Spin Decay for a Ball in Flight

Dave Tutelman  --  January 26, 2016

How fast does a spinning golf ball lose its spin in flight? That is what we examine here.

It started on Jan 25, 2016 in the Golf Teaching Professionals group on Facebook. Adam Young asked, "Anyone know what spin degradation rates are, typically? 500rpm/s?"

That's a good question. It was even phrased well. Most people would not have known offhand that a degradation rate would be in units of rpm per second. Since Adam's query also singled me out to look at the question, here's my take on it.

First of all, rpm per second is a very good try, but it isn't quite accurate. Rather than rpm/sec, we should be saying %/sec. That is, if you start with twice the spin, you'll have twice the rpm spin loss in the first second. This gives you an exponential decay, which makes intuitive sense.

Let's go with a percentage per second. But what percentage? How fast does the ball lose spin? When Frank Schmidberger and I built the TrajectoWare Drive software, we had to come up with a figure for this that matches reality. I went back to the source code for our ball flight algorithm, and deduced that we used a figure of 3.3% per second. Below, I'll talk about where that figure comes from.

As the discussion continued, Sue Ginter cited the TrackMan newsletter from October 2010, saying that the decay was about 4% per second. The excerpt from TrackMan is:
Spin rate: how many times the ball rotates per minute when leaving the clubface. This is independent of the orientation of the spin axis. Note that the spin rate drops during ball flight - typically 4% for each second.
That isn't exactly the same as 3.3%, but definitely close enough that we are confirming each other's numbers.


A ball in flight loses something like 3.3 to 4 percent of its spin for each second it is in the air.

In the six seconds of hang time typical on the PGA Tour, the initial spin is reduced to between 78 and 82 percent of its initial spin.

The rest of this article discusses the physical and mathematical basis of this conclusion.

Simple math

The simple physical reasoning behind the math runs something like this. The faster the ball spins, the more frictional force it generates. And more frictional force means more deceleration. (F=ma) So let's make the friction proportional to the existing spin rate. That means that the deceleration is proportional to the spin rate. There are a lot of processes that physicists and engineeers deal with that have the same property: temperature of two adjacent bodies starting with different temperatures, automotive shock absorbers, capacitors charging and discharging, a radioactive sample giving off radiation... lots of examples.

Consequently, the math for the example is well-known to physicists and engineers. It's an exponential function with a negative coefficient, of the form:
y  =  A e-t/T
A is a proportionality constant, and T is a coefficient called the "time constant" of the process. The significance of time constant T is that it is the time it takes for y to decay to 1/e of its original value. The graph at right shows what the exponential decay looks like. It shows e-t/T for three values of time constant T: a half second, one second, and two seconds. Each curve falls through the value .37 (which is 1/e) at its respective time constant: 0.5, 1.0, and 2.0.

When we plug the specifics into our generic exponential, we get a rather simple formula for spin at time t:

Spin(t) = Spin(0) e-t/30

When you look at Spin(1) (the spin after 1 second), you find that it is 96.7% of Spin(0), for a loss of 3.3%. And the way an exponential works, if you lose y% in time x, then you will lose the same y% in any time x along the time axis. That is, Spin(2)/Spin(1) will also be 96.7%. So will Spin(3)/Spin(2), etc.

When Frank and I developed TrajectoWare Drive, we did some careful checking against drive data, with spin of less than 4000rpm. Our production program loses accuracy pretty quickly above 4000rpm. We now have a modified algorithm in a prototype that deals with iron shots much better, and uses the same spin damping formula. But haven't done much validation against real data, and won't introduce it into a new version of TrajectoWare Drive until it has been properly tested. But I think this discussion has eliminated one possibility for an explanation of our high-spin problems: the spin decay assumptions. The fact that TrackMan reports a very similar decay (without qualifying the effective range) tells me we can't be too far off.

Are all balls the same?

Early in the discussion, Mike Duffey asked the very reasonable question:
I would guess that it might vary a fair amount between types of golf ball, mass distribution inside, surface material/design. Any thoughts on that part?
I agree with that assessment. From our F=ma model, the spin decay is angular deceleration -- negative acceleration. It is easy to see that a=F/m. And both F and m (well, "mass" in rotational problems is moment of inertia) are design variables in the golf ball.
  • The dimple pattern is going to affect the friction force between air and golf ball. Ball designers use different surface designs for exactly this purpose. And the friction force is what causes deceleration of the spin.
  • The distribution of mass will affect the moment of inertia of the golf ball. Remember, moment of inertia of any object is its resistance to angular acceleration, its desire to keep its spin exactly as it is right now. The higher the moment of inertia, the more slowly the ball will lose its spin.

    The cutaway drawings show the cross section of a 3-layer golf ball. The cover contributes very little mass, but there is plenty of mass in the outer mantle and inner core. Remember that moment of inertia is increased by moving the mass away from the center of rotation. So if the golf balls have the same mass (and all competitive golf balls are within a few grams of 46g), then the choice of material weights for the mantle and core would have a direct effect on decay of spin.
How much ball-to-ball difference would there be? I doubt the variation would be as much as 2:1 for real golf balls. That is still quite a bit of variation, but my personal guess is considerably less than that. The 30-second time constant that I am presenting is based on a single ball; I'm pretty sure it is the Titleist Pro-V1 of about 2006.

