Formula: Shock
Shock is the difference between the initial and final kinetic energy of the racquet.
The Shock formula is difficult but possible. The formula for kinetic energy is \(½M \times v^2\), and Shock is the change in kinetic energy from immediately before to immediately after impact, \(½M({v_1}^2 - {v_2}^2)\), so finding the mass center linear velocities before and after will give a formula for Shock. Apply Conservation of Angular Momentum, as with Torque, to get the angular impulse and the difference in angular velocities (\(w_1 - w_2\)) of the racquet before and after impact, then multiply (\(w_1 - w_2\)) by the distance from the axis of rotation to the mass center (\(r\)) to get (\(v_1 - v_2\)), which is one factor in (\({v_1}^2 - {v_2}^2\)). The tough part is getting the other factor, (\(v_1 + v_2\)). First, convert (\(w_1 - w_2\)) to a difference in linear velocity of the impact point (\(p_1 - p_2\)) by multiplying the angular velocity difference by the distance (\(d\)) from the axis of rotation to the impact point. Knowing this linear velocity difference of the impact point (\(p_1 - p_2\)), apply the Coefficient of Restitution formula: the difference in velocities of the impact point and ball after impact is equal to their velocity difference before impact, times the coefficient of restitution (which is a measure of the elasticity of the ball/racquet system). Solve for the linear velocity of the impact point before impact (\(p_1\)). Knowing that the linear velocities of the impact point and of the mass center are proportional in rotation and equal in translation, find the linear velocity of the mass center before impact (\(v_1\)) by multiplying by the ratio \(r \div d\). Note on this step. Since \(v_1 - v_2\) is known, and so is \(v_1\), \(v_2\) can be found. So now \(v_1 + v_2\) is known and the formula follows. Although the formula now looks hairy, that’s all there is to it.
In general, the total kinetic energy of a rigid body in plane motion (e.g. a swinging racquet) is the sum of:
- the kinetic energy associated with the linear velocity of the mass center (\(½Mv^2\)) and
- the kinetic energy associated with its rotation about the axis of rotation (\(½Iw^2\)).
The first component is due to body rotation, arm swing, etc., and is assumed to be the same for all racquets used by the same player under the two Benchmark Conditions. Over the course of the impact, the before vs. after difference in linear velocity of the mass center associated with this first component alone will be negligibly small anyway. Any difference will come as the racquet’s rotation slows down or speeds up during the impact. So the first component is disregarded, and the second component is what is concerned in the following derivation. However, the second component (rotational kinetic energy) is converted into kinetic energy associated with the linear velocity of the mass center so as to be expressible in joules. A complete legend of variables used follows after the derivation.
The first step follows from the Torque derivation:
\[ (v_1 - v_2) = \frac{rdb}{I} (s_2 - s_1) \]
\[ v_2 = v_1 - \frac{rdb}{I} (s_2 - s_1) \]
\[ p_1 - p_2 = \frac{d^2b}{I} (s_2 - s_1) \]
\[ s_2 - p_2 = c (p_1 - s_1) \]
\[ p_2 = s_2 - c (p_1 - s_1) \]
\[ p_1 - \left [s_2 - c(p_1 - s_1) \right] = \frac{d^2b}{I} (s_2 - s_1) \]
\[ p_1 = \left( \frac{1}{1 + c} \right) \times \left( s_2 + cs_1 + \left[ \frac{d^2b}{I} (s_2 - s_1) \right] \right) \]
\[ v_1 = \frac{r}{d}p_1 \]
\[ v_2 = v_1 - \frac{rdb}{I} (s_2 - s_1) \text{ (from above)} \]
\[ (v_1 + v_2) = \left( \left[ \frac{2r}{d(1+c)} \right] \times \left[ s_2 + cs_1 + \frac{d^2b}{I} (s_2 - s_1) \right] - \frac{rdb}{I} (s_2 - s_1) \right) \]
\[ Shock = \frac{M}{2}(v_1^2 - v_2^2) \]
\[ = \frac{M}{2} (v_1 + v_2) (v_1 - v_2) \]
\[ = \left( \frac{M}{2} \right) \left( \left[ \frac{2r}{d(1 + c)} \right] \times \left[ s_2 + cs_1 + \frac{d^2b}{I} (s_2 - s_1) \right] - \left[ \frac{rdb}{I} (s_2 - s_1) \right] \right) \times \left( \frac{rdb}{I} (s_2 - s_1) \right) \]
\[ = \left( \frac{Mr^2}{I} \right) \left( \left[ \frac{b(s_2 - s_1)}{1 + c} \right] \times \left[ s_2 + cs_1 + \frac{d^2b}{I} (s_2 - s_1) \right] - \frac{1}{2I} \left[ db(s_2 - s_1) \right]^2 \right) \]
Discussion
Impulse is a change in momentum, measured in units of momentum (\(Mv\)), and integrating all of these tiny changes over the time of impact (dwell time) gives \(\frac{1}{2}Mv^2\), which happens to be the formula for kinetic energy, and the dimensional recipe for the metric energy unit of the joule (kg·m²/s²). So Shock is the integral of impulse.
