Can the wind blow a golf ball into the
cup?
Dave Tutelman  Jan 6, 2009
At request of Tommy
Craggs  ESPN The Magazine
On January 6, 2009, I got a phone call from Tommy Craggs, a
factchecker for ESPN The Magazine. He was
following up on a "tall tale from sports", a story about golf.
A
gambler and golfer tells of a $30,000 putt he had in Las Vegas. It was
a twelvefoot putt, but he topped it so badly he thought it would only
roll three feet. With a strong helping wind, it rolled all the way to
the hole and in. The green was perfectly level, and yes the putt was
that
badly struck.
Tommy asked if the story was plausible, particularly the ball being
blown all the way to the hole. I
responded that it could definitely happen if the wind is strong enough.
The only questions are how strong is "strong enough", and
whether strong enough would ever happen in Las
Vegas while people were playing golf.
The computation of how strong the wind
has to be is interesting. Almost everything it requires is stuff a
college freshman would have to know to pass Physics 101. (The
wind resistance of a sphere is not firstsemester physics. Everything
else is. A couple of years ago, I tutored my son in freshman physics,
so I'm pretty familiar with the current contents of the course.)
Approach:
Here are the steps in our analysis:
 We need to know how fast the green is. Green speed is measured in a
unit called "stimp". Assume the green stimps at 10 or so. (That's
fast but not
PGA Tour fast. A lot of resort courses stimp at about 10  and this
took place at the old Desert Inn course, a classic resort setup.)
Compute from
that how much rolling friction a
golf ball will encounter.
 If the wind force exceeds the
friction force, then the ball, once set in motion downwind, will
continue to roll indefinitely. Find the amount of wind that will exert
enough force on a rolling golf ball to overcome the friction we just
computed.
We are not looking for an exact
answer, just a ballpark number. If the answer is 20mph then the story
is easily plausible, if 100mph then not so. The difference between
20mph and 21mph is not going to change the plausibility. Therefore, we
will keep the
computation simple and ignore minor
effects.
We will work in MKS units [meter/kilogram/seconds] to keep the formulas
easy, and convert to more usual golfing units where we want a feel for the numbers.
1. Find the rolling friction for a 10 stimp green.
A stimpmeter
is a nearlyfrictionless slanted ramp, 30" long (0.762m) and slanted at
an angle of 20°. You roll a golf ball down the track, imparting an
initial velocity to the ball. Then you measure how far the ball rolls.
That distance, in feet, is the stimp number, or speed, of the green.
So, if the green's stimp is 10, the ball will roll ten feet.
The
first step in finding the rolling friction is finding the speed of the
ball as it leaves the track and begins rolling on the green. The
geometry is a right triangle, with the hypotenuse .762m long at an
angle of 20° with the horizontal side. So, if the horizontal side is on
the ground, the upper corner is 0.261m above the ground. The
calculation is:
Height = .762 *
sin(20°) = 0.261
All
golf balls are very nearly the same weight, 46g. (Or, for our purposes,
0.046Kg.) Computing weight, which is the force exerted by gravity, we
get 0.451 Newtons.
Potential energy is weight (force) times height, or
Energy = .451 * .261
= 0.118 Joule
That potential energy is converted into kinetic energy by rolling down
the ramp.
Energy =
½ m v^{2}
= 0.118^{} = ½ * .046
* v^{2}
This equation is easily solved for velocity v, which is 2.26 m/s (or 5
miles per hour).
So
we have a ball going 2.26 m/s, and friction slowing it down. How much
friction will it take to stop the ball exactly 10 feet (3.05m) from the
end of the ramp? (Why ten feet? Because that's what a stimp of 10
means.) We can find the force from the old standby
F = ma
We
know the mass of the ball. We have to find the acceleration (actually,
the deceleration) due to friction with the grass. There is a wellknown
formula for velocity (v), distance (x), and acceleration (a):
v_{final}^{2} = v_{init}^{2}
+ 2ax
We
know that the initial velocity is 2.26m/s, the final velocity is zero,
and the distance is 3.05m.
0 = 2.26_{}^{2}
+ 2a * 3.05
Solving the equation for a, we find that the
acceleration is 0.837 m/s^{2}. Why negative?
Because the force is friction, which acts opposite to the motion. The force
decreases the velocity, so the
acceleration is a deceleration.
Now we can solve for the force.
F =
ma = .046
* .837 = 0.0385 Newton = .0087
pounds
That
is not a lot of force. It's about 1/7 of an ounce, or about 4 grams. I
guess that's about what we would expect from a fast green  not much
friction on the ball.
