From: whisnant@iastate.edu (Kerry L Whisnant)
Subject: Re: strings
Keywords: tension
Organization: Iowa State University, Ames, IA
Date: Wed, 15 Apr 1992 18:00:31 GMT

In article <1992Apr13.220627.7432@cbnewsm.cb.att.com> lfd@cbnewsm.cb.att.com (Lee Derbenwick) writes:
>In article <1992Apr12.184223.22668@seas.gwu.edu>, se020151@seas.gwu.edu (Hyok Lee) writes:
>> ...  By simple laws of physics, 
>> tighter strings do not contact the ball at a greater force than looser
>> strings.  F=ma; therefore, if the ball is accerated to the same speed, then
>> the normal force should be equal.  Looser strings stay in contact with the
>> ball longer, therefore impart more spin. ...
>
>The first sentence doesn't follow from "simple laws of physics".
>Let's take as a given the statement that "looser strings stay in
>contact with the ball longer".  (Other posters have referred to
>experimental evidence for this.)
>
>The average acceleration is delta-v/delta-t.  Since delta-v is
>stated to be the same for both cases, but delta-t is greater with
>looser strings, the average acceleration is _less_ with looser
>strings.  Therefore, since F=ma, the average force on the ball
>during contact is less with looser strings.

This isn't quite right either; delta-v is not the same since the ball
comes off faster with looser strings. However, the delta-t argument is
qualitatively correct since the difference in final velocities is
probably less than the difference in contact times.

>(A full calculation, to determine peak force, would also be
>interesting, but that requires a _lot_ more experimental data
>on ball and string deformation, etc.  So I'll have to make do
>with the average, here.)

In fact, you can solve this problem theoretically if you make some
simplifying assumptions:

1. The system is confined to one dimension.

2. The player exerts a constant force F on the racquet.

3. the strings obey Hook's Law (i.e., that the force of the strings on
the ball is -kx, where x is the distance the strings are stretched and
k is the "spring constant" that measures the stiffness of the strings;
to a good approximation, k will be proportional to the string
tension).

What you end up with is a straight-forward intermediate-level mechanics
problem with two bodies (ball and racquet). After some computation you get:

                                /                                 \
                               |           -1 /   F              \ |
time on racquet= t = sqrt(mu/k)| pi + 2 tan  | ------- sqrt(mu/k) ||
                               |              \mr vrel           / |
                                \                                 /

                           F     /               \
final ball speed=  vf = ------- |  t + sqrt(mu/k) | + vcm
                        mb + mr  \               /

                                   /               2  2\ 1/2
                                  |        2     mu  F  |     mu F
maximum force on ball= Fmax = sqrt| mu vrel  k + ------ |  +  ----
                                  |                 2   |      mr
                                   \              mr   /


where mb= mass of ball
      mr= mass of racquet
      mu= reduced mass of system = mb mr/(mb + mr)
      pi= 3.14159...
      vrel= relative speed between ball and racquet on initial impact      
      vcm= initial center of mass speed of system

From these equations it is clear that increasing k (i.e., the tension):

1. Decreases the time on the racquet
2. Decreases the final ball speed
3. Increases the maximum force on the ball

Thus tighter strings touch the ball a shorter time and give less
power, but the maximum force on the ball is greater.

We can use this result to guess what happens when top-spin is applied.
If the frictional force applied parallel to the ball's surface is
proportional to the normal force (the usual ansatz for simple physics
problems; the proportionality constant is the so-called coefficient of
friction) then the angular velocity (spin) applied to the ball should
be proportional to the velocity change; i.e., a tighter racquet which
gives less final speed also gives less spin.

However, if the frictional force increased faster than the normal
force (because the "grip" of the strings on the ball improved at
higher forces, i.e., the coefficient of friction is not constant but
is instead force-dependent), or if it depended on how fast the racquet
was scraped across the ball, then it could very well be that the spin
imparted would be greater for a tighter racquet. Thus the critical
question is the dependence of the frictional force vs. the normal
force. My guess is that the coefficient of friction increases with the
force applied (because the strings grip better), and that tighter
racquets would usually give more spin. But it would require some
experimental testing to know for sure.
-- 
Kerry Whisnant                  High Energy Physics Group
whisnant@iastate.edu            Department of Physics and Astronomy
whisnant@alisuvax.bitnet        Iowa State University, Ames, IA 50011
Newsgroups: rec.sport.tennis