By Jochen Voss, last updated 2012-02-18
The purpose of this text is to explain some physical background of ball juggling. While the number of hands and the timing of throws is abstracted away in the mathematical theory of juggling, these quantities play a central role in this text.
The present version of the text is work in progress and will hopefully be extended.
In this section we will have a look at a real-world juggling pattern to see what is going on. We will get timing information from the pattern, derive Shannon's first theorem about juggling, and finish this section by exploring how fast or slow you can juggle while keeping the same height of throws.
The basis of our analysis is a
sequence of pictures showing the
4444535344 with a time
resolution of 1/30 second. One of the pictures is shown at the right.
Table 1 gives a time-line of events in the
Table 1. This table summarises the sequence of events in the picture series. It gives the times and order of all throws and catches.
Our first aim is to measure the frequency of the throws. The linear least squares fit of the data from the table is t(n) = 0.229*n - 0.187, where t(n) denotes the time of throw number n. This is illustrated in figure 1. You can see that one beat lasts approximately 229ms but the right hand starts to lag behind a little bit in the second half of the sequence.
Figure 1. The linear least squares fit for the throwing times is t(n)=0.229*n-0.187. Thus the length of one beat is approximately 0.229 seconds.
In between the throws, the hands spend part of the time moving empty to catch the next ball. Another part of the time is spent holding or carrying the balls. This is illustrated in the next figure.
Figure 2. Graphical illustration of holding times: the upper band corresponds to the left hand, the lower band to the right hand. Gray strips indicate that the corresponding hand is empty. The unit on the horizontal axis is seconds.
At the beginning of the sequence the right hand is holding two balls (green and blue). After this initial phase there are some obvious balance considerations.
Combining these relations shows
and thus we get the following result, which was first derived by the famous mathematician Claude Shannon.
Theorem 1 (Shannon's first theorem). For a (not necessarily periodic) juggling sequence where each hand holds at most one ball at a time the following balance equation holds over each interval of time:
As an illustration we check that the theorem really holds for our example sequence of throws.
Table 2. Measured holding and empty times for the hands in the interval from t=0.100 to t=2.100.
Table 3. Measured holding and flight times for the balls in the interval from t=0.100 to t=2.100.
Example. For the juggling sequence described above and the 2 second time interval which starts at t=0.100 we get the holding and empty times given in table 2. The times for the balls are given in table 3. As you can see, the left hand is empty for 38.3% of the time (0.766s of the 2s interval) and the right hand is empty for 31.6% of the time (0.632s of 2s). The quantities from Shannon's first theorem are
Thus the left hand side of the equation in Shannon's theorem becomes (2.602+5.398)/(2.602+1.398) = 8 / 4 = 2 as expected.
Shannon's theorem as stated above does not require the juggling pattern
to be periodic. But when applied to patterns like
4 where all throws are the same height, things can be
Definition. A juggling pattern is called
uniform if each ball every time it is caught is held for the same
amount of time (the
holding time), each ball every time it is
thrown flies for the same amount of time (the
flight time), between
each pair of adjacent throws the hands are empty for the same amount of
empty time) and each hand holds at each instant of time
at most one ball.
Theorem 2 (Shannon's first theorem for uniform patterns). For uniform juggling patterns the following balance equation holds:
Proof. We reduce this statement to the general
version of Shannon's first theorem. Denote the holding time
by h, the flight time by f, the empty time
by e and the number of balls by b. Consider any
time interval of length h+f. During this interval each
of the balls flies for
f time units and thus the sum of
flight times is b⋅f (see
figure 3). Also, during this interval of time,
each ball is held for h units of time and thus the sum of holding
times is b⋅h. Finally each ball is thrown exactly
once during the time interval in consideration and thus the sum of empty
times is b⋅e. Using
theorem 1 we get
This completes the proof.
Figure 3. Timing diagram for a uniform 4-ball juggling sequence using 2 hands. One hand juggles the blue and yellow balls, the other hand the red and green ones. The horizontal axis is time, holding times are marked gray and flight times are marked in the ball's colour. The holding time is h=0.3, the flight time is f=0.6. The light gray boxes in the background mark two different time intervals of length h+f. The diagram illustrates that each time interval of this length contains equal amounts of accumulated holding, flight and empty times.
Good jugglers can vary the speed of a pattern to some extent while
keeping the height of their throws the same. The extremes are called
hot potato juggling (throwing the balls almost immediately after
catching them) and
lazy juggling (holding the balls as long as
possible). Since the height of the throw is determined by its flight time
(we will examine this relation below), keeping the height of the throws
the same implies that the flight time is not changed. How fast or how slow
can we juggle for a given flight time f?
Theorem 3 (speed of juggling for fixed flight times). For a uniform juggling sequence with evenly spaced beats and with h hands, b>h balls and flight time f, the length of a beat is between f/b and f/(b-h).
Proof. The limiting case for
juggling is the case where the holding time equals 0. Solving the
relation from theorem 2 for the empty time
Since the holding time is zero, the distance of one hand's throws is
f⋅h/b+0 = f⋅h/b.
Assuming that the beats for all h hands are evenly spaced we find
the length of a beat in
hot potato juggling with flight
time f to be f/b.
The limiting case for
lazy juggling is the case where the empty
time equals 0. From theorem 2 we get
and, since the empty time is zero, the length of a beat for
juggling with flight time f is
f/(b-h). This completes the proof.
Example. For the juggling sequence from
table 1 the average flight time of a
4 is approximately 0.6 seconds. Since the pattern has 4 balls
and two hands, the length of a beat for a uniform pattern would be between
0.6/4=0.15 seconds and 0.6/(4-2)=0.30 seconds. The length of the beats we
measured for the (non-uniform) pattern in
table 1 was 0.229 seconds (see
figure 1) which is almost exactly in the
middle of this interval.
During the flight times a ball's motion is governed only by the force exerted by earth's gravity (for our purposes here the friction of the air can be neglected). Therefore, the horizontal component of a ball's speed during flights in constant whereas the vertical component is subject to an acceleration of g=9.81m/s2 downwards. Solving the motion equations shows that, given the initial position x(0) and the initial speed v(0), the ball follows the following parabola (compare figure 4):
Figure 4. Jochen Voss, juggling three glow-balls. The marked ball is moving from the left to the right along the green parabola. The ball needs 0.567s to cover the designated section of its path, first raising 45.9cm and then falling 33.4cm.
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