Formula: Shoulder Pull
Shoulder Pull is the centripetal force, in Newtons, with the shoulder being a distance \(R\) (\(= 0.61 + r \div 100\)) meters from the racquet mass center. The Shoulder Pull force, measured in units of Newtons (1 Newton = 0.225 lb.) attributable to the racquet alone is derived from the Work formula and the general formula for centripetal force (centrifugal force).
Knowing the formula for Work, it’s an easy step to a formula for Shoulder Pull. Work is kinetic energy, whose formula is:
\[ KE = \frac{1}{2} Mv^2 \] (\(M\) is mass, \(v\) is linear velocity of the mass center.)
Shoulder Pull is centripetal force (equal and opposite to the centrifugal
force acting on the racquet), and the formula for centripetal (center seeking) force is:
\[ Centripetal\ force = \frac{Mv^2}{R} \]
(\(R\) is the distance from the shoulder to the racquet’s mass center, which is \(0.01r + 0.61\ meters\)). The distance from the shoulder to the axis of rotation is assumed to be 0.61 meters. It is the centripetal force due to the racquet alone that is of interest, so forget the arm mass.
\[ Work = \frac{M}{v}v^2 \]
\[ Work = \frac{Mr^2}{2} \times \left( \left[ \frac{1}{d(1+c)} \right] \times \left[ s_2 + cs_1 + \frac{d^2b}{I} (s_2 - s_1) \right] \right)^2 \]
\[ Shoulder\ Pull = \left[ \frac{2}{\left(\frac{r}{100} + 0.61 \right)} \right] \times Work \]
\[ Shoulder\ Pull = \frac{Mr^2}{0.01r + 0.61} \times \left( \left[ \frac{1}{d(1+c)} \right] \times \left[ s_2 + cs_1 + \frac{d^2b}{I} (s_2 - s_1) \right] \right)^2 Newtons \]
Dividing the Work formula by \(R\) and multiplying by 2, therefore, gives a formula for Shoulder Pull due to the racquet, in Newtons of force. One Newton = 0.225 pounds.
\(A_x\) | = | 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 |