Workin' on radian time

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Workin' on radian time

Lance Latham
Group -

In the U.S, a Sunday newspaper supplement known as
'Parade' magazine is commonly available. The mag
carries a column by Marilyn vos Savant, supposedly the
planet's smartest human. Since the column represents
her sole visible contribution to the progress of the
human race, one might question the claim, but
whatever.

In the column, Marilyn answers questions that are
intended to be 'clever'. The most recent Sunday issue
carried the following question:

What do the following clock times have in common?
12:00, 1:05, 2:11, 3:16, 4:22, 5:27, 6:33, 7:38, 8:44,
9:49, 10:55.

The answer is fairly obvious, but I'll leave it
unanswered here for those who wish to work on it.

The question does, however, lead to some more
interesting questions of possible interest to this
group.

Answer and additional questions below:






























First, the answer to the original question is that the
times represent moments at which the hour hand and
minute hand of an analog clock overlap, rounded to the
nearest minute.

A point of interest here is the observation that the
problem can be solved in a variety of ways.

One can approach the question as a variant of Zeno's
paradox of Hercules and the tortoise, for example, in
which the minute hand races along to the last point
where the hour hand was stationed, only to find that,
in the intervening time period, the hour hand has
moved forward again.

This approach leads to a solution by infinite series.

Another and shorter approach is to equate the
algebraic expressions for the time elapsed in catching
the hour hand, and solve for the angular position in
radians, then convert the result to decimal hours.

The problem also leads to some interesting extensions,
such as:

1. At what times do the hour, minute AND second hands
align? Do they ever align at all?

2. How does this problem translate to a 24-hour clock?
Does it translate at all?

3. Show a very simple model that permits calculation
of decimal hours for hour-minute hand overlap for an
arbitrary clock of N hours (and 60-minute hours). Does
a simpler model (equation) exist?

-Lance


Lance Latham
[hidden email]
Phone:    (518) 274-0570
Address: 78 Hudson Avenue/1st Floor, Green Island, NY 12183
 




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Re: Workin' on radian time

VictorEngel
Dear Lance and Calendar People,

> In the U.S, a Sunday newspaper supplement known as
> 'Parade' magazine is commonly available. The mag
> carries a column by Marilyn vos Savant, supposedly the
> planet's smartest human.

I think the claim is that she has the highest I.Q., not necessarily that she
is the smartest. And there is some controversy about her I.Q. as well. At
the very least, one needs to know how the various I.Q.s were derived.:
http://en.wikipedia.org/wiki/Marilyn_vos_Savant#What_is_her_IQ.3F


> In the column, Marilyn answers questions that are
> intended to be 'clever'.

... but which usually come off as being trite, in my opinion (ever notice
that "in my opinion" is right-handed on a standard keyboard?).

> 1. At what times do the hour, minute AND second hands
> align? Do they ever align at all?

Uhhh... Obviously, they align at midnight, by definition.
 
> 2. How does this problem translate to a 24-hour clock?
> Does it translate at all?
>
> 3. Show a very simple model that permits calculation
> of decimal hours for hour-minute hand overlap for an
> arbitrary clock of N hours (and 60-minute hours). Does
> a simpler model (equation) exist?

4. How about extending the problem to more hands and to
nonmultiples, such as the orbits of the planets?

By the way, my first approach to the problem would probably
be to use modulus arithmetic, noting that each hand has
a given speed, so each hand has a specific equation for
position (degrees as function of seconds after midnight):

h(t) = t/120
m(t) = t/10
s(t) = 6t

The question is equivalent to solving for h(t) = m(t) mod 360
and m(t) = s(t) mod 360.

Victor
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Re: Workin' on radian time

Lance Latham
RE:

... Marilyn vos Savant, supposedly
> the
> > planet's smartest human.
>
> I think the claim is that she has the highest I.Q.,

Lance replies:
I believe that's the actual claim. I was just
short-handing there. I don't put a lot of faith in
I.Q. scores in the near 100th-percentile. Anyway,
she's supposed to be intelligent.

>> > In the column, Marilyn answers questions that are
> > intended to be 'clever'.
>
> ... but which usually come off as being trite, in my
> opinion (ever notice
> that "in my opinion" is right-handed on a standard
> keyboard?).

Lance replies:
They generally run along the lines of 'what do these
words have in common?', with an answer based on
anagrammatic features or something else trivial. I'd
be more impressed if she had an answer for Riemann's
hypothesis.

> > 1. At what times do the hour, minute AND second
> hands
> > align? Do they ever align at all?
>
> Uhhh... Obviously, they align at midnight, by
> definition.

Lance queries:
Any other times? Why or why not?

> > 3. Show a very simple model that permits
> calculation
> > of decimal hours for hour-minute hand overlap for
> an
> > arbitrary clock of N hours (and 60-minute hours).
> Does
> > a simpler model (equation) exist?
>
> 4. How about extending the problem to more hands and
> to
> nonmultiples, such as the orbits of the planets?

Lance replies:
I considered extending the question to larger temporal
units but decided against it. Early clocks did things
like that, so it might warrant a look. The problems
start immediately, though with non-standard month
lengths and intercalation in annual, or other, cycles.

But it opens the door to a combined clock-calendar
that defines other units. There's plenty to play with
there.

 

> By the way, my first approach to the problem would
> probably
> be to use modulus arithmetic, noting that each hand
> has
> a given speed, so each hand has a specific equation
> for
> position (degrees as function of seconds after
> midnight):
>
> h(t) = t/120
> m(t) = t/10
> s(t) = 6t
>
> The question is equivalent to solving for h(t) =
> m(t) mod 360
> and m(t) = s(t) mod 360.

