Re: Azimuth Calculation

[Alfonso Pastor-Moreno]

Hi Brooke.Yes you are right.The links you proposed are easier to access.Thank you.About the Skylight Compass: It is a very interesting article.Thank you.Best regards.Alfonso Pastor-Morenogalvag...@gmail.comP.s Keith: Try with these links if they work for you.On 23 Jul 2024, at 23:57, Brooke Clarke  wrote:
  

  
  
Hi Alfonso:

These links may be easier to access:
https://www.academia.edu/44609415/Sun_Compass_The_Theory_behind_the_Sun_Compasses
https://www.academia.edu/85934045/The_Construction_of_the_Roman_Altitude_Cylinder_Sundials_Parts_1_2_3

PS Here is a Skylight Compass that uses the polarization of the sky.
https://prc68.com/I/S5807.shtml#Skylight_Compass
-- 
Have Fun,

Brooke Clarke
https://www.PRC68.com
axioms:
1. The extent to which you can fix or improve something will be limited by how well you understand how it works.
2. Everybody, with no exceptions, holds false beliefs.

 Original Message 


  
  
  To all the sundialists involved in this thread about Azimuth
Calculation.
  Hi all.
  Regarding the subject about Sun's Altitude vs Sun's Azimuth,
some years ago I wrote two papers that could be of your
interest. 
  They can be downloaded from Academia in the
following links.
  
  
  
Sun Compass. The Theory
  behind the Sun Compasses
  
  https://d1wqtxts1xzle7.cloudfront.net/65073791/Sun_Compass_Theory_APM_v1_3-libre.pdf?1606791440=&response-content-disposition=attachment%3B+filename%3DSun_Compass_The_Theory_behind_the_Sun_Co.pdf&Expires=1721729703&Signature=ZhdMFU6oekFXxpo2t4DrBxbQodbPCbeGW5MBADE7GPLveWD4eTSMcHIgVQ2dzxz0P5544CafvZuF7rpqGs4rYs5IjHI8pu-HHih-uPzjMwiZq8UFqN-u5jNA9iYfVSayCNCwPHBJEj1dFX-dMk2QeNPn2nzogFjiR2USnxhc3PDc9oLkYNZF-mvInQCK7m09~L4Y2tmotJEdScQjF2bhJQJQOKB~sCGWuhppzxWUcphyOTxpP7L1SXUjVYq2E~xDx5BEaOvGCAx2oqcefeqR9m2lLM-utgLYUwKL~X8KsheHDbiYqrYyecCRkrZXHTDTNoCeK7NZ3l9MuWtUSGQhmw__&Key-Pair-Id=APKAJLOHF5GGSLRBV4ZA
  
  
  
  
  
Construction of the Roman
  Altitude Cylinder Sundials by the Analemma of Vitruvius Parts
  1,2,3.
https://d1wqtxts1xzle7.cloudfront.net/90499756/Roman_Portable_Cylinder_Sundials_Construction_APM_Part_1_2_3-libre.pdf?1662003085=&response-content-disposition=attachment%3B+filename%3DConstruction_of_the_Roman_Altitude_Cylin.pdf&Expires=1721729833&Signature=LQuNqDY0cwMDWa1Y2aoN3ZpZgBIUHOdmGLGMvrRRYPEgoDcWN~bgiu7-ezw0ytBpC6RD0F~KLgXG9xF20nGz~NJl8WGGu---BFpi0141jPDFujWK2ppYJy6gAh-x0iwW0EtbEmiuSrCARU795xcZdVPvIvPl4YpknE-d8qyt8U6G~Mhm4j4h~mW-hTMocoOZurRXl~Qx0InCDd299OrKsFibLr6wBQXSfTK7k6b8Wh1msBMMEhWDF4IvjW9bKcZsJjG--NbViC95fdTftfHAihD3OKrTAeYcRYcOLjbRm7TxIMRsuv0ibbKAQXwITERssz3EKASlmRUGMOeXWwdtMw__&Key-Pair-Id=APKAJLOHF5GGSLRBV4ZA


About this subject in the NASS "The Compendium" 29(4)
  December 2022, and 30(1) – March 2023,  Robert L. Kellogg and
  myself, we wrote ex aequo an article entitled:
PORTABLE CYLINDER SUNDIALS CONSTRUCTION.


In the mentioned articles the subject 'Sun's
  Azimuth vs Altitude calculation" is treated extensively.
Maybe some of you could be interested in having a
  look to the articles.
Kind regards.

