Re: Azimuth and altitude

[Geoff Thurston]

Here is a small amendment to my previous message.

Although h0 becomes negative x does not so I should have written:

if the absolute value of A exceeds 90 degrees, then h0 and thus x becomes
negative. This is the reason for:
  If A < 90, h = x + h0
  If A > 90, h = x - h0

Geoff

On Thu, 18 Jul 2024 at 11:33, Geoff Thurston  wrote:

> Hi John,
>
> I have just been studying Jim's book and I have to admit that I was
> initially confused by the diagrams. However, after redrawing them for
> myself, I am convinced by Jim's argument and equations provided that
> azimuth is measured from the southern meridian. Referring to his equations:
>
> tan h0  = cos A / tan ϕ
>
> sin x = cos h0. sin d / sin ϕ
>
>
> From the first equation, we can see that h0 takes the same value for
> positive and negative azimuths which is what we would expect. However, if
> the absolute value of A exceeds 90 degrees, then h0 and thus x become
> negative. This is the reason for:
>   If A < 90, h = x + h0
>   If A > 90, h = x - h0
>
> Best wishes,
>
> Geoff
>
> On Wed, 17 Jul 2024 at 19:11, John Goodman via sundial <
> sundial@uni-koeln.de> wrote:
>
>> Diese Nachricht wurde eingewickelt um DMARC-kompatibel zu sein. Die
>> eigentliche Nachricht steht dadurch in einem Anhang.
>>
>> This message was wrapped to be DMARC compliant. The actual message
>> text is therefore in an attachment.
>>
>>
>> -- Forwarded message --
>> From: John Goodman 
>> To: Sundial List 
>> Cc:
>> Bcc:
>> Date: Wed, 17 Jul 2024 15:11:55 -0400
>> Subject: Re: Azimuth and altitude
>> Thank you for explaining. I can see how sine functions are symmetric
>> around 90 degrees but I still don't see how 90 degrees is an axis of
>> symmetry for measuring azimuth.
>>
>> If South is treated as 0 degrees instead of 180 degrees, then, as you
>> describe it, East could be -90 degrees and West would be +90 degrees. The
>> symmetry around South would then be Az < 0° or > 0° but still not the Az <
>> 90° or > 90° that was specified in Morrison's equations.
>>
>> 1. Calculate the sun's altitude when the declination is zero:  tan h0 =
>> cos A / tan phi
>>
>> 2. Calculate an auxiliary angle x from:  sin x = (cos h0 sin d) / sin phi
>>
>> 3. Calculate sun's altitude, h, for the azimuth and declination from:
>>
>> If A < 90, h = x + h0
>> If A > 90, h = x - h0
>>
>> NOTE:
>>
>> h = altitude
>> A = azimuth
>> d = solar declination
>> phi = latitude
>>
>>
>> On Jul 12, 2024, at 4:40 AM, Hervé Guillemet 
>> wrote:
>>
>> Hi John,
>>
>> If I understand your point, I think that the confusion comes from the
>> fact that for gnomonistthe origin of azimuth is South, and not North like
>>  it is for navigators (salesmen, pilots). In such case East is -90°, West
>>  is +90°, North is +/- 180°.
>> With this origin (South = 0°), there is a symetry in the trigonometric
>> circle : sin (pi - A) = sinA.  As an example sin 120° = sin 60° =
>> 0,866... For the same sine value, you have 2 angles. This is why you need
>>  to determine whether Az is > 90° or < 90°
>>
>> I hope that it answers your point.
>> Best regards Hervé Guillemet
>>
>> --
>> *De: *"John Goodman via sundial" 
>> @uni-koeln.de >
>> *À: *"Sundial List" @uni-koeln.de
>> >
>> *Envoyé: *Jeudi 11 Juillet 2024 20:03:10
>> *Objet: *Azimuth and altitude
>>
>> Diese Nachricht wurde eingewickelt um DMARC-kompatibel zu sein. Die
>> eigentliche Nachricht steht dadurch in einem Anhang.
>>
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>> text is therefore in an attachment.
>> ---
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>> <https://lists.uni-koeln.de/mailman/listinfo/sundial>lists
>> <https://lists.uni-koeln.de/mailman/listinfo/sundial>.uni-koeln.de/
>> <https://lists.uni-koeln.de/mailman/listinfo/sundial>mailman
>> <https://lists.uni-koeln.de/mailman/listinfo/sundial>/
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Re: Azimuth and altitude