Lance replies:
A Chinese Remainder approach, then. I did not try that
approach, and do not believe that it will lead to a
solution, but I'm willing to be convinced.

-Lance


Lance Latham
[hidden email]
Phone:    (518) 274-0570
Address: 78 Hudson Avenue/1st Floor, Green Island, NY 12183
 




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Re: Workin' on radian time

VictorEngel
In reply to this post by Lance Latham
Dear Lance,

> Lance queries:
> Any other times? Why or why not?

Noon.
 

> Lance replies:
> A Chinese Remainder approach, then. I did not try that
> approach, and do not believe that it will lead to a
> solution, but I'm willing to be convinced.

I don't believe there is a solution other than noon and midnight. The only
possible solutions would be the 11 times when the hour and minute hand align
(at which point the cycle repeats). These alignments occur exactly every
86400/22 seconds. At none of these times does the second hand align.
Therefore there is no time all three align except at the 12.

You might have better luck with a 13 hour clock (13-1 is not prime like 12-1
is), or by dividing a minute into a number of units that is a multiple of
11.

In general, I would guess that the fastest hand would have to move a
multiple of the quotient of the other two hands minus one times as fast as
the medium hand. I have not yet tested this idea.

Victor
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Re: Workin' on radian time

Lance Latham
RE:

---> > Any other times? Why or why not?
>
> Noon.

Lance replies:
The clock in the original question is a 12-hour clock,
so noon and midnight have the same representation of
12:00.  

> I don't believe there is a solution other than noon
> and midnight. The only
> possible solutions would be the 11 times when the
> hour and minute hand align
> (at which point the cycle repeats). These alignments
> occur exactly every
> 86400/22 seconds. At none of these times does the
> second hand align.

Lance replies:
One way to approach this problem is to calculate the
times when the hour and minute hands align, and look
at the value of the seconds unit. That might leave
open the question of whether there are other times
when all three hands align.

> In general, I would guess that the fastest hand
> would have to move a
> multiple of the quotient of the other two hands
> minus one times as fast as
> the medium hand. I have not yet tested this idea.

Lance replies:
Go for it. When everyone is exasperated or bored or
whatever, I'll post my results. This shouldn't be
rocket science, but it does have the quality of
suggesting more questions, which is always nice.

It should be assumed, of course, that the hands of the
analog clock are moving continuously and smoothly, not
jumping. That is, it really is continuous, analog
time.

-Lance


Lance Latham
[hidden email]
Phone:    (518) 274-0570
Address: 78 Hudson Avenue/1st Floor, Green Island, NY 12183
 




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the Point Re: Workin' on radian time

Brij Bhushan Vij
Lance & All:
>>I don't believe there is a solution other than noon
>and midnight. The only
>possible solutions would be the 11 times when the
>hour and minute hand align
>(at which point the cycle repeats). These alignments
>occur exactly every
>86400/22 seconds. At none of these times does the
>second hand align.
While this may be one area, needing more questions to be answered: What's
the point?
Even for 'sidereral astronomy' as I have attempted to hint, the question of
time count wrt its likage with arcAngle via my DEFINITIONS for 'New' time
unit - decimal or sidereal second; when linked with the New Length Unit (m')
as: *1/10^5th of Pi/180 or one-degree* automatically get resolved in fixing
the HOUR-Angle.
Thus, we have our existing working of 12/24-hourx100x100 units (2400000
decimalor sidereal seconds) LINKED to One-degreex100'x100" (arc-Angle) as
Earth spins around its axis during the day.
Regards,
Brij Bhushan Vij
(Monday, Kali 5107-W07-01)/265+D-151 (Tuesday, 2006 May 30H20:59(decimal) ET
Aa Nau Bhadra Kritvo Yantu Vishwatah -Rg Veda
Jan:31; Feb:29; Mar:31; Apr:30; May:31; Jun:30
Jul:30; Aug:31; Sep:30; Oct:31; Nov:30; Dec:30
(365th day of Year is World Day)
******As per Kali V-GRhymeCalendaar*****
"Koi bhi cheshtha vayarth nahin hoti, purshaarth karne mein hai"
Contact # 001(201)675-8548


>From: Lance Latham <[hidden email]>
>Reply-To: East Carolina University Calendar discussion List              
><[hidden email]>
>To: [hidden email]
>Subject: Re: Workin' on radian time
>Date: Tue, 30 May 2006 14:44:17 -0700
>
>RE:
>
>---> > Any other times? Why or why not?
> >
> > Noon.
>
>Lance replies:
>The clock in the original question is a 12-hour clock,
>so noon and midnight have the same representation of
>12:00.
>
> > I don't believe there is a solution other than noon
> > and midnight. The only
> > possible solutions would be the 11 times when the
> > hour and minute hand align
> > (at which point the cycle repeats). These alignments
> > occur exactly every
> > 86400/22 seconds. At none of these times does the
> > second hand align.
>
>Lance replies:
>One way to approach this problem is to calculate the
>times when the hour and minute hands align, and look
>at the value of the seconds unit. That might leave
>open the question of whether there are other times
>when all three hands align.
>
> > In general, I would guess that the fastest hand
> > would have to move a
> > multiple of the quotient of the other two hands
> > minus one times as fast as
> > the medium hand. I have not yet tested this idea.
>
>Lance replies:
>Go for it. When everyone is exasperated or bored or
>whatever, I'll post my results. This shouldn't be
>rocket science, but it does have the quality of
>suggesting more questions, which is always nice.
>
>It should be assumed, of course, that the hands of the
>analog clock are moving continuously and smoothly, not
>jumping. That is, it really is continuous, analog
>time.
>
>-Lance
>
>
>Lance Latham
>[hidden email]
>Phone:    (518) 274-0570
>Address: 78 Hudson Avenue/1st Floor, Green Island, NY 12183
>
>
>
>
>
>__________________________________________________
>Do You Yahoo!?
>Tired of spam?  Yahoo! Mail has the best spam protection around
>http://mail.yahoo.com
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Re: Workin' on radian time

Palmen, KEV (Karl)
In reply to this post by Lance Latham
Dear Lance, Victor and Calendar People

Here is my reply.

