Shollenbarger's Clock


by expectedvalue
Tags: clock, shollenbarger
expectedvalue
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#1
Jun16-06, 11:26 AM
P: 5
Shollenbarger's Clock:

Excluding 12:00, Is there ever a time on a clock where the second, minute and hour hands are equidistant from each other? (In other words, the three hands are all 120 apart? If yes, give the EXACT time(s) that this occurs. If no, prove that no such time exists. Good Luck!
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DM
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#2
Jun17-06, 01:30 PM
P: 352
-------------------
If you include 12:00
-------------------

The second hand could be on 12:00, the minute hand on 4:00 and the hour hand on 8:00

Time would be 8:20

--------------------
If you exclude 12:00
--------------------

I personally don't think it's possible.

If any four hour marks distance stand for 120° (each hour mark is 30°), then 12:00 must be included.

Thus no such time exists!
Parlyne
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#3
Jun17-06, 03:22 PM
P: 546
Quote Quote by DM
-------------------
If you include 12:00
-------------------

The second hand could be on 12:00, the minute hand on 4:00 and the hour hand on 8:00

Time would be 8:20
But, at 8:20:00, the hour hand of a clock doesn't point directly at the 8. In fact, it will be one third of the way from 8 to 9. Thus, at 8:20:00 the hands will not all be [tex]120^\circ[/tex] apart.

--------------------
If you exclude 12:00
--------------------

I personally don't think it's possible.

If any four hour marks distance stand for 120 (each hour mark is 30), then 12:00 must be included.

Thus no such time exists!
I think he was saying to exclude the time 12:00:00 from consideration, rather than the direction of pointing at the 12.

expectedvalue
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#4
Jun17-06, 05:15 PM
P: 5

Shollenbarger's Clock


Parlyne is correct. I wanted to exclude 12:00:00 from consideration as an answer, because all three hands are on top of each other. But one hand can be pointing at 12.

8:20:00 is not a valid solution:

From second to minute: 120°
From minute to hour: 130°
From hour to second: 110°
DM
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#5
Jun17-06, 05:29 PM
P: 352
What about:

hour hand on 4:00, minute hand on 12:00 and second hand on 8:00

Time = 4:00
expectedvalue
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#6
Jun17-06, 05:41 PM
P: 5
What about:

hour hand on 4:00, minute hand on 12:00 and second hand on 8:00

Time = 4:00
This time is: 4:00:40. This is not a valid solution:

From minute to hour: 119 2/3°
From hour to second: 119 2/3°
From second to minute: 120 2/3°

The hour hand cannot be at 4:00 or the minute hand at 12:00 when the second hand is at 8:00, because it would now be 40 seconds after 4:00, thus the hour hand and the second would have moved slightly.

Keep trying. (Fractions of a second are allowed as valid answers, if needed.)
DM
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#7
Jun17-06, 05:58 PM
P: 352
My last attempt of the day is:

Hour hand on 3:00

Minute hand on 0:59

Second hand on 0:00:39

Time = 3:59:39
expectedvalue
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#8
Jun17-06, 06:22 PM
P: 5
Time = 3:59:39
No, unfortunately 3:59:39 is not a valid answer:

From minute to hour: 121.925
From hour to second: 114.175
From second to minute: 123.9
geniusprahar_21
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#9
Jun18-06, 03:15 AM
P: 29
its not possible.......one way of proving it is assume the time as xhrs, yhrs, and z secs........and put appropriate conditions on them... then find angles and solve..........
DM
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#10
Jun18-06, 04:04 AM
P: 352
Quote Quote by geniusprahar_21
its not possible.......one way of proving it is assume the time as xhrs, yhrs, and z secs........and put appropriate conditions on them... then find angles and solve..........
This is a very messy explanation. "Put appropriate conditions on them" doesn't give us the reason.

What conditions?
gnomedt
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#11
Jun18-06, 01:38 PM
P: 38
I too believe it's impossible. I haven't finished the proof, but the idea is that I make the hour, minutes and seconds hands of the clock into unit vectors which are a function of time, then take dot products and set them equal (because the dot product of any two of the aforementioned vectors must be |1|*|1|*cos[120 deg], i.e. 4 hour-marks apart) and finally show that there's no way that that's possible. There might be an easier way, but I like vectors.

Now here's another brainteaser, if this one turns out to be impossible: what time on Shollenbarger's Clock will make the hands the closest to 120 degrees?
DaveC426913
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#12
Jun18-06, 03:35 PM
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P: 15,325
Correct me of I'm wrong, but no one said they have to be at exact integer times, right?

Don't these three hands line up at 120 degrees from each hour approximately 11 times in every 12 hour period? (I'll leave it to others to determine the exact times.)
Parlyne
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#13
Jun18-06, 05:09 PM
P: 546
Quote Quote by DaveC426913
Correct me of I'm wrong, but no one said they have to be at exact integer times, right?

Don't these three hands line up at 120 degrees from each hour approximately 11 times in every 12 hour period? (I'll leave it to others to determine the exact times.)
No. The hour and minute hands line up 11 times in each 12 hour period. The second hand will only be in line with these one of those 11 times.
Parlyne
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#14
Jun19-06, 12:30 AM
P: 546
I'm going to go out on a limb and say that there is no such time. To demonstrate, let's assume there is a time when the hands are all [tex]120^\circ[/tex] apart. Call the hour x, the minute y and the second z. Because of the format times are cited in, x and y must be integers.