[Geoff Thurston]

Hi John,

I have just been studying Jim's book and I have to admit that I was
initially confused by the diagrams. However, after redrawing them for
myself, I am convinced by Jim's argument and equations provided that
azimuth is measured from the southern meridian. Referring to his equations:

tan h0  = cos A / tan ϕ

sin x = cos h0. sin d / sin ϕ


>From the first equation, we can see that h0 takes the same value for
positive and negative azimuths which is what we would expect. However, if
the absolute value of A exceeds 90 degrees, then h0 and thus x become
negative. This is the reason for:
  If A < 90, h = x + h0
  If A > 90, h = x - h0

Best wishes,

Geoff

On Wed, 17 Jul 2024 at 19:11, John Goodman via sundial 
wrote:

> Diese Nachricht wurde eingewickelt um DMARC-kompatibel zu sein. Die
> eigentliche Nachricht steht dadurch in einem Anhang.
>
> This message was wrapped to be DMARC compliant. The actual message
> text is therefore in an attachment.
>
>
> -- Forwarded message --
> From: John Goodman 
> To: Sundial List 
> Cc:
> Bcc:
> Date: Wed, 17 Jul 2024 15:11:55 -0400
> Subject: Re: Azimuth and altitude
> Thank you for explaining. I can see how sine functions are symmetric
> around 90 degrees but I still don't see how 90 degrees is an axis of
> symmetry for measuring azimuth.
>
> If South is treated as 0 degrees instead of 180 degrees, then, as you
> describe it, East could be -90 degrees and West would be +90 degrees. The
> symmetry around South would then be Az < 0° or > 0° but still not the Az <
> 90° or > 90° that was specified in Morrison's equations.
>
> 1. Calculate the sun's altitude when the declination is zero:  tan h0 =
> cos A / tan phi
>
> 2. Calculate an auxiliary angle x from:  sin x = (cos h0 sin d) / sin phi
>
> 3. Calculate sun's altitude, h, for the azimuth and declination from:
>
> If A < 90, h = x + h0
> If A > 90, h = x - h0
>
> NOTE:
>
> h = altitude
> A = azimuth
> d = solar declination
> phi = latitude
>
>
> On Jul 12, 2024, at 4:40 AM, Hervé Guillemet 
> wrote:
>
> Hi John,
>
> If I understand your point, I think that the confusion comes from the fact
>  that for gnomonistthe origin of azimuth is South, and not North like it
> is for navigators (salesmen, pilots). In such case East is -90°, West is
> +90°, North is +/- 180°.
> With this origin (South = 0°), there is a symetry in the trigonometric
> circle : sin (pi - A) = sinA.  As an example sin 120° = sin 60° =
> 0,866... For the same sine value, you have 2 angles. This is why you need
> to determine whether Az is > 90° or < 90°
>
> I hope that it answers your point.
> Best regards Hervé Guillemet
>
> --
> *De: *"John Goodman via sundial" 
> @uni-koeln.de >
> *À: *"Sundial List" @uni-koeln.de
> >
> *Envoyé: *Jeudi 11 Juillet 2024 20:03:10
> *Objet: *Azimuth and altitude
>
> Diese Nachricht wurde eingewickelt um DMARC-kompatibel zu sein. Die
> eigentliche Nachricht steht dadurch in einem Anhang.
>
> This message was wrapped to be DMARC compliant. The actual message
> text is therefore in an attachment.
> ---
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> /
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>
>
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Re: Gnomon Gap Puzzle

[Geoff Thurston]

Happy New Year, Frank,

How about a T-shaped dial consisting of a vertical east dial backed by a
vertical west dial and sharing a solid sloping roof whose edges act as the
gnomons. The hour lines close to noon could be marked as a horizontal dial
on the "floor" of the dial.

I think that this would meet the noon gap requirement but I cannot think of
a location that demands it.

