Particularly troubling is the lack of prior consultation with the pros who will be using these balls. Which of them are in favor? True, it is rare to find any touring pros who might be called formally educated, so naturally they must expect their opinions to be ignored and their objections overruled. If they don’t want to play, there are plenty of ambitious youngsters eager to take their place. However, even though there might be no reason for courtesy or compassion, wouldn’t it be economically prudent to prolong the careers of the marquee players, and not to increase the already alarming rate of injuries among the recreational players?

The ITF has authorized a ball with a 15% greater diameter to be used “on an experimental basis.” The intention is that the bigger ball will meet more air resistance, therefore play will be slower. Fluffing the nap (felt covering of the ball) will increase diameter and drag, but apparently the intention of the ITF is to require ball manufacturers to mold a larger rubber core.

The larger diameter of the rubber core, even if the weight of rubber remains the same, will result in a higher rotational inertia for the ball. That means a “heavy” ball because players will be able to impart a lot of angular momentum (spin). Angular momentum is the product of the rotational inertia and the rotation speed, and the higher rotational inertia permits a much “heavier” ball at the same spin rate. High angular momentum of the ball on impact will aggravate Torsion (screwdriver twist on the handle), causing more stress on the arm of the receiver.

Another problem with bigger balls: they will radically change the game in the same way that the “spaghetti string” racquet did, by giving junkballers an edge. The ITF banned (retroactively) the spaghetti strings which imparted such extreme spin. The same reasoning should ban these balls.

Yet another problem with bigger balls: if the same ball weight (57 grams) is to be maintained, the rubber of the bigger ball must be made thinner to stretch over the larger surface. Thinner rubber means that the air will leak out easier, and higher air pressure will be needed to maintain the same ball bounce. These balls will go flat faster. They will also be less bouncy in actual pro-level play because of higher hysteresis loss from more air being compressed. These will be soft balls.

Presently, for professional tournament play, a ball must bounce more than 53 inches and less than 58 inches when released from a height of 100 in. That means that the coefficient of restitution for the ball itself (apart from the racquet) is between 0.73 and 0.76. It should be noted that the 100 in. drop height does not approximate the speed of a pro serve, so for testing the hysteresis loss from the bigger ball this test would be inadequate. Using softer balls, having a bounce at the low end of this range (low * c*), means higher Shock, Shoulder Pull, Work, Shoulder Crunch, Wrist Crunch, and Elbow Crunch for the players.

As you can see from the formulas, heavier balls (high ** b**) means both higher resultant forces from impact (Torque and Impulse Reaction),

Club players can take a lesson here, especially those who play on clay, where the balls get heavier as play goes on. **Change balls often** to protect your arm. Tennis balls are a bargain, so leave them on the court.

**Step One:** Select the racquet that is more head-light, i.e. which has its
balance point closer to the hand. This will mean the one with the lowest balance
spec. In our example, let’s say the most head-light is Blue. By adding weight
at the butt of the other racquet, Red, match its balance point to the most head-light,
Blue, using the formula for balance point shift:

* s = d**(

= shift in balance, in cm.s

A “point” is 1/8 inch, or 0.3175 cm. Points head-heavy (or head-light) are measured from the midpoint of the racquet’s length.

= distance from added weight to original balance point, in cmd

= amount of added weight, in gramsm

= original racquet weight, in gramsM

**Note: in the following discussion, the symbol * means multiply, /
means divide by, and ^ means to the power of (^2 means squared, or multiply that by
itself).**

**Example*** of balance shift calculation*: Red has balance
31.9 cm and Blue has balance 31.2 cm. Blue and Red are the same length and both
weigh 340 grams. The swingweight of Blue is 322 and Red is 330. The desired
balance point shift is 0.7 cm (31.9 – 31.2), so plug 0.7 into the left side of the
equation in place of

0.7 = 31.9*(* m*/(340 +

0.7/31.9 = 0.022 =

0.022*(340 +

7.48 + 0.022*

7.48 = 0.978*

From this calculation, you will need 7.6 grams at the butt end of Red to shift the balance of Red back to match the balance of Blue.

**Step Two:** Having matched the balance of Red to Blue, now you have to add some
weights to Blue in order to match Blue’s swingweight to the swingweight of Red, and also
to match Blue’s weight to the new total weight of Red, while keeping the match in
balance. This additional weight on Blue will be in two pieces, * a*
(headweight) and

* a* = headweight, in kg. — please note, this is kilograms, not
grams (a gram is a thousandth of a kilogram, so move your decimal point, e.g. 6 grams =
0.006 kg)

* b* = tailweight, in kg.

* x* = distance of headweight from the 10 cm axis, in
centimeters

* y* = distance of tailweight from the 10 cm axis, in
centimeters

r = distance, in cm, of the racquet’s balance point from the 10 cm axis of rotation. We
know r because we know the balance and the location of the axis [10 cm from the butt for
RDC swingweight measurements], so just subtract 10 from the balance spec and you have
r. The * r* in the formulas is not measured from the 10 cm
axis but from the axis specified by the player’s grip, so r is not

The four unknowns are * a*,

Equation (1): * a* +

The total added weight is the amount that will bring Blue up to the same weight as modified Red. Equation (1) follows from this requirement. We know the total added weight, so we can plug a number in there.

