Radar clock

[bonanova]

I found an old clock in the basement and determined that its minute and hour hands have lengths in the precise ratio of 2:1, the minute hand being the longer. Having nothing else to do, I mounted a miniature radar speed recorder on the tip of the hour hand and arranged for it always to point to the tip of the minute hand. How much time will elapse after the greatest negative (receding) speed has been recorded until the greatest positive (approaching) speed is recorded?

[Logophobic]

  Reveal hidden contents

Times of greatest receding speed:

12:10:55
1:16:22
2:21:50
3:27:17
4:32:44
5:38:11
6:43:39
7:49:06
8:54:33
10:00:01
11:05:28

Times of greatest approaching speed:

12:54:33
2:00:00
3:05:28
4:10:55
5:16:22
6:21:50
7:27:17
8:32:44
9:38:11
10:43:39
11:49:06

Time elapsed between greatest receding speed and greatest approaching speed:

43 minutes, 18 seconds

 

[jasen]

16,363636 minutes

  Reveal hidden contents

 

The hour and minute hands are superimposed only when their angle is the same.
in 12 hour, hour and minute hands superimposed 11 times.

so time between 2 superimposed is 12/11 hour = 65.454545... minutes

greatest negative (receding) speed when minute and hour hand form  -90 deg.
greatest positive (approaching) speed  when minute and hour hand form 0 deg.

because 1 circle is 360 deg, so :
time from negative to positive speed speed is

(12/11)/4 hour = 16,363636 minutes

 

 

[bonanova]

  On 1/16/2016 at 7:00 AM, jasen said:

16,363636 minutes

  Reveal hidden contents

 

The hour and minute hands are superimposed only when their angle is the same.
in 12 hour, hour and minute hands superimposed 11 times.

so time between 2 superimposed is 12/11 hour = 65.454545... minutes

greatest negative (receding) speed when minute and hour hand form  -90 deg.
greatest positive (approaching) speed  when minute and hour hand form 0 deg.

because 1 circle is 360 deg, so :
time from negative to positive speed speed is

(12/11)/4 hour = 16,363636 minutes

 

 

Expand  

@jasen, I think you're on the right track, but look more closely at when the speeds are greatest.

[DejMar]

  Reveal hidden contents

The greatest receding speed  occurs at 12:55 when the distance decreases by approximately 0.0959556 units since the preceding minute and the greatest approaching speed occurs at 11:06 when the distance increases by approximately the same amount. The time lapse between the two recordings is 10 hours and 11 minutes.

 

[bonanova]

  On 1/16/2016 at 8:45 PM, DejMar said:

  Reveal hidden contents

The greatest receding speed  occurs at 12:55 when the distance decreases by approximately 0.0959556 units since the preceding minute and the greatest approaching speed occurs at 11:06 when the distance increases by approximately the same amount. The time lapse between the two recordings is 10 hours and 11 minutes.

 

Expand  

@DejMar, that's not the result I get, but I also wonder

  Reveal hidden contents

Wouldn't the times of greatest approach and recession be mirror images? Are you assuming lengths of 2 and 1 for the hands, and does 0.0959556 represent a distance change per minute? That would be consistent with saying the distance decreases by that amount between 12:54 and 12:55, and increases by that amount between 11:05 and 11:06, which would restore the symmetry.

I get different times for the extremal speeds.

 

[DejMar]

In answer to you second question,

  Reveal hidden contents

I did not assume that the ratio of the hand lengths were 2 and 1 unit lengths . The ratio of the clock hands was stated in the problem. The unit length was simply taken to be the length of the hour hand with the actual length being unimportant.  The distance reported is only an approximation for the span between the two tips of the clock hands between those stated times. The actual calculated results were not exactly the same, and I truncated these results to the seven decimal places). The value for which the distance is equal does occur between each of those two minutes, thus there is some symmetry -- yet, I did not bother to discover when the calculated second or sub-second for each minute in which the points were equal, as the smallest unit of time I was concerned with was that of the minute. For each minute, I calculated the Cartesian-coordinates of the tip of the minute hand as travelling 6 degrees along the circumference of the larger circle of radius 2 and the same for the tip of the hour hand travelling 1/2 degree along the circumference of the smaller  circle of radius 1 sharing the same radial point. The distance between the two points was then calculated. The "speed" was assumed to be the difference in distance between each minute interval as time was a constant between each of those intervals. The minute intervals with the greatest receding and approaching distances were then assumed to fit the criteria in answering the problem.

 

[bonanova]

Solution:

  Reveal hidden contents

Consider the minute hand (red) fixed vertically. The hour hand (green), along with the clock face, then rotates CCW with time: the hour hand moves at a rate of 360^o every 12/11 hours. It moves directly toward or away from the tip of the minute hand at the points of tangency shown as black lines. When the length ratio is 2:1 the hour hand is 60^o either side of the minute hand, receding at the left position and approaching at the right position. Moving CCW, these positions are 240^o, or 2/3 of a rotation apart. That represents a time interval of (2/3)(12/11) hours or (2/3)(720/11) = 1440/33 = 43.6363... minutes.

Note the two of the critical positions occur at 10:00 and 2:00. So one answer is 4 hours. But that interval does not span adjacent critical positions. 43+ minutes is the shortest time interval between receding and approaching positioins.