An analog clock displays the time 3:15. What is the angle between the hours and the minutes hands?

An analog clock displays the time 3:15. What is the angle between the hours and the minutes hands?

homework goes to

7.5

Idk like 10?

7.5

Hour hand starts at pi/2 (3 o'clock) and moves at pi/6 radians per hour for 15 minutes. At 3:15, minute hand is at pi/2.

absoluteValue[pi/2 + (pi/6) * (15/60) - pi/2]

= 0.13 radians

7.5

Implying an analogue clock has to use hands.

What a shit question

Isn't it just zero? Big hand is on the three, little hand also on the three. Three minus three is zero.

wew

Hmm... The question is pretty ambiguous and the answer depends on how advanced you are in physics/math. I'm in 2nd year physics, so this might be overkill, but:

Is it analog or digital? This makes a big difference because if its analog it will operate based on newtonian mechanics, whereas if it was digital it would operate on quantum mechanics (electrons).

The size of the clock is important. The bigger the hands are the more the angle changes simply by having longer clock hands.

Lastly, how fast is the clock moving? Are we talking relativistic speeds? If so the angle will decrease proportionally to the square root of the speed of light squared divided by the energy, according to Einsteinian (relativistic) relativity.

Tricky question

Why is he wrong though? At exactly 3:15 shouldn't both the minute and hour hands point at exactly 45 degrees, assuming 12 o clock is the origio?

>whereas if it was digital it would operate on quantum mechanics (electrons)
w e w
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>The bigger the hands are the more the angle changes simply by having longer clock hands
oh okay you're just being a funny fuck l o l good 1 m8 got me real gud

I mean 90 degrees not 45

Most clocks don't move their hands in discrete steps, so at 3:15 the hour hand will have moved a quarter of the way from 3 to 4.

To answer this question in a particular way, let us consider a discrete model of the three hands of a clock's face, where all three hands of the clocks' face move from spot to spot in discrete steps. We may say first of all that each full turn of each hand represents a particular passage of time, and that the turns of each hand may therefore be compared to each other. we furthermore require that the second, the incrementing from one second to the next (or the previous, for discussion) is the base unit of our discussion. So that while the second hand makes 1/60th of a turn, the other two hands make much smaller turns in kind, and that the smallest of these, the corresponding turn of the hour hand, likewise serves as a basis for the number of legal positions that any hand may assume.

Now, consider the following:

1 full turn of the hour hand = 12 hours = 720 minutes = 43200 seconds

1 full turn of the minute hand = 1 hour = 60 minutes = 3600 seconds

1 full turn of the second hand = 1/60th hour = 1 minute = 60 seconds.

A little dimensional analysis and creativity allows us to immediately describe our discrete model in further detail. We have that the increment from one second to the next entails 1/60th turn of the second hand, or equivalently (simultaneously), 1/3600th turn of the minute hand and 1/43200th turn of the hour hand. Thus the hour hand advances through 43200 evenly spaced positions in teh course of one 12-hour cycle, while the other hands cycle around certain subsets of these positions, 3600 of them for hte minute hand, and just 60 of them for the second hand.

Let us continue this treatment.

Imagine in this treatment (that is, with these assumptions, this modeling of the movement of the hands of a clock face) that the numbers 1-12 on a clock face are instead replaced by their respective multiples of 3600, from 3600 (for 1) up through 43200 (for 12). such an image gives a helpful illustration of the positional and simultaneously angular system that we wish to treat of. The clock face has 43200 angular positions, and the hour hand from second-to-second simply increments from 1,2,3,... all the way up to 43200 at the end of a given 12-hour cycle. Meanwhile the minute hand cycles through the multiples of 12 in this exact same angular schema, and 12 times as fast, and the second hand cycles through the multiples of 720 over and over again.

We therefore represent 12:00:00 by (43200, 43200, 43200). At noon-midnight, the three hands coincide. At 12:00:01, we have (1, 12, 720).

At 3:00:00, we have (10800 , 43200, 43200).
At 3:05:00, we have (11100 , 3600 , 43200). We know this because the hour hand has incremented exactly 300 times, or 300 seconds, and the minute hand has incremented from "0", or 43200, in multiples of 12 over the same time. We also know that the minute hand is "exactly" at the "1/5/3600" position on the clock face. This insight makes the rest straightforward.
At 3:10:00, we have (11400 , 7200 , 43200).
At 3:15:00, we have (11700 , 10800 , 43200).

