Tennis Kinematics Transient Analysis: A Ball Spin & Racket Collision Description by Miltiadis A. Boboulos - HTML preview

From (Eq. 38) ∫ F_av dt and by assuming

t1

F_av = 1275 N
and t₁ - t₂ = t = 0.00274 sec
then S_x = 1275 . 0.00274 = 3.4935 N sec
But as the time of application of the force F _av is actually twice as much due to

the back shooting of the ball ricocheting from the string area, then

 

Sx max=1275 . 0.00274 . 2 = -6.987 N sec (Eq. 44)

 

9.4 The maximum ball speed when hitting the racket and the bouncing ball speed.

Above calculations have been based on maximum b l speeds when colliding with the racket, which is equal to 210
km/h or 58.3 m/sec [14]. While flying
within the maximum boundaries of the ^Y
court the ball can travel over ≥23 meters
provided it travels along a straight line m/ s e c
parallel to the length of the court and not U o > 5 8 , 3_{less that around 32 meters provided it U}

1

travels diagonally. The sp eed of the ball is Θ
reduced during the flight due to the air X resistance; therefore its initial speed R
should be higher [14]. ^{Fig. 14}

We calculated the actual speed of the ball based on the theory of an object (shell) flight in a resistance environment when shot at a certain angle from the surface plane. – Figure 14.

The equation for movement in resistance environment is:
X = cosθ₀ /b. h (Vo. cosθ₀ /V cosθ), (Eq. 45)
and from aerodynamics

V ²

 

b = R_x =C_x ρ₂ .S (Eq. 46)

 

where, Vo = Uo; V = U_1; R_x = sir resistance

ρ = air density = 0.125 (at normal temperature and humidity conditions and when measured at the level of the court);
S – cross sectional area of the ball (m²);
b₁ – ball diameter = 0.065 m;
Cx – air resistance coefficient, which depends on the shape ;
For spherical shape it is Cx = 0.14 for Re≤3.8 . 10⁵ = 380000
S – cross sectional area of the ball = 0.065²π/4 = 0.0033 m²
R_e = 69000.b₁.V(V =U₁)
forV =60m/secR_e = 69000.60.0.065 R_e = 269100< 380000

R
x

 

=

 

0

 

.

 

14

 

V²₁₆ .0.0033

 

R_x =0.000029.V =b.V²b =0.000029

 

The angle θ is unknown so from the equation

 

tg θ= tg^{θ g} (e^2ξ−1) (Eq. 47) _{0 2bV2cos θ}

 

0 0

 

We determine θ_o.

 

ξ=^bx (Eq. 48) _cosθ0

For X = 32 meters, which is the longe st possible flight of the ball over the court. From (19) at θo=12°-slanted initial trajectory – shooting at straight-forward aim angle
θ_o≤15° and for Vo=58.3 m/sec; tgθ_o=0.2126 and cosθ_o=0.9781:

0.000029.32

 

From ξ (Eq. 49) 0.9781

 

If we calculate for initial conditions:

 

^{9.81 (e 0.0019 1)tg θ= 0.2125} 2.0.000029 .58.3^{2 −}

 

.0.9781 (Eq. 50)

tg θ= 0.0206 ; θ= 1.178^o
Then from the equation for the movement of the ball in air environment we get:
V.cosθ= V_o cosθ_o e^-ξ (Eq. 51) θ = 1,178°; V_o=58,3 [m/sec]; ξ = 0,00095; 2ξ = 0,0019; e^-ξ =0,999
Vcos1,178° = 58,3 .cos12^°.5^-0,00095

V=^57,023 = (Eq. 52) _0,999

 

∠θ_o =O^o(U₁)V_o = 58,3 [m/sec] ; b = 0,000029 (ξ = 0,00095); X = 32m.

 

gx

 

²g22 ³

 

-y =x.tg^θ −_{3V0 cos3 θ2V0 cos2 θ 2} (Eq. 53) _{0 2}

 

0 o

 

Where e^2ξis determined in order of ξ rates. 9

 

,

 

81

 

.

