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Stopping and Turning back hands of time

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A peculiar Grandfather's clock has both its hour and minute hands
running in proper rates. Starting at 12 o'clock both runs clockwise.
But the hour hand stops when the minute hand passes over it.
And the minute hand turn its direction back when it is over the stopped
hour hand causing the hour hand to run clockwise again.
post-53237-0-63835300-1361389377_thumb.j
How long will it take for both hands to be on 12 o'clock again?
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Posted · Report post

A peculiar Grandfather's clock has both its hour and minute hands
running in proper rates. Starting at 12 o'clock both runs clockwise.
But the hour hand stops when the minute hand passes over it.
And the minute hand turn its direction back when it is over the stopped
hour hand causing the hour hand to run clockwise again.
How long will it take for both hands to be on 12 o'clock again?

24 hours

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Posted · Report post

If I'm reading this right...

23 hours. It would normally take 12 hours to get back to 12 o'clock, but the minute hand does one extra revolution each hour 1-11 each taking an hour for 11 more hours.

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Posted (edited) · Report post

I see the clock reads 10:12. Do we start now? Or do we start at 12:00?

Also, I'm a little unclear what happens when the minute hand and hour hand meet: (When you said the minute hand "passes over" the hour hand, I assume that the minute hand continues in its direction.)

* minute hand going clockwise (CW), hits moving hour hand--hour hand stops, minute hand continues CW

* minute hand going CW, hits stopped hour hand--hour hand starts, minute hand goes Counterclockwise (CCW)

* minute hand going CCW, hist moving hour hand--hour hand stops, minute hand continues CCW

* minute hand going CCW, hits stopped hour hand--hour hand starts, minute hand goes CW.

Edited by CaptainEd
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Posted · Report post

I see the clock reads 10:12. Do we start now? Or do we start at 12:00?

Also, I'm a little unclear what happens when the minute hand and hour hand meet: (When you said the minute hand "passes over" the hour hand, I assume that the minute hand continues in its direction.)

* minute hand going clockwise (CW), hits moving hour hand--hour hand stops, minute hand continues CW

* minute hand going CW, hits stopped hour hand--hour hand starts, minute hand goes Counterclockwise (CCW)

* minute hand going CCW, hist moving hour hand--hour hand stops, minute hand continues CCW

* minute hand going CCW, hits stopped hour hand--hour hand starts, minute hand goes CW.

That is the way I read the OP Cap'n.

But it may be the case that a forward-moving minute hand stops the hour hand, while a backward-moving minute hand starts the hour hand again. I think one of these interpretations precludes and answer, and that might make the OP as it stands a double puzzle.

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Posted (edited) · Report post

Tricky!

from the pictured position, it will take 2:38 more before both hands are pointing at 12. However, the minute hand will still be moving CCW. In one hour more, they will once again be pointing at 12, and both hands will be moving CW.

When both hands are together and moving CW,

it takes 1 hour 5 minutes + epsilon to meet again, when HH stops, MH continues CW.
One hour later, MH meets again and reverses direction (to CCW), while HH starts again.
55 minutes - epsilon later, they meet again, HH stops, MH continues CCW
one hour later, MH meets again and reverses direction to CW, while HH starts again.

In other words, in exactly 4 hours, the two hands moved exactly one sixth of the circle (that is, they are both pointing exactly at the number 2 more than last time.

So it would take 24 hours for the hands to get together at 12 with both hands moving CW. But only 23 hours for them to get together at 12, because that time, MH is still going CCW, and HH stops.

So, from a standing start--both hands at 12, both moving CW, it takes 23 hours to get back to both hands at 12 , 24 hours to do so with both hands moving CW. I agree with Rob's numbers.


BobbyGo may have envisioned the full cycle, but missed that last backward run.

If we are to tell how long from when the clock has HH past 10, MH at 12, we notice several things:
* HH must be moving--it only stops at a little after the odd hours, and exactly on the even hours; it's moving at other times
* MH must be moving CW--same condition
* Therefore it will be (3 hours - 22 minutes) before the hands are together at 12 and 12 (and that will be the 23 hour situation).

Edit: I agree, Bonanova, I'm not totally clear on the mechanism. A third interpretation might be that the MH reverses whenever it hits the HH. I'll take one of Y-san's Excedrins...