-----Original Message-----
From: East Carolina University Calendar discussion List
[mailto:[hidden email]]On Behalf Of Lance Latham
Sent: 30 May 2006 20:00
To: [hidden email]
Subject: Workin' on radian time


Group -

In the U.S, a Sunday newspaper supplement known as
'Parade' magazine is commonly available. The mag
carries a column by Marilyn vos Savant, supposedly the
planet's smartest human. Since the column represents
her sole visible contribution to the progress of the
human race, one might question the claim, but
whatever.

In the column, Marilyn answers questions that are
intended to be 'clever'. The most recent Sunday issue
carried the following question:

What do the following clock times have in common?
12:00, 1:05, 2:11, 3:16, 4:22, 5:27, 6:33, 7:38, 8:44,
9:49, 10:55.

The answer is fairly obvious, but I'll leave it
unanswered here for those who wish to work on it.

The question does, however, lead to some more
interesting questions of possible interest to this
group.

Answer and additional questions below:

<snip>

First, the answer to the original question is that the
times represent moments at which the hour hand and
minute hand of an analog clock overlap, rounded to the
nearest minute.

<snip>

The problem also leads to some interesting extensions,
such as:

1. At what times do the hour, minute AND second hands
align? Do they ever align at all?

KARL SAYS:
This is an interesting problem. They align at 12 o'clock and could align at other times. In 1/11 of 12 hours the second hand turns 60/11 times, which equals 5 5/11. Because the numerator 5 is not 1 and 11 is prime, there can be no additional triple alignments. In my next note you'll see a solution to a generalisation of this problem.

<snip>

3. Show a very simple model that permits calculation
of decimal hours for hour-minute hand overlap for an
arbitrary clock of N hours (and 60-minute hours). Does
a simpler model (equation) exist?

KARL SAYS:
I know of no such technique (or model) valid for all N or even just N=12.

However it can be done in reverse from a decimal to a 12-hour clock. This is because a decimal clock would have its hands align 9 times a day (for 10 decimal hours per day). The 36 times when the two hands are aligned, opposite or are at right angles occur at an intervals of 40 minutes exactly. This can be used to convert to 12-hour time. If there are 10 decimal hours in 12 hours, then this interval is 20 minutes.


Also of interest are the times of:
12:00 12:55 1:51 2:46 3:42 4:37 5:32 6:28 7:23 8:18 9:14 10:09 11:05.
I note clocks in shops and mail-order catalogues often say one of these times.


Karl


08(04(04
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Re: Workin' on radian time

Palmen, KEV (Karl)
In reply to this post by VictorEngel
Dear Victor and Calendar People

-----Original Message-----
From: East Carolina University Calendar discussion List
[mailto:[hidden email]]On Behalf Of Engel,Victor
Sent: 30 May 2006 22:03
To: [hidden email]
Subject: Re: Workin' on radian time


Dear Lance,

> Lance queries:
> Any other times? Why or why not?

Noon.
 

> Lance replies:
> A Chinese Remainder approach, then. I did not try that
> approach, and do not believe that it will lead to a
> solution, but I'm willing to be convinced.

I don't believe there is a solution other than noon and midnight. The only
possible solutions would be the 11 times when the hour and minute hand align
(at which point the cycle repeats). These alignments occur exactly every
86400/22 seconds. At none of these times does the second hand align.
Therefore there is no time all three align except at the 12.

You might have better luck with a 13 hour clock (13-1 is not prime like 12-1
is), or by dividing a minute into a number of units that is a multiple of
11.

KARL SAYS:
At 1/12 of the way through the clock cycle, the second hand would have rotated 60*13/12 times which is a whole number (65). So whenever the two hands are aligned the 'second' hand would point straight up. So no better luck!

VICTOR CONTINUES:
In general, I would guess that the fastest hand would have to move a
multiple of the quotient of the other two hands minus one times as fast as
the medium hand. I have not yet tested this idea.


KARL SAYS:
Suppose the long hand goes round N times faster than the short hand then the 'second' hand must go M times faster than the long hand. For all N-1 triple alignments to occur M must satisfy
M mod (N-1) = 1.
For N=12 we get M=1,12,23,34,45,56,67 etc.. .

Note that all N-1 triple alignments occur if M=N.

If N-1 is composite we can get some additional triple alignments with some other 'second' hand speeds. I've already shown this for N=10 and M=60 which aligns 3 times per clock cycle. To get T triple alignments per clock cycle where T is a divisor of N-1, one need only choose M such that M mod T = 1.

So Victor could get four triple alignments on his 13-hour clock if he were to have either 25 or 65 minutes in his hour. T=4 is particularly interesting because whenever the first two hands are aligned, the 'second' hand is aligned, opposite or perpendicular to them both. However, T=4 cannot occur with an even-number-hour clock.