Using these, we can express the angle at which each hand sits. The hour hand sits at [tex]\alpha = \left (\frac{x}{12} + \frac{y}{720} + \frac{z}{43200} \right)*360^\circ = 30x + \frac{y}{2} + \frac{z}{120}[/tex]. The minute hand sits at [tex]\beta = \left ( \frac{y}{60} + \frac{z}{3600} \right )*360^\circ = 6y + \frac{z}{10}[/tex]. And, the second hand sits at [tex]\gamma = 6z[/tex].

We set the second and minute hands [tex]120^\circ[/tex] apart by setting [tex]\beta = \gamma \pm 120[/tex]. This gives the condition [tex]y = \frac{59}{60} z \pm 20[/tex]. Since y must be an integer, z must be an integral multiple of 60/59. Thus, we can write [tex]z = \frac{60}{59}n[/tex] and [tex]y = n \pm 20[/tex], where n is an integer.

Finally, to put the hour hand in the correct position, we set [tex]\alpha = \gamma \pm 240[/tex]. This reduces to the condition that [tex]x = \frac{11}{59} n \pm \frac{23}{2}[/tex].

We can see that, for x to be an integer, n must be valued such that [tex]\frac{11}{59} n[/tex] is an odd half integer. But, since 59 is odd and 11 and 59 are relatively prime, the only way for [tex]\frac{11}{59} n[/tex] to be an odd half integer is if n is an odd half integer multiple of 59 - [tex]n = \frac{2m+1}{2} 59[/tex]. Since all factors in the numerator are odd, this can't be an integer. Thus, if n is an integer, x is not.

Thus, there are no x, y, and z satisfying the relationships above such that x and y are both integers.
geniusprahar_21
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#15
Jun19-06, 01:33 AM
P: 29
Quote Quote by DM
This is a very messy explanation. "Put appropriate conditions on them" doesn't give us the reason.

What conditions?
conditions like:
x,y,z should be integers...........or atleast x and y should be integers(if u want to accept decimal seconds). further 0<= x <12....... 0<= y <60 and 0 <= z < 60........with these conditions u'll never get an appropriate answer
ceptimus
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#16
Jun19-06, 05:43 AM
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P: 297
It gets close enough for practical purposes at

02:54:34.576
expectedvalue
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#17
Jun19-06, 08:46 AM
P: 5
I tried to clarify this earlier...

Decimal seconds are allowed. The seconds do not have to be integers. But give your answers with fractions.

It gets close enough for practical purposes at

02:54:34.576
Unfortunately, this is a brain teaser and not a practical problem. So, 'close enough' is not equidistant.
Jimmy Snyder
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#18
Jun19-06, 08:50 AM
P: 2,163
The answer is that the three hands never form 120 degree angles w.r.t each other. I have an inelegant solution. There are in fact, 22 times in a 12 hour period that the hour and minute hands are 120 degrees apart. It suffices to determine where the second hand is at those times. I did so and found that the second hand is never at the right place at the right time. Here are the results of a computer run. The first column is the time. The second column is the angle between the hour and minute hands. The third column is the angle between the minute and second hands. Take the first pair of lines for instance. Sometime between 00:21:49 and 00:21:50, the hour and minute hands were 120 degrees apart. However, at no time during that interval was the second hand near to 120 degrees from the minute hand. There is some symmetry in the data so that it turns out that with a little more analysis, you wouldn't need to check all 22 times. The closest approach is at 2:54:32 (and at 3 other times).

00:21:49 -119.9917 -163.1000
00:21:50 -120.0833 -169.0000

00:43:38 -239.9833 33.8000
00:43:39 -240.0750 27.9000

01:27:16 -119.9667 67.6000
01:27:17 -120.0583 61.7000

01:49:05 -239.9583 264.5000
01:49:06 -240.0500 258.6000

02:32:43 -119.9417 -61.7000
02:32:44 -120.0333 -67.6000

02:54:32 -239.9333 135.2000
02:54:33 -240.0250 129.3000

03:38:10 -119.9167 169.0000
03:38:11 -120.0083 163.1000

03:59:59 -239.9083 5.9000
04:00:00 120.0000 0.0000
04:00:01 119.9083 -5.9000

04:43:38 -119.9833 33.8000
04:43:39 -120.0750 27.9000

05:05:27 120.0250 -129.3000
05:05:28 119.9333 -135.2000

05:49:05 -119.9583 264.5000
05:49:06 -120.0500 258.6000

06:10:54 120.0500 -258.6000
06:10:55 119.9583 -264.5000

06:54:32 -119.9333 135.2000
06:54:33 -120.0250 129.3000

07:16:21 120.0750 -27.9000
07:16:22 119.9833 -33.8000

07:59:59 -119.9083 5.9000
08:00:00 240.0000 0.0000
08:00:01 239.9083 -5.9000

08:21:49 120.0083 -163.1000
08:21:50 119.9167 -169.0000

09:05:27 240.0250 -129.3000
09:05:28 239.9333 -135.2000

09:27:16 120.0333 67.6000
09:27:17 119.9417 61.7000

10:10:54 240.0500 -258.6000
10:10:55 239.9583 -264.5000

10:32:43 120.0583 -61.7000
10:32:44 119.9667 -67.6000

11:16:21 240.0750 -27.9000
11:16:22 239.9833 -33.8000

11:38:10 120.0833 169.0000
11:38:11 119.9917 163.1000


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