Required grip pressure
Dave Tutelman
 July 2, 2020
How tightly does a golfer have to
grip the club in order to make a
swing? I was asked this question by a reader, and the answer was more
interesting than either of us expected.
In early July of 2020, I got the following email from Chris Daigle.
I am a mechanical engineer and a golf hobbyist and I have been
searching for information but largely finding conflicting data.
I am interested to try and
quantify how much grip strength is needed to
swing a golf club. In my own experience, I feel a large pull
of the club when it reaches the bottom of the swing, so much so that I
can feel my hands react and grab the grip tightly as it approaches
impact with the ball. This seems to agree with a lot of
papers I have read but I cannot find consistent or useful data on swing
speed vs grip force applied. I know it can vary depending on
how one swings the club but I hear time and time again, faster swingers
generally have higher maximum grip strength capabilities than slower
swingers.
I would love to boil things down to a general guideline for myself and
other golf hobbyists so we can more clearly see the importance or
unimportance of grip strength as it limits or allows a certain swing
speed, primarily focusing on the driver of course.
Do you have any knowledge of this you can share? Like I have
said, I have read many papers, but most don't share swing speed of the
people they test and there is conflicting data I feel due to type of
swing, and measurement devices being used. I want to know how
much crushing force is being applied to that golf grip at its peak for
x speed swing.
In a subsequent note, he wrote:
I have attempted to very loosely estimate this just by calculating
centripetal force of an object rotating about a fixed point in space
but I know this is more complicated than that.
Chris, you're closer than you give yourself credit for. The calculation
of centripetal force is central to the whole thing. But there are other
factors as well.
Let's see what they are.
Overview
The
problem statement is, "How much grip pressure do I need for my swing?"
Golf instruction tells us that that we don't want more tension in the
hands and forearms than is absolutely necessary. But how much is
necessary? There's a traditional analogy, usually attributed to Sam
Snead, to
hold the club "like you were holding a baby bird in your hands". Poetic
for
sure, but is it realistic? Let's find out how much grip pressure you
need just to do the job of swinging your club at V
miles per hour. That has to be the bare minimum grip pressure that
works. And we don't want to use too much more than the bare minimum,
because tension in the hands and forearms interferes with release.
The first question we must answer is: at what point in the swing
does the maximum force requirement occur? The intuitive and obvious
answer is: just before impact, where the maximum angular velocity of
the club requires hands to exert the maximum centripetal force to hang
onto the club. We will proceed in this vein, and I believe it gives the
most useful
answer. But at
the end of this article, we will discover a point in the swing that may
call for an even larger hand force  though not one that involves
keeping
the club in the hands.
Given that assumption, there are three things we have to concern
ourselves
with:
 The
centripetal pull we have to exert just to keep the club on
its rotational path, and not let it fly out of our hands. This depends
on clubhead speed and some of the club's dimensions. This peaks out at
or near impact, which is why the pictures is shown just before impact.
 Any additional
force we need to continue
to
accelerate the clubhead
near impact.
 The hand forces
on the handle of the club to resist the forces
from #1 and #2.
In the body of the article, we will deal with them in this order. Then
we sanitycheck our results against some realworld data:
 What sort of grip force can the hands actually exert?
 What grip forces have actually been measured during
the downswing.
The results are interesting, to say the least.

Executive
summary
We will first find F,
the centripetal force the hands need to exert on the club to keep it
from flying down the fairway. Then we will discover that we can neglect
the possibility of an additional force to accelerate the clubhead in
the vicinity of impact. Finally, we look at the mechanisms by which
the hands can exert an axial force on the club shaft; they consist of
friction and taper. Applying these mechanisms to the formula for F,
we get the following expression for G,
the total force in pounds the hands exert on the grip.
G = 
0.00129 V^{2}
μ_{s}
R 

(M + m/2) [1  sin(a)] 
Where:
 V
is the clubhead speed, in miles per hour.
 μ_{s}
is the static coefficient of friction between the hands and the handle.
 R
is the distance from the midhands point to the center of mass of the
clubhead, in inches.
 M
is the mass of the clubhead, in grams.
 m
is the mass of the shaft, in grams.
 a
is the angle of taper of the grip where the hands are applying the
force.
Other topics covered include relating dynamometer readings to the
ability to apply grip pressure, and comparing our results to some
measured grip forces during the downswing.

Centripetal force
Separate out the
angular component
The
clubhead travels on a curved path. It requires a force
perpendicular to the path to keep it from flying off on a
straight line  thus flying completely out of the golfer's grip. This
perpendicular force is called a centripetal force.
Centripetal force depends on the curvature of the
path. Its usual formulation is:
Here the curvature is represented by the radius of curvature R.
But we have to remember that the clubhead's velocity is due to two
effects: the angular velocity of the club (hence the emphasis on
centripetal force) and the linear velocity of the hands' motion. On the
diagram (built on top of a photo from the Howard's Golf web site), we show
these two velocities, which add together to create the clubhead speed.
But only VR,
the velocity due to rotation, contributes to the centripetal force the
hands must exert. How can we separate out VL,
so only VR
contributes to the centripetal force?
For good quality swings, somewhere between 12% and 20% of the clubhead
speed depends on the linear hand speed VL.
Actually, that variation is mostly from study to study, not from golfer
to golfer; the results of any one study seem to come out within a range
of
a couple of percentage points across the golfers. So let's arbitrarily
use 15%, an 8515
split between speed due to rotation of the club and speed of the hands.
That means 85% of clubhead speed can be attributed to rotation of the
club, and contributes to the centripetal force. We will calculate the
centripetal force by using VR
and the obvious radius in the diagram. That system can be superimposed
on the linearly translating hands, but only the rotational component
gives the force exerted on the
hands system  which is what we are looking for.
A word about the radius R.
We are using the radius in the translating frame, so we are only
dealing with pure angular motion. That radius is from the center of the
hands (the "midhands
point") to the clubhead CG, so it is a few (3 to 4) inches
less than the length of the club itself.
Now we can find the force due to the clubhead speed V,
the clubhead mass M,
and the radius R.
F = 
M (.85 V)^{2}
R 
= 
0.723 M V^{2}
R 