  Alfonso Pastor-Moreno
galvag...@gmail.com
  


  On 22 Jul 2024, at 21:37,
guillemet.herve  wrote:

  


  


  HI John,
  
  
  It seems that the instrument you are
looking for is called a shepherd's dial. It is mobile,
you set the date, you hold it vertically and orientate
the horizontal equivalent gnomon towards the sun and it
gives you the time. It is calculated for a specific
latitude. 
  
  
  Best regards Hervé Guillemet 
  
  
  
Envoyé
  depuis mon appareil Galaxy
  






   Message d'origine 
  De : John Goodman via sundial
 
  Date : 22/07/2024 16:14 (GMT+01:00) 
  À : Fred Sawyer  
  Objet : Re: Azimuth Calculation 
  
  

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.--

Re: Azimuth Calculation

[Alfonso Pastor-Moreno]

To all the sundialists involved in this thread about Azimuth Calculation.
Hi all.
Regarding the subject about Sun's Altitude vs Sun's Azimuth, some years ago I 
wrote two papers that could be of your interest. 
They can be downloaded from Academia in the following links.

Sun Compass. The Theory behind the Sun Compasses
https://d1wqtxts1xzle7.cloudfront.net/65073791/Sun_Compass_Theory_APM_v1_3-libre.pdf?1606791440=&response-content-disposition=attachment%3B+filename%3DSun_Compass_The_Theory_behind_the_Sun_Co.pdf&Expires=1721729703&Signature=ZhdMFU6oekFXxpo2t4DrBxbQodbPCbeGW5MBADE7GPLveWD4eTSMcHIgVQ2dzxz0P5544CafvZuF7rpqGs4rYs5IjHI8pu-HHih-uPzjMwiZq8UFqN-u5jNA9iYfVSayCNCwPHBJEj1dFX-dMk2QeNPn2nzogFjiR2USnxhc3PDc9oLkYNZF-mvInQCK7m09~L4Y2tmotJEdScQjF2bhJQJQOKB~sCGWuhppzxWUcphyOTxpP7L1SXUjVYq2E~xDx5BEaOvGCAx2oqcefeqR9m2lLM-utgLYUwKL~X8KsheHDbiYqrYyecCRkrZXHTDTNoCeK7NZ3l9MuWtUSGQhmw__&Key-Pair-Id=APKAJLOHF5GGSLRBV4ZA


Construction of the Roman Altitude Cylinder Sundials by the Analemma of 
Vitruvius Parts 1,2,3.
https://d1wqtxts1xzle7.cloudfront.net/90499756/Roman_Portable_Cylinder_Sundials_Construction_APM_Part_1_2_3-libre.pdf?1662003085=&response-content-disposition=attachment%3B+filename%3DConstruction_of_the_Roman_Altitude_Cylin.pdf&Expires=1721729833&Signature=LQuNqDY0cwMDWa1Y2aoN3ZpZgBIUHOdmGLGMvrRRYPEgoDcWN~bgiu7-ezw0ytBpC6RD0F~KLgXG9xF20nGz~NJl8WGGu---BFpi0141jPDFujWK2ppYJy6gAh-x0iwW0EtbEmiuSrCARU795xcZdVPvIvPl4YpknE-d8qyt8U6G~Mhm4j4h~mW-hTMocoOZurRXl~Qx0InCDd299OrKsFibLr6wBQXSfTK7k6b8Wh1msBMMEhWDF4IvjW9bKcZsJjG--NbViC95fdTftfHAihD3OKrTAeYcRYcOLjbRm7TxIMRsuv0ibbKAQXwITERssz3EKASlmRUGMOeXWwdtMw__&Key-Pair-Id=APKAJLOHF5GGSLRBV4ZA

About this subject in the NASS "The Compendium" 29(4) December 2022, and 30(1) 
– March 2023,  Robert L. Kellogg and myself, we wrote ex aequo an article 
entitled:
PORTABLE CYLINDER SUNDIALS CONSTRUCTION.

In the mentioned articles the subject 'Sun's Azimuth vs Altitude calculation" 
is treated extensively.
Maybe some of you could be interested in having a look to the articles.
Kind regards.
Alfonso Pastor-Moreno
galvag...@gmail.com

> On 22 Jul 2024, at 21:37, guillemet.herve  wrote:
> 
> 
> HI John,
> 
> It seems that the instrument you are looking for is called a shepherd's dial. 
> It is mobile, you set the date, you hold it vertically and orientate the 
> horizontal equivalent gnomon towards the sun and it gives you the time. It is 
> calculated for a specific latitude. 
> 
> Best regards Hervé Guillemet 
> 
> Envoyé depuis mon appareil Galaxy
> 
> 
>  Message d'origine ----
> De : John Goodman via sundial 
> Date : 22/07/2024 16:14 (GMT+01:00)
> À : Fred Sawyer 
> Objet : Re: Azimuth Calculation
> 
> 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.---
> https://lists.uni-koeln.de/mailman/listinfo/sundial
> 
---
https://lists.uni-koeln.de/mailman/listinfo/sundial