Edited by CaptainEd
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@captainEd, the pic is not the start, both @ 12 o'clock is the start where the hands are moving clockwise, eventually the MH overtakes the HH causing it to stop but the MH keeps going until it reached the stopped HH again where then it reverses its direction (like bumping) while causing the HH to move again(resume running clockwise) . No double interpretations is intended here.

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If I'm reading this right...

23 hours. It would normally take 12 hours to get back to 12 o'clock, but the minute hand does one extra revolution each hour 1-11 each taking an hour for 11 more hours.

Both hands should point at 12
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@captainEd, the pic is not the start, both @ 12 o'clock is the start where the hands are moving clockwise, eventually the MH overtakes the HH causing it to stop but the MH keeps going until it reached the stopped HH again where then it reverses its direction (like bumping) while causing the HH to move again(resume running clockwise) . No double interpretations is intended here.

OK, TSLF, thanks.

23 hours.

At that moment, MH is moving counterclockwise. One hour later, MH meets again at (HH=12, MH=12). Analysis as given in my answer above

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There is a pattern here..

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There's more than meets the eye here.

I've gone back and forth and concluded a couple things.

The answer is not simply 23 hours.

The answer might be "a very long time, if ever."

I have some more work to do on this answer.

It's a very nice puzzle.

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@captainEd, the pic is not the start, both @ 12 o'clock is the start where the hands are moving clockwise, eventually the MH overtakes the HH causing it to stop but the MH keeps going until it reached the stopped HH again where then it reverses its direction (like bumping) while causing the HH to move again(resume running clockwise) . No double interpretations is intended here.

From the last part where the MH 'bumps' the HH making it move again. The MH travels CCW until it hits the HM again making it stop then makes a full revolution CCW till it 'bumps' the HH which starts moving again and the MH begins traveling CW?

There's more than meets the eye here.

I've gone back and forth and concluded a couple things.

The answer is not simply 23 hours.

The answer might be "a very long time, if ever."

I have some more work to do on this answer.

It's a very nice puzzle.

I agree now that I understand the OP a little more.

the answer will be a whole number of hours. The real problem is to find out how many times the HH stops which is the number of extra hours the MH goes around the clock.

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Here's what I think.


A normal clock has eleven times between midnight and noon when H and M coincide, counting only midnight or noon. They are equally spaced, and occur 12/11 hours apart. If this were a normal clock, we should add 1-hour waits at each of these and come up with 12+11 = 23 hours for the total elapsed time.

But if we start the clock, so to speak, with the hands leaving midnight, and we stop it with the hands reaching noon, then we should exclude them both and add only ten 1-hour waits, for a total elapsed time of 22 hours.

Both these answers are wrong. We don't have a normal clock.

When H commences motion at the 12/11 point [when the clock reads 1:05:27], M is moving backward. They next coincide 12/13 hour later [not 12/11 hours later - H and M are moving "toward" each other this time] when the clock reads 2:00:50.4. Note that 12/11 + 12/13 = 288/143 = ~2.014 is slightly greater than 2. After an hour wait, a cycle is complete. The cycle then repeats.

Disturbing. A small number of cycles no longer gets us to a reading of 12:00:00.

So, with M bouncing back and forth and H jumping hesitantly forward, when will the hands next be upward vertical? There are two possible answers. First, when a multiple of 288/143 is also a multiple of 12. These are the even-numbered points of coincidence, at the ends of complete cycles. Second, they could coincide mid-cycle. That is, when 12/11 plus a multiple of 288/143 is a multiple of 12.

[288/143]n = 12m -- or -- 12/11 + [288/143]n = 12m

Look at the first case, it's easier.

288n = 1716m is true when n=1716 and m=288.

The total elapsed time would be 12m + 2n = 6886 hours or just short of 287 days.

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Oh dear! My epsilons didn't cancel. Oops. Way to go, Bonanova!

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Here's what I think.

A normal clock has eleven times between midnight and noon when H and M coincide, counting only midnight or noon. They are equally spaced, and occur 12/11 hours apart. If this were a normal clock, we should add 1-hour waits at each of these and come up with 12+11 = 23 hours for the total elapsed time.

But if we start the clock, so to speak, with the hands leaving midnight, and we stop it with the hands reaching noon, then we should exclude them both and add only ten 1-hour waits, for a total elapsed time of 22 hours.

Both these answers are wrong. We don't have a normal clock.

