cp's OEIS Frontend

Consider an analog clock face to be a unit circle, with unit-length clock hands; the endpoints of the hands lie on the unit circle and form the vertices of a Clock Triangle inscribed within the circle.

The area within this Clock Triangle has maximum value 1.2990353071..., which occurs around 02:54:35 and at its mirror image around 09:05:25.

At time T seconds after 00:00:00, the clock hands are at angles

S (seconds hand) = T/60 * 360, (degrees)

M (minutes hand) = T/60/60 * 360,

H (hours hand) = T/60/60/12 * 360.

The clock cycle repeats every 12 hours = 43200 seconds.

The second 6 hours of the cycle is a mirror image of the first 6 hours.

The area within the Clock Triangle at any time is equal to

F(T) = abs(sin(H-M) + sin(M-S) + sin(S-H))/2.

(The derivation of this equation is not overly-complicated.)

The hour and minute hands are exactly 120 degrees apart at times

T = 14400/11*(3k+1) and T = 14400/11*(3k+2) for integer k.

There are 22 such times during every 12-hour cycle.

Empirically examining the relative extrema of F(T) near these 22 times, it is found that the largest F(T) occurs near T = 10475 (02:54:35), and near its mirror image T = 32725 (09:05:25).

Using Newton's iterative method to solve for Tmax in F'(Tmax) = 0,

Tmax = 10474.561690797181984...

F(Tmax) = 1.299035307107332...

Note: an equilateral triangle has area sqrt(3)*3/4 = 1.2990381056...

The inclusion of the second hand leads to solutions differing from those of A120500. Apart from the perfect match at midnight or noon, the other minima are characterized by small angles > 0 between the hour and the minute hands and a coincidence of the second hand with one of the other two hands. It turns out that the coincidence of the minute and second hands never leads to the smaller angle value. The exact times in seconds are given by A348759, with rounding to nearest second applied to determine the terms of this sequence.

The global minimum occurs at midnight or at noon, when the positions of all three clock hands coincide exactly. All other minima are characterized by the fact that the position of the second hand coincides exactly with the position of the hour hand. The neighboring situation in which the second hand coincides with the minute hand leads in all possible cases to a slightly larger angle between the minute and hour hand.

The three hands of an analog 12-hour clock are never exactly equidistant (mutually 120 degrees apart), but they are very nearly so at 22 times during the 12-hour clock cycle. These 22 cases occur at exactly a(n)*43200/1427 seconds after 00:00, according to the simple "midpoint" definition given below, although three other (more complicated) definitions of near-equidistance are also described here. 11 cases occur in the first 6 hours, and each has a mirror image case in the second 6 hours. For all n = 1..22, a(n) mirrors a(23-n).

There are at least four ways to define the precise times when each case attains its nearest equidistance. For each case, the four definitions generate four distinct solutions, which all lie within a fraction of a second from each other. The best pair of cases occur near 02:54:34.5 and its mirror image 09:05:25.5, which correspond to a(6) = 346 and a(17) = 1081. The second-best pair is a(11) and a(12). The third to 11th best are a(n) for n = 1, 5, 7, 2, 10, 8, 4, 9, 3 (and their mirrors 23 - n).

MIDPOINT (definition #1). The simplest definition first locates each of the 22 times when the hour and minute hands are exactly 120 degrees apart, then determines the nearest time (earlier or later) when the second hand is exactly midway between them and around 120 (not 60) degrees from each of them. The midpoint solution times occur at exactly a(n)*43200/1427 seconds after 00:00. The a(n) are found using the FORMULA below. Midpoint times are used as first approximations when calculating solution times for the other three definitions. A347040 gives standard clock times for a(n), rounded to the nearest second, in integer format hh:mm:ss.

LARGEST TRIANGLE AREA (definition #2). The best solution occurs when the triangle whose vertices are the endpoints of the three clock hands has maximum area. See A348637.

LEAST SQUARES (definition #3). The best solution occurs when the sum of the squares of the deviations of the angles of each pair of clock hands from 120 degrees is minimum. To find this time, express time T in units of 12-hours (T = 0 at 00:00 and T = 1 at 12:00), and express all clock hand angles A in units of 120 degrees (3 units = 360 degrees). In these units, the midpoint times are T(n) = a(n)/1427, and the three clock hand angles at any time T are

A(1) = hour-hand to second-hand angle = 3*(720-1)*T = 2157*T,

A(2) = minute-hand to second-hand angle = 3*(720-12)*T = 2124*T,

A(3) = hour-hand to minute-hand angle = 3*(12-1)*T = 33*T.

At any time near any T(n), all three A(j) will be near-integer, with their variances from 120 degrees represented by

V(j) = abs(A(j) - round(A(j))).

At exactly time T(n), these variations are all integer multiples of 1/1427, given by

V(3) = abs(33*a(n) - 1427*k)/1427,

V(1) = V(2) = V(3)/2,

where k is defined in the FORMULA below, and is equal to floor((3n-1)/2).

The sum of the squared variances to be minimized near time T(n) is

F(T) = V(1)^2 + V(2)^2 + V(3)^2.

The least-squares time t occurs when F'(t) = 0. The values of t are rational numbers.

LEAST VARIANCE (definition #4). The best solution occurs when the sum of the absolute values of the deviations of the angles of each pair of clock hands from 120 degrees is minimum. To find this time, use the same A(j) and V(j) as defined in the least-squares solution above, but instead minimize the sum of the variances

G(T) = V(1) + V(2) + V(3).

The least variance time u occurs when G'(u) = 0. The values of u are rational numbers.