All about Gear Effect - 3
Dave
Tutelman --
February 19, 2009
OK, let's review where we are...
The spin due to horizontal gear effect is given by:
| s |
=
|
|
≈ 16.4 Vb x |
|
(Equation 2 & 2a) |
And the spin due to vertical gear effect is given by:
| s |
=
|
|
≈ 25 Vb y |
|
(Equation 3 & 3a) |
Where the units are clubmaker-friendly units (at least in the USA) like
miles per hour, inches, and gram-centimeters-squared.
The horizontal-rotation moment of inertia Ih is
generally between 4000 and 5900 g-cm2 in
almost any 460cc clubhead. The vertical-rotation moment of inertia Iv
is 1/2 to 2/3 of Ih.
This results in very substantial spins, which can certainly
exceed 1000rpm and which exceed 2000rpm in some of our examples.
Now let's use this analysis to answer some of the questions
that arise about the practical application of gear effect.
Do weight
screws produce draws and fades?
In 2004, TaylorMade introduced the R7 driver with weighted screws to
allow changing the distribution of weight in the clubhead. They claimed
that choosing the right weights for the right holes could provide
trajectory control ranging from a 10-yards fade to a 10-yard draw.
Their 2008 model (the most recent at this writing) is the R7
Limited, which claims a 20-yard draw and 15-yard fade, for a
total of 35 yards of adjustability. And the concept (weight screws to
control trajectory) has created lots of imitators.
Figure 3-1
Figure 3-1 shows how the weight screws are supposed to work. The idea is to
move the center
of gravity. If you can move the center of gravity toward the heel, then
a center hit will behave the way a toe hit did before you moved the CG;
that is, the club will rotate around the [now heel-biased] CG and cause
hook spin. The difference is that the bulge is not starting the ball to
the right and giving it slice spin, so all the gear
effect behaves as a hook.
But can they actually do the job?
The question frequently comes up whether weight screws can really
control trajectory. The consensus of those who have looked hard at it
is: there isn't enough weight moved in the R7 or its imitators to do
anything noticeable. For instance:
- Tom
Wishon has repeatedly
held that it takes 30-40 grams to do anything worth doing here for 90%
of the world's golfers.
That's more than twice what TaylorMade moves in its screws.
- I
have heard an anecdote of a teaching pro with a science background who
conducted a blind test with the original R7. He found that what the
golfer believed about the weight configuration made a significant
difference; the actual weight configuration did not.
- I myself
have estimated the motion of the center of gravity due to the weight
screws at less than a tenth of an inch, and concluded that tiny a
difference would not matter.
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Figure 3-2
Now we have the tools to
evaluate this question analytically. The R7 Limited has three screws
totaling 18 grams: two 1g screws and a 16g screw. So the choices for
weighting are simple:
- 15g to the heel (draw bias).
- 15g to the toe (fade bias).
- 15g to the rear (neutral bias, but more loft at impact).
The reason I say 15g instead of 16g is that there is a 1g screw
remaining in the other holes, so only 15 grams are movable. The total
head weight (hence the club's MOI and swingweight) and the clubhead MOI
remain pretty much the same because the total weight in the screws is
unchanged and all three stations for the screws are roughly the same
distance from the clubhead CG.
We
can anlyze this as a 189-gram shell plus a 15-gram weight that we can
move to the heel or the toe. (I have not measured one. But Jeff Summitt advised me that
TaylorMade drivers have heads about 4 grams heavier than the industry
norm, giving it a 204g total head weight. But the numbers for gear
effect purposes will be indistinguishable from a 200g head.) The
spacing is close to a 4" range, so let us use
round numbers of 2" toward the heel or 2" toward the toe. Some simple
calculations show:
| Position of 16g
screw |
Position of center
of gravity |
| Heel |
0.15" toward heel |
| Center |
Center |
| Toe |
0.15" toward toe |
Fifteen hundredths of an inch is not much. But it is measurable, and
will produce some gear effect. Let's see how much.