Re: Azimuth Calculation

[guillemet.herve]

HI John,It seems that the instrument you are looking for is called a shepherd's 
dial. It is mobile, you set the date, you hold it vertically and orientate the 
horizontal equivalent gnomon towards the sun and it gives you the time. It is 
calculated for a specific latitude. Best regards Hervé Guillemet Envoyé depuis 
mon appareil Galaxy
 Message d'origine De : John Goodman via sundial 
 Date : 22/07/2024  16:14  (GMT+01:00) À : Fred Sawyer 
 Objet : Re: Azimuth Calculation Diese Nachricht wurde 
eingewickelt um DMARC-kompatibel zu sein. Dieeigentliche Nachricht steht 
dadurch in einem Anhang.This message was wrapped to be DMARC compliant. The 
actual messagetext is therefore in an attachment.---
https://lists.uni-koeln.de/mailman/listinfo/sundial

Re: Azimuth Calculation

[John Goodman via sundial]

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.--- Begin Message ---
Thank you for reworking your equations for me. I have to confess that I'm weak 
at math and matrices make me a little queasy. The spherical trigonometry of 
Morrison's solution is territory I feel safer exploring, once I get my 
quadrants under control.

> On Jul 20, 2024, at 5:44 PM, Steve Lelievre  
> wrote:
> 
> John,
> 
> Here's my second attempt.
> 
> Steve
> 
> >>
> 
> 
> 
> Sun position vector S is   where ω is hour angle, φ is latitude, and δ is 
> solar declination.
> 
> Azimuth (from south) obeys  so 
> Dividing right hand side by –cos δ and then rearranging  
> 

> Setting a = , b = 1, c =  gives
> 
> a cos ω – b sin ω = c
> 
> Use trig identity 
> 
> a cos ω – b sin ω ≡ R cos(ω + x) where R =  and 
> which converts our case to
> 
> R cos(ω + x) = c
> 
> which rearranges as
> 
> ω =  
> 
> or
> 
> ω = 
>  
> 
> To force correct quadrant
> 
> if γ > 90, use ω =360 − Term 1 – Term 2
> 
> if γ < −90, use ω = −Term 1 – Term 2
> 
> if γ = 0, use ω = 0
> 
> 
> 
> These rules are for latitudes north of Tropic of Cancer. I'll leave you to 
> figure out quadrant adjustments for other cases if you need them. 
> 
> 
> 
> 
> 
> 
> On 2024-07-20 9:13 a.m., John Goodman wrote:
>> No rush; thank you. I appreciate you applying your brain to my concerns!
>> 
>>> On Jul 20, 2024, at 12:11 PM, Steve Lelievre 
>>>   
>>> wrote:
>>> Oh, no! I think I may have my initial term for tan gamma inverted, so my 
>>> result would be an angle relative to EW.
>>> 
>>> I have to go out for the day now, but will check my calculation when I get 
>>> home.
>>> 
>>> Steve
>>> 

--- End Message ---
---
https://lists.uni-koeln.de/mailman/listinfo/sundial

Re: Azimuth Calculation

[John Goodman via sundial]

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.--- Begin Message ---
I look forward to seeing this article and the design of your volvelle!

What I would be ideal for me would be a portable device that would align with 
the sun (azimuth) and read off the time for the date at that location. If the 
altitude is measured too, I believe the date would be inferred too. Finding the 
orientation of the meridian is a problem if this device is frequently moved.

Have you seen this? http://www.dickkoolish.com/rmk_page/suncalc.html

> On Jul 20, 2024, at 4:43 PM, Fred Sawyer  wrote:
> 
> At the recent NASS conference in Vancouver I introduced a new variety of 
> azimuthal volvelle sundial based on the solution to this very question. At 
> some point it will appear in an article in The Compendium.
> 
> Fred Sawyer
> 
> 
> 
> On Sat, Jul 20, 2024 at 10:02 AM John Goodman via sundial 
> mailto: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 mailto:johngood...@mac.com>>
>> To: Alfred Galvagnon mailto:galvag...@gmail.com>>
>> Cc: Sundial List mailto:sundial@uni-koeln.de>>
>> Bcc: 
>> Date: Sat, 20 Jul 2024 10:01:38 -0400
>> Subject: Re: Azimuth Calculation
>> Thanks for the reference. My ultimate objective is to find the sun's hour 
>> angle on a given day, in a given location, when it reaches a selected 
>> azimuth. 
>> 
>> If I understand this paper, it's starting with equatorial coordinates and 
>> calculating an azimuth, which is sort of flipping my problem around.
>> 
>> > On Jul 19, 2024, at 2:56 AM, Alfred Galvagnon > > <mailto:galvag...@gmail.com>> wrote:
>> > 
>> > Maybe this article will help.
>> > 
>> > https://www.academia.edu/32880342/Non_current_ephemeris_for_approximated_calculations?source=swp_share
>> > 
>> > Alfonso Pastor Moreno
>> > galvag...@gmail.com <mailto:galvag...@gmail.com>
>> > ---
>> > https://lists.uni-koeln.de/mailman/listinfo/sundial
>> > 
>> 
>> ---
>> https://lists.uni-koeln.de/mailman/listinfo/sundial
>> 