Centripetal
force due to the shaft
That covers the force due to the clubhead. But the shaft has mass and
is rotating. So there must be a centripetal force required to keep the
shaft from flying off on its own  just like the clubhead. The
clubhead was easy, because we treated it as a point mass at a distance
from the midhands. But the shaft's mass is distributed more or less
uniformly from the midhands to the clubhead, and must therefore be
treated as a distributed mass.
You
won't miss anything but the math if you skip this note.
That turns this into a calculus problem  a simple
calculus problem, but one we need to formulate in calculus terms.
As per the diagram, we are going to integrate (add together) the
centripetal force due to all the infinitesimal lengths of shaft mass dm
between the midhands point and the tip.
We will make the assumption that the mass of the shaft (which we call m)
is uniformly distributed along its length. This is seldom the case, but
is close enough so the answer will not have a large error. Therefore,
we write:
Which gives us:
F = 
m
R 

R
∫
r=0 
v^{2}
r 
dr 
The velocity v
at distance r
from the midhands is proportional to r.
Which gives us:
F = 
m VR^{2}
R^{3} 

R
∫
r=0 
r dr 
Evaluating the integral and applying the range gives
F = 
m VR^{2}
R^{3} 

[ 
½ r^{2} 
R
]
r=0 
This is a reasonable and intuitively satisfying result. It says that
the force is half what it would be if the shaft's mass were a point
mass at R.
In fact, it is possible to generalize that a bit. If the mass of the
shaft is not uniformly distributed, the fraction is not one half, but
rather the distance from the hands to the balance point divided by R.
(Just so I don't forget how I derived it, I recorded the derivation in
the Appendix.)

Centripetal
force: the bottom line
Thus the centripetal force to keep the entire club from flying out of
the hands is:
F = 
0.723 V^{2}
R 

(M + m/2) 
That is great as far as it goes. But there remains the
small issue of measurement units. The equation demands a consistent set
of units, such as meters/kilograms/seconds (MKS) or feet/slugs/seconds.
But those are not the units that come to mind if you're an American
clubmaker or even American golf instructor. We are more comfortable and
intuitive in terms of:
 F
in pounds. (Good so far.)
 V
in miles per hour. (Not feet per second.)
 R
in inches. (Not feet.)
 M
and m
in grams. (Not slugs nor even pounds.)
That is just a complete scramble of units. Let's see if we can
find the conversion factors to make the units come out right. If we do
it well, it should just show up as a coefficient different from the
current 0.723.
Let's start by explicitly including a conversion multiplier to each of
our quantities. Since we want the force in pounds, let's use the
foot/slug/second system, since its unit of force is the pound.
Multiply
by ↓ 
to
convert ↓ 
into ↓ 
1.47 
miles per hour 
feet per second 
1/12 = 0.0833 
inches 
feet 
1/14594 
grams 
slugs 
F = 
0.723 * 1.47^{2} V^{2}
.0833
R 

(M + m/2)
14594 
Multiplying it out, we get our formula for centripetal force.
F = 
0.00129 V^{2}
R 

(M + m/2) 
This looks like the sort of formula we were hoping for. Let's compute
it for a few different clubs:
Club 
Driver
graphite shaft 
Driver
steel shaft 
5iron
steel shaft 
Club
length 
45.5 
44.5 
38 
R 
42 
41 
34.5 
Head
mass M 
200 
200 
256 
Shaft
mass m 
50 
115 
120 
Head
speed V ↓ 
Centripetal
force (pounds) ↓ 
10 
1 
1 
1 
20 
3 
3 
5 
30 
6 
7 
11 
40 
11 
13 
19 
50 
17 
20 
30 
60 
25 
29 
43 
70 
34 
40 
58 
80 
44 
52 
76 
90 
56 
66 
96 
100 
69 
81 
118 
110 
84 
98 
143 
120 
100 
117 
170 
130 
117 
137 
200 
140 
135 
159 
232 
150 
155 
182 
266 
This is certainly the same order of magnitude that the conventional
wisdom would lead us to expect. There is a graph of this table in the Appendix.

