# 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
 58,830 Vb C x Ih
≈  16.4 Vb x (Equation 2 & 2a)
And the spin due to vertical gear effect is given by:
s
 58,830 Vb C y Iv
≈  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.

## 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. 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.

## 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 3-3
The computation strategy is:
1. Find the size of the torque applied to the head by the ball.
2. Find the amount of clubhead twist during impact.
3. Find the torque in the shaft resulting from that size of twist.
4. 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:
s
 58,830 Vb C x Ih
(Equation 2)
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
 58,830 Vb C x Ih 1.68 2C
=
 49,417 Vb x Ih
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:
1. The torque on the clubhead due to impact with the ball - Tavg
2. 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. 