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The Secret of Jack Stulak's famous Golf Swing

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Stulak's Secrets of the Golf Swing

I was lucky enough to have Jack Stulak sit down with me and break down the secrets of his famous golf swing. He was able to explain, in the simplest of terms how to have a consistent swing every time. It is such a rarity for someone as famous as Jack, a legend in his own mind, to tell us common golfers how to have a great swing every time. Please make sure you thank Jack the next time you see him.

Physics Of A Golf Swing — Analysis

To begin the analysis for the physics of a golf swing, let's look at the strobe picture of Bobby Jones swing once more. To better illustrate the motion, I drew a red curve and blue curve. The red curve traces the motion of his hands at the grip location, throughout the swing. The blue curve traces the motion of the club head throughout the swing. The red curve is almost a perfect circle, as you can see. The blue curve is not a circle, but is spiral shaped. The largest radius of the (blue) curve occurs at the bottom of the swing, where the club is fully extended.

Now that we have a clearer picture of the motion involved, we can draw a schematic of the set up. Note that the physics of a golf swing will be treated here as a two-dimensional problem.

Where:

r is the radius of the golfer's swing

P is the grip location where the golfer is holding on to the club

G is the center of mass of the golf club

lG is the distance from P to the center of mass of the golf club

θ is the swing angle the golfer's arms make with the vertical

α is the angle the golf club makes with the golfer's arms

Next, we need to draw a free-body diagram of the golf club. This is the important object of study since it is the golf club that is reacting to "input" from the golfer, so it must be isolated from the system.

Where:

g is gravity, acting downwards. This value is equal to 9.8 m/s2

lclub is the length of the club

MP is the moment exerted on the golf club at P by the players hands

FPx is the horizontal force exerted on the golf club at P by the players hands

FPy is the vertical force exerted on the golf club at P by the players hands

G(x,y) is the position, in Cartesian coordinates, of the center of mass G. Note that this position is with respect to ground, and is a function of time.

We can express the position of the center of mass as follows:

Next, using Newton's Second Law we write the general force equation in the x-direction:

Where:

ΣFx is the sum of the forces in the x-direction

m is the mass of the golf club

aGx is the acceleration of the center of mass in the x-direction, with respect to ground

Note that

Thus we have

For simplicity, note the following notation:

For simplicity, it is also assumed that the golfer's arms rotate at a constant angular velocity. In other words, the angular acceleration is zero:

Similarly, by Newton's Second Law the general force equation in the y-direction is:

where ΣFy is the sum of the forces in the y-direction, and aGy is the acceleration of the center of mass in the y-direction, with respect to ground.

Note that

Thus we have

We must now write the general moment equation for rotation of a rigid body about its center of mass G.

Where:

ΣMG is the sum of the moments about the center of mass

IG is the moment of inertia of the golf club about its center of mass, about an axis pointing out of the page

αclub is the angular acceleration of the club

Note that

This is the angular acceleration of the golf club with respect to ground.

Thus, the moment equation becomes

To solve these equations we can substitute (1) and (2) into (3). The resulting equation is as follows:

This is a second order differential equation in terms of α. It can only be solved numerically. It will solved for the part of the swing where the wrists are uncocked, and subject to the following initial conditions at the release point, given as:

Note that α = 90° is a common (and approximate) angle that golfer's hold the club relative to their arms, prior to uncocking their wrists.

An additional condition is that MP = 0 since the wrists are allowed to rotate freely at the release point, and at every point in the swing thereafter.

The other unknown quantities in the equation such as r, m, lG, IG are constants based on the properties of the golf club and the parameters of the golfer's swing.

Using trial and error, one must input different wrist uncocking (release) angles, in the above equation (for θo), until α = 180° at θ = 180° (in the solution).

Physics Of A Golf Swing — Solution

The physics of a golf swing can be illustrated in the graph below. This graph shows a representative solution which gives good insight into how the golf club angle α increases as the swing angle θ increases.

For the above solution shown in the graph, I chose the following input values which I feel are fairly representative of the typical golf club and golf swing:

lG = 0.58 m

IG = 0.015 kg-m2

m = 0.300 kg

r = 0.5 m

Initial swing angle θoi (at start of swing) = 10°

Swing angle θo (just before release) = 57°

Initial club angle α (just before release) = 90°

So according to this particular model, the optimal release angle to uncock your wrists is θo = 55-60° .

Physics Of A Golf Swing — Closing Remarks

It is interesting to see that the most rapid increase of club angle α (shown in the graph above) begins at a swing angle of around θ = 120° (even though uncocking of the wrists occurs much earlier in the swing). This is because the club angle α increases slowly at first before reaching a rapid release rate. However, looking at the strobe picture of Bobby Jones' swing one might (mistakenly) conclude that the correct swing angle to uncock your wrists is at approximately 120°, simply because that is when the increase in club angle α becomes very noticeable. Thus, analyzing the physics of a golf swing using a dynamics approach, will yield more accurate information than if someone were to just observe a golfer's swing, and draw conclusions from that. The lesson here is that certain information can only come to the surface when the physics of the problem is examined carefully.

A final consideration of the physics of a golf swing is to see how fast the club head moves at the point of impact, at the bottom position (using the previous input values). Since the club head is moving horizontally (in the x-direction) at the bottom position, there is no vertical component of velocity (in the y-direction). Therefore, the velocity of the club head is given as:

Note, the above equation is derived by calculating dx/dt, and replacing lG with lclub.

Using a club length lclub = 0.7 m, r = 0.5 m, θ = α = 180°, dθ/dt = 15 rad/s, and dα/dt = 19 rad/s (as calculated from the solution), we get Vx = 31 m/s for the club head speed just before it hits the ball.

It is interesting that, in analyzing the physics of a golf swing, we see that the arms and wrists play a somewhat passive role, and yet a powerful hit results. So where does the swing energy come from? It comes from the muscles in the torso and shoulders which swiftly rotate the golfer's arms and club through the swing. Uncocking the wrists at the optimal swing angle means that, as much energy as possible is transferred to the club head (in the form of kinetic energy), just before it contacts the ball.