 

32

 

² 9,81.0,000029.32^{3 y =32.tg00 −} 2.58,3².1^{2 −} 3.58,3².1³ y = 0 –1,4770 – 0,0009 = -1,4779 m (Eq. 54) Provided that the initial speed V_o=20 m/sec, as:

 

R_c = b₁.V.69000 = 0,065.20.69000 = 89700; C_x = 0,5

 

ρ

 

From (Eq. 46) b = 0,0033.₂ = 0,0001

C_x values have been taken from the diagram indicating the C_x and K_e relation for a spherical body of an uniform surface – similar to the tennis ball – a study made in NASA’s aerodynamic tube.

From (Eq. 49)

 

tg

 

θ

 

= 0,2125

 

9,81

 

_{2.0,0001.58,32.0,9781}(e^0,065 (Eq. 55)

θ₀ = 12°
V_o=20 [m/sec]
b = 0,0001

ξ = ^0,0001.32 = 0,00327_0,9781

x= 32 [m ]
tgθ =0,021 ; θ = 1,203° Vcos 1,203° = V_o.cos12°.e^-ξ V.0,9998 = 20.0,9781.0,9967 V= 19,50[m/sec]

From (Eq. 54)

 

θ

 

0

= 0 ; y = 0
9
,
81

.

 

32

 

² 9,81.0,0001.32^{3 −} 3.20².1³2.20².1²

 

y = -12,557 – 0,027 = -12,53 [m]

From all equations given above and subsequent calculations the following conclusions were drawn up:
- For speeds close to the maximum speed of V=58.3 m/sec the speed variations resulting from the air resistance during the flight of the ball to the longest possible travel distance over the court, (which is approximately 32 meters) is quite insignificant. Therefore, we can assume for all further calculations that the speed of hitting the ball with the racket is the same as the speed of the ball when it collides with the racket when meeting it.
- For speeds close to V = 20 m/sec (when playing not so hard) the influence of the air resistance is reduced but the angle of the ball when falling on the court changes substantially. The flying speed for Vo=20 is reduced down to V=19.50, which for the purpose of the design of the test rig is also rather insignificant.
- From the last calculations above we also drew the conclusion that tennis players will find helpful. For horizontal flights from an initial height of y=1.47m [from (Eq. 54) above] the ball can fly a distance of 32 meters until it reaches the playground travelling along a nearly straight line provided it has been shot at a speed that is close or equal to 58.3 m/sec. Moreover, it would be better if it flied at a descending trajectory from the very beginning – shooting above the head from over 1.47 meters height.

9.5 Load analysis for the head and body of the racket and eventual oscillations and deformations.

As alread y mentioned in (35), (36), (37), when the ball meets the string the force F causes the tension reactive forces Fa and Fb in the strings. These forces result in a sudden compression load on the rim in the direction of the centre of the ellipse of the frame and also, a bending load in the direction of F and the body (head and handle) of the racket. If we selected one point M on the body of the racket then these forces would induce free oscillations with a restoration force of P = - CX, where P is the restoration force resulting from the elasticity of the construction.

The equation for the movement is:

 

mx’’ = -cx→ x’’+^c = 0_m

² If we substitute k =c/m, where k is the natural circular frequency, then x’’+k²x =0
Which represents a second order linear differential equation having constant

coeff icients. Its characteristic second order equation with constant coefficients is r +k^{2 2} = 0 and both roots are imaginary: r_1,2= ± ik.
The differential equation integral is:
X = C₁coskt+ C_2.k.t
X’ = -C₁k sinkt +C₂kcos t
C₁ and C₂ are integration constants.
If we substitute C₁ = Asinα and C₂ = Acosα and A and α are constant, then:

X = A(sinα . cos k.t + cosα . sin k.t) = A sin(k.t + α ). (Eq. 55)

Therefore, the point A moves harmonically. The constants A and α are determined using the initial conditions. When t = 0, X = X_o and V_x=V_x,o=X_o and having differentiated (Eq. 55) we receive:

Vx = x’ = Ak cos (kt+α) At t = 0 we get:

 

x

 

'

 

Xo = A sinα; X’o = Ak cosα ^o= Acos α_k

 

And for the amplitude we get:

 

A = x²_o+(^x'o )²; tg =^kxo , (Eq. 56) _{k x'o}

 

Where: T=2π/k is the harmonic frequency period

 

T = 2?^m (Eq. 57)_c

From equations (56) and (57) we found that the amplitude A and the initial phase depend only on the initial conditions Xo and Xo’ and from (27) we can see that the period T does not depend on the initial conditions (the property is called isochronism). This allows us to calculate T. To be able to do this it is sufficient to write down the differential equation for the movement as follows: X”+k²X = 0 and executing the

formula T =2?^m without integrating the equation. The differential equation is: _c

