Fwd: Vertical stereographic sundial

[Peter Mayer]

Hi Maciej,

Thanks for your recent post. Since no one seems to have replied to your 
plea in the final paragraph for software to draw a 
stereographic/Oughtred dial, here are some existing open software options:


* Gian Casalegna's _Orologi Solari_ will draw one: azimuth: 
horizontal/vertical: astrolabe


*Helmut Sondereggers' _Sonne_ will draw at least a horizontal 
stereograph: Azimuthal: Sun Altitude


* Valentin Hristov has written some elegant code for DeltaCad 
:www.math.bas.bg/complan/valhrist/az-ht.bas  [When I tried to re-run it 
a few minutes ago, it now seems to be stuck in a loop which I haven't 
had time to debug. I've attached an example which I created about 4 
years ago].


So: no need to create a stereograph by hand!

best wishes,

Peter


On 27/10/2022 4:02:34, ml...@interia.pl wrote:
CAUTION: External email. Only click on links or open attachments from 
trusted senders.


Dear All,

As proven in a separate tread on "Sun elevation tool/ horizontoscope", 
stereographic projection is

a fascinating geometric construction with great potential in gnomonics.

During last year I explored the idea of vertical stereographic sundial 
and it resulted in

a paper describing several stereographic sundial types:

- vertical stereographic sundial, for walls of any declination and for 
any latitude,
- four-sided stereographic vertical sundial, attractive for its 
“gnomon-less” form,
- double – stereographic & polar – vertical sundial for walls of any 
declination, a vertical reference to Oughtred’s horizontal design,

- folded pocket, double stereographic & polar – vertical sundial
- inclined, proclined and polyhedral stereographic sundials

The paper can be downloaded free of charge from Cursiva publishers 
website:


http://cursiva.pl/e-ksiegarnia/vertical-stereographic-sundial-properties-construction-and-related-instruments/ 



or from Academia.edu website:

https://www.academia.edu/88820719/Vertical_stereographic_sundial_Properties_construction_and_related_instruments 



Have fun with it !

Vertical stereographic sundials offer plenty of possibilities for 
sundial makers
and I believe might be a very interesting addition to existing sundial 
types.


As the geometric construction of vertical stereographic sundials with 
CAD software is time-consuming,

it would be great to include it in existing sundial design software.
I kindly ask sundial list members running such software, please 
consider it !


Regards,

Maciej Lose

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Peter Mayer
Department of Politics & International Relations (POLIR)
School of Social Sciences
http://www.arts.adelaide.edu.au/polis/
The University of Adelaide, AUSTRALIA 5005
Ph : +61 8 8313 5609
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e-mail:peter.ma...@adelaide.edu.au
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Typo correction in most recent post

[Michael Ossipoff]

# 5, in my most recent post, should say:

.

To determine the hour-angle of the mean-Sun:

.

1. Convert the minutes in the LMT to a fraction of an hour.

.

2. Subtract 12 hours from the LMT, or add 12 hours to the LMT…whichever
gives a number between 0 & 24.

.

3. Multiply the result by 15.

---

The reason for this correction is that I’d forgotten that hour-angle & time
are reckoned from different starting-points:

Hour angle is measured from the meridian (in the south)…

…& time is measured from the anti-meridian.   i.e. from when the Sun or the
mean-Sun is at the anti-meridian.
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EqT estimate without knowing when Eqt was zero

[Michael Ossipoff]

(I’d post this as a “reply” to my existing topic, but that thread doesn’t
show up in my display, & so it’s necessary to post this as a new topic.)

.

Yesterday I described a way of estimating EqT, which depended on the fact
that the EqT is zero on some known day.  But EqT can be also estimated
without knowing when it is zero.   …by use of Sidereal Time.

.

First in summary, & then with details for the summarized elements:

.

For some given time of day, in Standard-Time, convert it to local mean
time, by applying your longitude-correction.

.

1. Determine ecliptic-longitude as before. Even if not reliably, estimate
it as well as possible for a particular time of day. (Each half-day in
error for when an ecliptic-month starts only makes a 2 minute error in the
EqT estimate.)

.

2. As before, use Spherical Trigonometry’s Cosine Formula to determine the
Sun’s R.A., in degrees, from its ecliptic-longitude.

.

3. Determine the Sidereal Time (…the hour-angle of the Vernal-Equinox.
Hour-angle is measured westward around the Earth’s axis, from the meridian,
in the south.)

.

4. Subtract the Sun’s R.A. from 360, & add the result to the Sidereal Time
in degrees.

.

5. From the local mean time, determine the hour-angle of the mean Sun, in
degrees.

.

6. Subtract the hour-angle of the Sun from the hour-angle of the mean-Sun.
Multiply by 4, & that gives EqT.

.

Details, where needed:

.

2. The Spherical-Trigonometry Cosine Formula, for ecliptic-longitude & R.A:

.

To get the Sun’s R.A.:

.

Multiply the tangent of its ecliptic-longitude by the cosine of the
obliquity, & then take the inverse tangent of the result.

.

3. Sidereal-Time:

.

Sidereal Time, like Mean Time advances uniformly.

.

It takes us roughly 365.25 days to orbit the Sun.  That’s slightly longer
than a mean tropical year. That’s because the equinoxes come about 1/72 of
a degree to meet us partway, making our tropical year about 20 minutes
shorter than the sidereal year.  (Of course those year-lengths really
depend on from what part of our orbit they’re measured.)

