Might there be a formula to determine when watch hands coincide?

[ckirmser]

Summary:: Is there a formula to determine when watch hands overlap?

The other day, I was sitting on the can and, as I am won't to do, pondering matters of import to the known and unknown universe at large.

During the event in question, I happened to notice my watch - and I am having the most overwhelming feeling of deja vu; let me check my past posts...

Nope, nothing there. Hmm.

Well, anyway, I looked at my watch and I wondered if there might be a formula to determine when the minutes and hour hands would overlap.

The idea is based on an analog watch with the hands moving smoothly, not something that snaps to points on the dial. It can't be anything like, well, at 3:15, both hands are 90° along the dial, because at 3:15, the minute hand is, indeed, on the three, but the hour hand is 25% of the way towards four. So, they don't overlap. And, once the minute hand has moved towards that 25% point, the hour hand has not been idle; it's moved a little bit further along, too.

My mental musings have not resulted in anything satisfactory to solve this earth-shattering problem that will bring a utopian era of peace and good will to all.

[Moderator's note: Moved from a technical forum and thus no template.]

 

[mjc123]

Think: how many times does the minute hand go round the dial while the hour hand does it once? How many times must the minute hand "lap" the hour hand?

 

[ckirmser]

12 times. But, lapping the hour hand is not the question. I'm trying to see if it's possible to determine at what angles the hands overlap and, if so, what is the formula to figure those angles?

As an aside, this may not be the right forum for the question; it's not a homework question. Just something I thought of as I was sitting in the bathroom.

 

[Klystron]

I asked a similar question -- Why do clocks tick? -- of my jeweler/watchmaker grandfather while learning to read time from an analog watch. We disassembled a spring powered (wind up) clock and analysed the clockworks.

The positions of the watch hands are controlled by concentric cylindrical shafts connected to rotating gears. Depending on the mechanism one can determine coincident hand positions by counting and combining the gear teeth with attention to the gear actuators. Apparent smooth (continuous) hand motion and angles are largely illusions with this mechanism. Change of position of clockwork indicators is discrete and deterministic.

This encyclopedia entry contains a general overview.

 

[Merlin3189]

I don't think your answer of 12x is correct. The correct answer will help tell you what you want to know.

Don't worry about the HW heading, it's just the best place to put it to get the right people to look at it.

 

[mjc123]

12 times.

Is the answer to which question?

But, lapping the hour hand is not the question. I'm trying to see if it's possible to determine at what angles the hands overlap and, if so, what is the formula to figure those angles?

Yes it is. If the hour hand is overtaken n times in one circuit, what is the angle between successive positions of the hour hand when it is overtaken?

 

[osilmag]

The hour hand moves a distance of "5 minutes" every hour, i.e. there are 5 marks between 1 and 2. For every 360 degrees that the minute hand moves the hour hand moves 360/12 degrees.

 

[ckirmser]

Is the answer to which question?

Sorry, I misread your question. The minute hand will have gone around once; I was thinking of the second hand.

​

Yes it is. If the hour hand is overtaken n times in one circuit, what is the angle between successive positions of the hour hand when it is overtaken?

Perhaps, then, I misunderstood your question; I'm interpreting "lapping" as the passing of the hand. I'm not talking about the minute hand passing the hour hand, but when it coincides with the hour hand.

If, for instance, the hour hand is on 1, it will no longer be on 1 when the minute hand gets there. The hour hand will have moved fractionally towards 2. So, there is no overlap. The minute hand must move forward minutely - while at the same time, the hour hand is also moving forward - until it catches up at some point between the 1 and 2 markers.

 

[Merlin3189]

A rider: will it make any difference if there is a second hand? Where will all three hands line up?

 

[Janus]

The basic equation for the time period between objects rotating at different rates lining up is
1/(1/P1-1/p2)
Where P1and P2 are the periods involved.
thus for the Hour and minute hand to line up, it takes
1/(1/60-1/720)=~65.4545.. minutes between line ups. this happens 11 times every 12 hrs.
The sec hand passes the hour hand every ~1.034 minutes. or 696 times every 12 hrs.
All three hands only all line up with each other every twelve hrs at 12:00:00

 

[ckirmser]

The basic equation for the time period between objects rotating at different rates lining up is
1/(1/P1-1/p2)
Where P1and P2 are the periods involved.
thus for the Hour and minute hand to line up, it takes
1/(1/60-1/720)=~65.4545.. minutes between line ups. this happens 11 times every 12 hrs.

Well, cool beaners.

I thought the answer might involve some calculus because rates were involved.

Thanx, Janus!

The sec hand passes the hour hand every ~1.034 minutes. or 696 times every 12 hrs.
All three hands only all line up with each other every twelve hrs at 12:00:00

The three hands will line up only every 12 hours?

Hmm - OK, given some pondering, I guess that, as the second hand approaches the minute hand, as it sits over the hour hand, the minute hand will have moved off fractionally as the second hand coincides with it.

 

[kuruman]

The sec hand passes the hour hand every ~1.034 minutes. or 696 times every 12 hrs.