Force for
clubhead acceleration
The
next question we have to deal with is, how much additional force
does the golfer exert to keep the clubhead accelerating through
impact?
For
golfers who have acceleration going beyond impact, how are they
accelerating the clubhead? In particular how are they accelerating the
angular velocity? I single that out because linear acceleration along
the hand path requires forces along the hand path  at a right angle
to the major centripetal force we already have. Unless they are
major
forces indeed, they do not add noticeable magnitude to the grip force
required. The graph at the right shows how little effect a small force
has on the magnitude of the total force when added at a right angle to
a larger force. Both the added force and the resulting total are
normalized to a unit force. So even if we have a right angle force of half the centripetal force,
it only increases the total force by 12%. So unless we are looking for
rather precise results, we
can ignore forces along the hand path when looking for the total "pull"
on the hands.
Let's move on to forces that can cause angular acceleration. Many
good golfers do continue their angular acceleration right up
to and including impact. How do we know? Because biomechanics studies
show a positive net torque on the club, applied by the hands, even at
impact. Let's look at what two such studies tell us. 
These are graphs of the inplane ("alpha") torque during the swing up
until impact. They look different, but are really quite similar.
 The one on the left was posted by Dr YoungHoo Kwon
as part of a tutorial in the Golf Biomechanists group on Facebook in
2015; it may still be available there. It shows three curves:
 The hand couple.
 The moment of the hand force.
 The total torque, the sum of 'a' and 'b'.
 The one on the right is from a video screenshot of a video
by Dr Sasho MacKenzie. It shows the same three
curves, but with different names. The gray shaded area on the Kwon
graph is a translation of Kwon's notation to Sasho's.
The curves look different, but that difference can be
explained by the fact that Sasho's graph includes the backswing. If you
just look at the downswing, the curves are really very similar.
The
most significant similarity (important to us, for this discussion) is
the moment of the hand force at impact. It is about 50 Newtonmeters on
both graphs. Let us see what the moment of the hand force means.
This image shows Dustin Johnson at the moment of impact. His lead
(left)
arm is pulling up along its length. We don't know this for a
fact, but it is very likely because:
 The centripetal force at the speed he generates is
substantial  over 100 pounds.
 The mechanism to generate a force along the length of
the arm is much more powerful than is needed to generate a forward or
backward force along the hand path.
 As biomechanics studies show us, there is no need to
use forces other than along the arm, at least not in the vicinity of
impact.
So let's assume the line of force he applies to the grip is along the
left arm itself.
But that is not the line of the shaft! The club is lagging considerably
behind the line of force. In fact, that is where the angular
acceleration comes from. The "moment of force" that creates all that
clubhead speed is due directly to the torque represented by that [red]
force time the moment arm, the distance between the line of force and
the center of mass of the club. (Engineers call the center of mass the
"center of gravity", and it is almost the same place as what clubmakers
call the "balance point" of the club.)
The basics of physics say that a force on a body not through the center
of mass (CoM) generates a torque on the body. The torque is equal to
the magnitude of the force times the distance the line of force is from
the CoM. We can see that this torque will create angular acceleration
to increase the angular velocity of the club  which is exactly what
we should want to do. It is this moment of force, and not the hand
couple, that accelerates the club during the last 50100msec of the
swing.
I chose a picture of Dustin Johnson (as opposed to some other golfer)
because it gave me the most easily understood diagram; few other
golfers, if any, have the line of force leading the CoM by as much as
he does. This can look very different from golfer to golfer. I tried an
interesting calculation: What force would it
take to generate 50 Newtonmeters of torque given any particular
golfer's moment arm at impact. I estimated the moment arm for a
halfdozen pictures, and got anywhere from 55 to 220 pounds of force.
This rather neatly brackets the sort of numbers we got for centripetal
force. For professional golfers, we can expect the driver clubhead
speed to be 110125mph, giving a centripetal force of 85110 pounds.
What should that tell us? My conclusion is not tightly reasoned, but
seems (intuitively, to me, at least) to be a good working hypothesis.
Topnotch golfers apply all the centripetal force they can, and get their clubhead
speed from keeping the line of that force as far ahead of the club as
they can. The way a Dustin Johnson generates big clubhead speed is to
keep his pulling force further ahead of the club than other golfers can.
So I will
provisionally state that the force exerted by these guys is nearly the
centripetal force, but deflected backwards to move the extended line of
the force forward of the club's CoM. When you deflect a force, you turn
it into two component forces: a slightly smaller force in the original
direction, plus a component perpendicular to the original direction. We
can do a similar graph to the one above, to see how much of a
difference this makes in the force.
The graph at the right shows how much we have to multiply our original
force by to get back to the same centripetal component we had before we
deflected the force.
Even at 20° of deflection (what Dustin Johnson exhibits in the picture)
we only have to make up 6% of the force. That is small enough to be
almost negligible.
It is worth noting that the camera that took the picture is not aimed
perfectly
perpendicular to the swing plane. Therefore, any numbers will be
approximations at best. I used a plastic protractor on the computer
screen to determine that the angle was 20°. The graph says there is no
real requirement for precision  which is good, because there was no
real precision either.

The
bottom line for all this is that the centripetal force, all by itself,
is a pretty good approximation to the total force that needs to be
resisted by the golfer's grip pressure. It may be off by 510 percent
(probably less),
but it will give you a very good idea how much force is there to be
resisted. 
Grip pressure
At this point, we know
the force trying to pull the club out of our hands. The final step is
to find out how much grip pressure is needed to resist this force, how
much is needed to keep the club in our hands.
Friction
Now
we know almost everything we need in order to find the grip force. We
know that the force has to be enough to keep the club
from pulling out of the hands; the hands have to be able to apply the
full centripetal force, and maybe a few percent more. What is left is
friction.
Here is a view of the club's handle being held by a golfer wearing a
glove. The red arrows are the only force the golfer is able to apply
directly  squeeze the handle of the club. This action must be
enough to keep the club from being pulled out of the hands by
centrifugal force, which is the reaction force to the centripetal force
the golfer uses to keep the club rotating in a curved path.
But the red arrows, the force the golfer can apply, are all normal
(that is, perpendicular) to the direction of the centripetal force the
golfer wants
to apply. There is no component that can help in this case. How can a
normal force create a tangential force? The answer is friction. 
Here's
a diagram that, with only minor variation, is in every
physics and engineering text as "Figure 1" for the
chapter on friction. And it will be Figure 1 for us as well. What it
shows is:
 The green body rests on the gray surface, exerting
its weight as a force on the surface. The surface presses back (equal
and opposite reaction) with normal force N.
 Someone wants to pull the green body horizontally
along the surface, so they apply force P.
If there were no friction between the body and the surface, then the
body would accelerate according to P=ma.
 But there is friction. It shows up as a counterforce
F,
preventing the body from moving until P
exceeds the maximum F
that friction can impose.
That maximum F
depends on the "grippiness" of the two surfaces, which is called the
coefficient of static friction, μ_{s}.
It gives a very simple relationship:
F =
μ_{s} N
This seems to be the problem we are trying to solve. We have the
hands exerting forces all around the handle of the club (as in the
upper picture). We want those forces to prevent the club from flying
out of the hands under the pull of a centripetal force that could
exceed 100 pounds. So the problem would appear to be to minimize N
(the grip force) for a given F
(the centripetal force). In other words, we want a high friction μ_{s}
between
the handle and the hand or glove.
With this formulation, the goal we started with is N,
the sum of all the normal forces that the hands apply to the handle of
the golf club. Since we already know F,
we can find out how much N
is required from:
Grip
pressure (really total force on the grip) = N
= 
F
μ_{s} 