We will use equation #2a to compute the spin, and TrajectoWare Drive to
compute the hook distance.
In each case, we will assume a center hit.
Let's try computing the amount of hook for a heel-weighted driver, with
a variety of golfers. In each case, we will use a loft appropriate to the clubhead speed of the golfer.
The
golfer
|
Ball
speed
(mph) |
Hook
spin
(rpm) |
Hook
distance
(yards) |
| We'll start with the same golfer we used for the
sanity test of the analysis. In each case, we'll use the driver loft
that
gives the maximum distance according to TrajectoWare Drive, and see how
big a hook it gives us. |
150 |
369 |
12.5 |
Well, that was not enough to give us the 17.5
yard hook that TaylorMade claims.
Spin
and distance both increase with ball speed. How high a ball speed do we
need to give the advertised hook? |
168 |
413 |
17.5 |
| What
about the guy who is really the target for the advertising, the hacker
who needs to correct a bad slice? How much draw is he going to get
against his weak, 40-yard slice? |
120 |
295 |
6.0 |
What does this tell me?
- TaylorMade's claims can be achieved.
- But, in order to get that much hook/slice, you need a
clubhead speed in the neighborhood of 115mph. That's big hitting,
typical of a Tour player. (Last year, the average Tour clubhead speed
with a driver was about 112mph.)
- If you're the hacker with an 85mph clubhead speed and
a 40-yard slice, you will get back only 6 yards of that slice with
the R7 Limited. That 6 yards is not a major portion of your slice. More
important, it is probably less than the inconsistency of your launch
conditions; you probably won't even notice the 6-yard
improvement -- though it will still be there. Maybe that's what Wishon
meant
by 90% of golfers.
But can anybody actually hit the center repeatably
within a .15"
tolerance? After all, we only moved the CG only 0.15" away from the center
of the clubface.
It doesn't matter!
Remember, there is a bulge built into the clubface
to compensate for gear effect. What the weight screws do is bias the CG
relative to the bulge, not just on an absolute basis. The center of the
face has no bulge; if you hit it there, the CG bias will obviously
produce the simple hook in the table. As you move away from the
center, you get both gear effect and bulge -- which in combination
preserve the draw/fade bias built into a center-hit. So if you miss the
center, the shot will still finish about the same amount left of where
it would if you had the same impact with a neutral, unbiased driver.
Consider:
- If you hit it dead center, you will get the hook
suggested by the table above.
- If you hit it toward the toe, you will get more
hook than that amount of toe hit would give for a
neutral driver -- by roughly the
amount suggested by the table. So, after the bulge has its
say,
you will again wind up about that amount left of target.
- If you hit it toward the heel, the shifted CG will
give you less slice than gear effect would have
been for a neutral driver. So, after the bulge has its say, you will
again wind up about that amount left of target.
Yes I'm surprised, too. And I have to retract some things I've said in
the past.
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Does shaft
torque limit the gear effect?
If I get a torsionally stiff shaft, will that reduce the gear effect
compared with an easily twisted shaft? After all, we see those dramatic
videos of how much the head twists due to impact. Surely the shaft
torque will affect that.
Yes, it will affect those videos. But only a small portion of that
twisting occurs while the ball is still in contact with the clubhead.
So we have to compute whether that small amount of twisting can induce
enough torque in the shaft to compete with the torque that the ball is
applying to the clubhead due to the off-center hit.
Figure 3-3The
computation strategy is:
- Find the size of the torque applied to the head by the ball.
- Find the amount of clubhead twist during impact.
- Find the torque in the shaft resulting from that
size of twist.
- See whether the
shaft torque induced by that twist is a significant fraction of the
torque we found in step #1 above. If not, then shaft torque cannot
limit gear effect.