I get 60.0834 s = 1.0014 min. Here's how, formally using angular speeds $\omega_s=\frac{2\pi}{60}~\mathrm{rad/s}$ for the second hand and $\omega_h=\frac{2\pi}{12\times 3600}~\mathrm{rad/s}$ for the hour hand. For the first crossing
$$\frac{2\pi}{60}~(\mathrm{rad/s})\Delta t-2\pi=\frac{2\pi}{12\times 3600}~(\mathrm{rad/s})\Delta t~\rightarrow ~\Delta t=\frac{43200}{719}=60.0834~ \mathrm{s}$$ For the $n$th crossing,$$t_n=n\Delta t=\frac{43200 ~n}{719}~ \mathrm{seconds}$$
Note that this implies that the number of crossings is 719 as there are 43200 seconds in 12 hours. The fact that 719 is a prime number says that there can be no integer $\Delta t_n$ for $n<719$, i.e. all crossings before 719 occur at non-integer values of seconds. Since all three hands must line up after an integer number of seconds, no three-way line up can occur before $n=719$.

Edit: The original post was edited after @Janus posted #13. I modified the last equation to give an expression for the "clock" time after the $n$th crossing. All intervals are the same so $\Delta t_n$ is meaningless.

 

[Janus]

I get 60.0834 s = 1.0014 min. Here's how, formally using angular speeds $\omega_s=\frac{2\pi}{60}~\mathrm{rad/s}$ for the second hand and $\omega_h=\frac{2\pi}{12\times 3600}~\mathrm{rad/s}$ for the hour hand. For the first crossing
$$\frac{2\pi}{60}~(\mathrm{rad/s})\Delta t-2\pi=\frac{2\pi}{12\times 3600}~(\mathrm{rad/s})\Delta t~\rightarrow ~\Delta t=\frac{43200}{719}=60.0834~ \mathrm{s}$$ For the $n$th crossing,$$\Delta t_n=\frac{43200 ~n}{719}~ \mathrm{seconds}$$
Note that this implies that the number of crossings is 719 as there are 43200 seconds in 12 hours. The fact that 719 is a prime number says that there can be no integer $\Delta t_n$ for $n<719$, i.e. all crossings before 719 occur at non-integer values of seconds. Since all three hands must line up after an integer number of seconds, no three-way line up can occur before $n=719$.

Your right, I must have slipped up punching into the calculator. Double checking gives me:
1/(1/60 - 1/ 43200) = 60.083559... sec

 

[Janus]

Well, cool beaners.

I thought the answer might involve some calculus because rates were involved.

This same equation can be used to figure out when planet's make their closest approaches to each other. For instance, the Earth has an orbital period of 365.26 days, and Mars one of 686.98 days, Thus close approaches between Earth and Mars occur an average* of 1/(1/365.26-1/686.98) = ~ 780 days apart.
Or, given that the Moon orbits the Earth in 27.32 days relative to the stars, and the Earth orbit's the Sun in 365.26 days, also relative to the stars, then the Moon and Sun will be in opposition every 1/(1/27.32- 1/365.26) = 29.53 days, and thus this is the time between full Moons, on average.

*The fact that both the Earth's and Mar's orbits are not perfectly circular causes a slight a variation in this timing.

 

[ckirmser]

This same equation can be used to figure out when planet's make their closest approaches to each other. For instance, the Earth has an orbital period of 365.26 days, and Mars one of 686.98 days, Thus close approaches between Earth and Mars occur an average* of 1/(1/365.26-1/686.98) = ~ 780 days apart.
Or, given that the Moon orbits the Earth in 27.32 days relative to the stars, and the Earth orbit's the Sun in 365.26 days, also relative to the stars, then the Moon and Sun will be in opposition every 1/(1/27.32- 1/365.26) = 29.53 days, and thus this is the time between full Moons, on average.

*The fact that both the Earth's and Mar's orbits are not perfectly circular causes a slight a variation in this timing.

That never occurred to me, but, yes, you could look at planetary orbits as pseudo watch hands.

Interesting...

 

[kuruman]

That never occurred to me, but, yes, you could look at planetary orbits as pseudo watch hands.

Interesting...

Why "pseudo"? Isn't the Sun a real clock hand that defines the day to this day in every day life? Combine that with the clock hand of the Moon and you have lunar months and lunar years. People have been using celestial clocks for a long long time.

 

[ckirmser]

Why "pseudo"? Isn't the Sun a real clock hand that defines the day to this day in every day life? Combine that with the clock hand of the Moon and you have lunar months and lunar years. People have been using celestial clocks for a long long time.

Well, pseudo in that there are no actual, physical hands tying the planets to the Sun and, as you said, the orbits are elliptical; it's hard to get hands to trace an elliptical path on a watch.

Wouldn't the Sun be more an analog to the pin holding the hands, rather than one of the hands? I figure the "hands" to be the force of gravity, keeping the planets tied to the Sun; that the planets are just the tips of the "hands," rather than the whole thing.

 

[kuruman]

You are are thinking of a heliocentric model clock. A rotating hands type of clock makes sense if the observer is at rest with respect to the axis of rotation about which the hands rotate. We are at rest with respect to the Earth not the Sun, therefore we have to consider a geocentric model clock.

 

[ckirmser]

You are are thinking of a heliocentric model clock. A rotating hands type of clock makes sense if the observer is at rest with respect to the axis of rotation about which the hands rotate. We are at rest with respect to the Earth not the Sun, therefore we have to consider a geocentric model clock.

Well, yeah, I am.

if it's a geocentric clock, then this would have no relationship to a real clock, as the planets orbit the Sun, not the Earth. Unless your scope is only that of those objects orbiting the Earth, of course.

But, even then, the orbit of, say, the Moon, is also elliptical.