Hands are same length, what time is it?

sonhouse

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05 Apr '10 13:07

uzless
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08 Apr '10 02:33
Originally posted by AThousandYoung
Not true; the position of the hour hand between two numbers depends on the position of the minute hand. You would not confuse 9:30 with 6:45 because the minute hand points directly at the 6 in the first case and in the second the hour hand would almost be at 7.
By your logic, are there any example you can state where you can NOT tell the time, other than 12;00?

Perhaps I should have said Zero instead of all.
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08 Apr '10 02:367 edits
Originally posted by uzless
By your logic, are there any example you can state where you can NOT tell the time, other than 12;00?

Perhaps I should have said Zero instead of all.
You can tell 12:00 too.

I'm not sure. Maybe 5:32 vs 6:28? 1:52 vs 10:08?

Using that symmetry logic, there should be six pairs the times of which cannot be distinguished from the partner time.

The OP suggests 5:12 vs 2:26. This is not symmetric about the 12-6 axis...hmmm.
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08 Apr '10 02:43
Originally posted by sonhouse
This is a special LCD clock, it's a big screen so it's just an image of a clock, they are on the same plane.
My Aunt Mabel can make an atomic LCD clock by hand?!
sonhouse
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08 Apr '10 04:202 edits
Originally posted by AThousandYoung
My Aunt Mabel can make an atomic LCD clock by hand?!
She is very clever, Aunt Mabel🙂
I have a watch made by Casio that has a built in 60 khz receiver that decodes the signal broadcast by the WWVB NIST time standard transmitter near Fort Collins Colorado. here is a brief description:

WWVB continuously broadcasts time and frequency signals at 60 kHz. The carrier frequency provides a stable frequency reference traceable to the national standard. There are no voice announcements on the station, but a time code is synchronized with the 60 kHz carrier and is broadcast continuously at a rate of 1 bit per second using pulse width modulation. The carrier power is reduced and restored to produce the time code bits. The carrier power is reduced by 17 dB at the start of each second, so that the leading edge of every negative going pulse is on time. Full power is restored 0.2 s later for a binary “0”, 0.5 s later for a binary “1”, or 0.8 s later to convey a position marker. The binary coded decimal (BCD) format is used so that binary digits are combined to represent decimal numbers

The time code contains the year, day of year, hour, minute, second, and flags that indicate the status of Daylight Saving Time, leap years, and leap seconds.

So it seems Aunt Mabel only needs a radio...

BTW, any ham (my call is AI3N) knows about WWV, they also broadcast on 2.5, 5, 10, 15 and 20 Mhz and give time hacks once per minute. When I was on the Apollo at Goddard Space Flight center, my job was Apollo Tracking and Timing, the timing part was three atomic clocks at each downrange tracking site and at Goddard which kept the timing around the world to within 100 nanoseconds which allowed data to be transferred from one antenna to another due to the fact the Earth spins around and a given antenna site like Goldstone in Arizona and others, can only track for 7 or 8 hours and has to hand off the signal to another relay that can pick up the Apollo signals and get the data to Goddard in one piece. We used WWV signals as a rough in time hack, the 8th cycle of the 1000 hz time hack and at a particular phase angle of that 8th cycle to rough in the atomic clocks if they had to be replaced or recalibrated once a week. Great job BTW, held a moon rock in my hand!
sonhouse
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08 Apr '10 04:46
Congrats Iamtiger! 132 it is. I tried to cut and paste the solution from Plus magazine but it was full of formatting errors, didn't paste right. Look up Plus mag for their version of the proof that Iamtiger figured out!
PBE6
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08 Apr '10 13:39
I get the same answer as iamatiger. In the diamond-shaped grid mentioned in my previous post, there are 12 lines running in each direction (24 total). The total number of intersections is then 12*12 = 144. Of these, there are 12 points that run along the line theta(H) = theta(M), meaning that the time would be the same regardless of which hand is which. While unique, these times are not confusing so we exclude them, leaving 144 - 12 = 132 confusing times. Since each of these positions is mirrored on the other side of the theta(H) = theta(M) line, there are 132/2 = 66 clock positions which will be confusing.

Anther approach is to restrict your counting to those clock times that line on only one side of the theta(H) = theta(M) line (and not on it). In that case, the first line has 11 intersections, the second line has 10, and so on down to 1. The sum 1+2+3+4+5+6+7+8+9+10+11 = 66, again in agreement with iamatiger regarding the number of confusion clock positions.
PBE6
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08 Apr '10 14:052 edits
To find any intersection point on the grid, we simply note that the equation for the vertical slanted lines is:

y = 12x - (2*pi)(n-1)

where "n" is an integer from 1 to 12, and the equation for the horizontal slanted lines is:

y = (1/12)x + (pi/6)(M-1)

where "M" is an integer from 1 to 12. Equating the two, and after a little algebra, we get:

x = (2*pi/143)(12n + M - 13)
y = (2*pi/143)(n + 12M - 13)

with the pleasing symmetry found in iamatiger's solution.

