Physical principles for the golf swing

Forces

Dave Tutelman  --  March 5, 2020

Kinematics

We already have F=ma, what more do we need to understand how force produces motion?

People who can ask that question are the ones who most need to understand the real messages of kinetics and kinematics.
  • Position is not the same as motion (or velocity).
  • Acceleration is not the same as motion (or velocity).
Far too many laymen -- golfers, instructors, and TV personalities -- don't understand this. And they especially don't understand the second bullet, that acceleration and velocity can have different, even opposed, values. I have even met a fellow electrical engineer with whom I had to break off discussion because he refused to recognize that difference. (And he took enough physics course that this should not have been a problem.)

So please bear with me while I preach. At times, it may seem like I'm beating a dead horse, but without getting this distinction you will never understand some crucial aspects of the swing.

Introduction and examples

Because of F=ma we correctly equate force with acceleration. The force graph over time is the same shape as the acceleration graph over time. The mass is just a scaling factor; it tells us how tall that shape is, but it is still the same shape. So, in the discussion below, we will use "force" and "acceleration" almost interchangeably, except when there is a real need to distinguish between them.

Let's start with a very simple example of position, velocity, and acceleration. Here's an example I used for a more elementary tutorial look at the basics of physics. It shows position, velocity, and acceleration for both positive and negative acceleration. (Negative acceleration has its own name, "deceleration".)

Now we'll look at a live example of how acceleration and velocity are different. Here's a video of someone pushing a car for exercise.



He is turning force into velocity. But not directly! He is turning force into acceleration, which gradually turns into velocity. Since force is the same shape as acceleration (because of F=ma), we can sort of "see" the acceleration; it is pretty constant at the maximum this guy can exert on the car. So we have a constant acceleration. But, if we look at the wheels spinning, it is clear that they keep spinning faster and faster for at least the first 7 seconds. So we have a more-or-less constant acceleration and an increasing velocity.

Once you are comfortable with the tutorial and the live example, let's dig into what we can glean from the graphs we get from motion capture systems and biomechanics tech reports.

Velocity to position - slope

Let's look at how velocity can tell us about position. In order to do so, we'll take a quick journey back to grade school. Remember the problems you had to solve "DRT" problems from the formula
Distance  =  Rate * Time
That tells us how to go from a velocity (a rate) to a position (a distance from where we started). The way we do it is to add some distance for every unit of time the "system is running". How much distance per unit of time? The rate! That's what a rate is; it's something per unit of time. In this case, the something is distance. For instance, the rate may be 15 miles per hour. That means for every hour, we add 15 miles to the distance.

What does this look like on a graph? Here's a graph of a simplet DRT (Distance = Rate * Time) problem.

For elementary DRT  problems, the rate (velocity) was always a constant. On the graph, the velocity is the green curve, and it is a constant; it does not change with time.

The distance is a function of time. In fact, we know the exact funtion of time. It is our favorite formula, D=RT. As time increases, distance increases as a straight line. The slope of the line is the velocity. In fact, that is the definition of slope -- the rate of change with respect to time.

This tells me we could find the distance curve by laying together infinitesimal line segments, where the slope of each segment is equal to the velocity. Note that, in this case, the velocity is a constant. Therefore the slope is a constant. So we don't need to resort to a lot of tiny segments. If the slope is constant, then the curve on the graph is a straight line. And that is what we see here.
 

Velocity to position - area

There is another way of looking at velocity-to-position calculations. Looking at it from the slope point of view is fine if the velocity is constant, but we need something a little more powerful if the velocity varies. And, in most golf physics problems, the velocity does vary.

The diagram shows a velocity (the red curve on the graph) that varies with time. Suppose we want to find out how far we went during the interval from time=0 to time=100. All we know about converting velocity to distance is D=RT, but R (the rate, or velocity) keeps changing. How can we do D=RT if R is varying?