Coefficient
of friction for the grip
So what
is
the coefficient of friction for a golf grip? I found it hard to find a
really good answer. Here are two of the best resources I found.
 A really good introductory article on coefficient of friction. This
includes a table with μ_{s}
for a large variety of materials. This table introduces the notion 
which should have been obvious with a little thought  that a single
material does not have a coefficient of friction; it is a property of a
pair of materials meeting at a common surface. One of the lessons from
this article is that μ_{s}
for
"slippery" substances runs about 0.1 more or less, and for "sticky"
substances 1.0 or occasionally even more.
 A forum discussion in GOLFWRX. The
question is the one we have. I did not see a definitive, authoritative
answer, but most seem to be using 0.7 or so. (μ_{s}
for
rubber on rubber
is over 1.0, so 0.7 is certainly feasible.)
There are
obviously other considerations that will modify the 0.7 up or
down:
 Whether or not a glove is worn. I didn't see anything
in a technical discussion, but it seems from my personal experience
that μ_{s}
is
higher if a glove is worn than playing with bare hands.
 Wet or dry. Adding moisture in the form of sweat or
rain definitely reduces the coefficient of friction.
 Condition.
A brandnew grip tends to be soft and tacky, an old grip hard and
smooth. (There are some grip materials on the market today that I have
been unable to wear out or age out noticeably; they are made by Star
Grip and Pure Grips.)
To illustrate the importance of all this, let me cite a very fresh
anecdote  from today.
I just got home from the golf course, where I played a round in my
least favorite weather  hot and humid. I sweat a lot, enough so a
round like this can be dangerous to my health. But it is also a threat
to the integrity of my golf game, because a wet interface between hands
and handle has a much lower μ_{s}
than a dry one. And I find it difficult to wear a glove in such
weather, because the inside of the glove rapidly becomes an
uncomfortable swamp.
So I was playing without a glove, and
playing increasingly badly. My most common swing thought was, "Hang
onto the club because it is going to slip in your hands." After
completely duffing three shots on the fifteenth hole, I reached into
the bag and pulled out a lonely, seldomused glove, and pulled it onto
my left hand. Then I hit the best 3wood shot I had hit all day. I kept
taking the glove off and putting it back on, so I could have a
reasonable grip but not let the inside of the glove become gooey. And I
left it off for short game and putting, where the required forces are
low. That solved the problem. After my triplebogey on #15, I finished
bogey, par, par.
Hey, I'm in the middle of writing about this. I
should have known better. But it took me until a total meltdown on the
fifteenth to remember the importance of the glove in boosting the
coefficient of static friction.

Surface texture
It
is fun reading patents
and crazy ads in golf magazines, and one kind of offthewall
innovation in these places is new designs and materials for golf grips
and golf gloves. The point of many of these is to raise the coefficient
of static friction so high that you just don't have to worry about grip
pressure. For the most extreme of these inventions, the baby bird would
be quite safe, but the rules of golf would be seriously jeopardized.
Here are a few of the more obvious ones, but there are plenty of them.
 A
deep "tread" in the surface of the handle. This draws upon two effects.
(1) A rain tire gets its gripping power from the ability to channel
water into the grooves, allowing more intimate contact between the
raised part of the tire and the road. If the problem is not just
surface dampness but lots of water (like rain), this might also work
for a golf grip. (2) If you look at what causes friction on a microscopic level,
it is usually little irregularities in the surfaces grabbing one
another and even locking together until the force becomes too great to
resist. A deep tread goes this one better; it provide big irregularities
in the surface.
 A rubber imitation of the old leather wrap grip;
every
grip company makes one of these. Most of them also have holes in the
surface to imitate punched leather. The rationale is the same as the
reason above: provide depressions to channel off rain and sweat, and
provide some macroirregularity of the surface.
 Velcro!
I have seen (not recently) a matched grip and glove that gripped
together
with Velcro. The USGA and R&A disallowed it, of course.
 Nearly microscopic plastic ball and socket pairs on
matching grip and glove. Like Velcro, this has been disallowed.
 A glove
that could bring the coefficient of static friction up
to 1.5. It dates to 2016, and I have no idea whether it has been ruled
conforming, nor even whether conformance has been applied for.
 And the list goes on...
If you want amusement of this sort, go to any
patent archive web site
and search on "golf". I just tried this, and was treated to complete
plans, including circuit diagrams, for instrumenting a golf bag to warn
you if clubs are missing.

Taper
The picture shows two views (one natural and one
distorted) of a Golf Pride Tour Wrap grip, probably the most popular
grip of all time. The idea of tapering the handle almost certainly
originated from a desire to do better than just friction in letting
the hands apply a centripetal force.
Look at the exaggerated view, where we have increased the butt diameter
to artificially increase the amount of taper. I did this so the force
vectors can be seen. (This is really a 2D approximation, so we have
split N
into two forces, one on each side of the grip.) Note that the normal
force N,
which by definition is normal to the surface, is slanted in such a way
as to oppose F,
the centrifugal reaction force pulling the club out of the golfer's
hands. With a taper, N
has a component parallel to the axis of the shaft  the blue vector.
And the blue force is opposing the centrifugal reaction force, and
helping the friction force. That friction force is due to N
and a coefficient of friction between the hands and the handle.
If the angle of taper is a,
then the blue helping forces h
amount to:
h
= 2 N sin(a/2) ≅ N sin(a)
When we look at the true likeness of the Tour Wrap grip, we find that
the taper angle is only 4° at its steepest, so:
 The smallangle approximation above is certainly
valid for the Tour Wrap, and
 The total help given to the normal force N
by the taper is about 7%.
While 7% is not a huge gain, it is not trivial either. It is probably
worth tapering a grip to earn a 7% discount on the grip force we need
to apply. So, for a tapered grip, N
can be reduced to N[1sin(a)].
Research
and even simple observation tell us that most of the centripetal force
on the grip near impact is exerted by the lead hand. So we should focus
on the angle of taper under the lead hand. And in fact, that is where
the grip taper is almost always the strongest.
Another argument I once saw for a tapered grip is that it provides a
fixed
diameter the hands can impose to prevent any slippage at all. I don't
buy this, but I'm not an anatomist. Perhaps the hands can maintain a
diameter like that, but I have only heard of hands being able to impose
a diameter by exerting a force  like N.

