Dots on Golf Ball

So it all started when a question came in my mind. Why does a cricket ball swing? I set myself to the quest of understanding the entire mechanics. You just can’t help but wonder that how much of understanding it takes to explain a simple event. It is like pulling a single thread and eventually learning that you a pulling the entire fabric apart, so a metaphor could be the question for the thread and ignorance for the fabric.

Enough of the literature, now let’s take a moment and think about those little dimples on a golf ball. One might ask that why do we need them, surely the balls in other sports don’t bear such features. Well this is question we will be dealing with.

In golf a lot is dependent on the club speed (the speed at which the club hits the ball), the impact (position where the club hits the ball) and the range of the ball. The first two are characteristics of the player however the third one is dependent on the air and ball moving past it. A lot of studies have been done to study the forces encountered by the ball during its flight.


Let’s take a moment and understand the motion of the golf ball in air. It is simple projectile motion about which we have read in 11th standard but the only difference is the presence of air due to which drag force is imparted on the ball. Drag Force can be understood as the backward force experienced by you when you put your hand out of the car window.


So I read research by NASA where they studied the drag force in case of varying velocity of the incoming air (assuming that we are sitting on the ball).


****Explanation of the above figure:
1. The air flow is ideal and laminar and this is the simplest case in which no disturbances are generated.
2. If you increase the speed of air the air separates earlier from the surface giving rise to the vortex which leads to drag.
3. If the speed is further increased an alternating pattern of vortexes is formed
4. Again if you increase the speed, an early separated laminar layer is formed resulting in a lot of drag.
5. Now this is important case where due to turbulence the air is sucked in which reduces the drag.

Look at the picture given below and try to understand the relation between the drag and the vortex created by the flow of air over the smooth sphere. This is the case described in the point 4 above.




Drag force is given as F = (1/2)*p*V*V*A*C

Where, p = Density of air

V = Liner Velocity

A = Cross sectional Area of the face

C = Drag Coefficient
So we can see that drag is directly proportional to the drag coefficient for different values of velocity. Researches after the wind tunnel testing have already developed the Drag Coefficient vs Reynolds No. (It is a dimensionless number that gives the measure of momentum and the viscous force on a particular body, eg. Take a bucket of honey and shoot a bullet through it and then take water and shoot the same bullet through it, in which case do you expect the bullet to stop first? The answer intuitively is honey but in engineering we have a parameter of Reynolds number to explain it) graph to make our life easier.


But this is valid only for a smooth sphere and we all know that in reality there is nothing like a perfectly smooth surface. But then again! whats the point of putting small dots on the surface of the ball as it will increase the roughness of the surface? Well! as the researchers carried out experiments with rough spheres they observed an interesting thing. In a certain Reynolds range the drag coefficient of the rougher sphere are less than that of the smooth ones, it is show in the graph below:


Now let see whether the Reynolds Number developed when a golf ball is hit falls in that range:

So I googled  few things:

Kinematic Viscosity of Air (@ 20 C) = 1.36 * 10 ^ (-4) (ft^2/s)

Diameter of an average american golf ball = 42.7 mm


Average speed of the ball hit by an average golfer = 133 mph = 59.4563 m/s
Reynolds No. calculated by above data comes out to be 2.00935*10^5 (Re1)
Again! I found that minimum average speed imparted to the golf ball is 111 mph (= 49.6214 m/s) so the Reynolds No, for that is 1.6769*10^5 (Re2). When you mark these points on the graph shown above, they entirely falls in the zone where drag coefficient of golf ball is significantly lower than the smooth ones and nearly constant too (that makes calculations a lot easier).


Thus the Drag Coefficient came out to be ~0.25 which is exactly half of that in case of smooth sphere thereby halving the Drag force as well.

Theoretically it could be explained by using the fact that turbulent boundary layer developed sticks more to the surface of the ball than the laminar flow as show in the picture below.


Sorry for the long explanation and maths, here is a potato for you.


Now! I wonder what will happen to the speed of the moon had there been no craters on its surface (I guess nothing ‘coz there is no air in the space to impart drag) but that’s a story for a different time.
Till then, Stay Curious!
-By two Kam-akals (Pawan Kedia and Ujjwal Dubey)


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