@cobra91 I know that 3:16:21 don't work obviously from it says min hand 16 second hand 21 I wanted confirmation that the min and hour hand were exactly together at that time At 3:16:21.81, the hour and minute hands are between the '3' and '4' (and do line up exactly as you intended), but the second hand is between the '4' and '5'. I have always said the second hand will never be exactly at the right position but if you use 3:16:16 only 5 seconds away the hands appear together unless you use a magnifying glass. I had posted this earlier but for some reason it didn't show up.
Lost solution
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n9531l wrote:
Perhaps someone would like to consider the case of a real wristwatch, for which the second hand typically moves in discrete jumps. I have two of those, and for both, the second hand jumps once per tick. The Timex has five ticks per second, and the Bulova has six ticks per second. The minute and hour hands appear to move continuously, but presumably also jump once per tick in steps that are too small to see.
Assume all hands are straight up at midnight, and take as the start time the time close to Loyd's hand position that gives the least possible deviation from equal spacing. Is there ever a time when the deviation is less than it is just one tick (1/5 or 1/6 second) after the start time? If so, what is the first such time?
To solve this puzzle, you will have to define a measure for deviation from equal spacing, as I did previously, and include that definition as part of your solution.
Pure continuous hands movement is impossible in our quantum world.It must take discrete ticks.For the exact solution the puzzle creator must include the clock accuracy in his puzzle.
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#164
Who is ready for a new solution to Loyd's watch puzzle? I'm convinced that the puzzle as stated has no solution because of what was left unstated. So I will propose a modification of the puzzle statement, after which it will have an unambiguous solution. First, it is specified that the watch is one whose hands are all straight up every noontime and whose hands all move continuously. Next, the starting time is given as 2:54:35, so there's no need to try to guess it from the picture. Finally, the solver is asked to find the next time at which the angular separation between the hands is exactly the same as at the starting time, so there's no need to worry about what "appears" means.
I contend that there is one time which solves the puzzle as stated above. I will put off explaining the easy method of finding it to give those interested a chance to solve it on their own.
(Note: The solution will be the same if the clock ticks and the hands move at each tick in discrete steps.)
9:05:25
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#166
keju is first with the answer. I'll explain the solution method that occurred to me, and keju can say whether there's another way. Put another watch beside the first, and call it watch B. Watch B is identical to watch A in every respect except that it runs backwards. Its hands all move at their normal rates but counterclockwise. At noon all the hands of both watches point straight up. If a hand of watch A moves clockwise through a certain angle, the same hand of watch B will move counterclockwise through the same angle. So at a given time, the angular separation of all the hands is the same for both watches. When watch A gets to 2:54:35, watch B will be at 9:05:25, which is the solution.
Yup same method. Rotational mirror image. Only thing left is to show there is no other solution in between.
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#168
keju wrote:
Only thing left is to show there is no other solution in between.
That takes more work. In my case I had fortunately already done it 4 days ago with the exercise described in Comment #85. At that time I found the mirror-image answer without realizing its significance, and also found no other minima of M over the whole 12 hour range with the same value as at the starting time.
@n9531l I think we all agree that mathamatically even the watch pictured is not a perfect 120 degrees apart.If I am not mistaken M proves that the relation of the three hands position to each other is unique at any given time all through the twelve hour rotation. does M match 2:54:35 with 9:05:25 exactly? I think M will force you to say 12 hours....Again I have to say I couldn't have hoped to get such a well thought out elaboratly displayed set of solutions thanks. I googled watch mechanisms and this thread lost solutions came up on the page of sites.
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#170
@sameez1: Not sure I understand what you're asking. The angles between the hands at 2:54:35 are exactly the same as the angles between the hands at 9:05:25, as is clear from the backward running clock argument. M measures the deviation of the angles from exactly 120 degrees, so M has exactly the same value at both times (namely 0.8680555...). Of course M is not zero, because the angles are not exactly 120 degrees.
The maximum accuracy of that clock is half of a second for that given hands arrangement to be true.It is because the last tick of minute hand into 11 will have 30 second value.While the second hand currently in the last 30 second completion,thus those hands arrangement still make sense.Only if those hands seems a little bit shifted from 30 second notch that was caused by not accurate pinning of the hands.
For the accuracy of a quarter of a second those hands arrangement will be impossible.it is because the last tick of minute hand into 11 will have 15 seconds value while the second hand still pointing at more than 15 second into one minute completion.
and the answer is (maybe?) in about 30 seconds and a 1/4 tick, all hands equal distance apart at a reduced angle.
k2qs wins a chocolate fish.
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#173
Mr_Edt and k2qs need to say which clock they are talking about. Comment #166 was about a watch whose hands move continuously. Comment #159 mentioned two real wristwatches whose second hands move either 5 or 6 times per second. Various other clocks have been mentioned.
It would help if you say which comment you are responding to.
n9531l wrote:
Mr_Edt and k2qs need to say which clock they are talking about. Comment #166 was about a watch whose hands move continuously. Comment #159 mentioned two real wristwatches whose second hands move either 5 or 6 times per second. Various other clocks have been mentioned.
It would help if you say which comment you are responding to.
I refer to the original clock picture of the puzzle.It is to get a proper visualization to solve the problem.
I found out that all rotations fails. The best is still the first that I gave all the others are all worse.
But now I find that an mirroring the Original postition in the Y-axis give new prospects.
About 6 hours 10 minutes and 50 seconds plus a fraction of a second you get this time and clock.
about 5 and fraction minutes past Nine. ( about 5 minutes , 25 seconds and a fraction of a second.) This must be the solution. Because the angles of how the clock runs are before and after the hour excatly the same.( both 5 minutes and a fraction so the smal arm is just as must past the 9 as it is just before the 3 ) Now you can ad the fractions and gets the extra seconds. 
RubenHogenhout wrote:
6 hours 10 minutes and 50 seconds plus a fraction of a second
The analysis is very nice but what is the answer you would send in to the contest? This one?
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#177
RubenHogenhout wrote:
But now I find that an mirroring the Original postition in the Y-axis give new prospects.
Mirroring the hands across a vertical axis of the clock face is equivalent to having a clock that runs backwards. So you have indeed found the time at which the hand spacing is exactly the same as in the starting position. But how would you go about proving that this is the first time at which the spacing APPEARS to be the same?
I still believe the puzzle as stated has no solution, or several solutions, depending on the assumptions made by the solver.
n9531l wrote:
RubenHogenhout wrote:
But now I find that an mirroring the Original postition in the Y-axis give new prospects.
Mirroring the hands across a vertical axis of the clock face is equivalent to having a clock that runs backwards. So you have indeed found the time at which the hand spacing is exactly the same as in the starting position. But how would you go about proving that this is the first time at which the spacing APPEARS to be the same?
I still believe the puzzle as stated has no solution, or several solutions, depending on the assumptions made by the solver.
using this same idea couldn't you invert the clock in the mirror image for a sooner time? No I looked again it wouldn't work,seems you could use that as proof of no earlier time though.
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#179
@sameez1: No. That reflection puts the hands in a position that will never be reached unless they were misaligned to begin with.
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Perhaps someone would like to consider the case of a real wristwatch, for which the second hand typically moves in discrete jumps ...
Back in post 34, sameez1 considered a clock whose face was divided into 60 units, where every unit represents one tick. The second hand ticks once a second, the minute hand ticks once a minute, and the hour hand ticks once every 12 minutes.
Which is nothing like a real wristwatch.