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

## 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:

-K_{1 }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

K_{1}
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:

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.

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

e^{x}
e ^{ln(v)} |
= v = e ^{(
-K t + C )} |

v =
e^{C} 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 v^{2}
(eq. B-1)

Again, the first step is to get the

vs
on one side and the

ts
on the other.

Now we antidifferentiate each side
separately. A

v^{2}
in the denominator gives a very different answer from just a

v.

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.

Again we use the usual approach to evaluate

K and

C,
and find that our formula for spin

v
is

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.