Best wishes,

Geoff

On Tue, 1 Jan 2019 at 08:50, Frank King  wrote:

> Dear All,
>
> Here is a little Dialling Puzzle to start
> the New Year...
>
> We are all familiar with the term 'Noon Gap'.
> On a simple horizontal sundial with a plate
> gnomon, this is the gap on the dial plate
> between the two vertical faces of the gnomon.
>
> On the dial plate, there are two lines for
> 12 o'clock with the noon gap between.  Often
> this gap is left blank.  Sometimes there is
> a date or, perhaps, the maker's name.
>
> During the year just ended, I was asked to
> design a dial which had to fit in a rather
> unusual space.  After a little thought, I
> decided on a solution.  In this...
>
>  THE ENTIRE DIAL FITS INSIDE THE GNOMON GAP
>
> Question 1: What does the design look like?
>
> Question 2: Can this possibly look good?
>
> Question 3: What is the 'unusual space'?
>
> A Happy New Year to you all.
>
> Frank King
> Cambridge, U.K.
>
> ---
> https://lists.uni-koeln.de/mailman/listinfo/sundial
>
>
---
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Re:

[Geoff Thurston]

I used to subscribe to the  CALNDR-L mailing list but I gave it up because
I did not have time to consider the numerous fascinating postings. It seems
to be still available at:

http://myweb.ecu.edu/mccartyr/calndr-l.html

Geoff

2018-04-13 14:43 GMT+01:00 graham stapleton via sundial <
sundial@uni-koeln.de>:

> Diese Nachricht wurde eingewickelt um DMARC-kompatibel zu sein. Die
> eigentliche Nachricht steht dadurch in einem Anhang.
>
> This message was wrapped to be DMARC compliant. The actual message
> text is therefore in an attachment.
>
> -- Forwarded message --
> From: graham stapleton 
> To: "sundial@uni-koeln.de" 
> Cc:
> Bcc:
> Date: Fri, 13 Apr 2018 13:43:37 + (UTC)
> Subject: Existence of a Calendar Study Forum?
> (apologies for an off-subject posting - but who else is likely to know?)
>
> Does anybody know of any kind of group, forum, or organisation that
> studies calendars (either as a time system or tangible objects) in terms of
> their structure, mathematics, history, culture etc. It seems unlikely that
> there is, but it would be good to know.
>
>
> 
>  Virus-free.
> www.avast.com
> 
> <#m_-3292758373496658296_DAB4FAD8-2DD7-40BB-A1B8-4E2AA1F9FDF2>
>
> ---
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>
>
>
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Re: Hail to the Chief

[Geoff Thurston]

Dear Frank,

Can I help the POTUS? No.
Can I answer your question? Maybe but I am not used to working in imperial
units.

Consider a horizontal cross section of the gnomon at any height above the
dial plate. The shape of the section would be an ellipse with minor axis
equal to 0.5" aligned East-West. The major axis would be equal to 0.5"/sin
30 = 1" aligned North-South.

At noon, each section of the gnomon will cast a shadow as wide as its minor
axis, thus 0.5" wide. At 6 am or pm, each section will cast a shadow as
wide as its major axis, thus 1"  wide.

Best wishes,

Geoff



On 17 January 2018 at 18:04, Frank King  wrote:

> Dear All,
>
> I gather from the UK media that
> the recent medical check-up
> undertaken by the U.S. President
> involved some "cognitive tests"
> and he scored 30 out of 30.
>
> One of these tests requires the
> subject to mark in the hands of
> a clock at a specified time.
>
> It is, of course, reassuring to
> know that he is familiar with
> analog(ue) time measuring
> instruments so I would like to
> propose a test which is just a
> whisker more ambitious for next
> year...
>
> I would present the subject
> with a mock-up horizontal
> sundial equipped with a rod
> gnomon which has a circular
> cross-section and is about
> 0.5" in diameter.
>
> The dial would be properly
> set up for a latitude of
> 30 deg N. so the gnomon would
> slope at a 30 deg angle to the
> horizontal.
>
> I would then ask the subject to
> draw the shadow of the gnomon as
> it would be at 6am and 12 noon.
>
> Since the mock up would include
> a full set of labelled hour lines
> this should not be too much of a
> challenge BUT...
>
> Part of the required answer is
> that the shadow at 6am should be
> wider than the shadow at 12 noon.
>
> I would then ask by how much the
> shadow is wider at 6am than at
> 12 noon and to explain why.
>
> Can you help the President?
>
> Frank
>
> Frank H. King
> Cambridge, U.K.
>
>
> ---
> https://lists.uni-koeln.de/mailman/listinfo/sundial
>
>
---
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Re: Golden Ratio and Sundials

[Geoff Thurston]

Rod,

Thanks for the information. They look as if they have been been
attractively produced but in the UK, Issue 1 is available at £1.99 while
Issue 3 onwards costs £8.99.  So I shall not be subscribing £450 for a
complete set even if I had room on my shelves.