--- End Message ---
---
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Re: Azimuth Calculation

[Steve Lelievre]

John,

Here's my second attempt.

Steve

>>

Sun position vector S iswhere /ω/ is hour angle, /φ/is latitude, and /δ/ 
is solar declination.//


Azimuth (from south) obeys so //

Dividing right hand side by /–cos δ /and then rearranging

Setting /a/= , /b/ = /1/, /c/ = gives

/a cos ω – b sin ω = c/

Use trig identity

/a cos ω – b sin ω ≡ R cos(ω + x) /where/R = ///and

which converts our case to

/R cos(ω + x) = c/

which rearranges as

/ω = ///

or

/ω//= ///

//

To force correct quadrant

if /γ/> 90, use /ω//=/360 − Term 1 – Term 2

if /γ/< −90, use /ω//= /−Term 1 – Term 2

if /γ/= 0, use /ω/= 0


These rules are for latitudes north of Tropic of Cancer. I'll leave you 
to figure out quadrant adjustments for other cases if you need them.




On 2024-07-20 9:13 a.m., John Goodman wrote:

No rush; thank you. I appreciate you applying your brain to my concerns!

On Jul 20, 2024, at 12:11 PM, Steve Lelievre 
 wrote:


Oh, no! I think I may have my initial term for tan gamma inverted, so 
my result would be an angle relative to EW.


I have to go out for the day now, but will check my calculation when 
I get home.


Steve

---
https://lists.uni-koeln.de/mailman/listinfo/sundial

Re: Azimuth Calculation

[Fred Sawyer]

At the recent NASS conference in Vancouver I introduced a new variety of
azimuthal volvelle sundial based on the solution to this very question. At
some point it will appear in an article in The Compendium.

Fred Sawyer



On Sat, Jul 20, 2024 at 10:02 AM 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: Alfred Galvagnon 
> Cc: Sundial List 
> Bcc:
> Date: Sat, 20 Jul 2024 10:01:38 -0400
> Subject: Re: Azimuth Calculation
> Thanks for the reference. My ultimate objective is to find the sun's hour
> angle on a given day, in a given location, when it reaches a selected
> azimuth.
>
> If I understand this paper, it's starting with equatorial coordinates and
> calculating an azimuth, which is sort of flipping my problem around.
>
> > On Jul 19, 2024, at 2:56 AM, Alfred Galvagnon 
> wrote:
> >
> > Maybe this article will help.
> >
> >
> https://www.academia.edu/32880342/Non_current_ephemeris_for_approximated_calculations?source=swp_share
> >
> > Alfonso Pastor Moreno
> > galvag...@gmail.com
> > ---
> > https://lists.uni-koeln.de/mailman/listinfo/sundial
> >
>
> ---
> https://lists.uni-koeln.de/mailman/listinfo/sundial
>
>
---
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Re: Azimuth Calculation

[John Goodman via sundial]

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.--- Begin Message ---
No rush; thank you. I appreciate you applying your brain to my concerns!

> On Jul 20, 2024, at 12:11 PM, Steve Lelievre 
>  wrote:
> Oh, no! I think I may have my initial term for tan gamma inverted, so my 
> result would be an angle relative to EW.
> 
> I have to go out for the day now, but will check my calculation when I get 
> home.
> 
> Steve
> 
> 
> 
> On 2024-07-20 8:50 a.m., Steve Lelievre wrote:
>> 
>> On 2024-07-20 7:01 a.m., John Goodman via sundial wrote:
>>> My ultimate objective is to find the sun's hour angle on a given day, in a 
>>> given location, when it reaches a selected azimuth.
>> John, here's my attempt. I've never used this result so my calculation needs 
>> checking.
>> 
>> Cheers,
>> 
>> Steve
>> 
>> 
>> 
>> δ is solar declination
>> 
>> φ is latitude
>> 
>> ω is hour angle
>> 
>> γ is azimuth
>> 
>>  
>> Solar position vector S is
>> 