1. Find the torque applied by the ball.
Remember (from the analysis earlier in the article) that the ball's
momentum equals impulse.
mVb
= ∫F(t)dt
If we simplify things and say the force is constant over the duration of
impact, then we will underestimate the torque due to the ball. If the
force has a typical rise-and-fall variation, we will understimate the
maximum torque by a
factor of about 2. Let's do it that way to give shaft torque a chance.
mVb
= ∫F(t)dt =
Favg * .0005
So we can find the average force
Favg
= 2000 mVb
= 92 Vb
Since the torque about the center of gravity is the force times the
lever arm x
Τavg
= 92 xVb
(newton-meters)
Let's apply unit conversion factors so it works with inches, miles per
hour, and foot-pounds.
Τavg
= .771
xVb
(foot-pounds)
That is the torque applied by the ball to the clubhead.
2. Find the amount of clubhead twist during impact.
We know from equation 2 that:
That is the spin of the ball in rpm. Since the ball and the clubhead
are acting as a pair of gears during impact, the rotation rate of the
clubhead (also in rpm) should be the ball spin times the ratio of the
gear diameters. The effective radius of the clubhead rotation about its
CG is C
(see figure 1-3), giving a gear diameter of 2C.
The
diameter of a golf ball is 1.68". So
That is the rate of rotation of the clubhead when the ball leaves it.
If it had that rotation rate for the entire .0005 second of impact,
then the total rotation (in revolutions) during impact would be
| rotation (revs) =
49,417 Vb |
x
Ih |
.0005
60
sec/min |
= 0.412 Vb |
x
Ih |
But it is not rotating at this final rate for the entire duration of
impact. It builds up to that rate, gradually accelerating as the ball
does. For any plausible acceleration function, the actual rotation
during impact is half this amount, or
| rotation (revs) =
0.206 Vb |
x
Ih |
The rotation in degrees (which is what we need) is 360 times
this, or
| rotation (deg) =
74.2 Vb |
x
Ih |
3. Find the torque due to that shaft twist.
Now we know how much the head rotates during impact. We can turn
rotation into shaft torque by remembering that the "torque
rating" τ of a shaft is the number of degrees the
shaft will
twist with an applied torque of one foot-pound. So the torque is the
degrees of shaft twist divided by the torque rating.
| Tshaft |
= |
rotation(deg)
τ |
= |
74.2 xVb
τ Ih |
4. Compare shaft torque with impulse torque
We now know:
- The torque on the clubhead due to impact with the ball - Tavg
- The torque on the clubhead from the shaft due to twist - Tshaft
Let's compare them. We'll look at their ratio.
Τavg
Tshaft |
= |
.771 xVb
74.2
xVb / (τ Ih)
|
= .0104
τ Ih |
The effect of shaft torque will be greatest if we allow
the clubhead to twist by using a low-MOI clubhead. So let's select from
the low end of our driver
data: 4000g-cm2. The ratio becomes 41.6τ.
Suppose we take a very "low torque" (that is, torsionally stiff) shaft,
to emphasize the effect of the shaft. A driver shaft with a torque
rating of 2° is stiff indeed! Even with this stiff a shaft, the ratio
is 83.
What does that mean? It means that the torque due to the ball
striking the clubhead is 83 times as great as the torque
supplied by the shaft to oppose the clubhead twist. In other words, the effect of shaft torque is negligible.
If we move away from drivers, it
might be possible for shaft torque to influence gear effect for a
hybrid with a small (low-MOI) head and a larger-tip steel shaft with a
torque rating well below 2°. I doubt that the influence would be
anywhere near 50% of the spin, but it might be measurable -- unlike the
case for a driver.
Bottom line: No matter how we
stack the deck in favor of shaft torque, the driver head's moment of
inertia supplies about 99% of the resistance to twist, and the
shaft only about 1%. The shaft torque is negligible in resisting
gear
effect.
Last modified - Feb 19, 2009
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