multiplied by -- so they cancel out.
  3. The path is horizontal, but the force is vertical. So there is no component of the force in the direction of the path. No work is done while traversing path C.
  4. The force is upward but the path is downward. The distance the ball moves is -y, and that minus sign is important It means that negative work is being done on the ball. Whatever extra work was done getting the ball well above the upper shelf is given back when the ball is put down on the shelf (along path D).
These are the basic ways the path can be something different from straight up. (What about a curved path? It is composed of small segments of the style of path B.) And none of them matter. The work is "adjusted" for each path so that the final work done is just Wy on that path, and a total work of Wh.

So we can state that the potential energy of any object of weight W at a height h above our zero reference is exactly Wh.

Spring

A spring, so matter what its size, shape, or intended use, is a device that can exert a force against being deflected, that force being proportional to the amount that it is deflected. Its behavior can be described very concisely by its "spring constant" K, which is the force divided by the deflection. K is the same value at any deflection, since the force and deflection are proportional. (Of course, this breaks down once a spring is deflected beyond its useful range, but for this example we'll assume we are staying in the useful range.)

We have a coil spring with spring constant K. If we compress the spring an amount x, how much potential energy have we given to the spring?

Because of the definition of the spring constant, we know that, for any x,
F = Kx
So the force changes as we press the spring. All we have to do to calculate work is add up all the tiny contributions of the force at some value of x, times the tiny distance to the next x we look at. Adding together miniscule pieces like this is the job of integral calculus, which gives us a very simple formula. (If you ever took integral calculus 101, this one was covered in the first week.)

Work  =  F dx  =  Kx dx

Work  =  ½ Kx2

The potential energy of the compressed spring is the work done to compress it, so the potential energy is ½Kx2

Kinetic energy

As we noted earlier, a mass at speed has energy due solely to its velocity. This form of energy is called kinetic energy, and it is given by the formula:

KE  =  ½ mv2

Where does this come from? Look at it this way. In order to move a weight upwards a distance Wh, we had to do work equal to Wh. That work turned into potential energy. That suggests a potential for somthing, it could potentially become something, maybe some other form of energy. So let's see what would happen if we tipped the ball off the upper shelf and let it fall to the lower shelf. The falling ball has the potential to acquire some velocity v. Now let's remember what "conservation of energy" means. Energy cannot be created nor destroyed. All that potential energy has to turn into kinetic energy, or the work done lifting the ball (potential energy) would not be conserved. The ball's mass m moving at a velocity v has to have the same energy as the potential energy was before the ball was pushed off the shelf.

When we do the math, we find that the formula for kinetic energy has to be the one given above, or energy is not conserved. So, reiterating once again:

KE  =  ½ mv2

For those who are interested, here's a proof that conservation of energy leads to our formula for kinetic energy. It's here mostly to show the beauty and consistency of physics, but you don't need to know it to understand golf swing biomechanics.

We'll start out by finding the velocity of the ball v when it reaches the lower shelf.

v  =  at

This is just the relationship of acceleration and velocity. If you're confused by this, better go back and look at the basics. It's important.

v  =  gt                  (eq 1)

where g is the acceleration of gravity. But we don't know t. Let's find it. We know that distance is the integral of velocity over time, so

h   =   g dt   =   ½ g t2

We can solve this for t, and plug back into equation 1.

t   =   sqrt (
2h

g
)

v   =   gt   =   sqrt ( 2hg )             (eq 2)

But we know already that

PE   =   Wh   =   mgh

because the weight is the "mg force". And we can solve equation 2 for hg, which we will substitute back into potential energy. This will give us the kinetic energy, because we started out with conservation of energy -- the two energies must be the same. Solving equation 2 for hg:

hg   =   ½ v2

Plugging back into E=mgh we get

KE   =   ½ mv2

Q.E.D.

A couple of golf examples

Before we go further, let's look at some simple questions about golf that might be answered by what we know now. In both cases, we don't have enough knowledge yet to get the real-world answer, but we can certainly answer a simpler question. It also happens to be the question so often framed by people who know just enough physics to be dangerous.
  1. How much clubhead speed can we get from gravity working on the clubhead?
  2. How much clubhead speed can we get by recovering shaft bend as a "kick velocity"?

Gravity

We are going to do a simplified analysis of how much of our clubhead speed comes from gravity. We are simplifying in that we are only considering gravity's action on the clubhead. There is also a small contribution from the weight of the shaft and a significant contribution from the weight of body parts (mostly the arms). But we don't have the tools yet to do those estimates. This would require both a knowledge of angular energy and the dimensions and inertial properties of some significant body parts.

We will solve the problem by finding:
  • How much potential energy does the clubhead have at the top of the backswing?
  • If all that potential energy is converted to kinetic energy by the time the club gets to impact, how fast does that mean it is traveling?