Best wishes,

Geoff

On 24 June 2017 at 12:26, rodwall1...@gmail.com 
wrote:

> Hi Geoff and Frank,
>
> My book also shows this.
>
> By The Way, the book is one of a series called the Mathematical World by
> National Geographic. Where in Australia the 1st book. The Golden Ratio. The
> mathematical language of beauty. Was AUD $2. $2 to get you to purchase the
> 1st book. And then you may want to purchase the series of books. Of course
> they will not $2.
>
> Regards,
>
> Roderick Wall.
>
> - Reply message -
> From: "Frank King" 
> To: "Geoff Thurston" 
> Cc: "Michael Ossipoff" , "Sundial Mailing List" <
> sundial@uni-koeln.de>
> Subject: Golden Ratio and Sundials
> Date: Sat, Jun 24, 2017 8:16 PM
>
> Dear Geoff,
>
> Many congratulations on your proof...
>
> When I set the puzzle, I thought three things:
>
>  1. I am really setting this for Geoff to solve.
>
>  2. He will certainly solve it and will probably
> be the first to publish.
>
>  3. His proof will either match mine or be more
> elegant.
>
> I was right on all three counts.  Your proof is
> just what I had in mind.  Once you spot those
> two triangles it is obvious that they are
> similar and the rest comes out in the wash!
>
> Very best wishes
>
> Frank
>
> ---https://lists.uni-koeln.de/mailman/listinfo/sundial
>
>
---
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Re: Golden Ratio and Sundials

[Geoff Thurston]

Frank,

I think the most elegant proof that the diagonal to side ratio in a
pentagon equals phi is as shown in the attachment.

Geoff



On 23 June 2017 at 08:08, Frank King  wrote:

> Dear All,
>
> Referring to the Golden Ratio and Sundials, Donald
> Snyder wrote:
>
>   I see nothing obvious except ... trivial
>   possibilities.
>
> Try Googling   Dodecahedral Sundial  and you will
> see many examples.  Here is one chosen at random:
>   http://stretchingtheboundaries.blogspot.co.uk/2012/09/
> dodecahedral-sundial.ht
> ml
>
> The faces are all regular pentagons and the ratio of
> the distance between any two non-adjacent vertices
> and the length of a side is the golden ratio.
>
> Exercise for the reader:
>
>   Come up with a simple proof of this!
>
> In some (slightly contrived) sense, a regular
> pentagon incorporates 25 instances of the
> Golden Ratio, so a Dodecahedron incorporates 300
> such instances.
>
> Frank H. King
> Cambridge, U.K.
>
> ---
> https://lists.uni-koeln.de/mailman/listinfo/sundial
>
>
---
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Re: Sundials that come with 'baggage'.

[Geoff Thurston]

Just a guess, but might San 1790 be a misinterpretation of F.an 1790 (an
abbreviation of Fecit anno 1790). This would then match Mrs Gatty's
description pretty well.

Geoff

On 20 June 2017 at 10:53, Patrick Powers 
wrote:

> Hello all,
>
> It’s always interesting to look a little further into an old sundial –
> especially when it is one that is one of the lesser-known ones and is of
> some age too.
> This is true of a lovely dial set up in 1790 in Grantham, UK. [Grantham 
> appeared
> as early as 1086 in the Domesday Book and was of course the place where
> Isaac Newton was first educated].
>
> The dial of interest is a West Declining Dial which was partially
> ‘restored’ in 1968 but which restoration has left a number of questions
> behind.
>
> I would be interested to hear of any others’ comments about the various
> examples of restoration ‘drift’ and other oddities that are evident on this
> dial.
> My initial summary about the dial can be found via a link on the *SunInfo*
> Web Page (www.bit.ly/suninfo) or more directly at http://www.ppowers.com/
> grantham.htm.
>
> What might the letters ‘San’ mean?, is the motto unique in the UK? and why
> were the declination lines not restored too?, all come to mind.
>
> [image: image]
>
> Patrick Powers
>
>
> ---
> https://lists.uni-koeln.de/mailman/listinfo/sundial
>
>
>
---
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Re: Capuchin and Regiomontanus dials

[Geoff Thurston]

Michael,

I seem to recall that sec^2(x)=1+tan^2(x)

Therefore sec^2(lat).sec^2(dec)=(1+tan^2(lat)).(1+tan^2(dec))

=1+tan^2(lat)+tan^2(dec)+tan^2(lat).tan^2(dec)

=(1+tan dec tan lat)^2 + (tan dec - tan lat)^2

I guess that this relationship, which is just a variant of sin^2+cos^2=1,
should have been known to the dial designer.