Grip
force: the bottom line
Club 
Driver
graphite shaft 
Driver
steel shaft 
5iron
steel shaft 
Club length 
45.5 
44.5 
38 
R 
42 
41 
34.5 
Head mass M 
200 
200 
256 
Shaft mass m 
50 
115 
120 
Head speed V ↓ 
Total grip
force (lbs) ↓ 
0 
0 
0 
0 
10 
1 
1 
2 
20 
4 
4 
6 
30 
8 
10 
14 
40 
15 
17 
25 
50 
23 
27 
39 
60 
33 
39 
57 
70 
45 
53 
77 
80 
59 
69 
100 
90 
74 
87 
127 
100 
92 
108 
157 
110 
111 
130 
190 
120 
132 
155 
226 
130 
155 
182 
265 
140 
180 
211 
308 
150 
207 
242 
353 
The total normal grip force, summed (or integrated) over the area of
glovetogrip contact, is in this table. If it looks remarkably like
the table of centripetal force earlier in the article, that is no
coincidence. It is a table for the same clubs at the same clubhead
speeds; the only difference is that the entry is the total grip force
instead of the centripetal force.
That transformation, from centripetal force to the total force the
hands exert on the grip, boils down to a simple one:
 Divide the centripetal force by the static
coefficient of friction, and
 Multiply by [1sin(a)],
where a
is the angle of taper under the lead hand.
This
table is computed using a coefficient of friction of 0.7 and a taper
angle of 4°, both of which should be fairly typical. With those
numbers, the total force is a third larger than the centripetal force.
There is a graph of this table in the Appendix. 
Measured human
grip strength
Up to this point, we have been modeling the force
needed to swing the golf club.
Now let's consider the force
available. How much grip
strength can the golfer actually apply, and is it enough?
Finally, in the next section, we will compare our results to
experimental studies of the force
actually exerted in a golf swing.
What is grip
pressure and why do we care?
I don't know whether
you noticed, but every time I started to get specific  for instance,
to compute anything  I talked about force
instead of pressure.
In fact, they are different, and the answers to any of the quantitative
questions have answers of force, not pressure. Let's look at the
difference so we can understand why all the answers are forces.
In physics and engineering, pressure is the force
per unit area.
If you double the area but keep the pressure constant, then you double
the force. That is very important in hydraulic and pneumatic
mechanisms  it is the basis of how they work  but it is also an
important distinction here. For
instance, we measure force in pounds, but pressure is PSI,
pounds per
square inch. (In the metric system, force is Newtons, and pressure is
Newtons per square meter, also called Pascals.)
Here is a picture that may make things clearer. (I can hope so,
anyway.) It shows two rectangular prisms with their dimensions and
other physical properties. They are the same weight and volume. Let's
see what pressure each one exerts on the surface it stis on.
 The red shape sits on a 2x2 inch "footprint"; that's
4 square inches. Pressure is force per unit area. So its 4 pounds of
weight apply a pressure of 4 pounds divided by 4 square inches, or 1psi
(pounds per square inch) of pressure.
 The blue shape sits on a 1x1 inch "footprint"; that's
1 square inch.
So its 4 pounds of weight apply a
pressure of 4 pounds divided by 1 square inch, or 4psi of pressure.
. 
So
now we know a bit about what grip pressure is. We already know
the force the hands have to exert on the grip. Can't we convert that to
grip pressure by dividing by the area where the hands meet the handle?
There are several things standing in the way of coming up with a useful
number that way.
 There are big hands and little hands, and a whole
spectrum of sizes in between. We would have to quantify the size of the
hand before we had an area we could divide by.
 There are interlock grips, overlap grips, tenfinger
grips, and other more obscure grip styles. This provides another
variability in the area of the
hands on the handle.
 So far, we are assuming constant pressure wherever
the hands meet the handle. That is an incorrect assumption. Here is a
pair of hands after executing a
golf swing, taped with Pressurex Film,. (It comes from a
study we will discuss below.)
The thing to note
here is the extreme variation in the amount of pressure on various
parts of each hand. The left thumb and the pad just under the third,
fourth, and fifth fingers show a lot of pressure, while the tip of the
right forefinger shows almost none. Also, the right hand (the trail
hand for this test) sees far less pressure in the golf swing than the
left. In order to compute the total force
exerted, we would have to integrate (sum up) the force at each spot
times the size of the spot.
So there is no single number representing the
pressure. If you insist on a single number, you might choose the
average pressure, or the maximum, or the average on each hand, or...
But they would all give different numbers, and then you'd have to
explain which number you chose and why. No, there is no single number
that usefully represents the pressure. So let's stick with total hand
force.
Why should we care
about grip pressure (or force)?
Just grab the club, tightly enough and then some, and go after the ball
with it. The reason is that a tight grip and the tense forearms
required to produce it has a bad effect on the shot. at the very least,
it reduces clubhead speed. It is also easily argued that it has the
same effect as trying to manipulate the club with the hands; it hurts
consistency and accuracy.
Why should it reduce clubhead speed? That doesn't make sense. A tighter
grip should let you use more force to swing the club. In fact, our
graphs show the force increasing with clubhead speed.
Let's start with the graphs. Don't confuse cause and effect. The
increase of maximum force with clubhead speed is not because the
additional force produces the speed. Rather, it is because additional
clubhead speed requires additional centripetal force to prevent the
club from flying out of the hands. The force did not produce the speed,
but is needed because the speed is there.
Now let's look at why a tight grip might reduce clubhead speed.
Instructors know it does; here's one of many videos around supporting the
notion. The reason boils down to a debate
in the biomechanics world about whether hand torque is
positive or negative at impact. The large majority of biomechanists
have concluded that the hand torque coming into impact is negative;
that is, it is slowing the club down. The club is moving so fast that
the hands can't keep up with it. If that is true, then any effort the
hands try to exert late in the downswing is counterproductive; since
the hands can't keep up, they will slow the clubhead as it approaches
impact. The best we can do is relax our hands, wrists, and forearms as
much as possible. A corollary to this advice is to use as little grip
force as we can get away with, because applying grip force with the
hands will tense up the grip, make it less like the ideal frictionless
hinge it should try to be.
Which brings me to the real value of hand strength. Between writing the
previous
paragraph and this one, I took a lunch break and caught up with a
biomechanics forum where I hang out. What was freshly posted there
came from a past long drive champion (Monte Scheinblum, 1992
US champion). It was so spoton for this point, let me splice together
a couple of sentences from the thread so it
stands on its own:
The player has to be capable of
withstanding a great deal of outward force from the clubhead’s momentum
near impact, which is why I believe my ability to create high speeds is
greatly attributed to my 100^{th} percentile hand
strength.
I responded:
Yes, but cause and effect get a
little muddied here. The
hand strength, rather than helping create the speed, allows you to hang
onto the club. BUT... Yeah, it also helps create the speed, because
your hands, wrists, and forearms can be more relaxed when exerting the
force necessary to hang on.
Monte's reply was that he agreed 100%.
The
significance is that Monte can use a much smaller fraction of his
strength capability to prevent the club from flying off. With all that
hand strength left in reserve, he can relax his forearms, wrists, and
hands at a higher speed than you or I could while exerting the force
necessary
to hang onto the club.
Having established that hand strength is
a producer of clubhead speed  but in a surprising and unexpected way
 let's look at what hand strength is for typical golfers and then how
that relates to the grip force that needs to be applied.

