Sixteen and Four Elevenths Minutes Past Three

It is three o'clock. The hands are a right angle apart. When is the next moment they lie exactly on top of one another?

Not 3:15. At 3:15 the long hand has reached the 3, but the short hand left it fifteen minutes ago and is now a quarter of the way to the 4. The answer is $16 + \tfrac{4}{11}$ minutes past three, which is 3:16:21.8181 and so on forever, because elevenths do not terminate.

One number does the work

Measure angles clockwise from twelve, in degrees, and let $t$be minutes elapsed after three o'clock. The long hand covers 360 degrees in 60 minutes and the short hand covers 360 degrees in 720 minutes, so

The gap starts at 90 degrees and closes at the difference of the two rates:

Everything about this problem lives in that 11. Ninety degrees at five and a half degrees a minute is

and $\tfrac{4}{11}$ of a minute is $\tfrac{240}{11} = 21.8181$ seconds. Substituting back into equation (1), both hands read $\tfrac{1080}{11} = 98.1818$ degrees, which is the check that the answer is a coincidence and not merely a root of something.

Why 3:15 is so persuasive

At $t = 15$ equation (1) gives 90 degrees for the long hand and 97.5 for the short one. Seven and a half degrees of daylight, which on a dial of ordinary size is about two millimetres at the tip and easy to miss.

The reason the error is so natural is that the reflex answers a question about the longhand only. "When does the minute hand reach the 3" has the answer 3:15 and is a well-posed question. The problem asked about both hands, and the short hand has been moving the entire time at half a degree a minute, which over fifteen minutes is exactly the 7.5 degrees left over.

The same calculation at every hour

At $h$ o'clock the gap is $30h$ degrees, so the coincidence in that hour is at

For $h = 1$ that is 5 minutes 27.27 seconds; for $h = 3$ it reproduces equation (3); for $h = 6$ it is 32 minutes 43.64 seconds. The one to notice is $h = 11$, where equation (4) gives exactly 60 minutes. The coincidence that "belongs" to the eleven o'clock hour happens at twelve o'clock, so there is no moment strictly between eleven and twelve when the hands agree.

Eleven times, and per what

That missing coincidence is the whole reason for the following, which is stated wrongly more often than not. Consecutive coincidences are separated by

Twelve hours is 720 minutes, which is exactly 11 of those intervals, so there are eleven coincidences per twelve hoursand therefore twenty-two per day. The common phrasing "the hands overlap eleven times a day" is wrong by a factor of two.

The count is easier to feel than to prove. In twelve hours the long hand laps the dial twelve times and the short hand once, so the long hand gains eleven full laps on the short one, and each gained lap is one overtaking.

That is also where the elevenths in the answer come from. The two hands have speeds in the ratio 12 to 1, and every question about their relative position is answered by the difference of those numbers rather than by either one of them. Eleven is $12 - 1$, it divides nothing about a clock face, and so it produces the recurring decimals that make 3:16:21.8181 look like a strange answer to a tidy question. Any pair of hands with speeds in the ratio $n$ to 1 would give $n-1$ coincidences per cycle and fractions with $n-1$ underneath.

Where on the dial they meet

The coincidence positions are the angles $360n/11$ for $n = 0, 1, \dots, 10$, evenly spaced by 32.727 degrees. Since 32.727 is not a whole number of the dial's 30-degree hour steps, only $n = 0$ lands on a numeral. Every other coincidence happens between two marks, which is another reason the wrong answer feels right: people expect the hands to meet at something.

The same machinery answers the neighbouring questions without any new ideas. Write the signed gap as $g(t) = 90 - \tfrac{11}{2}t$. Setting $g = -180$ gives the moment the hands point opposite ways, $t = 270/5.5 = 49\tfrac{1}{11}$ minutes past three, and setting $g = -90$ gives the next right angle at $t = 180/5.5 = 32\tfrac{8}{11}$. Oppositions also happen twenty-two times a day, interleaved with the coincidences.

The assumption that holds it all up

Equation (1) assumes both hands move continuously. That is how a sweeping mechanical movement behaves, and it is not how many quartz movements behave. If the long hand jumps once a minute and sits still in between, then at every instant it is at a whole multiple of 6 degrees, while the short hand at those instants is at $90 + t/2$ with $t$ an integer. Equality would need $6t = 90 + t/2$ with $t$ a whole number, and $180/11$is not one. On such a clock the hands never coincide exactly after three o'clock at all.

One more consequence of the elevenths, for anyone who owns a watch with a second hand. All three hands coincide only at twelve o'clock exactly. The second hand joins the other two only when $720n/11$ is also a multiple of $60/59$ minutes, which forces $708n = 11k$, and since 11 does not divide 708 it must divide $n$. So $n = 0$ or $n = 11$, and both mean midnight or noon.

Sources and further reading

• The family of dial problems — Clock angle problem
• The rates in equation (1) — Angular velocity
• The subtraction in equation (2) — Relative velocity

The answer was confirmed by exact rational arithmetic and then found again without any formula: a scan of a full twenty-four hours at steps of a tenth of a second located the first crossing after three o'clock at 16.363636 minutes, counted exactly 11 coincidences in twelve hours and 22 in twenty-four, and measured the gap at 3:15 as 7.5 degrees.