Geoff

On 15 May 2017 at 16:32, Michael Ossipoff  wrote:

> Thanks for the Regiomontanus slide.
>
> Then the original designer of that dial must have just checked out the
> result of that way of setting the bead, by doing the calculation to find
> out if
> squrt((1+tan dec tan lat)^2 + (tan dec - tan lat)^2)) = sec lat sec dec,
> as a trial-and error trial that?
>
> Or, I don't know, is that a trigonometric fact that would be already known
> to someone who is really experienced in trig?
>
> ---
>
> What's the purpose of the lower latitude scale, on the dial shown in that
> slide?
>
> 
>
> When I described my folded-cardboard portable equatorial-dial, I
> mis-stated the declination arrangement:
>
> Actually, the sliding paper tab (made by making two slits in the bottom of
> the tab, and fitting that onto an edge of the cardboard) is positioned via
> date-markngs along that edge. The declination reading, and therefore the
> azimuth, is correct when the shadow of a certain edge of the tab,
> perpendicular to the cardboard edge on which it slides, just reaches the
> hour-scale on the surface that's serving as a quarter of an Disk-Equatorial
> dial.
>
> Actually, that dial was intended as an emergency backup at sea, where
> there would always be available a horizon by which to vertically orient the
> dial.
>
> The use of a plumb-bob for that purpose was my idea, because, on land
> there often or usually isn't a visible horizon, due to houses, trees, etc.
> Maybe, in really flat land, even without an ocean horizon, even a
> land-horizon could be helpful, but such a horizon isn't usually visible in
> most places on land.
>
> But then, with the plumb-line, it's necessary to keep the vertical surface
> parallel to the pendulum-string, and keep the pendulum-string along the
> right degree-mark, while making sure that the declination-reading is right,
> when reading the time.
>
> ...Four things to keep track of at the same time.   ...maybe making that
> the most difficult-to-use portable dial.
>
> With the Equinoctical Ring-Dial, the vertical orientation, about both
> horizontal axes, is automatically achieved by gravity, so only time and
> declination need be read.
>
> And, with a pre-adjustable altitude-dial, only the sun-alignment shadow
> and the time need to be read.
>
> With my compass tablet-dials, one mainly only had to watch the compass and
> the time-reading. Of course it was necessary to hold the dial horizontal,
> without a spirit-level, but that didn't keep them from being accurate.
>
> Michael Ossipoff
>
>
>
> On Sun, May 14, 2017 at 5:18 PM, Fred Sawyer  wrote:
>
>> Michael,
>>
>> See the attached slide from my talk.  All the various dials work with a
>> string of this length.  They vary simply in where the suspension point is
>> placed.  The pros and cons of the various suspension points were part of my
>> presentation.
>>
>> Fred Sawyer
>>
>>
>> On Sun, May 14, 2017 at 4:40 PM, Michael Ossipoff > > wrote:
>>
>>> When I said that there isn't an obvious way to measure to make the
>>> plumb-line length equal to sec lat sec dec, I meant that there' s no
>>> obvious way to achieve that *with one measurement*.
>>>
>>> I was looking for a way to do it with one measurement, because that's
>>> how the use-instructions say to do it.
>>>
>>> In fact, not only is it evidently done with one measurement, but that
>>> one measurement has the upper end of the plumb-line already fixed to the
>>> point from which it's going to be used, at the intersection of the
>>> appropriate latitude-line and declination-line.
>>>
>>> That's fortuitous, that it can be done like that, with one measurement,
>>> and using only one positioning of the top end of the plumb-line.
>>>
>>> But of course it's easier, (to find) and there's an obviously and
>>> naturally-motivated way to do it, with *two* measurements, before
>>> fixing the top-end of the plumb-line at the point where it will be used.
>>>
>>> The line from that right-edge point (from which the first horizontal is
>>> drawn) to the point where the appropriate latitude-line intersects the
>>> vertical has a length of sec lat.
>>>
>>> So, before fixing the top end of the plumb-line where it will be used
>>> from, at the intersection of the appropriate lat and dec lines, just place
>>> the top end of the plumb line at one end of that line mentioned in the
>>> paragraph before this one, and slide the bead to the other end of that
>>> line.   ...to get a length of thread equal to sec lat.
>>>
>>> Then, have a set of declination marks at the right edge, just like the
>>> ones that are actually on a Regiomontanus di