Hand force
of real golfers
Chris
Daigle, who started me off on this,
pointed me to a video
that discusses the hand force in a golf
grip and exercises to improve it. Hand force is measured by an
instrument called a dynamometer. You squeeze it and it measures the
force you apply, much like a scale measures weight. In fact, exactly
like a scale measures weight.
You
grip the dynamometer between your four fingers and the base of the
thumb. The fingers press down with some amount of force. (In the
picture, it is 100 pounds for the combined force of all four fingers.)
The base of the thumb presses upward with the same force. We know it
is the same force because F=ma and the dynamometer isn't going
anywhere. It isn't accelerating, so the net force must be zero.
With this combination of forces, the dynamometer will indicate 100
pounds of grip force.
(Actually, it isn't exactly the same force. That hand also has to cover
the weight of the dynamometer itself, or it will fall to the floor. But
it is orders of magnitude less than the grip force.)

Here is a table of the grip strength distribution in
some population, I'm not sure what. The source is the user manual for a JAMAR brand
dynamometer.
Here are a few things that jump off the chart at me:
 It
is not terribly surprising that the right hand is stronger than the
left. You would expect the dominant hand to be stronger, and
righthanders constitute a substantial majority of the population.
 This RHLH
relationship is not what you would want to see. The picture of grip
pressure on the gloves say that most of the gripping force is done by
the lead, or nondominant hand. We will see how much more later, but it
is really a dramatic difference. Near impact, almost the whole grip
force is applied by the left hand.
 For men, the age for highest
leadhand grip strength is 3539 years old. At that age, the average
male has a 113pound grip in the lead hand. At age 3034, the average
female can generate 68 pounds of left hand grip strength.
 In the quote above, long drive champion Monte
Scheinblum claims to have 100^{th}percentile
grip strength. Let's assume this translates to the 3sigma point on the
bell curve, which would be the top 0.1% of the population. At age
3539, the average leadhand strength is 112.9 pounds and the standard
deviation (one sigma) is 21.7 pounds. So Monte would have a leadhand
strength of 178 pounds (112.9+3*21.7).

Are
these numbers  the measured grip strength  consistent with the
required grip force for the clubhead speeds we know golfers attain?
Let's look at three data points, using a graphiteshaft driver. We will
use a split between lead hand and trail hand at impact of 955; I got
that proportioning from measurements which we will discuss in the next
section. That means full strength for the left hand and 5/95=0.053
strength for the right.
Golfer 
Grip strength
L + .053 R 
Corresponding
clubhead speed 
Comments 
37 year old
male
(avg strength) 
119 lb 
114 mph 
This is tour
clubhead speed.
So the average golfer has
enough strength 
33 year old
female
(avg strength) 
72 lb 
89 mph 
The average
female golfer
probably isn't close to this.
But LPGA tour does better. 
Monte Scheinblum
assumed 37yo
(3sigma strength) 
188 lb 
143 mph 
In 1992, when
Monte was
champion, 143mph would have
been enough. Not today. 
At first blush, this would appear to be a successful sanity test. But
notice that there is not a lot of margin, a lot of "headroom". The
golfer needs to use 8090% of their grip strength to keep the club from
flying off. Think about how much effort you have to put in to get a
dynamometer to 8090% of your maximum reading. That does not square
with relaxes forearms and wrists. You may be able to hang onto the
club, but it is unlikely that you can get the club to the speed you are
otherwise capable of.
I conclude that the grip pressure is just too high for the grip
strengths in the table. How can we resolve this dilemma?
Let's remember that the grip strengths in the table are dynamometer
readings. Let's also remember how dynamometer readings relate to the
forces the hands are exerting. Remembering this, let's look at an
oversimplified view of the forces involved.
On the left, we see the grip pressure represented by six forces of 20
pounds each, all normal to the surface of the grip. Oversimplified for
tutorial purposes, not because this is a useful model of the grip. What
we have is a total normal force of 120 pounds available to create
friction that will exert a centripetal force.
On the right we have the same six forces, but we are going to analyze
them differently. We are looking for the net force in a vertical
direction
represented by the yellow dashdot line. So we have to resolve each
force into its components and find the component parallel to the yellow
line. That turns out to be 40 pounds downward (red vectors) balancing
out 40 pounds upward (blue vectors).
It is the latter model that is closer to how a dynamometer works. The
fingers press down on one side, the thumb pad presses up on the other,
and the instrument measures the pressure in between. The number that
pops up is the red numbers in the diagram on the right. (That is
numerically the same as the blue numbers; the upward and downward
forces are equal, or the dynamometer would accelerate off somewhere.)
So the dynamometer is reporting only a third of the total normal force
the hand can exert. Of course, the diagrams oversimplify reality; the
factor is probably less than three. But it is certainly more than two,
because the hand must exert both the upward and the downward forces on
the grip. So the dynamometer understates the total normal force the
hand can exert by a factor of between two and three, probably closer to
two.
That makes more sense. It means that the golfer is exerting less than
half his or her grip strength to keep the club from flying off, which
makes for less speedrobbing tension. Of course, the golfer might still
use too tight and too tense a grip  but at least we know now it is
not necessary to do so, and not advantageous either.