2. Find the wind to provide enough force to exceed rolling
friction.
The force
of wind on any body is:
F =
½ ρ v^{2}
A C_{d}
where:
ρ =
density of the air. Use 1.3 Kg per cubic meter
for this number.
v = velocity of the wind.
A = area that the body projects to the wind.
C_{d} = the "coefficient of drag" which depends on
the shape of the body.
The golf ball is a sphere of 43mm diameter. (We ignore
the dimples
for this calculation.) That gives a circular area of 0.00145 square
meters seen by the wind. The
coefficient
of drag of a sphere is 0.47. So the equation above for wind
force becomes:
F =
½ * 1.3 * v^{2}
* .00145 * .47
But
we know from step #1 that the force of the wind has to be more than the
force of friction if a ball set in motion will continue to roll across
the green indefinitely. And have already calculated that friction force to be .0385 Newtons. So the equation becomes:
.0385 =
½ * 1.3 * v^{2}
* .00145 * .47 = .000443 v^{2}
Solving for wind speed v, we get 9.3 meters per second, or
about 21 miles per hour.
Bottom line
Twentyone miles per hour is
a bit windy to be
playing golf  more than a twoclub wind  but I'm sure all of us
have played in more wind than that.
However, that isn't the whole
story. Wind speed varies with height; the wind is stronger aloft. Our
calculations showed what the wind has to be within an inch or two of
the ground. But, when we talk about how strong the wind is, we usually
mean anywhere from 5 feet (where we live as humans) to 30 feet (the
height where some weather stations report their readings). Using a mathematical
approximation,
we see that the wind speed at these heights can be up to 1.5 times the
wind that is affecting the putt. So the story of the windblown putt may
require as much as 32mph wind speed, as felt by the golfer or mentioned
in a weather report.
Is the story still possible at 32mph? Well,
golf is
played under such conditions. I played today in nearly that much wind
(New Jersey in January  brrrr!) The Scots and the Irish play in that
much wind all the time. But how likely is it in Las Vegas?
A
trip to the National
Weather Service
web site shows us that the monthly peak gust in Las Vegas ranges from
58 to 90mph every
month of the year. So a 32mph gust when the putt was struck is
not at all unlikely. This point of view was reinforced when I shared an
early draft of this article with my friend Bob Doyle,
also a golfer and retired engineer. In his reply, he mentioned, "I
played a Vegas course when the wind was gusting to 50. My two
sons and I gutted it out."
More improbable is that the gust is
perfectly aligned with the putt. But the wind has to be aligned
somewhere, so this may certainly have happened. Given a 4inch hole and
a 12foot putt, there is 1 chance in 230 that
the wind will be aligned well enough to blow the putt into the hole. In
fact, our chances become a little better when we take into account that
the wind only has to make a 9foot putt; according to the story, the
golfer's mishit got the ball three feet closer without any aid from the
wind. That's 1 chance in 180.
Again, not outrageous. Combining all the coincidences, we have an event
of low probability that:
 There are golfers on the course in weather that produces
 A 32mph or higher gust,
 Which is aligned in a 1 out of 180 direction,
 With $30,000 riding on the result of the putt.
But there is nothing in the physics that says it can't happen. And, if
you watch sports (or anything else) long enough, just about anything
that can happen will happen.
And what about the bottom line for you?
Did you remember enough physics? Would you have been able to answer
Tommy if he had asked you? Did you at least follow the solution as you
read it?
If it makes you feel any better, I got the answer wrong the first try.
I made a numerical error in the simplest part, computing the
crosssectional area of the golf ball. I didn't notice that until a day
later.
Minor
effects we ignored in the computation
 The stimpmeter track's friction is small, but not zero. We
treated it as zero.
 Some
of that kinetic energy from the stimpmeter is rotational (the ball is
rolling). This is much less than the linear motion of the ball, so we
ignore it.
 There is probably a small amount of energy lost at the
transition between the track and the ground. We ignore it.
 Rolling
friction on grass is not perfectly independent of speed. But, with a
green as fast as a 10 stimp, we can consider it the same at any speed.
 Dimples
on the golf ball change its aerodynamic properties. Most of this change
is the lift, next is highspeed drag, and lowspeed drag is a distant
third place. So we will ignore the dimples for this calculation.
 The
ball is initially going downwind, which reduces the relative speed of
the wind and thus the force from the wind. But, as the ball slows down,
the computed wind speed gets ever closer to the actual speed relative
to the ball. So the computed wind speed will ideally just let the ball
continue to creep, not stop altogether.
Last updated  Aug 4, 2009
