# Racquet Power and the Ideal Racquet Weight

By Rod Cross

What is the ideal weight of a racquet that will give the maximum possible power out of the racquet? Most elite players use a moderately heavy racquet, around 340 grams or so, but is that the best weight for maximum power or is 340 grams a compromise between power and control or maneuverability? A simple calculation suggests that the best weight, in terms of racquet power, should be the geometric mean of the ball weight and the weight of a player’s arm.

## CALCULATING IDEAL RACQUET WEIGHT

By “geometric mean,” I mean the following. Take any two numbers, say 4 and 16. The average or mean of these two numbers is (4 + 16)/2 = 10 which is half way between 4 and 16 in the sense that 10 - 6 = 4 and 10 + 6 = 16. The geometric mean of 4 and 16 is the square root of (4 × 16) = square root of 64 = 8, which is also half way between 4 and 16 in the sense that 8 is twice as big as 4 and 16 is twice as big as 8. Similarly, the geometric mean of 2 and 18 is the square root of (2 × 18) = 6. In this case, 6 is 3 times bigger than 2 and 18 is 3 times bigger than 6.

Figure 1. Assuming no energy loss,
maximum energy is transferred from one ball to another if they are both the same mass. |

Example 1: A Tennis Ball. A tennis ball weighs 57 grams. An arm weighs about 2000 grams (4.4 pounds). The average of these two weights is 2057/2 = 1028 grams, which is way too heavy for a racquet. The geometric mean gives a more sensible answer for racquet weight. Multiply the numbers together and you get 57 × 2000 = 114,000 square grams. Now take the square root of 114,000 which is 337.6 grams. That is the geometric mean, and it may be the ideal weight in terms of racquet power (assuming your arm weighs 2000 grams). A 337.6-gram racquet is 5.92 times heavier than a 57-gram tennis ball, and a 2000-gram arm is 5.92 times heavier than a 337.6-gram racquet.

Example 2: A Baseball Bat. In case you think the answer here is purely coincidental, consider a baseball bat. A baseball weighs 144 grams. You swing with two arms weighing, say, 4000 grams. Multiply these together and you get 144 × 4000 = 576,000. The square root of 576,000 is 759 grams (27 ounces). That seems about right for a baseball bat. If your two arms weigh 6000 grams, then the calculation suggests a 929-gram bat (33 ounces).

## GETTING ENERGY INTO THE RACQUET

There is actually some interesting physics behind this geometric mean calculation. In order to serve as fast as possible I need a heavy racquet in order to get as much energy as possible out of my arm and into the racquet. I figured that out last year by throwing different objects and measuring how fast I could throw them. I can throw a 57-gram tennis ball at 20 meters/seconds, I can throw a 144-gram baseball at 16 meters/seconds and I can throw a 730-gram steel ball at 12 meters/seconds. Younger people than I can throw faster but the proportions remain about the same.

Figure 2. If one ball is very much heavier
than the other (in this case a little more than five times), the heavy ball retains more energy. In this example the heavy ball retains about 46 percent of its initial energy. The lighter ball has a greater velocity but less energy than the ball in Figure 1. |

Everyone can throw a light ball faster than a heavy ball, but the heavy ball always ends up with more energy than the light ball. The kinetic energy of each ball is mv2/2 where m is the mass of the ball and v is the ball speed. In my case, this amounts to 11.4 Joule for the tennis ball, 18.4 Joule for the baseball and 52.6 Joule for the steel ball. If I wanted to smash something like a rock by throwing a ball at it, then I’d be better off throwing a heavy steel ball than a light steel ball. It’s the same with a racquet. I can’t swing a heavy racquet as fast as a light racquet but I can get more energy into a heavy racquet. Imagine serving with a 20-gram racquet. You could swing such a racquet really fast, but it wouldn’t pack any punch at all since it’s not heavy enough and doesn’t have enough energy. Using this argument, I need a racquet that is as heavy as I can swing to get as much energy as possible into the racquet.