Thus the angle between the hour hand and the minute hand is simply the difference of 11700 and 10800, which is 900, with respect to 43200. This amounts to 900/43200 turns, which simplifies to 9/432 or 3/144 or 1/48 turns. Since 360 is highly composite, it is easy to express this in terms of degrees, for example. 1/48th of 360 is just 360/48 = 15/2 = "7.5 degrees", as other anons have already pointed out. These amount to a few different ways of answering the OP's question, but we've described a rather sophisticated discrete model that can be turned toward any other related question.

Is that a valid assumption? The problem doesn't say really.

But if it is, then it moves a quarter of a twelfth around the clock, which is 7.5 degrees

Using the above, we can push the analysis of hte discrete model a bit futher. We know that for two hands to coincide at a given second (or moment), their two coordinates must have the same values. When does this happen in the discrete model, generally speaking?

At 12:00:00, we have that all three hands coincide at "top-dead-center", to borrow a term from mechanics and engineers.

The minute hand and the second hand only coincide once per hour, at the top of a given hour. Because each time that the minute hand reaches a multiple of 720 (one of the sixty legal positions of the second hand), the second hand is always at 43200, starting a new minute. Thus the two may only coincide exactly at the top of an hour.

The hour hand and the second hand only coincide once in every 12-hour cycle, being the noon-midnight 12:00:00 endpoints which end and begin such cycles. This is true because each time that the hour hand reaches a multiple of 720 (once every 12 minutes), the second hand is always at 43200, starting a new minute, as before.

What of the hour hand versus the minute hand? The two have ten opportunities to coincide in the /body/ of a 12-hour cycle, not counting the noon-midnight endpoints when they do. Consider that after 12:00:00, the minute hand does not lap the hour hand from 12:00:01 through 1:00:00. Further, from 11:00:00 through 11:59:59, the minute hand has yet not lapped the hour hand, except again at 12:00:00. So apart from the noon-midnight endpoints, no lapping happens in the bodies of these "top" hours.

the other ten moments where the minute hand laps the hour hand actually occur in between the seconds of our discrete model. They can be reasoned as follows: 1:05:27-28, 2:10:54-55, etc (see pic related). Yet it never happens in our discrete model that the minute hand and the hour hand occupy the same position, except at noon-midnight. We therefore also say that the minute hand and the hour hand only coincide once every 12-hour cycle.

>Most clocks don't move their hands in discrete steps
name three

From this it also readily follows that at no other time in a 12-hour discrete cycle do all three hands coincide at some other position, for the hour hand is always elsewhere (1-11, or 3600-etc) apart from noon-midnight, at every other possibility of the minute and second hand coinciding. In the discrete model, if you wish to speak of any two hands coinciding, you must at least be at the top of an hour, or otherwise at noon-midnight.

We can even entertain what it means for two positions to be antipodal, and then pose/rehash the related set of questions along the lines of the above. If two unequal positions are antipodal, then the result of subtracting the smaller of the two from the larger is always exactly 21600. By simple analogy, consider the 1-12 on a standard clock face: we say that 12 and 6 are antipodal points, 1 and 7 antipodal, and so on. The difference by subtracting the smaller from the larger in every case is 6. Simply multiply 6 by 3600 to give our new base, 21600 -or in a sense, exactly half of the cycle.

Are any two of the hands ever antipodal? Why or why not?

>the one in your high school English class
>the one next to your mom's bed
>the one at the strip club your mom works at

>how do clocks work i dunno lol

>grad school in a nutshell

depends on the clock really. the question does not specify how the hour arm behaves

Consider the minute hand versus the second hand. (Again,) the only times at which the two have a chance to be antipodal are the legal positions of the second hand, or the multiples of 720. But each time that the minute hand is at one of these positions, the second hand is at 43200 starting a new minute. Thus the only situations once per hour where the two hands are antipodal are the x:30:00 situations, at the bottom of any one given hour, when the minute hand is at bottom-dead-center (21600), while the second hand is of course at 43200. Incidentally notice that the larger minus the smaller is of course 21600, fitting our definition of antipodal positions in the discrete model.