We can solve the problem by slicing time into small pieces, so small that the velocity is pretty close to constant over the small piece. The diagram shows the slices as rectangles, whose width is the time slice and whose height is the velocity at that particular time. Let's look at a few things about the rectangles:
  1. The velocity (the red curve) does not change a lot over a single rectangle.
  2. We have chosen a rectangle height that tries to intersect the velocity in the center of the top line of the rectangle -- sort of equals out the velocity above and below the rectangle.
  3. Now we will say, "OK, let's ignore the small change in velocity over a single rectangle, we'll consider it to be a constant velocity. The magnitude of the velocity is the height of the rectangle."
  4. Let's look at the distance traveled over an individual time slice. Suppose the width of the slice (as a time) is t0 and the height (which is a velocity) is v0. Then v0t0 is the distance traveled during t0, because D=RT.
  5. Let's look again! v0t0 is not just the distance. It is also the area of that rectangle. Remember your geometry; the area of a rectangle is base times height.
Bottom line: if we add together the v0t0 for all the individual rectangles, we get:
  • A very good approximation of the distance traveled, and also...
  • A very good approximation of the area under the velocity curve.
What that means is that the distance is equal to the area under the curve. That should be true no matter how you find the area. Of course, you have to be careful about units of velocity and time, and keep the area in those units -- but if you do, then the area is the distance traveled.

Also worth noting: as you reduce t0 (and therefore increase the number of slices you'll have to add up), the more accurate your approximation of both distance and area. That's what a lot of computer usage is in biomechanics research -- adding up little increments of D=RT to get a total distance.

Examples of position, velocity, and acceleration: clubhead [speed] in downswing.

Acceleration to velocity

Acceleration has the same relationship to velocity that velocity does to position. Acceleration is the rate of change of velocity. If you have a graph of velocity, acceleration is the slope of the graph -- just as velocity is the slope of the position graph.

Please please please remember that acceleration is not velocity. Too often I see golf instructors confuse acceleration and velocity, and wind up with ridiculous conclusions. One of the more common confusions sounds something like, "You say there is a backwards force on the club? But then it would go backwards, and we both know that doesn't happen." Let's deconstruct this and see what is really being said.
  • A backwards force will produce an acceleration in the direction of the force. - Perfectly true.
  • A backwards acceleration means the club would start going backwards. - Ummm, no! That is a statement about velocity, not acceleration. A backwards acceleration means the club will begin to slow down. If it is applied for a long enough time, then it will start to go backwards.
Yes, there are situations where a reverse acceleration will look like an instantaneous reversal of direction. If the force is large enough and the mass is small enough, then the acceleration will be so big that our eyes will miss the slowing-down phase altogether. Let's look at a couple of examples.
  1. An elite tennis player returns a stroke. The ball arrives at 50mph and the player hits it back at 75mph. (Yes, those are pretty typical speeds.) The ball is on the racquet face for only 5msec. When you do the f=ma math, that takes a force of 143lb. Given that the tennis ball weighs only 57g (that's about 2oz), 143 pounds of force changes its direction so fast it looks instantaneous. It isn't instantaneous; the first 2msec of the 5msec impact it is slowing down.
  2. A freight train hits the brakes to stop. The brakes cause a backward acceleration, but the velocity keeps going forward for a while before the train finally stops. From a typical speed of 55mph, it takes almost three minutes to come to a halt -- with a backward acceleration the whole time. The problem is that the train has a huge mass and the brakes can apply only so much force, so the backward acceleration is pretty small.

You probably knew this already, but backward acceleration, or negative acceleration, is called "deceleration". I have been calling it "backward acceleration" to make a point.

If you want to better understand force, acceleration, velocity, and position, you should review the calculations for examples #1 and #2 above. I have the detailed solutions to both at the end of this section.

