Non digital, traditional watch, how many times its minutes hand comes exactly over hours hand and after how?

Question

much time each time?

Answer 1

This happens every 1 hour, 5 and 5/11ths of a minute. In 12 hours, there are PRECISELY 11 crossings (same simultaneous pointings) of minute and hour hands.

How can this be shown? "Let me count the ways."

Thank you, zanti3, for confirming what I just wrote. There are in fact so many ways of showing this, that I hardly know where to begin. But let me do so, anyway.

1. First, there is the obvious explicit demonstration method:

Take such an analogue watch, borrowing your aged grandparent's 21st birthday wristwatch or analogue kitchen clock if need be. Arrange the hands to both point to '12.' (This is the time 12:00.) Twirl them, and simply count how many times they overlap until you have 12:00 again. You'll find that they simply passed one another 11 times in 12 hours, and therefore each overlap time is separated by exactly 1 hour, 5 and 5/11ths of a minute from the next one, before or after.


2. For those not yet convinced, (a) either repeat this with more careful intervening observation, or (b) do it as a thought experiment.

2(a) Repeat what was done in (1), but note carefully the first couple of overlaps following 12:00. It might seem at first that they overlap or line-up again at 1:05, but if you look carefully, it's a small amount later. Twirl the hands again: it's now at 2:10 and so much more that it's almost 2:11. If the first line-up was at 1:05 and "a bit," the second must be at 2:10 and "2 bits." Aha, a pattern emerges; it will carry on until 11:55 and "11 bits" --- but that clearly has to be 12:00 again. Therefore, 11 "bits" is 5 minutes, so that each "bit" is in fact 5/11 minutes.

Therefore, 1 hour, 5 and 5/11ths of a minute pass between each overlap or line-up.

2(b) This way entirely discards the physical watch, and considers the problem as a thought experiment, as pioneered by good old Isaac, of course. Argue as follows:

After 1 hour, the minute hand has come back to point to '12,' but the hour hand has moved on to '1.' (Amazing!) That means that the hour hand travels angularly at exactly 1/12th the speed of the minute hand.

AHA, IT'S THE TORTOISE AND THE HARE, or XENO'S (OR ZENO'S?) PARADOX!:

Let the minute hand move on for 5 mins. It will now point at the '1.' But in that time, the hour hand will have moved on to point exactly 5/12ths of a minute ahead.

Let the minute hand move on for 5/12ths of a minute; the hour hand will have moved on 5/(12)^2 minutes, etc., etc.

Now in the corresponding Xeno's (or Zeno's?) Paradox, it was argued that the Hare could never catch the Tortoise. But this was because at least some Greeks could not contemplate that the sum of an infinite (geometrical) series could converge to a finite answer, and that there could be values of distance reached in this (unequal) race that would lie beyond ... what?: NOT infinity, but beyond a FINITE sum of an infinite number of terms, each decreasing in a specified way.

In Xeno's (Zeno's?) case, where the tortoise, say had a 100 yard start, and the hare ran 10 times faster, the convergent (or passing) point was at 111.1111... yards.

Note that this implies that 1/10 + 1/(10)^2 + 1/(10)^3 + ... = 1/9.

Now the pedestrian way to show this is of course to simply sum the geometric series. However, this is quite unnecessary. Decimal arithmetic shows it must be so: ANYONE can simply add 0.1 + 0.01 + 0.001 + 0.0001 + ...; clearly it's 0.1111..., which you all recognize as the decimal representation of 1/9. QED.

This obviously works for ANY "integer fractional" geometric series (and in fact for "non-integer fractional" geom. series, though that may be harder to see).

In the case at hand, consider 1/12 + 1/(12)^2 + 1/(12)^3 + ... Work in "12imals", i.e. decimal-like fractions in base 12: they're simply 0.1 + 0.01 + 0.001 + ... all over again, summing once more to 0.1111..., which is the representation of 1/11 in "12imals." (Think about it.)

And THAT'S why you can see that this thought experiment yields 1 and 1/11th of an hour for the "overlap time" once again.

There are many more ways of demonstrating this kind of repeated overlap or line-up on what are effectively circular race-tracks. I'll finish with one very important application, in Astronomy:

3. Planets in "circular orbits" around the Sun. (O.K., they're not really circular, but let's consider them so for simplicity.)

Planets go around the Sun in orbits with different periods, measured relative to the stars --- their so-called Sidereal Periods. But we can't directly measure such periods, as the measurements we can ourselves make are necessarily performed only on this rotating, orbiting ball we call the Earth. However, we CAN measure when the Sun (more precisely the "anti-Sun" or midnight meridian), the Earth, and the planet of interest "line-up", i.e. overlap as in some giant celestial clock. (In practice, it's necessary in general to interpolate between two suuccessive night's measurements, with the planet just on either side of the midnight meridian by small but measurable amounts.) That gives us P_overlap, otherwise known as the Synodic Period for the planet concerned.

By considering a chase around the celestial race track of the kind we considered in the case of the clock, or alternatively realizing it's all about relative angular speeds, and what it takes to get one full rotation ahead or behind, one can establish the following formula:

1/P_overlap = 1/P_fast - 1/P_slow.

For an exterior planet (Mars, Jupiter, ...) the Earth has the shorter period, i.e. P_fast is 1 year. For an interior planet, Earth has the longer period, so P_slow is 1 year. (Astronomy books often labour to establish TWO different formulae, but the particular formulation here covers BOTH cases at once!) It was by using such formulae that Copernicus actually established what the orbital periods of the classical planets were, from their "overlap" (or "synodic") periods.

So: to those who say "How could you even think of this clock problem?" or "What's the point?", remember that out of what Copernicus did came Kepler's Three Laws of Planetary Motion, Newton's explanation, the flowering of modern science that then followed, and ultimately today's technological society.

Thus, if someone hadn't been interested in when the "hands" of a celestial clock overlapped, you wouldn't be sitting here reading this today.

Live long and prosper.

Answer 2

Start at noon. Both hand are pointed at 12, overlapping. At 12:01 the minute hand is moved off the 12. 59 minutes later it is 1pm and the hour hand is pointing at the one and the minute hand is again on the twelve. The one is 5 minutes away from the 12. So in 5 minutes the hands will again overlap...however, the hour hand does move along towards the two over the period of an hour.

So the answer to your first question is straight forward: once an hour the two hands overlap. The second one is tougher. At 1:05 the two may over lap, but not say at 7:35, more like 7:38. You've got to figure how much the hour hand moves to cover that 5 minute space on the watch dial...60 minutes/5=12 min.

SO at 1:05, 2:10, 3:16, 4:21, 5:27, 6:32, 7:38, 8:43, 9:49,10:54, 11:59 ansd 12. You can figure out the minutes in between.

Just a timely guess

Answer 3

Ya' know! That's what I like about you... You know which men you want to seduce, and then turn into a pathetic simpering whimpering empty shelluva' man, within a very short time... (By the way...where do you jog? For you...I'll wear spandex, and make sure my sweet oz is ready and willing and waiting for you, as I temporarily suck in my stomach, and increase my jogging pace to fool you into thinking I'm a "God of stamina and strength," before I hide shortly thereafter, and wheeze breathlessly from that momentary "spurt" of jogging effort designed to get you all "aroused...")

Answer 4

Spock is right - it's 11 times every 12 hours.

(Ah, Doctor, I see you have expanded on your answer....)

Answer 5

Once every hour I suppose

Answer 6

why do you want to know something like that?