Likewise, the hour hand and second hand can only be antipodal at the multiples of 720, and agan the hour hand advances to one of these sixty positions once every 12 minutes. Each time, the second hand is of course at 43200 and by the same logic as the above, the only time in a 12-hour cycle at which the hour hand and the second hand are antipodal is thus 6:00:00, or (21600, 43200, 43200). We notice immediately that this situation entails one, distinct from the above case, where the /minute hand and the hour hand/ are antipodal, at 6 o'clock, and not at noon-midnight.

Are there any other times when the minute hand and the hour hand are directly antipodal? We have every expectation that after an ugly bruteforcing of the converse of the above "never coincides" logic, that we would find that the answer is no. But the fact is that we haven't checked yet. Still, to wrap up this treatment of the discrete model as far as we care to push it, we propose to do exactly that.

Finally it is of course absurd to speak of /all three/ of the hands where any two are antipodal each to the other. The closest analogue would be to identify the situation where each hand is as close to a 120-120-120 degree arrangement as possible, whether exactly in the discrete model or by closest approximation.

This is exactly why I am developing a condiration of a "discrete" model (most probably a pure math abstraction), to be followed immediately by the consideration of the continuous model (gear-ratios that are actually practicable from an engineering point of view?)

I made a thread about this topic recently (I am not the OP I'm just taking the opportunity to develop my thoughts) and one poster flatly indicated that "that is not how clocks work", in reference to my discrete model. It seems to me that any piece of engineering (timepieces) must approximate the above models to within a certain tolerance in order to be a useful clock, however.

The discrete model is most likely to be correct. I do not know of any clockwork which is not discrete motion. Unfortunately without knowing the gearing inside the particular clock it's kind of impossible to know the exact position of the hands, because even in apparently continuous hand movement you'd have to first know how often the hair spring induces motion. Probably there are only a practical few choices on all this, though. I'd guess the spring oscillates either once every half second, quarter second, or eighth of a second. But this is just a guess.

Another instance where continuous analysis is a degenerate case of discrete analysis. (With apologies to Zeilberger.)

The discrete model does not apply to the modern electronic (and old electro-mechanical) silent clocks that neither 'tick' nor jump. The modern version runs on a tiny synchronous motor followed by a normal gear train and the second hand moves continuously. This is a 'true' analog clock (still radio-controlled). I have one in my office and I like it a lot. It reminds you that time silently 'flows'.

Consider the time frame from midnight until (but not including) noon.
- When will the hour hand and the minute hand share a common position?
- When will they form a straight line?
- How often could you interchange the two pointers and still get a valid clock reading?

Your language seems to suggest a blend of discrete-and-continuous, depending on the hand, but I can't be sure based on the mechanical description. But I've seen clocks with the second hand moving in a smooth continuous motion, while I also noticed that the other hands did not "jump" at a minute's start/end, so that I assumed just by eyeballing it that the actual mechanical principle (which must necessarily approximate one of the two above mathematical models in order to be a decent Western timepiece).

You have also rephrased certain questions that I've already asked and answered, especially the first one. This redux of the same questions rhetorically suggests that in the mechanical thing that you're talking about, the answers are somehow different. If so, how? The third question also seems to me to be in one sense trivial in that if you perfectly interchange the hour hand and the minute hand (and their roles) at any one moment in time, and if in this interchange you still understand which is meant to be doing which, then you get the same correct clock reading. The trouble is in which hand moves 12 times as fast as the other hand, and whether that information carries over in such a theoretical swap or not.

the sentence fragment at the end of this first paragraph should be concluded with "is continuous for all three hands".

pi()/24 radians. The minute hand is at 0 (unit circle), and the hour hand is 1/4 of 1/12th the clock away.

1 - At h(12/11) hours with h = 0..10, 11 solutions.
2 - (6/11) hours later, also 11 solutions.
3 - Surprisingly frequent. It happens about every 5 minutes and 2.1 seconds. The exact moments are at (n*12)/(11*13) hours with n = 0..(11*13)-1 (-1 because 12:00 would be the exluded noon). So 143 occurrences in the given interval, first at 00:00:00, last at 11:54:57.9

>decent Western timepiece
Here in the East time has no hiccup.