Aunt Mabels clock shows 132 situations you can’t see what time it is. How do we calculate these 132 possibilities?
Take two normal clocks. Let the angle between the hour hand and the vertical be φ1 and between the minute hand and the vertical be α1 for the first clock and for the second clock φ2 and α2 respectively. The hour hand covers in 1 minute an angle of 6°, the minute hand covers in that time ½°. Let the time clock1 shows be a1 minutes past n1 o’clock and let clock 2 point to a2 minutes past n2 o’clock. So we can write:
φ1 = 6a1 and
α1 = 30n1 + ½ a1
(the minute hand covers 30° in 1 hour, so 30n1° in n1 hour and ½ a1° in a1 minutes). We obtain for the second clock:
φ2 = 6a2 and
α2 = 30n2 + ½ a2.
Aunt Mabels clock will confuse us if
φ1 = α2 and
φ2 = α1.
This gives
6a1 = 30n2 + ½ a2 and
6a2 = 30n1 + ½ a1.
From the first equation:
a1 = 5n2 + (1/12)a2 (i).
Fill in into the second equation:
6a2 = 30n1 + (5/2)n2 + (1/24)a2 or
a2 = (720/143)n1 + (60/143)n2 (ii).

With i and ii we can calculate all possibilities.
An example: take n1 = 4 and n2 = 8. ii gives a2 = 23,50 and i gives a1 = 41,96. Aunt Mabels clock can’t decide between 41,96 minutes past 4 and 23,50 minutes past 8.

Remark 1: You can easily find the time the hands are on top of each other:
a = (60/11)n.
Remark 2: You can uniquely determine φ when you know α.
φ = 12(α – 30n), n such that 0 ≤ α – 30n < 30 .

## Clock confusing

The confusing clock.

Aunt Mabels clock shows 132 situations you can’t see what time it is. How do we calculate these 132 possibilities?

Take two normal clocks. Let the angle between the hour hand and the vertical be φ1 and between the minute hand and the vertical be α1 for the first clock and for the second clock φ2 and α2 respectively. The hour hand covers in 1 minute an angle of 6°, the minute hand covers in that time ½°. Let the time clock1 shows be a1 minutes past n1 o’clock and let clock 2 point to a2 minutes past n2 o’clock. So we can write:

φ1 = 6a1 and

α1 = 30n1 + ½ a1

(the minute hand covers 30° in 1 hour, so 30n1° in n1 hour and ½ a1° in a1 minutes). We obtain for the second clock:

φ2 = 6a2 and

α2 = 30n2 + ½ a2.

Aunt Mabels clock will confuse us if

φ1 = α2 and

φ2 = α1.

This gives

6a1 = 30n2 + ½ a2 and

6a2 = 30n1 + ½ a1.

From the first equation:

a1 = 5n2 + (1/12)a2 (i).

Fill in into the second equation:

6a2 = 30n1 + (5/2)n2 + (1/24)a2 or

a2 = (720/143)n1 + (60/143)n2 (ii).

With i and ii we can calculate all possibilities.

An example: take n1 = 4 and n2 = 8. ii gives a2 = 23,50 and i gives a1 = 41,96. Aunt Mabels clock can’t decide between 41,96 minutes past 4 and 23,50 minutes past 8.

Remark 1: You can easily find the time the hands are on top of each other:

a = (60/11)n.

Remark 2: You can uniquely determine φ when you know α.

φ = 12(α – 30n), n such that 0 ≤ α – 30n < 30 .

Hub Boreas