Summarising the solution:

If the fastest hand turns M times faster than the medium speed hand,
the medium speed hand turns N times faster than the slowest hand
and there are T alignments of all three hands in one clock cycle, then the following all apply:
(1) T is a divisor of N-1.
(2) M is one greater than a multiple of T.
(3) T is the greatest divisor of N-1 that satisfies (2).


Karl

08(04(04
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Re: Workin' on radian time

VictorEngel
In reply to this post by Lance Latham
Dear Karl, Lance, and Calendar People,

I remind the list members that my Septimal clock at

http://the-light.com/cal/veseptimal.html

has hands that line up twice per hour.

Victor
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Re: Workin' on radian time

Palmen, KEV (Karl)
Dear Victor Lance and Calendar People

-----Original Message-----
From: East Carolina University Calendar discussion List
[mailto:[hidden email]]On Behalf Of Engel,Victor
Sent: 31 May 2006 14:13
To: [hidden email]
Subject: Re: Workin' on radian time


Dear Karl, Lance, and Calendar People,

I remind the list members that my Septimal clock at

http://the-light.com/cal/veseptimal.html

has hands that line up twice per hour.

KARL SAYS: I think N=M=49

Then T=48 and so the do hands align twice per hour, but it would be next to impossible to distinguish two consecutive aligned times.

If we were to have N=M=7, all three hands would be aligned once every four hours and the slowest and fastest hands would be aligned once every half an hour. Furthermore whenever these two hands are aligned the other (medium) hand will be a multiple of 45 degree relative to these two aligned hands.

Karl

08(04(04
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Re: Workin' on radian time

Lance Latham
In reply to this post by Palmen, KEV (Karl)
Group -

Since it's possible that not everyone has had an
opportunity to work on this yet, I won't spoil it by
revealing a full set of answers, but I'll release a
solution for the original problem. That solution,
applied to the 24-hour clock, leads to the general
equation to which I referred earlier.

Here goes.

The 12-hour clock represents a 12-hour period. There
are 2*pi radians in the circle of the clock face, so
the hour mark '1' is at an angle:

   theta = 2*pi   = pi
           ----    ---- radians
            12       6

from the position marked '12'.

The angular speed of the hour hand, H, is therefore:

  rH = pi/6 rad/hr

Measured in minutes, rH = pi/360 rad/min.

The corresponding angle for the minute hand, M,
represents 5 minutes, so the angular speed of M is:

   rM = pi/6 * 1/5 rad = pi/30 rad/min

There are several methods for proceeding to a
solution. Perhaps the easiest one goes like this. One
wishes to solve for the exact time when H and M
overlap for the first case, 1:05. Draw a straight line
that represents the distance 'd' around the clock
face.

  '12'           '1'
   P0             P1         P2
   |--------------|-----------|-------------> d

The distances do not have to be to scale. One knows
that the distance (P1 - P0) = pi/6 radians. The
distance (P2 - P1) can be called 'x'. Initially, set M
to P0 and H to P1, to represent the time 1:00.

One wishes to advance M at rate rM from P0, and H at
rate rH from P1 until M overtakes H at point P2. The
difference 'x' represents the time past 1:05 consumed
by H while M catches up.

A solution for 'x' can be had by setting up 2
equations. Since M and H must both arrive at P2
simultaneously, they spend the same amount of time 't'
to do so. Thus one may write:

   t = x / (pi/360) for H

and

   t = (pi/6 + x) / (pi/30)

using the identity:

   time = distance / rate

for both M and H, using their respective distances and
rates.

One then has the equation of the two, which can be
solved for 'x', which turns out to be:

   x = 5*pi/330  or pi/66

Thus the entire distance (P2 - P0) is:

   d = pi/6 + x

which turns out to be 0.571199 radians.

Since 1 radian = 6/pi hours, the conversion from
radians to decimal hours is trivial, yielding the
result:

   D = (6/pi) * d = 1.09090909... hours

which can then be easily converted to HMS format to
yield the final result of:

   1:05:27, rounded to the nearest second.

This is the exact moment (to the nearest second) when
M overtakes H and aligns with it.

Some hints for further solution follow below.















1. Work the second case for clock time 2:11. From the
result, one should be able to deduce a simple formula
for calculating the H-M match times for any hour 'h'
for a 12-hour clock. (Variable 'h' has values
1,2,...11 for a 12-hour clock.)

2. Work the first case for a 24-hour clock, i.e., when
do H and M align on a 24-hour clock, using the method
described, starting with H at '1', M at '12'.

3. From the result, one should be able to deduce a
simple formula to predict decimal hours, D, from 'h'
for an N-hour clock.

Good hunting!

-Lance


Lance Latham
[hidden email]
Phone:    (518) 274-0570
Address: 78 Hudson Avenue/1st Floor, Green Island, NY 12183
 




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Re: Workin' on radian time

VictorEngel
In reply to this post by Lance Latham
Dear Karl,

If you have defined what N and M are, I think it was lost. The last email I
have from you explains that a subsequent email would give your solution. I
did not receive another post from you until this one.