You can use the same methods to go from acceleration to velocity that we did to go from velocity to position. For instance:
  • Use V=aT the way we used D=RT for position (distance) problems.
  • Find the area under the acceleration-vs-time curve. Sometimes you can just calculate it, if the curve is a shape whose area we know. Much of the time, you will need to break it down into small slices of time, then add the areas of those slices together.
Now a couple of simple numerical examples. Let's go back and solve the two examples above, the tennis stroke and the frieght train.
  1. The tennis stroke - an elite tennis player returns a ground stroke. What is the force applied by the racquet to the ball, averaged over the duration of impact?
    • We can look up typical numbers, and find that:
      • The typical ground stroke in a pro tennis match is struck at 75mph, and loses speed to air resistance and friction with the ground before it reaches the other player; let's use 50mph as the remaining speed.
      • The ball is on the face of the racquet for about 5msec.
      • A tennis ball has a mass of 57g.
    • So a good ground stroke turns a velocity of -50mph into +75mph in 5msec.
    • That's a change in velocity of 125mph. (75mph - (-50mph)) in 5msec.
    • Let's get useful units. We'll use International System (SI) units. So 125mph becomes 56m/s. (I used Calc98 to convert units. I strongly recommend it.) 57g becomes 0.057Kg, and 5msec becomes 0.005sec.
    • "Average acceleration" means we can assume the acceleration is constant over the duration of impact. It isn't constant, but if we took an average, the number would come out as if it were that constant value. So we will find acceleration a.
    • We know that V=aT. Since we know V=56m/s and t=0.005s, the equation becomes 56=0.005a
    • Solving for a, we get  a=56/.005=11,176m/s2
    • We can plug this, along with the tennis ball's mass, into F=ma, and come up with F=.057*11,176
    • Now we just have to do the computation and find that F=637N=143lb.
    It should not come as a shock that applying 143 pounds of force to a 2-ounce ball would certainly look like an instantaneous change of direction for the ball. In fact, it took 5msec, but that is too fast for the eye to distinguish.
  2. The freight train - a freight train applies the brakes. How long does it take to stop?
    • Again, let's start by looking up typical numbers. I was hoping for a time to stop, but all I could find was a distance, so we'll have to do some calculating to find the time.
      • A freight train going a typical 55mph takes between 1 and 1.6 miles to stop.
    • Get our numbers into consistent units. This time we'll use the American system of feet and seconds. (We don't have enough information for mass or force, so we won't have to worry about slugs and pounds.)
      • 55mph = 81ft/s
      • A mile is 5280ft. We'll pick a nice round 7000ft, which is roughly midway between 1.0 and 1.6 miles.
    • What equations might we have that would give us more information?
      • V=at. But we only have V; we have neither a nor t.
      • D=½at2. But we only have D; we have neither a nor t.
      • So we have 2 equations in 2 unknowns. We should be able to solve them, but it could take some work.
    • There are certainly ways of solving such sets of equations. But if we look carefully, we can see a quick solution.
      • The equation for V is simply at.
      • But there is an at in D=½at2=½(at)t So we can plug in V for at and get D=½Vt. We already have values for D and V, so...
      • 7000=½*81t. We solve for t and get t=7000*2/81=173sec. That is nearly 3 minutes (180sec)!
    • Now we can go back to basics, V=at. We know both V and t now.
      • 81=173a, which solves to 0.47ft/s2, or 0.32mph/s.
    • If you didn't see the shortcut that I saw, there are not-too-difficult ways to solve small equation sets like this. They justrequire more work. Sometimes (often) insight saves work. But it's usually a good idea to go back and check -- make sure your insight was correct. In my case, I plugged my values for a and t back into the original equations. Indeed, they did balance as they should have, so my work was correct.

    The bottom line is that our train loses about a third of a mph every second. Starting at 55mph, it takes almost three minutes of backward acceleration to get down to a standstill.


Optional Math Note

If you know a little bit about calculus, this note may give you a more quantitative insight into position, velocity, and acceleration. If you don't know calculus, just skip over it.

