A Time puzzle to solve...

We know that both H and M hands are constantly moving, which means that at 1:05:27.2727 seconds the centre of M moves over the centre of H, but it is only for an instant. We have also established that M moves at 12x the velocity of H, but neither are ever still.

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Originally Posted by ian

Is your analogue clock a Mondial one as used by the Swiss railways?
In those the minute hand jumps forward 6 degrees as the second hand reaches the 12 o'clock position, but the hour hand creeps forward at a regular rate of 30 degrees per hour. Given that, the hour hand is never precisely above the hour hand between 1 and 2 o'clock.
So the question is moot.

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This is not true. My opening statement of this post means that the question is indeed not moot for a Mondial clock. In this case, M moves at an obviously much faster speed than in a regular clock, but in both cases the M hand is moving across H – just at a different speed. That means there is indeed an instant in time when the hands are perfectly centred together – it is just briefer by the ratio of the speeds of the two different M hands..

If I knew the velocity of M (when it moves, or even how long it takes to move) in a Mondial clock of choice I believe I could work out that precise instant of time to any choice of decimal places. :D

Ive been working 2 days straight serving cruise ship people, Iam tired and as long as my eyes are closed before 1AM comes I couldnt care :D

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Originally Posted by FenceFurniture

in those additional 3 seconds H has moved another (0.1 * 3)/12° or 0.025°,
hence a further ~¼ of a second (actually 0.272727 to 6 decimal places, but let's round it to 0.3) has to be added to the time for Swiss precision.

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So, to be even more precise we can say it occurs at 5 5/11 minutes past 1? :D

Slightly different way, using simultaneous equations

Using T to represent the time in minutes the rotation of the hour hand on the clock face is

H = T / (60 x 12) + 1/12

1/12 representing the starting position of 1pm

Minute hand is just M = T/60. One revolution every 60mins.

When aligned M = H, so T/60 = T/ (60×12) + 1/12

This solves to T = 60/11
Giving
M = 1/11

This is 5.4545... minutes in decimal or 5min 27 and 27/99 seconds

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Originally Posted by taz01

Slightly different way, using simultaneous equations

Using T to represent the time in minutes the rotation of the hour hand on the clock face is

H = T / (60 x 12) + 1/12

1/12 representing the starting position of 1pm

Minute hand is just M = T/60. One revolution every 60mins.

When aligned M = H, so T/60 = T/ (60×12) + 1/12

This solves to T = 60/11
Giving
M = 1/11

This is 5.4545... minutes in decimal or 5min 27 and 27/99 seconds

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That's basically the way I did it

On my antique mantle clock, the hands would be at 6:30 ..... because the clock is buggered and the hands are always at 6:30.

Another project.

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Originally Posted by FenceFurniture

If I knew the velocity of M (when it moves, or even how long it takes to move) in a Mondial clock of choice I believe I could work out that precise instant of time to any choice of decimal places. :D

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Hi Brett

to assist.
the minute hand on a Swiss Railway Mondaine clock (I shouldn't trust my very faulty memory when it comes to spelling) -- jumps forward 6 degrees over 1/16th of a second when the second hand reaches the 12 o'clock position. The second hand jumps forward 6 degrees every second.

BTW, the vibration frequency of the quartz crystal found in most analogue clocks is 32.768 kHz -- thousand cycles per second. So if you are wishing to be precise with the time the minute and hour hands are precisely aligned, you will need to factor in the quartz crystal vibration frequency.

Later.

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Originally Posted by ian

BTW, the vibration frequency of the quartz crystal found in most analogue clocks is 32.768 kHz -- thousand cycles per second. So if you are wishing to be precise with the time the minute and hour hands are precisely aligned, you will need to factor in the quartz crystal vibration frequency.

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The original question doesn't limit the type of analog clock to a quartz movement, unless one interprets 'regular' in the question to mean as regular as a quartz movement as distinct from a 'regular' 12 number clock face.

If the latter, then it includes clocks with mechanical movements, the accuracy of which depend upon the quality of manufacture of the gears and springs etc. If so, isn't it impossible to calculate the alignment of the hands on a given mechanical clock without knowing further details of its deviation from a theoretical standard equivalent to a quartz movement as the mechanical hands may be ahead of or after the hands on a quartz movement?

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Originally Posted by GraemeCook

On my antique mantle clock, the hands would be at 6:30 ..... because the clock is buggered and the hands are always at 6:30.

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Confirms the old comment that even a broken clock is correct twice a day.

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Originally Posted by 419

If so, isn't it impossible to calculate the alignment of the hands on a given mechanical clock without knowing further details of ....................................................................................................................

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Ummmm, I actually don't think I care.



It's just a bit of fun....no need to join in bogging it down in pedantry.

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Originally Posted by FenceFurniture

Ummmm, I actually don't think I care.



It's just a bit of fun....no need to join in bogging it down in pedantry.

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Sorry, I must have missed the bits in this thread that weren't pedantic.

Well you didn't miss this, and it's most definitely not pedantic.

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I reckon it's around 1.05 pm.

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Like I say, it's just supposed to be a bit of fun.

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Originally Posted by FenceFurniture

Like I say, it's just supposed to be a bit of fun.

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Fair enough.

For a bit of fun, could you explain how your not pedantic calculations and assessments of contributors' various answers apply equally to predicting the position of the hands on a mechanical analogue clock of unknown manufacture and unknown accuracy?

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Originally Posted by 419

Fair enough.

For a bit of fun, could you explain how your not pedantic calculations and assessments of contributors' various answers apply equally to predicting the position of the hands on a mechanical analogue clock of unknown manufacture and unknown accuracy?

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No. Because I couldn't be fagged. :;