Victor

> -----Original Message-----
> From: East Carolina University Calendar discussion List
> [mailto:[hidden email]]On Behalf Of Palmen, KEV (Karl)
> Sent: Wednesday, May 31, 2006 8:31 AM
> To: [hidden email]
> Subject: Re: Workin' on radian time
>
>
> Dear Victor Lance and Calendar People
>
> -----Original Message-----
> From: East Carolina University Calendar discussion List
> [mailto:[hidden email]]On Behalf Of Engel,Victor
> Sent: 31 May 2006 14:13
> To: [hidden email]
> Subject: Re: Workin' on radian time
>
>
> Dear Karl, Lance, and Calendar People,
>
> I remind the list members that my Septimal clock at
>
> http://the-light.com/cal/veseptimal.html
>
> has hands that line up twice per hour.
>
> KARL SAYS: I think N=M=49
>
> Then T=48 and so the do hands align twice per hour, but it
> would be next to impossible to distinguish two consecutive
> aligned times.
>
> If we were to have N=M=7, all three hands would be aligned
> once every four hours and the slowest and fastest hands would
> be aligned once every half an hour. Furthermore whenever
> these two hands are aligned the other (medium) hand will be a
> multiple of 45 degree relative to these two aligned hands.
>
> Karl
>
> 08(04(04
>
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r-sec & 11-Hour Clock Re: Workin' on radian time

Brij Bhushan Vij
In reply to this post by Lance Latham
Lance, group:
>>Since it's possible that not everyone has had an
>opportunity to work on this yet, I won't spoil it by
>revealing a full set of answers, but I'll release a
>solution for the original problem. That solution,
>applied to the 24-hour clock, leads to the general
>equation to which I referred earlier.
It is NOT difficult to achieve what Lance has in mind - BUT to what
advantage? Our minds are set to use 24-hour clock work and 2 Pi radian
imbedded into our 'conceptions'. For any New adventure, we must be clear
that any system can be made to work so long as we are 'linking' the *NEW
TIME UNIT WITH THE NEW LENGTH UNIT* for our clock mechanisms to work for
astro-calendaric calculations! I show this as:
Radian Time: 86400/22 sec =3927.273 sec =65.455 min =1hr 5m 27s.273
[1s=9,192,631,770 cs-133]
1. 65.455 min =1hr 5m 27s.273 2. 130.91min =2h 10m 54s.6
3. 196.365 min =3h 16m 21s.9 4. 261.82 min =4h 21m 49s.2
5. 327.275 min=5h 27m 16s.5 6. 392.73 min =6h 32m 43s.8
7. 458.185 min =7h 38m 11s.1 8. 523.64 min =8h 43m 38s.4
9. 589.095 min =9h 49m 05s.7 10. 654.55 min =10h 54m 33s
11. 720.005 min =12-HOURS [Note: the difference of 0.3 sec is due to
TRUNCATION].
The 11-Hour Clock can have HOUR-angle =43200/11 atomic-seconds (or r-sec).
     Thus, r-sec can be defined as: *1/110000th  of the atomic day =0.7855 s
or (72120374745 cs-133~]. If Earth were to remain 2Pi ‘radians’, the
HOUR-angle would be 32.7273-degree.
If a REFORM of 'time & calendar is desired' our thoughts must be practical
from view point of a man-on-street and that: what this costs to tax-payer?
Regards,
Brij Bhushan Vij
(Tuesday, Kali 5107-W07-02)/265+D-152 (Wednesday, 2006 May 31H10:88(decimal)
ET
Aa Nau Bhadra Kritvo Yantu Vishwatah -Rg Veda
Jan:31; Feb:29; Mar:31; Apr:30; May:31; Jun:30
Jul:30; Aug:31; Sep:30; Oct:31; Nov:30; Dec:30
(365th day of Year is World Day)
******As per Kali V-GRhymeCalendaar*****
"Koi bhi cheshtha vayarth nahin hoti, purshaarth karne mein hai"
Contact # 001(201)675-8548


>From: Lance Latham <[hidden email]>
>Reply-To: East Carolina University Calendar discussion List              
><[hidden email]>
>To: [hidden email]
>Subject: Re: Workin' on radian time
>Date: Wed, 31 May 2006 06:37:05 -0700
>
>Group -
>
>Since it's possible that not everyone has had an
>opportunity to work on this yet, I won't spoil it by
>revealing a full set of answers, but I'll release a
>solution for the original problem. That solution,
>applied to the 24-hour clock, leads to the general
>equation to which I referred earlier.
>
>Here goes.
>
>The 12-hour clock represents a 12-hour period. There
>are 2*pi radians in the circle of the clock face, so
>the hour mark '1' is at an angle:
>
>    theta = 2*pi   = pi
>            ----    ---- radians
>             12       6
>
>from the position marked '12'.
>
>The angular speed of the hour hand, H, is therefore:
>
>   rH = pi/6 rad/hr
>
>Measured in minutes, rH = pi/360 rad/min.
>
>The corresponding angle for the minute hand, M,
>represents 5 minutes, so the angular speed of M is:
>
>    rM = pi/6 * 1/5 rad = pi/30 rad/min
>
>There are several methods for proceeding to a
>solution. Perhaps the easiest one goes like this. One
>wishes to solve for the exact time when H and M
>overlap for the first case, 1:05. Draw a straight line
>that represents the distance 'd' around the clock
>face.
>
>   '12'           '1'
>    P0             P1         P2
>    |--------------|-----------|-------------> d
>
>The distances do not have to be to scale. One knows
>that the distance (P1 - P0) = pi/6 radians. The
>distance (P2 - P1) can be called 'x'. Initially, set M
>to P0 and H to P1, to represent the time 1:00.
>
>One wishes to advance M at rate rM from P0, and H at
>rate rH from P1 until M overtakes H at point P2. The
>difference 'x' represents the time past 1:05 consumed
>by H while M catches up.
>
>A solution for 'x' can be had by setting up 2
>equations. Since M and H must both arrive at P2
>simultaneously, they spend the same amount of time 't'
>to do so. Thus one may write:
>
>    t = x / (pi/360) for H
>
>and
>
>    t = (pi/6 + x) / (pi/30)
>
>using the identity:
>
>    time = distance / rate
>
>for both M and H, using their respective distances and
>rates.
>
>One then has the equation of the two, which can be
>solved for 'x', which turns out to be:
>
>    x = 5*pi/330  or pi/66
>
>Thus the entire distance (P2 - P0) is:
>
>    d = pi/6 + x
>
>which turns out to be 0.571199 radians.
>
>Since 1 radian = 6/pi hours, the conversion from
>radians to decimal hours is trivial, yielding the
>result:
>
>    D = (6/pi) * d = 1.09090909... hours
>
>which can then be easily converted to HMS format to
>yield the final result of:
>
>    1:05:27, rounded to the nearest second.
>
>This is the exact moment (to the nearest second) when
>M overtakes H and aligns with it.
>
>Some hints for further solution follow below.
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>1. Work the second case for clock time 2:11. From the
>result, one should be able to deduce a simple formula
>for calculating the H-M match times for any hour 'h'
>for a 12-hour clock. (Variable 'h' has values
>1,2,...11 for a 12-hour clock.)
>
>2. Work the first case for a 24-hour clock, i.e., when
>do H and M align on a 24-hour clock, using the method
>described, starting with H at '1', M at '12'.
>
>3. From the result, one should be able to deduce a
>simple formula to predict decimal hours, D, from 'h'
>for an N-hour clock.
>
>Good hunting!
>
>-Lance
>
>
>Lance Latham
>[hidden email]
>Phone:    (518) 274-0570
>Address: 78 Hudson Avenue/1st Floor, Green Island, NY 12183
>
>
>
>
>
>__________________________________________________
>Do You Yahoo!?
>Tired of spam?  Yahoo! Mail has the best spam protection around
>http://mail.yahoo.com
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Re: Workin' on radian time

