Math Question, involves clocks...?

Question

The time is now 12:00 PM. When is the next time the clock hands will be on top of each other. All 3 hands are involved. (Hour, minute, and second hand)

And is there a formula to solve this, 'cause i think i have to have that.

Answer 1

There is a formula for it, but it is not directly solvable. The minute hand travels 360º in 60 min, and the hour hand travels 360º in 12*60=720 min. The position of the hour hand is given by t*(360/720) or t/2, for t = minutes. The position of the minute hand is given by t*(360/60) or 6*t; But after 360º the angle starts at zero again; this is expressed by the modulo function; a mod b is the remainder from dividing a by b. In our case, the position of the minute hand is (6*t) mod 360; the hands overlapping equation is then

(6*t) mod 360 = t/2, or t = 2 * (6*t) mod 360.

This is a discontinuous function that can be solved by successive approximation.

A plot of this equation is shown here:

http://img89.imageshack.us/img89/968/clockhandska9.png

The times where the hands overlap are the point where the dashed line (t/2) intersects the solid lines (sloping portion only). The time for the first overlap is 65.41 minutes after start, or 1:05:24.6.

EDIT: There is a simpler formula. It is derived as follows:

At the top of the hour, the hour hand point exactly to the hour. Set the hour be n, 1 ≤ n ≤ 11; The minute hand moves at 6º per min. Say during the time it takes for the minute hand to reach the hour hand, the hour hand moves dº. Therefore the minute hand takes 5*n * d/6 minutes to reach the hour hand. The hour hand moves .5 deg/min, therefore the time for it to move d is 2*d minutes.

5*n + d/6 = 2*d

5*n = d(2-1/6) = d*(11/6)

d = (30/11)*n

The total travel of the minute hand is 30*n + (30/11)*n = (360/11)*n

At 6º per min, this gives in minutes (60/11)*n.

The time readings for each hour are the n : 5.454*n

n.........Time
1..........1:5.45
2..........2:10.91
3..........3:16.36
etc


EDIT:

I just read that you need to include the second hand as well. The time calculated from the above formula occurs when the second hand points to 12 on the clock. To include the movement of the second hand, we have to add to the motion of the minute and hour hands an amount equal to T/60, where T is the new position of the hour and minute hands, since T/60 is how much the minute hand has to move. Thus the new T = old T + T/60. The old T is n*(60/11). So

T = n*(60/11) + T/60; T*(1 - 1/60) = n*(60/11)

T = n*(60/11) / (1 - 1/60)

The new times come out n : 5.547*n

n.................Time

1......1:05.547 or 1:05:32.82
2......2:11.094 or 2:11:05.64
3......3:16.641 or 3:16:38.46
4......4:23.188 or 4:23:11.28
etc.

Answer 2

I assume all the hands move continuously and since a circle is 360 degree.

Since the hour hand move 360 degree every 12 hours, ie, it moves 1/120 degree every second.
Since the minute hand move 360 degree every hour, ie, it moves 1/10 degree every second.
Since the second hand move 360 degree every minute, ie, it moves 6 degree every second.

Given that the 12:00pm is position 0 and time 0.

So, your question becomes at what time, all the hands will at the same position?

I have tried to plug that into excel. I don't think they will be at the same position until 12:00:00am. 01:05:05 seems to be reasonable, but the time for the minute hand pass the hour hand, and the time for the second hand pass the minute hand is different.

I am not sure if this is the answer you are looking for, but I hope this will help you think.

Answer 3

10 minutes if the 12-hour device is used. If the time exhibits 3 digits, then the 1st and final would desire to be equivalent for a palindrome sequence. the middle one purely alterations each and every 10 minutes, so it takes 10 minutes between palindromes. If the 24-hour device is used, the shortest time is 28 minutes (23:32 ? 00:00).

Answer 4

01:05.05 AM

I don't know about a formula.. just took an 'ol clock and spun the hands around.

Answer 5

actually 00:01:01 Cause the short hand moves with every rotation of the long one

Answer 6

1:06:06