When H commences motion at the 12/11 point [when the clock reads 1:05:27], M is moving backward. They next coincide 12/13 hour later [not 12/11 hours later - H and M are moving "toward" each other this time] when the clock reads 2:00:50.4. Note that 12/11 + 12/13 = 288/143 = ~2.014 is slightly greater than 2. After an hour wait, a cycle is complete. The cycle then repeats.

Disturbing. A small number of cycles no longer gets us to a reading of 12:00:00.

So, with M bouncing back and forth and H jumping hesitantly forward, when will the hands next be upward vertical? There are two possible answers. First, when a multiple of 288/143 is also a multiple of 12. These are the even-numbered points of coincidence, at the ends of complete cycles. Second, they could coincide mid-cycle. That is, when 12/11 plus a multiple of 288/143 is a multiple of 12.

[288/143]n = 12m -- or -- 12/11 + [288/143]n = 12m

Look at the first case, it's easier.

288n = 1716m is true when n=1716 and m=288.

The total elapsed time would be 12m + 2n = 6886 hours or just short of 287 days.

Great, the number 143 is there bonanova...thanks for the solution ,best one.

HH pattern : long run + stop + short run + stop
65.45454545454550 min + 60 min + 55.38461538461540 min + 60 min = 4.01398601398601 hrs
32.7272727272727 deg + 0 deg + 27.6923076923077 deg + 0 deg = 60.4195804195804 deg
For smallest whole number of revolutions Rev = 60.4195804195804 deg x N/360
Rev = 24 cycles N=143 patterns Time= 4.01398601398601 hrs x 143 = 574 hrs

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If I understood the statement of the problem right:

1). The hour hand changes its state from stopped to moving clockwise, or from moving to stopped every time the minute hand passes over it.

2). The minute hand continues moving in the same direction as before when it passes over the moving hour hand and reverses its direction when it passes over the stopped hour hand.

The full cycle is:


(12/11 + 12/13 +2) hours.

To get to an exact hour mark that cycle must be multiplied 11*13=143 times

(12/11 + 12/13 +2)*143=574

Now 574 has a remainder of 10 when divided by 12. Or it can be viewed as remainder = -2.

Thus it takes 6 of those "exact hour" cycles to get to 12:

574*6=3444.

Remembering that on the last hour of the cycle the minute hand is making a backward circle while the hour hand stays put, thus repeating the same time:

3444 - 1 = 3443 hours. After that much time the hour and the minute hands will meet at 12.

There is another possibility for intermittent exact hour mark meetings.

Let the clock run to its first stop at 12/11 hours, then start adding full cycles. This way we need a multiple of 13 full cycles where we may run into an exact hour mark meeting. Playing with modular arithmetic we find that starting at 12/11 after 65 cycles an exact hour mark meeting will occur:

12/11 + (12/11 + 12/13 +2)*65 = 262.

That again gives the remainder of 10 (-2) when divided by 12 and we need 6 of those big cycles to get to the 12-hour mark:

262*6 = 1572 hours. When the hour and minute hand meet at 12 again.

Here during the last round of the last cycle the hour hand is moving, so we may not subtract an hour. Still it is less than 3443, so that should be the answer. There is no point in trying to mix the 574 and 262- hour cycles since they have the same remainder from the division by 12.

1572 hours.

Edited by Prime
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Every 60.4195804195804 deg both hands are aligned and cw. If a divider is used, 143 arc steps along the circle will point one of its arms to 360 deg..

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Every 60.4195804195804 deg both hands are aligned and cw. If a divider is used, 143 arc steps along the circle will point one of its arms to 360 deg..

I see, I mixed up the adavance of the hour hand with the elapsed time for each cycle. And I calculated period incorrectly for the second part in my previous post.

Still, after the corrections, I get a different answer. My hour and minute hands will meet a lot sooner.

Why rejecting the occurences where the clock hands meet at 12 in a mid-cycle?

I see nothing in the OP prohibiting that. In fact, the clock hands will meet again at 12 on a mid-cycle before they do so after a whole number of cycles.

1.The full cycle is (12/11 + 12/13 +2) = 574/143 hours elapsed time.

The hour hand advances (12/11+12/13) = 288/143 hours

After 143 such cycles, the hour hand advances 288 hours, which is divisible by 12. (288/12 = 24).