## GETTING ENERGY INTO THE BALL

However, I also need to get as much energy as possible out of the racquet and into the ball. In theory I could get all the energy of the racquet into the ball if the racquet weighed the same as the ball. That’s what happens in pool or billiards when two balls collide and when they are identical in weight. If ball A collides head-on with ball B, and if ball B is initially at rest, then ball A comes to rest and ball B takes off with the same speed as ball A had before the collision. All the energy of ball A is transferred to ball B (Figure 1). If ball A was heavier than ball B, then A would transfer some of its energy to B and keep some for itself, which means that A would continue on its path at a reduced speed after the collision (Figure 2). A racquet or a baseball bat doesn’t come to a complete stop when it hits a ball because it is heavier than the ball, and so is the arm used to swing the racquet or the bat. Clearly, what is needed is some sort of compromise where the racquet is heavy enough to get a lot of energy out of the arm into the racquet and light enough to get a lot of energy out of the racquet and into the ball.

Arms and racquets are complicated objects and it is not easy to come up with a simple formula to work out the ideal racquet weight in terms of the energy transfer. It is much easier to think about colliding balls. Suppose that a heavy ball collides with a light ball and an eighth of the energy of the heavy ball is given to the light ball. If the heavy ball collides with a medium weight ball, then the heavy ball will give more of its energy to the medium weight ball, say half of it. If the medium weight ball then collides with the light ball and gives half its energy to the light ball, the light ball will end up with a quarter of the initial energy of the heavy ball. In this example, of using an intermediate weight ball, there is a factor of 2 increase in the energy transferred to the light ball compared with a direct collision between the heavy and light ball. The ideal weight of this intermediate ball, to transfer as much energy as possible to the light ball, turns out to be the geometric mean of the weights of the heavy and light balls (Figure 3). That’s how I got the above formula for the ideal racquet weight. The physics here also helps to explain why I can hit a tennis ball a lot faster than I can throw it, by using a racquet that is lighter than my arm but heavier than a tennis ball. The effort on my part is the same when I throw or hit a ball but the racquet improves the efficiency of the energy transfer from my arm to the ball.

Figure 3. How much energy is transferred from one ball to another in a collision depends on their relative weights. If the aim is to get the most energy from a heavier to a lighter ball, then using an intermediate weight ball with a mass equal to the geometric mean of the heavy and light ball will optimize the transfer of energy and, thus, the velocity of the lighter ball. The examples show that using intermediate balls with masses less or more than the geometric mean result in less energy and velocity transfer to the light ball. |

## HOW ARMS AND RACQUETS ARE LIKE POOL BALLS

Arms and racquets are not the same as spherical balls, but they behave in the same general way. In fact, they behave in almost exactly the same way. For example, when I throw a ball, I keep the ball in my hand for about 0.2 seconds as I accelerate my arm and then I release the ball. In some sports like beach volleyball, the ball is thrown up and is served by hitting it with the hand or fist. In that case, we can regard the serve as a collision between the arm and the ball, or we can regard the serve as a collision between a heavy ball (the arm) and a light ball (the volleyball). Regardless of whether the ball is hit or thrown, it turns out that a light ball can be hit or thrown faster than a heavy ball, but a heavy ball ends up with more energy than a light ball. At least, that is the result when the ball being hit or thrown is lighter than the arm as it is in all ball sports (apart from the shot put).

If a light ball hits a heavy ball then it works the other way. That is, as the heavy ball gets heavier its energy decreases. Maximum energy is transferred during a collision when both balls are the same weight. Throwing can be regarded as a relatively long collision between the arm and a ball. Hitting is a much shorter collision, lasting much less than 0.2 seconds, but the energy transferred from the arm to the ball will be the same if the person doing the throwing or the hitting exerts the same effort in swinging the arm. The action of the arm when serving with a racquet is not a lot different from throwing the racquet, so the energy transferred from the arm to the racquet will increase as the weight of the racquet increases.

Serving with a racquet is like throwing it and throwing it is like hitting it with your hand in terms of the final speed and energy of the racquet. The effective mass of the hand is about half the actual mass of the arm, but the effective mass of the racquet, at the spot where the ball is hit, is about half the actual mass of the racquet. (This is explained in the book *The Physics and Technology of Tennis,* by H. Brody, R. Cross and C. Lindsey, available from the USRSA.) This changes the numbers slightly but the general idea is still the same. That is, the racquet needs to be lighter than the arm and heavier than the ball in order to generate maximum power. There are other factors that influence racquet power, such as the actual impact point on the strings, the actual balance point and the swingweight, but the weight of the racquet is the main factor.

See all articles by Rod Cross

**About the Author**

Rod Cross retired in 2003 as an honorary member of the Sydney University staff and continues to work on the physics of sport and forensic physics. He is a physicist and co-author of *The Physics and Technology of Tennis* available from the USRSA.

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