Palmen, KEV (Karl)
In reply to this post by Lance Latham
Dear Lance and Calendar People

Lances explanation seems to be unnecessarily complicated.
Can any calendar person make it simpler.

I would NOT regard any conversion to decimal time based upon it as simple.


Here's my simplification:

Firstly, one can use the speed of the slowest hand as the unit speed.
Then in the 12 hour clock. The hour hand travels at speed 1 the minute hand at speed 12. The speed in this unit gives the number of revolutions in one clock cycle (12 hours in this case).

The angle between the hour hand and the minute hand changes at their relative speed which is 11 and that's why they meet 11 times in one clock cycle (of 12 hours). 1/11 of 12 hours is 01h 05m 27s to the nearest second or 01h 05 27s 3/11s exactly. This exact value enables you to work out all 11 values exactly:

01:05:27 3/11
02:11:54 6/11
03:17:21 9/11
04:22:49 1/11
05:27:16 4/11
06:32:43 7/11
07:38:10 10/11
08:43:38 2/11
09:49:05 5/11
10:54:32 8/11

For the 24-hour clock the relative speed is 23 in units of the hour hand speed. So the interval is 24/23 hour = 01h 02m 36s 12/23s exactly.


For the problem of finding how many times T that all three hands of a clock whose 3 hands meet at least once and whose hands travel at speeds 1, N and M*N, I get the answer:

T is the largest divisor of N-1, such that M is one greater than a multiple of T.  
This is equivalent of saying that T is the largest common divisor of N-1 and M-1 (only just realised that!).

We get for example the following:

N  M   T
12 60  1
10 60  1
13 60  1
13 65  4
13 70  3
13 25  12
10 10  9
10 100 9
12 144 11
24 70  23

Karl

08(04(04


-----Original Message-----
From: East Carolina University Calendar discussion List
[mailto:[hidden email]]On Behalf Of Lance Latham
Sent: 31 May 2006 14:37
To: [hidden email]
Subject: Re: Workin' on radian time


Group -

Since it's possible that not everyone has had an
opportunity to work on this yet, I won't spoil it by
revealing a full set of answers, but I'll release a
solution for the original problem. That solution,
applied to the 24-hour clock, leads to the general
equation to which I referred earlier.

Here goes.

The 12-hour clock represents a 12-hour period. There
are 2*pi radians in the circle of the clock face, so
the hour mark '1' is at an angle:

   theta = 2*pi   = pi
           ----    ---- radians
            12       6

from the position marked '12'.

The angular speed of the hour hand, H, is therefore:

  rH = pi/6 rad/hr

Measured in minutes, rH = pi/360 rad/min.

The corresponding angle for the minute hand, M,
represents 5 minutes, so the angular speed of M is:

   rM = pi/6 * 1/5 rad = pi/30 rad/min

There are several methods for proceeding to a
solution. Perhaps the easiest one goes like this. One
wishes to solve for the exact time when H and M
overlap for the first case, 1:05. Draw a straight line
that represents the distance 'd' around the clock
face.

  '12'           '1'
   P0             P1         P2
   |--------------|-----------|-------------> d

The distances do not have to be to scale. One knows
that the distance (P1 - P0) = pi/6 radians. The
distance (P2 - P1) can be called 'x'. Initially, set M
to P0 and H to P1, to represent the time 1:00.

One wishes to advance M at rate rM from P0, and H at
rate rH from P1 until M overtakes H at point P2. The
difference 'x' represents the time past 1:05 consumed
by H while M catches up.