Equation (2): * a**(

Equation (2) follows from the requirement that balance remain unchanged. No change in balance means equilibrium about the original balance point, which is now the same for the two racquets. Equilibrium means that the sum of moments about the balance point is zero. Each moment is the weight times its lever arm, and the lever arm in each case is the distance of the weight from the original balance point. The distance for

Equation (3):

The total added swingweight (right side of the equation) is the difference between the original swingweight of Blue and the swingweight of Red after modification to match its balance to Blue, so subtract to find this difference and plug in the number. Red’s post-modification swingweight will usually be approximately the same as its original swingweight since the addition of the weight at the butt end to match the balance will not add appreciable swingweight due to its short distance (10 cm) from the axis of rotation. Ten grams at the butt would be necessary to add 1 swingweight unit. Swingweight is mass (in kg), times distance (in cm) squared, this distance being from the mass element to the axis of rotation you are considering, so you can figure out how much swingweight you have added by the tailweight.

Equation (3) follows from the definition of swingweight: the sum of the products of all
the mass elements and the squares of their distances from the axis of rotation. Each
added weight will contribute some swingweight about the 10 cm axis. The total added
swingweight about the 10 cm axis from the two additional weights * a*
and

Equation (4): [I’ – M’*r’^2] – [I – M*r^2] = * a**(

Three equations are not enough to solve for four unknowns, so this fourth one is necessary. Although it might look scary/boring/impossible, it is only high school math, and it isn’t a lot to ask that you know how to do this. Equation (4) follows from application of the Parallel Axis Theorem to Blue, before and then after the addition of weights * a* and

Here are the definitions of variables used in the following derivation of Equation (4):

I = swingweight of Blue before modification, in kg.cm^2

I’ = swingweight of Blue after modification, in kg.cm^2

Ic_{1} = moment of inertia about an axis of rotation through the balance point of
Blue before modification, in kg.cm^2

Ic_{2} = moment of inertia about an axis of rotation through the balance point of
Blue after modification, in kg.cm^2

M = mass of Blue before modification, in kg.

M’ = mass of Blue after modification, in kg.

r = distance from axis of rotation to balance point of Blue before modification, in cm.

r’ = distance from axis of rotation to balance point of Blue after modification, in cm.

The Parallel Axis Theorem tells us that * before* modification of
Blue, its swingweight about the 10 cm axis (I), which we have from measurement, was equal
to the sum of its swingweight about an axis through its balance point (Ic

I = Ic

and therefore, by algebra:

Ic_{1} = I – M*r^2

Applying the Parallel Axis Theorem to Blue after modification:

I’ = Ic_{2} + M’*r’^2

and therefore:

Ic_{2} = I’ – M’*r’^2

The difference between Ic_{1} and Ic_{2}, the swingweights about the
balance point (mass center) axis before and after modification, will therefore be:

[I’ – M’*r’^2] – [I – M*r^2].

This difference, which constitutes the left side of Equation (4) can be determined
because all of the terms in the expression are already known: I’ is the
post-modification swingweight of Red, which you are trying to match, and you know this; M’
is the new weight of Red, which you are also trying to match, and you already have this
from the balance modification that’s been done; r’ is the new distance of the balance
point of Blue from the 10 cm axis of rotation after modification, which will be the same
as the original r because we are keeping the balance of Blue the same now that Red has
been matched to it; I is the original measured swingweight of Blue; M is the original
measured weight of Blue; and r is the original measured distance from the 10 cm axis used
for swingweight measurements to the balance point of Blue (i.e., it’s the balance spec of
Blue, minus 10, because 10 cm is the distance from the butt end to the axis of rotation
used by the RDC, and balance is measured from the butt). Ic_{2} – Ic_{1}
is therefore a known quantity. It is the increase in the swingweight (also known as
moment of inertia, or rotational inertia) of Blue about an axis of rotation going through
Blue’s balance point, or mass center.

This known increase is equal to the sum of the contributed swingweights about the mass
center axis due to the added weights * a* and

Ic_{2} – Ic_{1} = * a**(

Ic_{2} – Ic_{1} = [I’ – M’*r’^2] – [I – M*r^2] from before,
so

[I’ – M’*r’^2] – [I – M*r^2] = * a**(

Voila, Equation (4). The left side is can be found from measurements, so plug
these known values in (see discussion above) and you will have some number for the left
side instead of the variables there. With the left side now reduced to a number, now
we have four equations involving only four unknowns, and it is therefore possible (but not
easy) to solve for the four unknowns — * a*,

A high-end pocket calculator that can solve a system of four simultaneous equations is what you’ll need to do this job, such as the Texas Instruments TI-86.

Formula for shift in balance: * s = d**(

= shift in balance, in cms

= distance from added weight to original balance point, in cmd

= amount of added weight, in gramsm

= original racquet weight, in gramsM

Equation (1): * a* +

Equation (2): * a**(

Equation (3): * a**

Equation (4): [I’ – M’*r’^2] – [I – M*r^2] = * a**(

*Unknowns in Equations 1 – 4: a, b,
x, y*

= headweight, in kg. — please note, this is kilograms, not grams (a gram is a thousandth of a kilogram, so move your decimal point, e.g. 6 grams = 0.006 kg)a

= tailweight, in kg.b

= distance of headweight from the 10 cm axis, in centimetersx

= distance of tailweight from the 10 cm axis, in centimetersy

*Other symbols/constants:*

r = distance from axis of rotation (10 cm) to balance point of Blue before modification, in cm: r = Blue balance spec – 10

r’ = distance from axis of rotation to balance point of Blue after modification, in cm

I = swingweight of Blue before modification, in kg.cm^2

I’ = swingweight of Blue after modification, in kg.cm^2: I’ = Red swingweight after balance matching.

M = mass of Blue before modification, in kg

M’ = mass of Blue after modification, in kg. M’ = Red weight, after balance matching.

*Measured Specs:*

r_{1} = balance of Blue

r_{2} = balance of Red

M_{1 }= weight of Blue

M_{2} = weight of Red

I_{1} = swingweight of Blue

I_{2} = swingweight of Red