Measurement
of grip pressure
Let's move on to actual observations of grip force during the golf
swing. In 2007, Erin Schmidt
published a PhD thesis on the subject (Schmidt, Erin R. (2007): Measurement of grip force and
evaluation of its role in a golf shot. Loughborough
University. Thesis. https://hdl.handle.net/2134/13125).
Most of the measurements were
done by electronic pressure sensors, specifically arrays of sensors on
the gloves and on the grip of the club.
For
us though, the point is not the details of the instrumentation, but
what we can learn about grip pressure during the golf swing. Here is
Schmidt's Figure 82, which I have annotated to identify things
described in the caption and text of the source document. The graph is
the total grip force (the sum
of all the glove sensors) vs time for the swing of a professional
golfer. Not
just any professional, but a Ryder Cup team member, so someone who is
really good.
I looked at this and noticed several things, some which rang true and
others that made me say, "Hmmm!"
 The
force provided by the right hand, especially in the downswing where the
forces are high, is a very small fraction of the total force,
just a shade over 5% near impact.
 The total force near impact is only 160N, or 36
pounds. Hmmm!
In order to exert the centripetal force to hold onto the club at a
115mph (typical pro tour) driver speed, we would need 120 pounds, more
than three times what Schmidt reports.
 The maximum force occurs
about 80msec before impact, and there is a dip in total force just
before impact  exactly where we would have expected the maximum force
exerted. Hmmm!
Let's discuss all three.
1. Very
little right hand force
We
should already know this, even without seeing Schmidt's graph. If you
watch the tour pros play on TV, you will have noticed some big hitters
that actually let go with their right hand at or just after impact. For
instance, Vijay Singh, Fred Couples, and Phil Mickelson. (I don't know
if any golfers from the most recent generation do, but these guys did
even at their prime.) That is zero percent right hand at impact, even
more extreme than the 5% we used in our calculations earlier. There is
even a video by Clay Ballard explaining
how and why this works.
The
fact that Ballard has bothered to make a video on the subject suggests
this is not
known and practiced by everybody. Good golfers do it, but many
beginners and duffers do not. Here is a set
of force traces
like the one above, but for sixteen of the amateur golfers in the
study. They are presented in order of increasing handicap; the handicap
index is in parentheses on the graph. While there is not a perfectly
uniform
increase of percent right hand force with handicap, you can notice a
trend. And the two highest index golfers in the figure have righthand
contributions of more like 30% than 5%.

2. Not
enough total force
At impact, 36 pounds is not nearly
enough to keep the club from flying out of our hands; that would be
more like 120 pounds. How is it even possible to measure such a low
number?
Remember that Schmidt had two different types of
measurements: sensors on the gloves and sensors on the grip. The
separatehands graph above
was made from glove measurements. If we compare it with graphs
from the grip sensors, we will find a similar shape but a very
different
scale.
Unfortunately,
Schmidt did not capture the swing of the Ryder Cupper using the grip
sensors, just the glove sensors. So I can't compare the two instruments
with the same swing. As my "plan B"I have chosen one trace from a
display of grip sensor data of multiple golfers.
I chose the trace that looked to my eye to be most like the shape of
the professional. It turned out to be a 6handicap golfer. (The
multiple traces on the graph reflect a different point Schmidt was
making: the swing of any one golfer is a "signature" that doesn't vary
much from swing to swing.)
In this trace, the force at impact is
about 500N, or 113 pounds. That is much closer to the 120 pounds we
need to keep the clubhead from flying off. The difference might be
experimental, or it might be due to a very understandable difference in
clubhead speed between the singledigit amateur and the Ryder Cup team
professional. Depending on which explanation is correct, there is a
factor of 3.1 or 3.3 between the glove sensors and the grip sensors.
For the rest of our interpretation, let's assume a compromise value of
3.2.
We still ought to try to understand why there is a
difference between the glove sensor and the grip sensor. Schmidt has
noted differences in the area covered by the sensors. That is a
credible explanation; if you look at the pictures of the glove and the
grip above, the grip has a much larger percentage of the area covered.
So the glove might be applying forces to the grip through a surface
other than the sensor itself. I suspect that could completely explain
the difference.
So why use an instrumented glove at all, if its
reported force is a factor of more than 3 on the low side? The answer
is detail. Only with the glove sensors can you assign a force to each
finger, and even a LHRH assignment may be less than 100% reliable from
an instrumented grip.
We will need this full understanding as we go on to the next "Hmmm!"

3. Maximum
force 80 milliseconds before impact
The
maximum total force for the professional golfer occurred about 60% of
the way through the downswing, a full 80msec before impact. That is
when the torque accelerating the club is changing from the hand couple
to
the inertial moment of the force. The centripetal force being exerted
by the hands is a lot lower than at impact, because:
 The angular velocity is so much lower than impact.
This is the major reason.
 The
radius of curvature is larger than it will be at impact. Looked at a
different way, a much higher percentage of the angular motion of the
club is due to the rotating left arm than is the case at impact, where
most of the angular velocity is due to wrist hinging..
So
I was very surprised to find a greater total force on the grip 80msec
before impact than
that required to counter centripetal force at impact. Where is this
force coming from? It certainly isn't a "hang on" force to keep the
club from flying away.
All I could think of was the hand couple,
perhaps at its maximum before it gave way to inertial release. Let's
check this out. Fortunately, we have some actual measurements to help
us:
 Schmidt has included the forces exerted by the
individual fingers. This is necessarily recorded from the glove
sensors, so we will have to multiply things by 3.2 when we are done.
 To
quantify the hand couple, there is another paper, by Koike, measuring
the hand
forces on the grip during the downswing. These forces are reported as
the individual hand forces and the individual hand couples.
Let's
start with the individual finger forces of the left hand. Here is
Schmidt's graph of the finger forces in Newtons, which are from the
same professional swing as the hand forces shown earlier. It is very
clear where the forces peak, as well as the "slot" just before impact.
The significant forces are:
 Middle finger: 60 N
 Thumb: 40 N
 Ring finger: 25 N
 Little finger: 15 N
 Palm:
though significant, we will ignore it. If you look at the sensors on
the glove, the palm sensors would be on the side when gripping a
club (not top and bottom); the force would be almost all crossplane
and not inplane.
 Index finger: too small to bother with, at about 5 N.
Schmidt
has a similar graph for the right hand. But, as we have already seen,
the righthand forces are smaller by an order of magnitude  so much
smaller that we are not going to bother doing the detail with them that
we are doing with the left. Accounting for the right hand will only add
6% or so to the totals, at least for the professional swing we are
analyzing.

Let's
see what that looks like on a force diagram. Here are the significant
forces, as identified above. The index finger exerts a very small
force, and we can see that the palm is pressing perpendicularly to the
other forces.
Let's see what the net force and torque the left
hand is exerting on the club. The forces, of course are in red and
labeled. The green line is a "ruler", with tick marks every 25mm
(roughly one inch). We see that the finger forces are on the 0, 25, 50,
and 100mm ticks.