Differential calculus

Let's describe position as a number x. We will restrict ourselves in this note to positions only in the x-axis. If you think about it a bit, that means that velocity and acceleration can only have components in the x-axis. First, let's look at what velocity and acceleration are in terms of differential calculus. Then we'll look a bit at the implications for integral calculus and differential equations.


In terms of
differential calculus
In terms of
geometry
Position
x
The position itself
Velocity
dx

dt
  or   x'
The slope of the
position curve
Acceleration
d2x

dt2
  or   x"
The slope of the
velocity curve

Integral calculus

As you would expect if you remember that integration is anti-differentiation, you can go from acceleration to velocity to position by integrating at each step. If velocity x'=v and acceleration x"=a, then:

v = ∫ a(t) dt

x = ∫ v(t) dt = a(t) dt dt

Differential equations

If you get into making or solving mathematical models for biomechanics, you will spend much time solving the differential equations for the F=ma of your particular model. (Also the torque equivalent, but we haven't gotten there yet.) That is what mathematical models turn out to be; sets of differential equations.

Let's try a very simple example. The air drag on a golf ball is proportional to the square of the velocity of the ball, and opposite in direction to the velocity. So for motion in the x-axis:
F = - K x' 2
since x' is the velocity. We need the minus sign because the force is opposite the velocity in direction. Now let's remember that F=ma. So, substituting x" for acceleration, this becomes:
F = m x" = - K x' 2

This is a differential equation that we can solve to get the motion of the ball (assuming that air drag is the only force acting on the ball -- not often the case, but we're keeping it simple here). We can make it look more like our usual idea of a differential equation with just a little algebraic manipulation:

x" = - K

m
x' 2or
   perhaps  
x" +
K

m
x' 2  =  0

Curved motion

Remember that force is a vector quantity. That means that it might not be pointed in the same direction as velocity, nor directly opposed to the velocity. For example, this picture shows a force at right angles to the velocity vector. There is no component of the force that is either accelerating nor decelerating the magnitude of the velocity. But the force is real, and it will impart a real acceleration to the gray ball. That acceleration will be one to change the direction of the velocity, directing it more downward.
Let's look at a picture of this situation over time. This picture shows the initial force at a right angle to the initial velocity. The force is downward, and it continues to be downward.

Remember, velocity is the accumulation of acceleration over time. If there were no force, then the path would continue horizontally (the dotted line). But the force, and thus the acceleration, is downward, so the velocity will accumulate a downward component. The original rightward component of velocity is still the same, but there is a growing downward component of velocity as well. The result is the parabolic path shown.

This is not a made-up example; it is very real and even common. If you throw or shoot a ball horizontally, it will retain its horizontal velocity (ignoring aerodynamic drag), but will be acted on by a downward force -- gravity. In a vacuum, the path would be a parabola just like this one, and for just the same reasons.
Now let's look at a more interesting and complex example. Suppose the force changes direction as the velocity changes direction, so that the force is always perpendicular to the path. But we will let the force remain the same in magnitude; only the direction changes. What does this do to the path of the ball?
  • Since the acceleration is always at right angles to the path, the magnitude of the velocity never changes, just the direction.
  • That means the picture is always the same, except rotated through some number of degrees. Therefore the curvature of the path is always the same.
What we have just described is the ball moving in a perfect circle at a constant speed.
A real example of this behavior is twirling a ball on the end of a string. We've all done this at one time or another.

The string is exerting a force that accelerates the ball toward our hand (the center of the circle). That force is called the centripetal force. Since we know that every force has an equal and opposite reaction, the ball exerts an equal and opposite force on the end of the string, called centrifugal force. (Note that it is a real force, but most informal mentions of "centrifugal force" are not real at all. They refer to an apparent force that pushes the ball or other body outward. Bear in mind that there is no need for a force to explain why the body wants to move in a straight line rather than a curved path.)

The magnitude of the centripetal force F needed to move mass m at velocity v in a circle of radius R is:

F  =
m v2

R




Last modified --  Oct 2, 2024