Proportional or square law?

There is a potential problem with the assumption of deceleration proportional to spin. The frictional force is closely related to classical aerodynamic drag. It's not exactly the same -- it is tangential motion with no windward nor leeward component. But we should look at any difference in the mathematical model between this and drag. And there is a difference. The drag force is not just proportional to the velocity; it is proportional to the square of the velocity. So let's see if this matters, and how much.

I did a quick bit of math (below) to see how much it matters if decay is proportional to the square of spin, instead of just proportional to spin. I found the equation for spin vs time with this new assumption. It is

Spin(t)  =   Spin(0)
t/T   +  1

where T is the time constant that would give an exponential curve with the same initial decay rate (that is, at t=0).

I plotted the proportional (exponential) and square law curves out to almost 10 seconds of hang time. (On the PGA Tour, the typical hang time is about 6 seconds, and seldom exceeds 8 seconds.) The two curves differ in the way you would expect, but perhaps by less of an amount than you would expect. If they are chosen to start decaying at the same rate, then the square-law retains its spin a little better than the proportional curve. That is because a friction that varies with the square of the spin falls off faster than friction that is proportional to the spin. So, as the ball slows down, the friction decreases more, the deceleration decreases more, and there is less slowing of the spin.

This graph leads me to the conclusion that, for practical purposes, it doesn't matter whether the decay is based on a proportional or square law frictional force; the curves are too close to make any noticeable difference in ball flight, nor bite upon landing. At 6 seconds (the PGA Tour typical), the difference in spin is less than 2%. If we assume this is on a spin of 3000rpm, that is a total difference of 50rpm. Yes, TrackMan could measure that (they claim an accuracy of 15rpm), but it is a remarkably small difference.

Solution of the differential equations

No, I'm not assuming the average reader cares to see the differential equations or their solutions. But some might -- and anyway, I'd like to archive it here so I can refer to it again if I have to. Who knew I'd still be solving differential equations in 2016? The last college course I took in DiffEq was in 1959.

(a) Decay assumed proportional to spin ("proportional")

The decay of the spin is due to the frictional force of the air on the spinning ball. In linear terms, if we "unrolled" the ball to a flat surface, we are talking about F=ma, where the force is friction and the acceleration is really deceleration of the surface of the ball -- corresponding, of course, to decay of the spin rate.

Let's assume that the frictional force is proportional to the speed of the air over the ball: that is, proportional to the spin rate. In that case, our basic F=ma equation becomes:

-K1 v  =  m a  =  m dv/dt

The minus sign recognizes that the force is negative, decelerating force. The last step recognizes that acceleration is the first derivative of velocity. That is, acceleration is the rate of change of velocity with time.

With little algebra, and combining K1 and m into a single constant K:

dv/dt  =  -K v                         (eq. A-1)

The first step in solving a simple Diff Eq like this is to get the vs on one side and the ts on the other. With basic algebra:


 =  -K dt

Now we can integrate (well, really "antidifferentiate") each side separately. And we'll add the arbitrary constant C that a Diff Eq solution generally requires.

ln ( v )  =   -K t  +  C

We would like to find v (the spin). Right now it's inside a natural logarithm, but that's not hard to fix. Let's consider each side to be the exponent of e. So, if the whole side is "x", we're going to make it ex

e ln(v)   =  v  =  e ( -K t  +  C )

v  =  eC  e-Kt

If we use the usual approach to evaluate C, and use a time constant T instead of K, we find that our formula for spin v is

v(t)  =  v(0)  e-t/T

So the decay is exponential. It is well known to second-year college students in physics and engineering that any effect that grows or decays proportionally to its current value follows an exponential curve. Our result is no surprise.

(b) Decay assumed proportional to square of spin ("square law")

Let's repeat the above derivation, but we will assume that the frictional force is proportional to the square of the velocity. So equation A-1 becomes:

dv/dt  =  -K v2                         (eq. B-1)

Again, the first step is to get the vs on one side and the ts on the other.


 =  -K dt

Now we antidifferentiate each side separately. A v2 in the denominator gives a very different answer from just a v.


 =   -K t  +  C

This time, we will invert each side, instead of raising it to a power of e. At the same time, we'll multiply both sides by -1.

v  =   1

Kt   +  C

Again we use the usual approach to evaluate K and C, and find that our formula for spin v is

v(t)  =   v(0)
Kt   +  1

K is chosen to give the correct initial decay rate. Since this is not a true exponential, it would be incorrect to talk about K being related to the "time constant" of decay, because time constant is specifically a property of an exponential decay. But if we relate it to an exponential that has the same initial rate of decay, then K=1/T with T being the time constant of the related exponential.

v(t)  =   v(0)
t/T   +  1

Last modified - Jan 29, 2016