 

mX” = G - P

 

Dying oscillations occur when a restoration force P and a resistance force

R are available (internal resistance).
R = λV
λ -coefficient of proportionality
V –velocity
R – acceleration in the opposite direction
The differential equation is:
mx’’ = P+R
mx’’+cx+λV = 0
If we substituted K2=c/m and 2h=λ/m, then

x’’+2hx’+kx = 0 (Eq. 58) The characteristic equation has the following roots:

 

r _1,2 = -h± h²−k² (Eq. 59) Three alternatives are possible depending on the sign of the determinant Δ= h²-k²

 

r

 

1,2

 

= -h

 

±

 

i
k

 

2_h2 −

 

a) Low resistance alternativeΔ<0 and k>h. The roots are complex:

 

The common integral is:

x
=
^−ht (c₁ coskt k² − h² +c₂ sin+ k² −h²

_e ) (Eq. C₁ =A sin α; C₂ = A cos α

x = A_e^−ht sin(t k² −h² + α (Eq. 61) where A and α are the integration constants.
A and α are determined by the initial conditions for the movement.
The movement is not periodical:

T₁ = ^π (Eq. 62) k² −h²

The resistance available increases
the period. When h<<k, T₁<<T (Low _X
resistance does not practically change
the period of oscillations) [16]. _Vo>0

b) High resistance alternative

 

Xo

 

Δ>0 and k<h. The roots of the ^{Vo<0
characteristic equation are real t}and negative. The common _{Fig. 16 integral is:}

 

x =C₁.e^t (−h+ h² +k² )+C₂e^−t(h+ h² −k² ) (Eq. 63) When t→∞, X decreases monotonously and tends to zero. The movement in the point is aperiodical.

c) The third alternative is when Δ=0 i.e. k=h. In
this case the roots of the characteristic equation
are equal, i.e. r_1,2=-h._{Fig. 15}

The integral of the differential equation is:

X = C₁ e^-ht +C ₂ te^-ht (Eq. 64) The movement in this case is aperiodical as in Fig. 15.
With reference to all proved above for the rackets the parameters that are

essential are the real values for the coefficients λ and C as well as the value of friction resistance. Therefore, the mass of the string area is increased for the purpose by providing some additional mass. Additional mass or cut-outs could be provided in the head of the racket [16].

9.6 Loading the frame and body by the horizontally acting force. Deformation estimations.

The critical load on the racket results from the maximum value of the load applied by the ball moving (in both directions) at maximum speed. From Figure 3 and the experimental data we can determine the spring constant C of the racket body:

C = 10.835(N)/2.25(mm) = 4.82 N/mm (Eq. 65)

 

and from Figure 1 C = 2.717 N/mm.

If we go back to Figure 4 the load arra ngement for the body of the racket (head + handle) will look as shown underneath in Figure 10:
E – modulus of elasticity. For steel E = 2.1 . 10⁶ ; J – geometric inertia moment (cm⁴ ); P – loading force; l – beam length; tg α_A = -Pl²/2EJ; EJ = Pl³/3fA
The deformation f_A resulting from the collision is given by the expression:

f_A = -Pl³/3EJ. (Eq. 66)

The differential equation is EJy’’=
Where, M is the bending momentum; m is the mass per unit length of the racket and M/l = m.
EJδ⁴ y/δ⁴x = mδ² y/δt²
If we substitute EJ/m = k⁴ then
EJ/m . δ⁴ y/δ⁴x + δ² y/δ²t = 0

⁴ k . . δ⁴ y/δ⁴x + δ² y/δ²t = 0,
where: Y(x,t) = X(x).T(t)
Then: X(x)= C₁ cosωx + C₁ cosωx sinωx + C₃ chωx + C₄ shωx
t(t) = Acospt + Bsinpt
As the actual values for E and m are different for different designs and materials

of rackets we can make the calculation using the experimentally established constant

C₂ = 4.82 N/mm from (65) (for an actual racket C₁ = 2.717 N/mm.
f^max.c=F=F/ C_1, (Eq. 67) but F = 1275 N (from Eq. 33)
f^max = 1275/4.82 = 264.5 mm for Figure 2 and
f^max