A solution for 'x' can be had by setting up 2
equations. Since M and H must both arrive at P2
simultaneously, they spend the same amount of time 't'
to do so. Thus one may write:

   t = x / (pi/360) for H

and

   t = (pi/6 + x) / (pi/30)

using the identity:

   time = distance / rate

for both M and H, using their respective distances and
rates.

One then has the equation of the two, which can be
solved for 'x', which turns out to be:

   x = 5*pi/330  or pi/66

Thus the entire distance (P2 - P0) is:

   d = pi/6 + x

which turns out to be 0.571199 radians.

Since 1 radian = 6/pi hours, the conversion from
radians to decimal hours is trivial, yielding the
result:

   D = (6/pi) * d = 1.09090909... hours

which can then be easily converted to HMS format to
yield the final result of:

   1:05:27, rounded to the nearest second.

This is the exact moment (to the nearest second) when
M overtakes H and aligns with it.

Some hints for further solution follow below.















1. Work the second case for clock time 2:11. From the
result, one should be able to deduce a simple formula
for calculating the H-M match times for any hour 'h'
for a 12-hour clock. (Variable 'h' has values
1,2,...11 for a 12-hour clock.)

2. Work the first case for a 24-hour clock, i.e., when
do H and M align on a 24-hour clock, using the method
described, starting with H at '1', M at '12'.

3. From the result, one should be able to deduce a
simple formula to predict decimal hours, D, from 'h'
for an N-hour clock.

Good hunting!
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Re: Workin' on radian time

Palmen, KEV (Karl)
In reply to this post by Lance Latham
Dear Lance and Calendar People

-----Original Message-----
From: East Carolina University Calendar discussion List
[mailto:[hidden email]]On Behalf Of Lance Latham
Sent: 30 May 2006 20:00
To: [hidden email]
Subject: Workin' on radian time


3. Show a very simple model that permits calculation
of decimal hours for hour-minute hand overlap for an
arbitrary clock of N hours (and 60-minute hours). Does
a simpler model (equation) exist?

KARL SAYS:
It just occurred to me that one could imagine a hand on the a clock that is the reflection of the minute hand in the hour hand. This imaginary hand would go round that clock anticlockwise ten times every twelve hours, which is once in a kind of decimal hour.

Its speed is twice the speed of the hour hand minus the speed of the minute hand. In units of the hour hand speed the result is 2-12= -10.

Karl

08(04(04
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Re: Workin' on radian time

VictorEngel
In reply to this post by Lance Latham
Dear Karl, Lance, and Calendar People,

I said:

> You might have better luck with a 13 hour clock (13-1 is not
> prime like 12-1
> is), or by dividing a minute into a number of units that is a
> multiple of
> 11.
>
> KARL SAYS:
> At 1/12 of the way through the clock cycle, the second hand
> would have rotated 60*13/12 times which is a whole number
> (65). So whenever the two hands are aligned the 'second' hand
> would point straight up. So no better luck!

I didn't mean to imply to retain 60 minute hours. Of course, that would need
to be adjusted.
 
> So Victor could get four triple alignments on his 13-hour
> clock if he were to have either 25 or 65 minutes in his hour.

Right.

Karl Continues:

> Summarising the solution:
>
> If the fastest hand turns M times faster than the medium speed hand,
> the medium speed hand turns N times faster than the slowest hand
> and there are T alignments of all three hands in one clock
> cycle, then the following all apply:
> (1) T is a divisor of N-1.
> (2) M is one greater than a multiple of T.
> (3) T is the greatest divisor of N-1 that satisfies (2).

That is an eloquent way of putting it.

Victor
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Re: Workin' on radian time

VictorEngel
In reply to this post by Lance Latham
Dear Karl and Calendar People,

I said:

> Dear Karl, Lance, and Calendar People,
>
> I remind the list members that my Septimal clock at
>
> http://the-light.com/cal/veseptimal.html
>
> has hands that line up twice per hour.
 
Karl replied:

> KARL SAYS: I think N=M=49
>
> Then T=48 and so the do hands align twice per hour, but it
> would be next to impossible to distinguish two consecutive
> aligned times.

I don't agree, or at least I don't think it matters much. The alignment
positions
are in the same position that the small hand of a 24 hour clock would be in
at the
same time. Every two hours, the positions would be at the 12 familiar clock
face
positions. Every 30 minutes, they would line up at these positions divided
in half
twice. We're already familiar with the 12 clock face positions, and
visualizing halves
comes naturally.

Victor
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Re: Workin' on radian time

VictorEngel
In reply to this post by Lance Latham
Dear Karl and Calendar People,

I originally came up with my Septimal Clock to keep track of the moon phases
in convenient 49th days. It occurs to me that 49th days can be tracked using
a closk with 50 rather than 49 units. If each of 3 hands goes 50 times as
fast as another and the slowest hand rotates once in a day, then all the
hands line up every 49th day.

Victor
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Re: Workin' on radian time

Lance Latham
In reply to this post by Palmen, KEV (Karl)
RE:

> Lances explanation seems to be unnecessarily
> complicated.
> Can any calendar person make it simpler.
>
> I would NOT regard any conversion to decimal time
> based upon it as simple.

Lance replies:
The original solution is relatively straightforward,
but not trivial. The work leads to results that are
much simpler.

The conversion to decimal time is a multiplication.

>
> Here's my simplification:
>
 ...1/11 of 12 hours is 01h 05m 27s to the
> nearest second or 01h 05 27s 3/11s exactly.


Lance replies:
Karl is on the right track here to the simple solution
for the 12-hour clock.