.

Because of the Sun’s apparent eastward movement with respect to the fixed
stars, due to our orbital motion around it, in a sidereal year the fixed
stars will have been perceived to go around us one more time than the Sun
has.  In a sidereal year, the stars will have gained 360 degrees over the
mean Sun.

.

So, in a day they’ll gain about (360/365.25) degrees, or about .9856 degree.

.

On the day of the Autumnal Equinox, if it occurs at Local Mean Time (LMT)
midnight, the Vernal Equinox will be on the meridian at midnight, because
the equinoxes are opposite.

.

So, how many days since the Autumnal Equinox, & how many hours since
midnight? Each hour since midnight advances the Vernal-Equinox 15 degrees
westward. Each day since the Autumnal Equinox avances the Vernal Equinox
about .9856 degree westward.

.

Of course if the Autumnal Equinox occurred some hours later than midnight
LMT, then at any later date & time the Vernal Equinox’s hour-angle will be
advanced 15 degrees less for each of those hours later.

…& of course the opposite if it’s earlier.

.

In that way, you can determine hour-angle of the Vernal Equinox at any time
& date.

.

5. Hour angle of the local mean sun:

.

It’s 15 degrees for every hour past noon local mean time.

,

I haven’t tried this method yet, but its error is probably about the same
as the other method I described.

.

I guess most of the error comes from the approximation of the Solar
Ecliptic Longitude via the Indian National Calendar.

.

That calendar seeks to make each of its ecliptic-months start as close as
possible to the actual astronomical ecliptic-month, & so I wouldn’t expect
large cumulative errors to accumulate.

.

The length of an Indian National Calendar ecliptic-month could be off by up
to half a day at most. Each half-day of error makes a 2 minute EqT error.  But
the calendar’s errors are intended to not accumulate. As I described for
October 29th, the EqT error was only 1 minute.
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Re: Paris sundial

[Bill Gottesman]

Is this real or a concept? The shadows don’t move in the rest of the image.
- Bill

On Sun, Oct 30, 2022 at 7:14 AM Alexei Pace  wrote:

> Interesting one in Rue Montmartre
> https://www.instagram.com/reel/Cj0hS2JIPQV/?igshid=YmMyMTA2M2Y=
>
> On Wed, Oct 26, 2022, 11:58 Werner Riegler  wrote:
>
>> Dear John,
>>
>> I know this object as “Horizontoscope".
>> http://www.horizontoscop.com/eng/index_eng.html
>>
>> I bought one recently after reading about it in one of Helmut
>> Sonderegger’s articles, where he gives the math of it.
>>
>> https://www.herzog-forsttechnik.ch/wp-content/uploads/2022/08/Sonnenkompass_Flyer-2022.pdf
>>
>> It’s quite interesting how the hyperbolic surface makes the image
>> independent of the height between the observing eye and the device.
>>
>> There is a german wikipedia entry.
>> https://de.wikipedia.org/wiki/Horizontoskop
>>
>> Maybe someone from the sundial list can produce an english entry with the
>> theory. It’s a nice device !
>>
>> best regards
>> Werner
>>
>>
>>
>>
>>
>> On 26 Oct 2022, at 02:45, John Pickard  wrote:
>>
>> Good morning,
>>
>> Has anyone come across this dial-related device?
>>
>>
>> https://picclick.co.uk/ARCHITECT-TOOL-Window-SUNLIGHT-SUN-ELEVATION-Enraf-144741549298.html
>>
>> Cheers, John.
>>
>> Dr John Pickard.
>>
>> ---
>> https://lists.uni-koeln.de/mailman/listinfo/sundial
>>
>>
>> ---
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>>
>> ---
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>
>
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Paris sundial

[Alexei Pace]

Interesting one in Rue Montmartre
https://www.instagram.com/reel/Cj0hS2JIPQV/?igshid=YmMyMTA2M2Y=

On Wed, Oct 26, 2022, 11:58 Werner Riegler  wrote:

> Dear John,
>
> I know this object as “Horizontoscope".
> http://www.horizontoscop.com/eng/index_eng.html
>
> I bought one recently after reading about it in one of Helmut
> Sonderegger’s articles, where he gives the math of it.
>
> https://www.herzog-forsttechnik.ch/wp-content/uploads/2022/08/Sonnenkompass_Flyer-2022.pdf
>
> It’s quite interesting how the hyperbolic surface makes the image
> independent of the height between the observing eye and the device.
>
> There is a german wikipedia entry.
> https://de.wikipedia.org/wiki/Horizontoskop
>
> Maybe someone from the sundial list can produce an english entry with the
> theory. It’s a nice device !
>
> best regards
> Werner
>
>
>
>
>
> On 26 Oct 2022, at 02:45, John Pickard  wrote:
>
> Good morning,
>
> Has anyone come across this dial-related device?
>
>
> https://picclick.co.uk/ARCHITECT-TOOL-Window-SUNLIGHT-SUN-ELEVATION-Enraf-144741549298.html
>
> Cheers, John.
>
> Dr John Pickard.
>
> ---
> https://lists.uni-koeln.de/mailman/listinfo/sundial
>
>
> ---
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>
>
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