A Time puzzle to solve...

[FenceFurniture]

Looking at the face of a regular analogue clock, at what time between 1pm and 2pm is the minute hand precisely positioned over the hour hand?

No cheating, and you must supply a proof of your answer!

[419]

Can't calculate it precisely (I never went past Form 2 maths and wasn't any good at that), but I reckon it's around 1.05 pm.

Reasoning is that the hour hand has to move through 30 degrees (1/12th of 360 degrees) on a 12 hour clock face for each hour while the minute hand has to move through 360 degrees each hour, so 360/30 =12. An hour has 60 minutes so divided by 12 = 5, thus 1 pm + 5 = 1.05pm.

[rwbuild]

1:06:06 pm because the minute hand advances 1 minute exactly for every 1 full rotation of the second hand plus 6 sec for each complete rotation of the second hand

[419]

1:06:06 pm because the minute hand advances 1 minute exactly for every 1 full rotation of the second hand plus 6 sec for each complete rotation of the second hand

I might be missing something here (and quite likely am), but wouldn't that result in 1:05:06?

Admittedly, I'm wedded to my brilliant deduction that it's 1:05 as the base.

[ian]

Looking at the face of a regular analogue clock, at what time between 1pm and 2pm is the minute hand precisely positioned over the hour hand?

No cheating, and you must supply a proof of your answer!

It largely depends on the mechanism controlling the clock hands.

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.

[mic-d]

I did it this way

IMG_9968.jpg

if the answers not clear at the bottom it is 1:05.45pm
and I see I forgot to write in 'and the hour hand moves at 0.5º/min' when I did my neat rightup

[rwbuild]

I might be missing something here (and quite likely am), but wouldn't that result in 1:05:06?

Admittedly, I'm wedded to my brilliant deduction that it's 1:05 as the base.

No because at 1Pm the minute hand is at 12 and hour hand at 1 so when the minute hand moves to exactly the 1 position the hour hand has also moved proportionately past the 1

[FenceFurniture]

All I can say is none of the answers are right so far. The time will obviously have to be between 1:05 and 1:10 though.

And it's not a Mondial clock, because that would take the fun out out of it.

[FenceFurniture]

Extracting the correct parts of answers:

the hour hand has to move through 30 degrees (1/12th of 360 degrees) on a 12 hour clock face for each hour while the minute hand has to move through 360 degrees each hour, so 360/30 =12.

Correct. But if the minute hand (M) has already moved though 30°, how far has the hour hand (H) moved as well? Obviously the M catches up to the H before 1:10. Perhaps Trial & Error might get you closer to the accurate (as opposed to "around") time? Having said that, there is a very precise formula that can be derived.

[mic-d]

All I can say is none of the answers are right so far. The time will obviously have to be between 1:05 and 1:10 though.

And it's not a Mondial clock, because that would take the fun out out of it.

Can you point out the error in my maths please FF? if you need it converted to seconds 1:05.45pm would be 1:05:27pm

[FenceFurniture]

Can you point out the error in my maths please FF? if you need it converted to seconds 1:05.45pm would be 1:05:27pm

Ah! I read your answer as 45 seconds past 1:05. So in that case your answer is indeed correct Mick! It's actually closer to 1:05:28 because the seconds are 27.73.

As follows:

Every second the M hand moves 0.1°, and the H hand moves 1/12 of that, or 0.083333 repeater°
So if H starts at 30° and s is the seconds until M = H then
30 + 0.083333 * s = 0 + s * 0.1
Multiply both by 10 ► s = 300 + s*0.83333

or ► s = 300 + s/12
Multiply both by 12 ► 12s = 300*12 + s
so ► 11s = 3600
so ► s = 3600/11
= 327.73 seconds later
or 1:00 plus 327.73 seconds which is 1:05 and 27.73 seconds.

[mic-d]

Ah! I read your answer as 45 seconds past 1:05. So in that case your answer is indeed correct Mick! It's actually closer to 1:05:28 because the seconds are 27.73.

As follows:

Every second the M hand moves 0.1°, and the H hand moves 1/12 of that, or 0.083333 repeater°
So if H starts at 30° and s is the seconds until M = H then
30 + 0.083333 * s = 0 + s * 0.1
Multiply both by 10 ► s = 300 + s*0.83333

or ► s = 300 + s/12
Multiply both by 12 ► 12s = 300*12 + s
so ► 11s = 3600
so ► s = 3600/11
= 327.73 seconds later
or 1:00 plus 327.73 seconds which is 1:05 and 27.73 seconds.

If I account for my rounding error, it is actually 1:05:27.2727272727...pm. You need to check your maths 3600/11

Cheers
M

[FenceFurniture]

If I account for my rounding error, it is actually 1:05:27.2727272727...pm. You need to check your maths 3600/11

Cheers
M

Yeah, a misread of the calculator.

[ian]

By brute force -- and assuming the clock mechanism allows for continuous movement of the minute hand, and 1 second (= 6 degrees) jumps by the second hand ...

At 1:05
the minute hand has moved to 30 degrees around the clock face
the hour hand has moved 30/12 (= 2.5 degrees) from its starting position at 30 degrees, so it's total movement around the clock face is 32.5 degrees -- the minute hand is clearly behind the hour hand.

at 1:06
the minute hand has moved a further 6 degrees, so it is 36 degrees around the clock face,
the hour hand has moved to 33 degrees around the face -- clearly the minute hand is ahead of the hour hand.

at 1:05:30
the minute hand will be at a position 33 degrees around the clock face,
the hour hand will have moved to a position [30 + (30/60 x 5.5)] = 32.75 degrees around the clock face, marginally behind the minute hand, but can you determine the difference in the angle?

at 1:05:24
the minute hand will be at a position 32.7 degrees around the clock face,
the hour hand will be positioned at [30 + (30/60 x 5.4)] = 32.7 degrees


DONE

.
.
.
.

I hope

[FenceFurniture]

at 1:05:24
the minute hand will be at a position 32.7 degrees around the clock face,
the hour hand will be positioned at [30 + (30/60 x 5.4)] = 32.7 degrees


DONE

.

...not quite

M moves 360° in 60 minutes or 3600 seconds, so 0.1° per second.
H moves at 1/12 that velocity
After 324 seconds (at 1:05:24):
M will have moved 32.4° from 0°, and be in position 32.4° but
H will have moved 1/12 of that (2.7°) from 30° position to be at 32.7°.

That means that M is still 0.3° (or 3 seconds) behind H. That means the time has to be at least 1:05:27
However, following on from that, 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.