Following the First Law of Thermodynamics, we note that an impact of a ball with a racquet is an adiabatic process (negligible heat transfer), so the change in the internal or potential energy of the particles of the racquet is equal to the sum of the work done on the racquet by the player and by the ball. More background on the change in kinetic energy: If you drop something, its mass (\(M\)), accelerated by gravity (\(g\) = 9.81 m/s²), goes a certain distance (\(h\)) before hitting the ground. Mass × acceleration is force (Newton’s Second Law), and force × distance equals energy (in joules), so the kinetic energy of the object right before it hits the ground is \(Mgh\). Just before impact, the object had a certain velocity \(v\) due to its acceleration by gravity, and it therefore had a kinetic energy of \(\frac{1}{2}Mv^2\). At the bottom of its path its kinetic energy is zero because its velocity is zero, and all of the kinetic energy it used to have has been turned into potential or internal energy. If it rebounds, some of this potential energy becomes kinetic energy again, the amount recycled into kinetic energy being determined by the elasticity of the collision (coefficient of restitution). What is not recycled but remains in the racquet as an increase in its internal energy is Shock.
| \(A_x\) | = | Impulse Reaction, the translational force acting at the axis of rotation due to impact, in Newtons. Note that when \(d\) = \(q\) (\(q\) is the distance from the axis of rotation to the center of percussion), the expression within the second set of parentheses becomes zero. |
| \(a\) | = | linear acceleration of the mass center, in m/s² |
| \(b\) | = | mass of the ball, in kg |
| \(c\) | = | coefficient of restitution of the racquet/ball system |
| \(d\) | = | distance from the axis of rotation to the impact point, in cm |
| \(e\) | = | the distance from the axis of rotation to the tip |
| \(F\) | = | force applied at mass center, in Newtons |
| \(I\) | = | moment of inertia (swing weight) of racquet, in kgf/cm² |
| \(I_5\) | = | moment of inertia (swing weight) of racquet at 5cm from the butt, in kgf/cm² |
| \(I_7\) | = | moment of inertia (swing weight) of racquet at 7cm from the butt, in kgf/cm² |
| \(I_{10}\) | = | moment of inertia (swing weight) of racquet at 10cm from the butt, in kgf/cm² |
| \(I_a\) | = | moment of inertia (swing weight) of racquet at distance \(a\) from the butt, in kgf/cm² |
| \(M\) | = | mass of the racquet, in kg |
| \(m\) | = | mass in kg |
| \(ω\) | = | angular velocity of racquet, in radians/s |
| \(p\) | = | linear velocity of impact point, in m/s |
| \(r\) | = | distance in cm from mass center (balance point) to axis used in the stroke |
| \(s\) | = | ball velocity, in m/s (positive is away from player) |
| \(s_1\) | = | velocity of ball before impact, in m/s |
| \(s_2\) | = | velocity of ball after impact, in m/s |
| \(T\) | = | torque at axis of rotation, in Nms |
| \(t\) | = | dwell time, or duration of impact, in seconds |
| \(v\) | = | linear velocity of the mass center, in m/s |
| \(v_1\) | = | linear velocity, just before impact, of racquet mass center, in meters/second |
| \(v_2\) | = | linear velocity, just after impact, of racquet mass center, in meters/second |