The three clock arms problem revisited

This is a problem that Slvador thought about when we were 7th graders. He wanted to come up with an equation to find when the three arms of a clock intersect. Slvador would you please put the equation?

Anyways, today I was trying to approach this problem graphically, and I found something interesting. The probability that the three arms would intersect is very low, except when the time is 12:00!

I will tell you how! First I assumed the following:

1- The thickness of the three arms is zero, just like a line in a 2D plan.
2- The arms are continuously in motion (quartz clock), they do not do the tick-tack thingy, and we have such a clock actually.

Ok, now we can note that for a single cycle of the hours arm, there will be 12 cycles of the minutes arm and 720 for the seconds arm. This can be graphically represented on a 2D plan, with the cycle being represented on the Y axis, while the time is represented on the X axis in seconds. Starting with the seconds arms, we will have 720 parallel lines that represent the motion of the seconds arm motion like saw teeth, and 12 parallel lines representing the motion of the minutes arm, also like a saw teeth, and finally one line that represents the hours arm.

Of course, the seconds arm will intersect with the hours arm 720 times, and with the minutes arm the same amount of times. The minutes arm will intersect with the hours arm 12 times.

Now let’s think about it, what are the chances that 11 points of those 1452 points could represent the solution for the system? Please note my assumptions and answer. I think it is very low, but we need to try to see if we can find any points that can solve the system of equations that is represented graphically above.

Posted in Uncategorized

2 thoughts on “The three clock arms problem revisited

  1. If you took y axis as angle, then there will the saw tooth graphs for each (saw tooth means look like this : /|/|/|/|/|)

    I don’t think the three clocks every meet except at 12:00. to prove it, i will use my old beautiful equation for the meeting of the hours and minutes which is

    60/11 * hour = minutes of the meeting

    for e.g if we want to find out when do they meet between the 2:00 and 2:59, so hour = 2 , so 60 / 11 * 2 = 10.909090909090909090909090909091
    so at 2 and 10 minutes and 0.90909090909090 * 60 seconds the hours and minutes meet
    now at 0.9090909090 *60 = 54.545454545454545454
    so the seconds will be really far from the two meetings

    I can easily do that for the 12 hours and prove they only meet at 12:00

  2. I meant by cycle on the y-axis that 0 to 360 i on the Y axis! and the saw tooth is more like / / / / / / /

    yea, that is what I ment by very low probability, they never meet, except at 12:00!

Leave a Reply

Your email address will not be published. Required fields are marked *