# Another clock problem

Mark Gonsoulin

Greenhorn

Posts: 1

posted 12 years ago

At 12 midnight.

Assuming the hands don't jump.

The s-hand will meat the others ca. one minute later.

But the m-hand moved more way than the h-hand, and keeps moving faster.

It will get close to the h-hand at about 1 pm, to be more precise at 1:05.

At 1:05:05 the s-hand reaches that position (nearly) too, but the m-hand moved a bit more away, than the h-hand.

This repeats every 1:05:05.

If you calculate the values:

You see the hands would meet again at 11:55:55 and then at 12:60:60 which is 01:01:00.

Assuming the hands don't jump.

The s-hand will meat the others ca. one minute later.

But the m-hand moved more way than the h-hand, and keeps moving faster.

It will get close to the h-hand at about 1 pm, to be more precise at 1:05.

At 1:05:05 the s-hand reaches that position (nearly) too, but the m-hand moved a bit more away, than the h-hand.

This repeats every 1:05:05.

If you calculate the values:

You see the hands would meet again at 11:55:55 and then at 12:60:60 which is 01:01:00.

Pete Kirkham

Greenhorn

Posts: 3

posted 12 years ago

assuming continuous movement:

position of second hand (360 degrees in 60 seconds) = t * 360 / 60 = t * 6;

position of minute hand (360 degrees in 60 * 60 seconds) = t * 360 / (60 * 60) = t / 10;

position of hour hand (360 degrees in 60 * 60 * 12 seconds)= t * 360 / (60 * 60 * 12) = t / 120;

hands meet when

(t*6) % 360 = (t / 10) % 360 = (t / 120) % 360;

or

(t*6) = (t / 10) - a * 360 = (t / 120) - b * 360;

where a, b are integers

(t*6) = (t / 10) - a * 360 ; => 59 t + 3600 a = 0;

(t*6) = (t / 120) - b * 360; => 719 t + 43200 b = 0;

gives two equations that we can eliminate t in:

42421 t + 2588400 a = 0

42421 t + 2548800 b = 0

2588400 a - 2548800 b = 0

so

a = (2548800 / 2588400) b;

=>

a = (708 / 719) b;

so b increments in multiples of 719, and a in 708

back to:

59 t + 3600 a = 0;

gives

t = (708 * 3600 / 59) k, k = {0,1,2...};

t = 43200 k;

every 43200 seconds, or 12 hours.

position of second hand (360 degrees in 60 seconds) = t * 360 / 60 = t * 6;

position of minute hand (360 degrees in 60 * 60 seconds) = t * 360 / (60 * 60) = t / 10;

position of hour hand (360 degrees in 60 * 60 * 12 seconds)= t * 360 / (60 * 60 * 12) = t / 120;

hands meet when

(t*6) % 360 = (t / 10) % 360 = (t / 120) % 360;

or

(t*6) = (t / 10) - a * 360 = (t / 120) - b * 360;

where a, b are integers

(t*6) = (t / 10) - a * 360 ; => 59 t + 3600 a = 0;

(t*6) = (t / 120) - b * 360; => 719 t + 43200 b = 0;

gives two equations that we can eliminate t in:

42421 t + 2588400 a = 0

42421 t + 2548800 b = 0

2588400 a - 2548800 b = 0

so

a = (2548800 / 2588400) b;

=>

a = (708 / 719) b;

so b increments in multiples of 719, and a in 708

back to:

59 t + 3600 a = 0;

gives

t = (708 * 3600 / 59) k, k = {0,1,2...};

t = 43200 k;

every 43200 seconds, or 12 hours.