EDIT: Just trying to simplify things.

Taking the solution one step further towards usefulness and legibility, we can convert the hand positions above to times quite easily. Since the numbers above represent the rotation of the hands around the clock face, we multiply by the appropriate factor to convert them into times.

To get the hour, we divide "x" by 2*pi and then multiply by 12 to get:

hour = 12x/(2*pi) = (12/143)(12n + M - 13)

This number will give you a fractional component of the hour which represents how far the hour hand has traveled towards the next number on the clock. For simplicity, you can simply use the integer part and leave the fractional component for the minute hand.

To get the minute, we divide "y" by 2*pi and then multiply by 60 to get:

minute = 60y/(2*pi) = (60/143)(n + 12M - 13)

Again, this number will give you a fractional component of the minute which represents the seconds and fractional seconds. For simplicity, you can simple use the integer part and leave the fractional component off.

Using these formulas, I generated the following list of confusing times (rounded to the nearest minute) on Excel, each paired with its confusing partner:

1 - 1:00 … 12:05
2 - 2:01 … 12:10
3 - 2:06 … 1:10
4 - 3:01 … 12:15
5 - 3:06 … 1:16
6 - 3:11 … 2:16
7 - 4:02 … 12:20
8 - 4:07 … 1:21
9 - 4:12 … 2:21
10 - 4:17 … 3:21
11 - 5:02 … 12:25
12 - 5:07 … 1:26
13 - 5:12 … 2:26
14 - 5:17 … 3:26
15 - 5:22 … 4:27
16 - 6:03 … 12:30
17 - 6:08 … 1:31
18 - 6:13 … 2:31
19 - 6:18 … 3:31
20 - 6:23 … 4:32
21 - 6:28 … 5:32
22 - 7:03 … 12:35
23 - 7:08 … 1:36
24 - 7:13 … 2:36
25 - 7:18 … 3:37
26 - 7:23 … 4:37
27 - 7:28 … 5:37
28 - 7:33 … 6:38
29 - 8:03 … 12:40
30 - 8:08 … 1:41
31 - 8:13 … 2:41
32 - 8:18 … 3:42
33 - 8:23 … 4:42
34 - 8:29 … 5:42
35 - 8:34 … 6:43
36 - 8:39 … 7:43
37 - 9:04 … 12:45
38 - 9:09 … 1:46
39 - 9:14 … 2:46
40 - 9:19 … 3:47
41 - 9:24 … 4:47
42 - 9:29 … 5:47
43 - 9:34 … 6:48
44 - 9:39 … 7:48
45 - 9:44 … 8:49
46 - 10:04 … 12:50
47 - 10:09 … 1:51
48 - 10:14 … 2:51
49 - 10:19 … 3:52
50 - 10:24 … 4:52
51 - 10:29 … 5:52
52 - 10:34 … 6:53
53 - 10:39 … 7:53
54 - 10:44 … 8:54
55 - 10:50 … 9:54
56 - 11:05 … 12:55
57 - 11:10 … 1:56
58 - 11:15 … 2:56
59 - 11:20 … 3:57
60 - 11:25 … 4:57
61 - 11:30 … 5:57
62 - 11:35 … 6:58
63 - 11:40 … 7:58
64 - 11:45 … 8:59
65 - 11:50 … 9:59
66 - 11:55 … 11:00
uzless
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10 Apr '10 15:541 edit
Originally posted by forkedknight
How precise are we getting here? At 12:05, technically you could tell because the hour hand should be 2.5 degrees off of 12, but practically speaking it would be impossible to tell between that and 1:00

*edit* My math was slightly wrong, but my point stands.
uzless
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10 Apr '10 16:005 edits
Originally posted by PBE6
To find any intersection point on the grid, we simply note that the equation for the vertical slanted lines is:

y = 12x - (2*pi)(n-1)

where "n" is an integer from 1 to 12, and the equation for the horizontal slanted lines is:

y = (1/12)x + (pi/6)(M-1)

where "M" is an integer from 1 to 12. Equating the two, and after a little algebra, we get:

x = :35 … 6:58
63 - 11:40 … 7:58
64 - 11:45 … 8:59
65 - 11:50 … 9:59
66 - 11:55 … 11:00
Not sure how we are saying 11:55 and 11:00 (among others) are indistinguishable.

Draw two clocks.