Elapsed time after 143 cycles is 574 hours. And, because the last step of the cycle the hour hand stayed put for exaclty one hour, there was a meeting at the same spot one hour prior. Thus after 573 hours from the start the hour and minute hand meet at the 12-hour mark.

2. The hour and minute hands also meet at exact hour mark after 65 full cycles plus 12/11 of an hour. (12/11 + 12/13 + 2)*65 + 12/11 = 132.

It so happens that 132 is also divisible by 12. 132/12 = 11

Thus the first meeting at 12 hour mark will take place exactly after (12/11 + 12/13 + 2)*65 + 12/11 = 262 hours. That is a lot sooner than 573 hours calculated in step 1.

All that is relatively easy to figure out using regular fractions and modular arithmetic. But it does not hurt to verify using the angles in degrees in decimals as calculated by TSLF.

The first step of the cycle the hour hand travels (12/11)*360/12 = 32.72(72).... degrees.

On the third cycle step, the hour hand travels (12/13)*360/12 = 27.69231... degrees.

Making the full cycle: 32.7272(72) + 27.692308 = 60.41958...

After 65 full cycles plus the first step of the next cycle the hour hand travels 60.41958.. * 65 + 32.7272(72) = 3960 (exactly). Which also happens to be divisible by 360. 3960 / 360 = 11.

And so all 12-hour mark meetings' elapsed times are as following:

262 + 574*n

263 + 574*n

573 + 574*n

574 + 574*n

The first meeting is after 262 hours.

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Every 60.4195804195804 deg both hands are aligned and cw. If a divider is used, 143 arc steps along the circle will point one of its arms to 360 deg..

I see, I mixed up the adavance of the hour hand with the elapsed time for each cycle. And I calculated period incorrectly for the second part in my previous post.

Still, after the corrections, I get a different answer. My hour and minute hands will meet a lot sooner.

Why rejecting the occurences where the clock hands meet at 12 in a mid-cycle?

I see nothing in the OP prohibiting that. In fact, the clock hands will meet again at 12 on a mid-cycle before they do so after a whole number of cycles.

1.The full cycle is (12/11 + 12/13 +2) = 574/143 hours elapsed time.

The hour hand advances (12/11+12/13) = 288/143 hours

After 143 such cycles, the hour hand advances 288 hours, which is divisible by 12. (288/12 = 24).

Elapsed time after 143 cycles is 574 hours. And, because the last step of the cycle the hour hand stayed put for exaclty one hour, there was a meeting at the same spot one hour prior. Thus after 573 hours from the start the hour and minute hand meet at the 12-hour mark.

2. The hour and minute hands also meet at exact hour mark after 65 full cycles plus 12/11 of an hour. (12/11 + 12/13 + 2)*65 + 12/11 = 132.

It so happens that 132 is also divisible by 12. 132/12 = 11

Thus the first meeting at 12 hour mark will take place exactly after (12/11 + 12/13 + 2)*65 + 12/11 = 262 hours. That is a lot sooner than 573 hours calculated in step 1.

All that is relatively easy to figure out using regular fractions and modular arithmetic. But it does not hurt to verify using the angles in degrees in decimals as calculated by TSLF.

The first step of the cycle the hour hand travels (12/11)*360/12 = 32.72(72).... degrees.

On the third cycle step, the hour hand travels (12/13)*360/12 = 27.69231... degrees.

Making the full cycle: 32.7272(72) + 27.692308 = 60.41958...

After 65 full cycles plus the first step of the next cycle the hour hand travels 60.41958.. * 65 + 32.7272(72) = 3960 (exactly). Which also happens to be divisible by 360. 3960 / 360 = 11.

**I guess this is the closest when both are almost at 12' because (10.9090909090909 rev = 60.41958041958040deg X 65 / 360) : acceptable as 11.

Nice find thanks. I take the more exact (24.0000000000000 rev = 60.41958041958040deg X 143/360 deg) out of my excel tabulation results

And so all 12-hour mark meetings' elapsed times are as following:

262 + 574*n

263 + 574*n

573 + 574*n

574 + 574*n

The first meeting is after 262 hours.

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This is the answer: Good one Prime! Thanks.

After 65 full cycles plus the first step of the next cycle the hour hand travels 60.4195804195804

* 65 + 32.7272727272 = 3960 (exactly). Which also happens to be divisible by 360. (3960 / 360 = 11 rev.) I was mistaken in my last post this one is OP fine.

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