>> so
>> 

>> Rearrange:
>> 

>> Above equation is of form:
>> 

>> Using trig identities:
>> 
 where  and 
>> in which case
>> 

>> or
>> 

>> or
>> 

>>  
>> We have to watch out for quadrant in inverse functions.
>> 
> -- 
> https://www.gnomoni.ca 
> https://www.youtube.com/@gnomonica
> ---
> https://lists.uni-koeln.de/mailman/listinfo/sundial
> 

--- End Message ---
---
https://lists.uni-koeln.de/mailman/listinfo/sundial

Re: Azimuth Calculation

[Steve Lelievre]

Oh, no! I think I may have my initial term for tan gamma inverted, so my 
result would be an angle relative to EW.


I have to go out for the day now, but will check my calculation when I 
get home.


Steve




On 2024-07-20 8:50 a.m., Steve Lelievre wrote:



On 2024-07-20 7:01 a.m., John Goodman via sundial wrote:

My ultimate objective is to find the sun's hour angle on a given day, in a 
given location, when it reaches a selected azimuth.


John, here's my attempt. I've never used this result so my calculation 
needs checking.


Cheers,

Steve


/δ/is solar declination

/φ /is latitude

/ω/is hour angle

/γ/is azimuth

Solar position vector S is

so

Rearrange:

Above equation is of form:

Using trig identities:

where and

in which case

or

or

We have to watch out for quadrant in inverse functions.


--
https://www.gnomoni.ca
https://www.youtube.com/@gnomonica
---
https://lists.uni-koeln.de/mailman/listinfo/sundial

Re: Azimuth Calculation

[John Goodman via sundial]

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.--- Begin Message ---
This is a lot of math for me to digest at once! 

One sanity check I like to use is setting the azimuth to due south (whether 
180° or 0°).  In this case, latitude, and declination don't make any difference 
and the hour angle should be zero.

Having written these equations, can you readily see if this sanity check holds 
true?

> On Jul 20, 2024, at 11:50 AM, Steve Lelievre 
>  wrote:
> 
> 
> On 2024-07-20 7:01 a.m., John Goodman via sundial wrote:
>> My ultimate objective is to find the sun's hour angle on a given day, in a 
>> given location, when it reaches a selected azimuth.
> John, here's my attempt. I've never used this result so my calculation needs 
> checking.
> 
> Cheers,
> 
> Steve
> 
> 
> 
> δ is solar declination
> 
> φ is latitude
> 
> ω is hour angle
> 
> γ is azimuth
> 
>  
> Solar position vector S is
> 

> so
> 

> Rearrange:
> 

> Above equation is of form:
> 

> Using trig identities:
> 
 where  and 
> in which case 
> 

> or
> 

> or
> 

>  
> We have to watch out for quadrant in inverse functions.
> 
> ---
> https://lists.uni-koeln.de/mailman/listinfo/sundial

--- End Message ---
---
https://lists.uni-koeln.de/mailman/listinfo/sundial

Re: Azimuth Calculation

[Steve Lelievre]

On 2024-07-20 7:01 a.m., John Goodman via sundial wrote:

My ultimate objective is to find the sun's hour angle on a given day, in a 
given location, when it reaches a selected azimuth.


John, here's my attempt. I've never used this result so my calculation 
needs checking.


Cheers,

Steve


/δ/is solar declination

/φ /is latitude

/ω/is hour angle

/γ/is azimuth

Solar position vector S is

//

so

Rearrange:

Above equation is of form:

Using trig identities:

where and

in which case

or

or

We have to watch out for quadrant in inverse functions.
---
https://lists.uni-koeln.de/mailman/listinfo/sundial

Re: Azimuth Calculation

[John Goodman via sundial]

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.--- Begin Message ---
Thanks for the reference. My ultimate objective is to find the sun's hour angle 
on a given day, in a given location, when it reaches a selected azimuth. 

If I understand this paper, it's starting with equatorial coordinates and 
calculating an azimuth, which is sort of flipping my problem around.

> On Jul 19, 2024, at 2:56 AM, Alfred Galvagnon  wrote:
> 
> Maybe this article will help.
> 
> https://www.academia.edu/32880342/Non_current_ephemeris_for_approximated_calculations?source=swp_share
> 
> Alfonso Pastor Moreno
> galvag...@gmail.com
> ---
> https://lists.uni-koeln.de/mailman/listinfo/sundial
> 

--- End Message ---
---
https://lists.uni-koeln.de/mailman/listinfo/sundial

R: RE: Azimuth calculation/Wall declination

[sun.di...@libero.it]

May I suggest you to have a look at Orologi Solari ?