> For the 24-hour clock the relative speed is 23 in
> units of the hour hand speed. So the interval is
> 24/23 hour = 01h 02m 36s 12/23s exactly.

Lance replies:
And for the 24-hour clock.

Running these out to the nearest second will show that
only for 12:00 is it possible to align hour, minute
and second hands exactly. If one tolerates some error
(it cannot be greater than 30 s), then alignment could
be claimed for 1:05:05, for example. An error in terms
of radian angle is most useful, I believe.

Since Karl has figured out the basic simplification,
namely that each clock has a characteristic constant
that determines the clock times for H-M alignment, it
only remains to relate the two clocks by a general
equation, which is:

   D = h * (1 + 1 / (N - 1))

where 'h' is the integer hour and N is the number of
hours on the clock face. That is, if N = 12, N-1 = 11,
and
 
   D = h * (1+ 1/11) = h*1.090909...

Check this using h = 11, which gives D = 11 + 11/11 =
12.

-Lance


Lance Latham
[hidden email]
Phone:    (518) 274-0570
Address: 78 Hudson Avenue/1st Floor, Green Island, NY 12183
 




__________________________________________________
Do You Yahoo!?
Tired of spam?  Yahoo! Mail has the best spam protection around
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Re: Workin' on radian time

Sepp Rothwangl
In reply to this post by Lance Latham
Group,

doesn't this idea calendrically end up in an all planet alignment, or
the GREAT YEAR CONJUNCTION, which was aim of the AD count adjustment?

For me it sounds very familiar ...  ;=)

Servus
Sepp

Am 31.05.2006 um 15:37 schrieb Lance Latham:

> Group -
>
> Since it's possible that not everyone has had an
> opportunity to work on this yet, I won't spoil it by
> revealing a full set of answers, but I'll release a
> solution for the original problem. That solution,
> applied to the 24-hour clock, leads to the general
> equation to which I referred earlier.
>
> Here goes.
>
> The 12-hour clock represents a 12-hour period. There
> are 2*pi radians in the circle of the clock face, so
> the hour mark '1' is at an angle:
>
>    theta = 2*pi   = pi
>            ----    ---- radians
>             12       6
>
> from the position marked '12'.
>
> The angular speed of the hour hand, H, is therefore:
>
>   rH = pi/6 rad/hr
>
> Measured in minutes, rH = pi/360 rad/min.
>
> The corresponding angle for the minute hand, M,
> represents 5 minutes, so the angular speed of M is:
>
>    rM = pi/6 * 1/5 rad = pi/30 rad/min
>
> There are several methods for proceeding to a
> solution. Perhaps the easiest one goes like this. One
> wishes to solve for the exact time when H and M
> overlap for the first case, 1:05. Draw a straight line
> that represents the distance 'd' around the clock
> face.
>
>   '12'           '1'
>    P0             P1         P2
>    |--------------|-----------|-------------> d
>
> The distances do not have to be to scale. One knows
> that the distance (P1 - P0) = pi/6 radians. The
> distance (P2 - P1) can be called 'x'. Initially, set M
> to P0 and H to P1, to represent the time 1:00.
>
> One wishes to advance M at rate rM from P0, and H at
> rate rH from P1 until M overtakes H at point P2. The
> difference 'x' represents the time past 1:05 consumed
> by H while M catches up.
>
> A solution for 'x' can be had by setting up 2
> equations. Since M and H must both arrive at P2
> simultaneously, they spend the same amount of time 't'
> to do so. Thus one may write:
>
>    t = x / (pi/360) for H
>
> and
>
>    t = (pi/6 + x) / (pi/30)
>
> using the identity:
>
>    time = distance / rate
>
> for both M and H, using their respective distances and
> rates.
>
> One then has the equation of the two, which can be
> solved for 'x', which turns out to be:
>
>    x = 5*pi/330  or pi/66
>
> Thus the entire distance (P2 - P0) is:
>
>    d = pi/6 + x
>
> which turns out to be 0.571199 radians.
>
> Since 1 radian = 6/pi hours, the conversion from
> radians to decimal hours is trivial, yielding the
> result:
>
>    D = (6/pi) * d = 1.09090909... hours
>
> which can then be easily converted to HMS format to
> yield the final result of:
>
>    1:05:27, rounded to the nearest second.
>
> This is the exact moment (to the nearest second) when
> M overtakes H and aligns with it.
>
> Some hints for further solution follow below.
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
> 1. Work the second case for clock time 2:11. From the
> result, one should be able to deduce a simple formula
> for calculating the H-M match times for any hour 'h'
> for a 12-hour clock. (Variable 'h' has values
> 1,2,...11 for a 12-hour clock.)
>
> 2. Work the first case for a 24-hour clock, i.e., when
> do H and M align on a 24-hour clock, using the method
> described, starting with H at '1', M at '12'.
>
> 3. From the result, one should be able to deduce a
> simple formula to predict decimal hours, D, from 'h'
> for an N-hour clock.
>
> Good hunting!
>
> -Lance
>
>
> Lance Latham
> [hidden email]
> Phone:    (518) 274-0570
> Address: 78 Hudson Avenue/1st Floor, Green Island, NY 12183
>
>
>
>
>
> __________________________________________________
> Do You Yahoo!?
> Tired of spam?  Yahoo! Mail has the best spam protection around
> http://mail.yahoo.com
>
>
Sepp Rothwangl

www.calendersign.com
Y -669; CEP - 244274
Anno-Domini-Hoax 2006-06-01
=======================

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You need to consider also the one, who forgets where the way is going.
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The web-net is the best device to catch the ICHTHYS:

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