It's a few minutes to 8.45pm, Finnish time, and the clock hands indicating minutes and hours are almost on top of each other. At the exact moment where the hand indicating minutes catches the hand indicating hours, where will the hand indicating seconds point?
Originally posted by talzamir It's a few minutes to 8.45pm, Finnish time, and the clock hands indicating minutes and hours are almost on top of each other. At the exact moment where the hand indicating minutes catches the hand indicating hours, where will the hand indicating seconds point?
Note that the value of Fract must be betwee 0 and 1, so Int has different constraints for each of the above:
A2 : 4 <= INT <= 11
B2 : 0 <= INT <= 7
A2a: 0 <= INT <= 3
B2a: 8 <= INT <= 11
To think about seconds we need to define:
Fract*60 = Int_M + Fract_S
And the angle of the Seconds hand is:
Fract_S * 360
Now, by stepping through the valid values of INT for equations A2, B2, A2a and B2a we should be able to get the times when the hour and minute hand are 120 degrees apart and see what all the angles are and whether we have found a solution, this is where I would try a few calcs out in excel, but I am in a house over christmas where the computer does not have any working spreadsheet app on it. Oh well, I'll try to download star office tomorrow, await the next post.
Worked it out, the answer seems to be no, as when the hours and minutes are 120 degrees apart, the seconds are a different number of 11ths of a degree from an integer than the hours and minutes are, except for times like 4:00 when the seconds and minutes are zero.
However neither this answer, nor the original question, take into account the fact that the hours, minutes and second hands all move in steps rather than continuously, if that is taken into account there MAY be an answer.
Hmm, Assuming any clock time, from 0 to 12 hours is an integer number of second ticks, whereby:
The second hand moves 260/60 = 6 degrees per tick
The minute hand moves 360/(60*60) = 1/10th of a degree per tick
The hour hand moves 360/(60*60*12) = 1/120 th of a degree per click
So, the minute hand moves 1/10 - 1/120 = 11/120ths of a degree with respect to the hour hand, per tick
So, for the minute hand to move 120 degrees ahead of the hour hand takes (120*120)/11 ticks (approx 20 minutes) but since 11 is prime with respect to 120 this means that the minute hand is only an exact multiple of 120 degrees ahead of the hour hand after 11 such cycles. i.e. every 120*120 seconds (4 hours). So the minute hand is 120 degrees ahead of the hour hand only at 4:00, 8:00 and 12:00, and at these times the hour and the seocnds hand are in the same place.
Therefore, regardless of whether the hands move smoothly or in jerks, the hand of a clock face never divide it into thirds.
Interesting question and answer. I agree the clock face is not perfectly divided into three equal segments.
However (and there is usually a however) if we allow some degree of tolerance, I can find a number of number of solutions where it is "pretty close". I have defined "pretty close" as being within one half of one degree of the perfect 120 degree angle.
The closest times are generally clustered around 2:54:34, 9:05:25 and 5:49:09. There are other times if you increase the tolerance to 1 degree, but these 3 are the closest.
If we define a 'tick' as the smallest unit of movement for the second hand (i.e. the number of ticks per second) then the 'tick' that yields the closest result is 89 ticks per second, at the times 2:54:34 + 50/89 ticks and 9:05:25 + 39/89 ticks. This has the smallest difference between the largest and smallest angles, at approximately one quarter of 1 degree. On a sufficiently small clock, I believe such a difference would not be visible to the naked eye.
Using 89 ticks/second, at 2:54:34:50 the hour hand is at the angle 87.288°, the minute hand is at the angle 327.456° and the second hand is at the angle 207.371°. The maximum angle is 120.085° and the minimum angle is 119.832°.
P.S. Using 16 ticks per seconds (and multiples thereof) yields a similar time and angles, albeit with a slightly higher difference.
This part :
"The maximum angle is 120.085° and the minimum angle is 119.832°."
should have read:
"The maximum difference between angles is 120.085° and the minimum difference between angles is 119.832°." An 'angle' is defined as the position of the hand on the clock, using a 360° circle where 0° is equivalent to 12 o'clock. The difference between angles speaks for itself but is the difference between the angles of two differing clock hands.
Can you get clocks where, as the second hand moves round, the hour and minute hand stay completely still and then they both move (minute by 6 degrees, hour hand by 1/10th of a degree) when the second hand gets to 0?
If you can get such clocks this question has two exact answers.
I suppose you might be able to get them, because less energy might be spent moving the hands so the battery might last longer. Additionally the mechanism might be simpler.
If you state that as your assumption, then you will have solved the puzzle. However, I haven't ever seen any such clocks - although I admit I am not a clock expert by any definition! Can you imagine looking at a clock where the time is 12:59:59 and the hour hand is still pointing at 12? That would be really confusing.
Originally posted by andrew93 If you state that as your assumption, then you will have solved the puzzle. However, I haven't ever seen any such clocks - although I admit I am not a clock expert by any definition! Can you imagine looking at a clock where the time is 12:59:59 and the hour hand is still pointing at 12? That would be really confusing.
I don't think it would be all that confusing. As I stated, the hour hand moves when the minute hand moves, so the hour hand would be 59/60ths of the way to 1.