It includes two different tools for wall declination measurement.

It could maybe help you.

Greetings.

Gian

http://digilander.libero.it/orologi.solari



 
Messaggio originale
Da: ka...@karonadams.com
Data: 01/08/2011 17.36
A: "Bill Gottesman", 
Ogg: RE: Azimuth calculation/Wall declination

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-->

THIS is the kind of thing I have been looking for! The kinds of techniques that 
make the complex math work simply.
 
Cathedrals were built a thousand years ago by people who had little more 
literacy than needed to do their jobs. None of them knew what a sine or cosine 
or tangent was. They knew how ratios worked and they knew the practicalities of 
geometry.
 
Thanks for the great tool. As it happens, I need to measure the declination of 
a wall and this helps a lot!
 

Karon Adams
Accredited Jewelry Professional (GIA)
You can send a free Rosary to a soldier!
www.facebook.com/MilitaryRosary
www.YellowRibbonRosaries.com
 



From: sundial-boun...@uni-koeln.de [mailto:sundial-boun...@uni-koeln.de] On 
Behalf Of Bill Gottesman
Sent: Monday, August 01, 2011 10:55 AM
To: sundial@uni-koeln.de
Subject: Re: Azimuth calculation/Wall declination
 
About 10 years ago I worked out a simple method to measure wall declination 
using just a carpenter's square and an accurate watch.  The methods is 
described here http://www.precisionsundials.com/wall%20declination.pdf, and a 
simple windows program that does all the calculations for you is here 
http://www.precisionsu

RE: Azimuth calculation/Wall declination

[karon]

THIS is the kind of thing I have been looking for! The kinds of techniques that 
make the complex math work simply.

 

Cathedrals were built a thousand years ago by people who had little more 
literacy than needed to do their jobs. None of them knew what a sine or cosine 
or tangent was. They knew how ratios worked and they knew the practicalities of 
geometry.

 

Thanks for the great tool. As it happens, I need to measure the declination of 
a wall and this helps a lot!

 

Karon Adams

Accredited Jewelry Professional (GIA)

You can send a free Rosary to a soldier!

www.facebook.com/MilitaryRosary

www.YellowRibbonRosaries.com

 

From: sundial-boun...@uni-koeln.de [mailto:sundial-boun...@uni-koeln.de] On 
Behalf Of Bill Gottesman
Sent: Monday, August 01, 2011 10:55 AM
To: sundial@uni-koeln.de
Subject: Re: Azimuth calculation/Wall declination

 

About 10 years ago I worked out a simple method to measure wall declination 
using just a carpenter's square and an accurate watch.  The methods is 
described here http://www.precisionsundials.com/wall%20declination.pdf, and a 
simple windows program that does all the calculations for you is here 
http://www.precisionsundials.com/walldeclination.exe.  When I tested it out 
many years ago, I believe it gave results repeatable at different times and on 
different days to a few tenths of a degree.

-Bill

On 7/31/2011 9:57 AM, Andrew Theokas wrote: 

Fellow dialists:

 

I am using the following well known formula to calculate the sun’s azimuth for 
a particular time and location:

 

Azimuth= tan-1(sin H/(sin φ*cos H – cos φ*tanδ)

 

where 

H= Sun’s hour angle

φ= the latitude - 42.3 degrees

δ is the sun’s declination - 18.62 degrees

 

The location is in Boston, USA or 42.3 degrees N and 71.04 degrees west

 

I am using the azimuth-azimuth approach to find the declination of a wall found 
here:

 

http://www.mysundial.ca/tsp/wall_declination.html

 

the time the measurement was made was 11:18 am (daylight savings time is in 
effect)

 

I can easily calculate that the azimuth with respect to the wall is 26.8 
degrees.

 

Here is the problem: using two other independent methods I find that the wall’s 
declination is 20 degrees East.

 

So 26.8 degrees – Sun’s Azimuth should equal about twenty degrees.

 

But, using the above equation I cannot get an Azimuth value to work. One place 
where I might be in error is the value of the Hour angle which I compute to be 
about –16 degrees.

 

But you can also find the Hour Angle on line here at 
http://pveducation.org/pvcdrom/properties-of-sunlight/sun-position-calculator

 

Where might I be going wrong?

 

Many thanks for a reply!

