## 03 March 2010

### Turning clocks upside down

Apparently this is a puzzle blog now.

This morning at 9:30 I picked up my watch and turned it upside down. It appeared to read 4:00. The hour hand was actually in the position it would be at at 3:30, of course. But the minute hand was pointing straight up, so the time must be on the hour. Since I could easily tell that the hour hand wasn't pointing directly to the right, I suppose my brain interpreted it as 4:00 instead of 3:00. Of course I did not think any of this consciously, but only reconstructed it after the fact; my thought process was more like "hey, 9:30 upside down looks like 4:00. that's weird."

Is it ever possible to interpret the hands of a clock, turned upside-down (i. e. rotated by 180 degrees), unambiguously as a time? (Fudging like my brain apparently did this morning does not count.)

Eric said...

Both the hour hand and the minute hand indicate minutes. Rotating 180 degrees changes the interpretation of the minute hand by 30 minutes, but it has no effect on the interpretation of the minutes being indicated by the hour hand. So, I'd say it's not possible.

Anonymous said...

Interpret the dial as a (looping) infinitely divisible strip of length 60. At time a:b, the hour hand is at position

h = 5a + b/60

and the minute hand is at position

m = b

Turning upside-down, we add or subtract 30 to or from both of these numbers. (signs independently)

h' = 5a + b/60 +- 30
m' = b +- 30

So, can there be a':b' with

5a' + b'/60 = 5a + b/60 +- 30
b' = b +- 30

5a' + b/60 +- 2 = 5a + b/60 +- 30

5a' +- 2 = 5a +- 30

The right hand side is divisible by 5, but the left one isn't, so there's never any way to pick an integer a' to make this work out.

jbb said...

Each hand travels in a circle, so its position is parametrized by S^1. Thus the set of all possible times throughout the day is a path along the torus S^1 x S^1. This is a (12,1) curve (I'm thinking 12 times through the hole in the bagel, once around the long way).

Rotating the watch corresponds to rotating each S^1 factor. Thus we rotate the bagel, exchanging front to back, and then roll it through its hole, exchanging inner radius with outer radius.

The first move sends the "time-curve" to itself, while the second moves it entirely off itself (because the curve only passes around once the long way, and so it's never behind itself, if you know what I mean). Hence no valid "upside-down" times.

Of course, this is just a different way of saying exactly what was said above by unapologetic.

A related (and similarly solved) problem is to assume the hour and minute hands are indistinguishable, and then ask how often you can determine the time of day from their (unordered) positions.

Anonymous said...

As an aside, I think it approximately works out to 5.5 hours behind (or 6.5 hours ahead). In India, we commonly say that flipping your watch shows you GMT. (India is GMT+5.5). :)

dogstar30 said...

It's not exact, but 10:49/9:54 is really close.

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