Re: The duration of the year

[Geoff Thurston]

Doug,


   1. You could do what the Egyptians and the Greeks did 1500 years
   earlier. Use a mural quadrant in the meridian to establish the greatest and
   least altitude of the noon sun ( from this you could calculate the orbital
   inclination and your latitude) and mark the mid altitude. You then sit and
   count the days between each instant that the midday sun attains this
   altitude in spring. You could refine the measurement by noting the altitude
   for several days and interpolating to determine the moment of equinox.
   After just a few years you would determine that the length of the tropical
   year is about 3651/4 days and the longer you observed the more precise this
   figure would become. You could also use an equatorial ring to make the
   observations which should make direct observation of daytime equinoxes
   possible but this would be more difficult to construct accurately.
   2. If you were equipped with a transit telescope and a means of
   recording transits to the second, then you could adopt the same procedure
   as the ancients but your results would converge more rapidly on the true
   length of the tropical year.

Best wishes,

Geoff

On 18 February 2017 at 18:07, Douglas Bateman <
douglas.bate...@btinternet.com> wrote:

> Given that this group has experts on the calendar and the earth’s orbit, I
> have a couple of questions.
>
> 1. Assuming that I was living a 1000 years ago, and had unlimited time
> watching the sun and stars (and *without prior knowledge*) how would I
> notice that each year was growing by about a quarter of a day?
>
> 2. Assuming that in 1850s I had access to a good transit telescope, and a
> reasonable clock (daily errors about 1 second a day), how would I refine
> the quarter of a day into several decimal places?
>
> These questions have been prompted by a debate in horological circles that
> the astronomers in the 1800s could have benefited by having a clock that
> was better than a second a month. My own view is that the 1 second a day
> was adequate because the clock is only put to use for *differential*
> measurements in time between frequent ‘clock stars’ each night and the
> transits of interest. Neglecting cloudy periods for the sake of argument.
>
> I look forward to receiving good advice.
>
> Regards, Doug
>
>
>
> ---
> https://lists.uni-koeln.de/mailman/listinfo/sundial
>
>
>
---
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Re: Unusual bi-annual sundial

[Geoff Thurston]

Thank you, Frank, for that comprehensive analysis of the problem. However,
I wonder if the errors might be masked by the 32 arc minute  solar penumbra.

Best wishes,

Geoff

On 19 January 2017 at 16:33, Frank King  wrote:

> Dear John,
>
> I wondered when someone would spot that there is a
> whole can of worms waiting to be opened here...
>
> > Won't the factors that necessitate the addition
> > of a leap day prevent this alignment from
> > happening at exactly 11/11 11:11 every year?
>
> Quite so.  No doubt you looked at the time-lapse
> video and spotted that the circle of light DIDN'T
> properly centre itself on the Great Seal of the
> United States.  This is surely only one step less
> sinful than being disrespectful to the US flag?
>
> OK, take a deep breath and see what we are up
> against...
>
> First we need to be clear what is meant by the
> time 11:11?  I assume this is clock time in
> Anthem, Arizona, and a little research suggests
> they are on Mountain Time there and that they
> don't observe Daylight Saving.  [Just think how
> the whole scheme could be wrecked if they did
> go over to Daylight Saving and the clocks didn't
> go back until after 11 November!]
>
> To me, their interest is at 18:11:00 UTC but that
> is a detail.
>
> The big difficulty is that, at this exact time of
> day, the solar declination varies with the leap
> year cycle and there is a steady drift.  As a
> result both the solar altitude and solar azimuth
> vary from one year to the next.  Let's see by
> how much...
>
> I'll take it that the Geographical Coordinates
> of Anthem are:
>
>33° 51' 15" N 112° 7' 30"
>
> Using GCstudio I determined the following data
> for 10 years starting in 2016, a leap year:
>
>   2016  -17°41'09"  +36°25'01"  +161°40'45"
>   2017  -17°37'11"  +36°28'55"  +161°39'53"
>   2018  -17°33'13"  +36°32'52"  +161°39'05"
>   2019  -17°29'12"  +36°36'55"  +161°38'33"
>   2020  -17°41'38"  +36°24'36"  +161°41'11"
>   2021  -17°37'47"  +36°28'23"  +161°40'14"
>   2022  -17°33'48"  +36°32'21"  +161°39'31"
>   2023  -17°29'52"  +36°36'14"  +161°38'36"
>   2024  -17°42'18"  +36°23'55"  +161°41'16"
>   2025  -17°38'23"  +36°27'48"  +161°40'23"
>
> The four columns show: year, declination, alt, az
> as they are at Anthem at 11:11:00 Mountain Time
> on 11 November in the 10 years shown.
>
> Take declination first.  You see that starting in
> 2016 the declination gets about 4 minutes less
> negative on successive years until there is a
> sudden jump back which is A LITTLE TOO BIG.
> This sets the pattern.  We become less negative
> until 2024 when there is another jump.
>
> The jumps back over-compensate because the tropical
> year is slightly less than 365.25 days.
>
> You will see that the solar altitude increases by
> just under 4' a year before falling back just over
> 12' in a leap year.  You will see that even in this
> little table the range of altitudes is about 11'
> and this will be noticed by careful observers.
>
> The azimuth varies too of course but by not so
> much and its main effect is to make you have to
> worry about just how to align the slabs.
>
> OK, what should they have done?
>
> Well one approach is to settle on the 2016 figures
> and note that over the next 36 years the data for
> 2016 will be somewhere near the middle.  After
> that the drift will become more noticeable but the
> designer will probably be dead and won't care.
>
> Things gradually get worse and worse until The
> Great Correction over the years 2096 to 2004
> when the omission of a leap year in 2100 will
> reverse some of the damage.
>
> Most people know that the Gregorian Calendar
> was an improvement over the Julian Calendar but
> almost all readers of this list will live their
> entire lives enduring pure Julian Drift.
>
> This is a massive imposition and we should all
> be lobbying for a much better 33-year Calendar
> originally designed by Omar Khayyam in 1079,
> long before John Dee and others rediscovered
> it.  This was over 500 years before Pope
> Gregory's tinkering in 1582.  Why didn't
> Pope Gregory do a proper job then?
>
> That's a long story but the result is that we
> are lumbered with an unhelpful calendar which
> is, I suppose, upward-compatible with its
> predecessor.
>
> I share the view that "upward-compatibility is
> the business of deliberately not putting right
> someone else's mistakes".
>
> Many apologies.  Another rant I fear!
>
> Very best wishes
>
> Frank
>
> Frank King
> Cambridge, U.K.
>
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Re: Leap Second Quiz Question

[Geoff Thurston]

Happy New Year to one and all,

The University Clock has gained 0.5 seconds against UTC in the past week
but is now 2.0 seconds ahead of UTC because of the leap second. If Frank
removed coins in accordance with the LWG for 0.5 seconds, then the clock's
rate would be rectified but the clock would still be 2 seconds fast next
week. Therefore, Frank should apply a LWG of 2.5 seconds and adjust again
when the clock is showing the correct time.

Geoff



On 1 January 2017 at 12:30, Frank King  wrote:

> Dear All,
>
> I hope you all enjoyed the extra second
> in bed this morning and that your alarm
> clock didn't go off one second early.
>
> Here is an easy question to start off
> the New Year...
>
> Every Sunday at 08:00 I check the first
> stroke of the hour-bell of the University
> Clock against a radio-controlled UTC clock.
>
> If it is slow I add coins to the tray on
> the pendulum.  If it is fast I remove
> coins.  My formula for the required
> adjustment includes a figure for:
>
> Last Week's Gain [LWG]
>
> Here are my recent observations:
>
>   25 December   clock 0.5 seconds fast
>1 Januaryclock 2.0 seconds fast
>
> Is the appropriate figure for LWG:
>
> a) 0.5 seconds
> b) 1.5 seconds
> c) 2.5 seconds
>
> Frank
>
> Frank H. King
> Keeper of the University Clock
> Cambridge, U.K.
>
>
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Re: Sundial Puzzle Corner

[Geoff Thurston]

I like David's solution and I am in awe of his stone-cutting capability.
However, to return to your problem where the ellipse has been returned
without any markings, why not draw an ellipse of the same specified
dimensions on paper with the axes already marked on it? This could then be
matched to the "close-to-perfect" slate ellipse to locate the axes and the
centre. It probably would not be necessary to go to the trouble of drawing
a complete ellipse. A polygonal approximation might work.