Let's resolve this into a net force and a couple.
The net force is easy. It is 60 N upward. (15+25+6040)
It is harder to know where that net force is, but we will need to in
order to find the equivalent couple. We separate out the upward and
downward forces and find where each group would be as an equivalent
single force. The downward forces are easy; there is only one of them,
the 40N force. We treat the upward forces as if they were weights, and
find the center of gravity of those weights. That turns out to be 64mm
from the downward force. So the two solid red vectors are the
equivalent upward and downward forces. They are:
 60 N upward force.
 40*0.064 = 2.56 Nm clockwise couple.
We
mustn't forget that this set of glove sensors gives a total force
roughly 3.2 times smaller than the grip sensor.) So, multiplying by
3.2, the force and torque are more likely to be:
192 N upward and 8.2 Nm in a
direction to release the club.
Do we have a sanity test for this number? It turns out that we do.

A
Japanese team headed by S. Koike has published several papers over the
years, describing the results of studies involving an instrumented grip
to determine the hand forces and hand couples during the swing. Here is
an excerpt from the abstract of their 2006 paper (S. Koike, H.
Iida, H, Shiraki, M. Ae, An
Instrumented Grip Handle for Golf Clubs to Measure Forces and Moments
Exerted by Each Hand During the Swing Motion,
Engineering and Sport 6, 2006.)
An
instrumented grip handle was designed to simultaneously measure the
forces and moments exerted by each hand on the handle during golf
swing. Eleven pairs of strain gages were attached on the surface of an
aluminum bar inserted under separated grip covers... A professional
golf player participated in this study and performed golf swings with
several clubs.
Here is a graph of the key results in Koike's study. I
have shown our lefthand crossshaft force and torque on the graph. The
red dots
are the values we just computed from the Schmidt measurements, and the
green dots are
Koike's measurements.
The good news is our numbers are within a factor of two of Koike's
numbers. The bad news is:
 Our force is low by about a third and our couple low
by about a half, compared with Koike's numbers.
 The numbers for the
right hand are not in the same ballpark as Koike's. We wrote off the
right
hand forces measured by Schmidt as too small to consider. Koike has the
right hand comparable to the left hand.
Let's look at each of these problems.
The
first is not really that much of a problem. The factor of 3.2 was the
result of eyeballing the grip sensor measurements when comparing them
against the glove sensor measurements. We were further hampered by not
having a grip sensor graph of the Ryder Cup professional's swing, so we
were operating by guesswork. We compared the glove measurements of a
top tour professional with the grip measurements of a six handicap
amateur. So the actual factor might be quite different.
An
even more likely explanation is our assumption that the grip sensor
measurements were an accurate representation of the total grip force.
But look at the instrumented grip. It is likely that there is
unmeasured pressure here, just less unmeasured pressure than the glove
sensor measurements. If we accounted for everything, we might get
numbers close to what Koike did.
The second discrepancy, the near absence
of any significant righthand influence, is much harder to explain.
Perhaps Schmidt got it right and Koike did not. Perhaps they both got
it right for the population of golfers they tested. I don't know.

Conclusion
We first
found F,
the centripetal force the hands need to exert on the club to keep it
from flying down the fairway. Then we looked at the mechanisms by which
the hands can exert an axial force on the club shaft. They consist of
friction and taper. Applying these mechanisms to the formula for F, we get the
following expression for G,
the total force in pounds the hands exert on the grip.
G = 
0.00129 V^{2}
μ_{s}
R 

(M + m/2) [1  sin(a)] 
Where:
 V
is the clubhead speed, in miles per hour.
 μ_{s}
is the static coefficient of friction between the hands and the handle..
 R
is the distance from the midhands point to the center of mass of the
clubhead, in inches.
 M
is the mass of the clubhead, in grams.
 m
is the mass of the shaft, in grams.
 a
is the angle of taper of the grip where the hands are applying the
force.
This is not necessarily the largest force the hands exert on the
grip during the downswing, but it is the largest one where loss of the
club would be the
consequence of not having enough hand strength. The crossshaft force
exerted
earlier in the downswing  at the beginning of release  may actually
call
for an even larger total force. Whether it does varies considerably
from golfer to golfer. But, for that force, the hands push and pull
instead of squeeze.
Appendix
Shaft with a
nonuniform mass distribution
Math stuff. You won't miss much if you
don't bother reading this. But
if you're an engineer, you might enjoy it.
This is the derivation of the centripetal force of the shaft, where the
mass
distribution is arbitrary. The centripetal force turns out to depend
only on where
that distribution places the balance point of the shaft. We will refer
to the diagram
we used for the same calculation with a uniform distribution.
Let us begin by exploring the balance point. We define the distance
from midhands to the balance point as B,
and the arbitrary mass density along the shaft as D(r).
B
is calculated as:
B = 
1
m 

R
∫
r=0 
r
D(r) dr 
Every engineer should know this off the top of their head. You did
dozens of homework problems in school based on this formula. Remember
this; we'll get back to it at the end.
Now let's take our original formula for the centripetal force on the
shaft, slightly reformulated for an arbitrary D(r).
Specifically, we know that dm=D(r)dr.
F = 
R
∫
r=0 
v^{2} dm
r 
= 
R
∫
r=0 
v^{2} D(r) dr
r 
Plugging in v=(V/R)r
from our earlier work, we get:
F = 
R
∫
r=0 
r^{2} VR^{2}
D(r) dr
R^{2}
r 
= 
VR^{2
}
R^{2} 

R
∫
r=0 
r D(r) dr

Now let's get clever and multiply both numerator and denominator by the
shaft mass m.This
should not change anything.
F = 
m VR^{2
}
R^{2} 
( 
1
m 

R
∫
r=0 
r D(r) dr ) 
We should recognize the part inside parentheses as the balance point
distance B.
So the centripetal force becomes:
That is, the centripetal force is a fraction of the force from a point
mass at the
tip of the shaft (the first factor), that fraction being B/R.
If the shaft has a uniform mass distribution, then the balance point is
in the middle and B/R
is one half  the result we got earlier.
Graph of
centripetal force vs clubhead speed
Graph of grip
force vs clubhead speed
Last
modified  July 23, 2020