Here is part of a typical swing sequence page from Golf Digest magazine, one of many they have published of Tiger Woods' swing over the years. The clubhead's height at the top of the backswing is the yellow arrow h.  Tiger is 6'2" tall, and some measurement of the picture itself says that h is no more than 6'9" or 81". Let's do this analysis using non-metric units of the foot/slug/second system; we'll do the next example in metric.

Wait! Slugs? What is that? You perhaps thought mass would be in pounds? OK, force and mass are not the same thing; if they were, then F=ma would be meaningless. In every system, there are units of force and different units of mass. In the system we use in the USA, pounds and ounces are units of force, not mass. So, working from F=ma, the unit of mass would be the unit of force divided by the acceleration of gravity -- 32 feet per second per second. That unit is called a slug. In other words, a mass that weighed one pound would be a mass of 1/32 slugs.

OK, let's proceed. The contribution of the gravity on the clubhead is the loss of potential energy as the clubhead moves from the top of the backswing to impact. (Well, not exactly "loss"; it is converted to kinetic energy, which is the point of this problem.) The potential energy change depends only on the mass m of the clubhead and difference in height h. That was the point of our look at several paths going from the lower shelf to the upper; only the change in height matters, not the path it takes to get there.
  • The mass m of the clubhead is 200 grams (the mass of most driver heads) converted to slugs, which is .0137 slugs.
  • The weight of 200g of mass is 0.44 pounds. That's the force exerted by gravity on the mass.
  • The height difference h is the 81" we identified earlier, minus about 2" between the ground and the CoM of the clubhead at impact. That's 79", which we need to express as 6.6 feet (because the consistent system of units requires feet as the measure of distance, not inches).
So we multiply height and force together, and find the potential energy loss is 2.9 pound-feet. Conservation of energy says the clubhead gains exactly that same amount of kinetic energy. So let's set up that equation.

Kinetic Energy  =  ½ mv2  =  2.9

2.9  =  ½ * .0137 v2

Solving for v, we get:

v  =   sqrt (
2 * 2.9

.0137
)  =  20.6 ft/sec  =  14 mph

That's out of a clubhead speed over 120mph (this is Tiger swinging the driver). It is certainly non-negligible, but we aren't getting a lot of "oomph" from gravity working on the clubhead.

Shaft bend

The shaft of a golf club functions as a spring. We can bend it (for instance, clamp the grip and deflect the clubhead) and expect that:
  • It will offer a resistance, a force, against the bending.
  • It will spring back when we release the bending.
Let's treat it as a spring and compute how much clubhead speed we might possibly get from it if it bends back during the swing then perfectly times its straightening to occur at impact. We should recognize this as another example of potential energy being converted to kinetic energy. This time it is a spring rather than gravity, but the principle is the same. And, as promised, this time we will use the MKS (meter/kilogram/second) system of units.


Here is a driver shaft being deflected by an amount D at the start of the downswing. Most swings have the maximum deflection later than this, so let's use D to mean the maximum deflection of the shaft during the swing before impact. To find the clubhead speed contribution from shaft kick, we must:
  • Compute the potential energy represented by the shaft bend.
  • Equate the kinetic energy of the clubhead to the potential energy we just computed, and find the velocity that energy would require.
We know that the potential energy of a spring is given by ½ Kx2. Obviously the spring deflection x is D. Measurements of shaft bend during the downswing tell us that, for a golfer (not a long drive competitor) with a reasonably well-fit shaft, the deflection of the driver shaft almost never exceeds 10cm, and is considerably less for most golfers. So let's start with x=10cm.

How about the spring contstant K? I have a basement full of clubmaking tools, so I went down to the basement and measured a bunch of rather stiff shafts -- that is, springs with a high K. (We're going with a high-end K and x, because we would like to see the maximum clubhead speed advantage we might get from shaft flex.) I found that a deflection of 1cm (or 0.01m) gave a force of about 3.5 Newtons. That gives a K (force divided by deflection) of 350N/m. So...

Potential Energy  =  ½ Kx2  =  ½ 350 * .12  =  1.75 Nm

We need to find the clubhead speed representing 1.75Nm of kinetic energy possessed by a driver clubhead of 200 grams (0.2Kg). BTW, there is a name for the energy of 1 Nm; it is called a "Joule".

Kinetic Energy  =  ½ mv2  =  1.75

1.75  =  ½ * .2 v2

Solving for v, we get:

v  =   sqrt (
2 * 1.75

.2
)  =  4.18 m/sec

For the benefit of us Americans and the units we have some intuition for, that is about 9.4mph.

Actually, that is not the way it works. In the early 1990s, TrueTemper (the shaft manufacturer) found out a few interesting facts using their ShaftLab instrumentation.
  1. The highest kick velocity they measured in years of testing was about 11mph. (That is quite consistent with our result.)
  2. For some reason, the kick velocity did not seem to show up as added clubhead speed. (We will find out why toward the end of this golf swing biomechanics tutorial. We're not ready yet to understand the reason. In fact, TrueTemper did not understand the reason back in 1994. It wasn't until 2016 that the paper was published that showed what was really going on.)



Last modified -- Dec 3, 2022