On 31 October 2016 at 08:22, David  wrote:

> I cut my own slate ellipses, often up to 3cm thick. In the process of
> marking out the ellipse I will have drawn both the major and minor axes. I
> make a small indentation with a metal scribe at both ends of both axes as
> well as the centre point. These indentations are deep enough to be
> noticeable but shallow enough to be erased easily on the cleaning-up
> operation after the eventual painting/gilding of the finished inscription.
> The indentations can be made more prominent by painting them with a spot of
> light-coloured acrylic paint. After the ellipse has been cut out (which I
> do with a series of short straight cuts from a bench-mounted, water-fed
> circular saw, followed by a hand-held water-fed cylindrical abrasive drum)
> the axes are easily drawn through the still-visible marks.
> If the job of drawing and cutting out has to be done by another person,
> such as at the stonemason's yard, they could be asked to leave similar
> marks to help you with the alignment of the inscription.
> David Brown
> Somerton, Somerset, UK
>
>  On 30/10/2016 21:29, Karl Billeter wrote:
>
>> On Mon, Oct 31, 2016 at 08:24:04AM +1100, Karl Billeter wrote:
>>
>>> On Sun, Oct 30, 2016 at 02:37:04PM +, Frank King wrote:
>>>   ...
>>>
 Almost the first task is to find
 the centre and the axes.  Clearly
 you cannot fold a slate in half
 and the traditional way to proceed
 is to put a large sheet of paper
 over the slate and crease it down
 all round the rim.

 You then cut round the crease and
 attempt to follow your procedure!

>>> Wrap the slate with a reflective strip (smooth, shiny plastic? thin
>>> polished
>>> metal?).  Playing around with a laser should find the focii.
>>>
>> You could also try balancing the slate on rod to find the centre but it's
>> probably too hard to get the required accuracy and it might be a little
>> risky!
>>
>> K
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>>
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Re: Sundial Puzzle Corner

[Geoff Thurston]

Frank,

Thanks for the puzzle. I think that these instructions might work if you
ventured to the antarctic circle during the southern winter and then
trecked to a position such that your latitude is greater than 90-Dec (so
that the midnight sun is visible) but less than 90. The shadow of the stake
would then be longest and point to true north at true solar midnight.

Geoff



On 28 October 2016 at 06:58, Frank King  wrote:

> Dear All,
>
> I have been looking at new U.K. educational
> website which has a whole category devoted
> to sundials.  Early on, there is section
> "Finding True North".  See:
>
>   http://wiki.dtonline.org/index.php/Finding_True_North
>
> This is what it asserts:
>
>   The Sun can be used to find True North
>   quite simply by placing a vertical stake
>   in the ground and noting which direction
>   the longest shadow points to.
>
> You may now take a two-minute break while
> you recover from rolling on the floor in
> a state of helpless laughter.
>
> When you have recovered, ponder this puzzle:
>
>   Where on the planet would you have to
>   be, and at what time of year, for these
>   instructions to give the correct result?
>
> Moral:
>
>   It is better to learn from other people's
>   mistakes than from your own.
>
> Frank King
> Cambridge, U.K.
>
> P.S. The home page of the website is:
>
>   http://wiki.dtonline.org/index.php/Category:Sundials
>
> It won't take long before you notice other
> erroneous assertions :-)
>
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Re: Using analemmatic sundials for determining sun exposure times

[Geoff Thurston]

Thanks for the mental exercise, Frank.  I think that Willy might squeeze
another 15 seconds of illumination at the end of December owing to the
rapid change in the EoT. A useful graph showing the rate of change of EoT
is provided by the USNO at:

http://aa.usno.navy.mil/faq/docs/eqtime.php

Geoff Thurston

On 14 June 2016 at 13:01, Willy Leenders  wrote:

>
> See on my website: http://www.wijzerweb.be/analemmatischengels.html
> The maximum time (12 hours) of sunlight on a vertical wall of any
> direction is above the Arctic Circle on a day when the midnight sun shines
>
> Willy Leenders
> Hasselt in Flanders (Belgium)
>
> Visit my website about the sundials in the province of Limburg (Flanders)
> with a section 'worth knowing about sundials' (mostly in Dutch):
> http://www.wijzerweb.be
>
>
> Op 14-jun-2016, om 10:29 heeft Frank King het volgende geschreven:
>
> Dear Dan,
>
> The approach that you describe does
> indeed show you how long the sun can
> theoretically shine on a vertical
> wall, for a given solar declination,
> provided (as you say) that you know
> the times the sun rises and sets.
>
> Now a simple exercise...
>
> What is the maximum amount of time
> the sun can shine on a vertical wall?
>
> You can choose the latitude and you
> can choose the orientation of the
> wall and you can choose the solar
> declination.
>
> Frank King
> Cambridge, U.K.
>
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