 

Andrew Theokas

 

 

 
 
---
https://lists.uni-koeln.de/mailman/listinfo/sundial
 
---
https://lists.uni-koeln.de/mailman/listinfo/sundial

Re: Azimuth calculation/Wall declination

[Bill Gottesman]

  
  
About 10 years ago I worked out a simple method to measure wall
declination using just a carpenter's square and an accurate watch. 
The methods is described here http://www.precisionsundials.com/wall%20declination.pdf,
and a simple windows program that does all the calculations for you
is here http://www.precisionsundials.com/walldeclination.exe. 
When I tested it out many years ago, I believe it gave results
repeatable at different times and on different days to a few tenths
of a degree.

-Bill

On 7/31/2011 9:57 AM, Andrew Theokas wrote:

  

  Fellow dialists:
   
  I am using the following well known formula to calculate
the sun’s azimuth for a particular time and location:
   
  Azimuth= tan-1    (sin
  H/(sin φ*cos H – cos φ*tanδ)
   
  where 
  H= Sun’s
hour angle
  φ= the latitude - 42.3 degrees
  δ is the sun’s declination -
  18.62 degrees
   
  The location
is in Boston, USA or 42.3 degrees N and 71.04 degrees west
   
  I am using
the azimuth-azimuth approach to find the declination of a
wall found here:
   
  http://www.mysundial.ca/tsp/wall_declination.html
   
  the time the
measurement was made was 11:18 am (daylight savings time is
in effect)
   
  I can easily
calculate that the azimuth with respect to the wall is 26.8
degrees.
   
  Here is the
problem: using two other independent methods I find that the
wall’s declination is 20 degrees East.
   
  So 26.8
degrees – Sun’s Azimuth should equal about twenty degrees.
   
  But, using
the above equation I cannot get an Azimuth value to work.
One place where I might be in error is the value of the Hour
angle which I compute to be about –16 degrees.
   
  But you can
also find the Hour Angle on line here at http://pveducation.org/pvcdrom/properties-of-sunlight/sun-position-calculator
   
  Where might
I be going wrong?
   
  Many thanks
for a reply!
   
  Andrew
Theokas
   
   

  
  

---
https://lists.uni-koeln.de/mailman/listinfo/sundial



  

---
https://lists.uni-koeln.de/mailman/listinfo/sundial

RE: Azimuth calculation

[Roger W. Sinnott]

Andrew:

 

I think your numbers make sense.

 

You didn’t mention the date, but I suspect your calculation was for July 30, 
2011.  For this date at 11:18 a.m. EDT, and the lat/long of Boston, I find the 
following values ---

  

   Sun’s declination:   +18.5 degrees

   Sun’s azimuth:133.4 degrees (from north through east)

 

When you say the wall “declination” is 20 degrees east (from due south), this 
is the same as saying that the normal to the wall faces azimuth 160 degrees 
(measured from north through east).

 

Then, you measured the Sun’s azimuth to be 26.8 degrees (east) with respect to 
the wall, implying that the Sun’s azimuth is 160 – 26.8 = 133.2 degrees.  This 
is rather close to the 133.4 that I got independently above. 

 

Some reference books define azimuths as the angle from due south, while others 
define it as the angle from true north, so it is important to keep these 
straight.

 

Roger 

 

 

From: sundial-boun...@uni-koeln.de [mailto:sundial-boun...@uni-koeln.de] On 
Behalf Of Andrew Theokas
Sent: Sunday, July 31, 2011 9:57 AM
To: sundial@uni-koeln.de
Subject: Azimuth calculation

 

Fellow dialists:

 

I am using the following well known formula to calculate the sun’s azimuth for 
a particular time and location:

 

Azimuth= tan-1(sin H/(sin φ*cos H – cos φ*tanδ)

 

where 

H= Sun’s hour angle

φ= the latitude - 42.3 degrees

δ is the sun’s declination - 18.62 degrees

 

The location is in Boston, USA or 42.3 degrees N and 71.04 degrees west

 

I am using the azimuth-azimuth approach to find the declination of a wall found 
here:

 

http://www.mysundial.ca/tsp/wall_declination.html

 

the time the measurement was made was 11:18 am (daylight savings time is in 
effect)

 

I can easily calculate that the azimuth with respect to the wall is 26.8 
degrees.

 

Here is the problem: using two other independent methods I find that the wall’s 
declination is 20 degrees East.

 

So 26.8 degrees – Sun’s Azimuth should equal about twenty degrees.

 

But, using the above equation I cannot get an Azimuth value to work. One place 
where I might be in error is the value of the Hour angle which I compute to be 
about –16 degrees.

 

But you can also find the Hour Angle on line here at 
http://pveducation.org/pvcdrom/properties-of-sunlight/sun-position-calculator

 

Where might I be going wrong?

 

Many thanks for a reply!

 

Andrew Theokas

 

 

---
https://lists.uni-koeln.de/mailman/listinfo/sundial

Re: Azimuth calculation

[Fabio Savian]

Hi Andrew,

have you correct H for the longitude and EoT?