For 1155 the hour hand would be ALMOST to the 12 and the minute hand would be on the 11....BUT,

For 11 oclock the hour hand would be on 11 and the minute hand would be DIRECTLY on the 12.

There should'nt be confusion between these two times. I think the problem with this solution is that it rounds to the nearest minute, thereby forcing the hour hand to move or stay put when, in reality, it shouldn't.

This is why I said earlier i didn't think there were any times where the hands were interchangeable other than 12:00

If we're ignoring the gear ratio on the clock hands then ok, 66 it is.
PBE6
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12 Apr '10 14:54
Originally posted by uzless
Not sure how we are saying 11:55 and 11:00 (among others) are indistinguishable.

Draw two clocks.

For 1155 the hour hand would be ALMOST to the 12 and the minute hand would be on the 11....BUT,

For 11 oclock the hour hand would be on 11 and the minute hand would be DIRECTLY on the 12.

There should'nt be confusion between these two times. I thin ...[text shortened]... e other than 12:00

If we're ignoring the gear ratio on the clock hands then ok, 66 it is.
The confusing times are rounded to the nearest minute. At no point did I ignore the gear ratios. If you want the exact times, feel free to solve the equations yourself.
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12 Apr '10 15:55
Originally posted by uzless
Not sure how we are saying 11:55 and 11:00 (among others) are indistinguishable.

Draw two clocks.

For 1155 the hour hand would be ALMOST to the 12 and the minute hand would be on the 11....BUT,

For 11 oclock the hour hand would be on 11 and the minute hand would be DIRECTLY on the 12.

There should'nt be confusion between these two times. I thin ...[text shortened]... e other than 12:00

If we're ignoring the gear ratio on the clock hands then ok, 66 it is.
PBE6 is correct, the approximation is what's confusing you.

Think of the clocks at 11:55 and 11:00. You're correct that the hours hand at 11:55 is not exactly to the 12, but what happens if a few fractions of a second before 11:00? The minutes hand is ALMOST to the 12. A similar reasoning for the hands on the 11 (which are also changing) and you realize that they MUST overlap at some point. It's like the hermit going up and down the mountain and occupying the same spot at the exact same time on each trip.
uzless
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12 Apr '10 19:151 edit
Originally posted by Palynka
PBE6 is correct, the approximation is what's confusing you.

Think of the clocks at 11:55 and 11:00. You're correct that the hours hand at 11:55 is not exactly to the 12, but what happens if a few fractions of a second before 11:00? The minutes hand is ALMOST to the 12. A similar reasoning for the hands on the 11 (which are also changing) and you realize t ing up and down the mountain and occupying the same spot at the exact same time on each trip.
I understand what you are saying. It makes sense what you are saying. But only if when you look at the clock, the actual real time is NOT exactly at a minute. Any time that is just fractions just before or just after the exact minute will in fact cause the confusion as you suggest.

I was assuming that when you looked at the clock it actually was that minute, neither just before or just after fractionally.

I misread the OP..he said how many "moments" are there, versus what I thought as "minutes"
uzless
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12 Apr '10 19:16
Originally posted by PBE6
The confusing times are rounded to the nearest minute. At no point did I ignore the gear ratios. If you want the exact times, feel free to solve the equations yourself.
ha, you know better than anyone that even a 1971 calculator has more math skillz than I
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12 Apr '10 22:40
It depends on how the hands move. I have seen clocks where the minute hand moves in minute increments almost instantaneously, and then stays still until the next minute. Usually in such clocks, the hour hand moves when the minute hand moves, by 1/60th of an hour at a time.

With such a clock we would have (using hour positions) 11:55: minute hand at 11, hour hand at 12+5/60ths vs 11:00, minute hand at 12, hour hand at 11, these are not equivalent positions, and if we go to 11:59 the hour hand goes wrong.

In fact, with such a clock there are no duplicate times at-all. The closest we get is 2:36 which looks pretty much like 7:13, but the minute hand in 2:36 is at 36/5 = 7+1/5 and the hour hand in 7:13 is at 7+13/60ths which is 1/60th wrong.

The problem is only any good if both hands move smoothly at a constant speed. The suggestion that this clock had a "large lcd" display was bad, because the resolution of the screen would prevent smooth movement.
sonhouse
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12 Apr '10 23:53
Originally posted by iamatiger
It depends on how the hands move. I have seen clocks where the minute hand moves in minute increments almost instantaneously, and then stays still until the next minute. Usually in such clocks, the hour hand moves when the minute hand moves, by 1/60th of an hour at a time.

With such a clock we would have (using hour positions) 11:55: minute hand at 11, ...[text shortened]... arge lcd" display was bad, because the resolution of the screen would prevent smooth movement.
It's an HD screen, 4000X4000 pixels🙂