You may get a confirmation of your calculation with GnomoLab of Sundial Atlas.
In www.sundialatlas.eu click GnomoLab in the right column.
Click on the orange asterisk beside latitude to digit the coordinates.
I use your data but they are not accurate enough to locate the building.
Pan and zoom the map to get the building.
With GnomoLab you can evaluate the declination of the wall, get the parameters 
of the dial and watch it with the shadow in real time.

ciao Fabio

Fabio Savian
fabio.sav...@nonvedolora.it
Paderno Dugnano, Milan, Italy
45° 34' 10'' N   9° 10' 9'' E
GMT +1 (DST +2)<>---
https://lists.uni-koeln.de/mailman/listinfo/sundial

Re: Azimuth calculation

[Thibaud Taudin-Chabot]

hundred meters you can easily use the latitide/longitude as if it is a 
rectangular system. If the distances are in the regions of several 
kilometers/miles it will be a bit more complex.


At 16-11-2000 11:03 -0200, you wrote:
-Original Message/Oorspronkelijk bericht--

Hello Friends

It's been a long, long time since I last disturbed you with my novice
questions.
I was just acculating credits do be entitled to ask the following
question
that has more to do with navigation than any other thing:

a) If I am using UTM coordenates, what is the easiest way to calculate
the bearing from point A to point B in the chart;
b) Same questions if  am using latitude and longitude

When using UTM I have come accross a solution that works but I must
confess I hate it because I don't think it is elegant and it takes a lot
of
time so I am sure there must be a better solution.

For the UTM it is much simpler because I can always create a Pythagorean

triangule whose sides are the difference of Northing and Easting of the
points,
so I have three sides and an angle. Now, if I make the origin point the
origen
of a Cartesian system I can find the Azimuth adding together the angle I
found
plus 0, 90, 180 or 270 if the destination point is on the first, second,
third or
fourth quadrant.

I does work, but there must be a simpler way to do it.

If what I have are the geographic coordinates it takes me much more time

with the spherical triangles and I am never sure I've done the right
thing.

I've read some books on celestial navigation and position astronomy. I
can see
the solution is there, but it does not seem I have the expertise to
convert all those
useful information in a simple formula for this specific calculation.

And I feel very unhappy when I have to fill out a couple of pages with
ugly
calculations just to find out an azimuth I can easily find with a GPS or
with
a protactor and a map!

Best regards

- fernando





--
Fernando Cabral Padrao iX Sistemas Abertos
mailto:[EMAIL PROTECTED]  http://www.pix.com.br
Fone Direto: +55 61 329-0206mailto:[EMAIL PROTECTED]
PABX: +55 61 329-0202   Fax: +55 61 326-3082
15º 45' 04.9" S 47º 49' 58.6" W
19º 37' 57.0" S 45º 17' 13.6" W


-
Thibaud Taudin-Chabot
52°18'19.85" North  04°51'09.45" East
home email: [EMAIL PROTECTED]
(attachments max. 500kB; for larger attachments contact me first)

Re: Azimuth calculation

[Richard Langley]

The Earth has a flatenning of about 1 in 300.  So this is the rough size of
maximum distance errors you can expect by assuming the Earth to be a sphere
rather than an ellipsoid.  For example, the distance between Washington
D.C. and Los Angeles is 3,711 km (2,004 nautical miles) assuming the earth to
be a sphere with a radius of 6371 km but 3,719 km (2,008 nautical
miles) using the best fitting (WGS 84) ellipsoid.
-- Richard Langley

On Thu, 16 Nov 2000, Fernando Cabral wrote:

>Hello Richard
>
>Minutes after sending my previous message (should this always happen?) I found 
>a
>page on the Internet which seems to give me what I need.
>
>I have checked it with some coarse calculations. Since I don't have my GPS
>here with me, I can't compare the results I got with the GPS'.
>
>Nevertheless, they seemed right to me.
>
>Previously I had been using two formulae given by J. Meeus in his book
>"Astronomical Algorithm" (if my memory is of any avail), one of which
>he says is not very precise. They both have agreed very well with
>the results given by the GPS. With the less precise I have obtained
>results within 0,3% and with the more precise numbers are the same.
>
>Now, in this page by Fiona Vincent, it seems I found a still simpler
>way to calculate both distances and azimuth.
>
>Can you give me an idea on what magnitude of error should I expect from them
>(perhaps due to the Earth flattening)?
>
>Thank you.
>
>- fernando
>
>[Image]
>
>
>
>
>
>Richard Langley wrote:
>
>> On Thu, 16 Nov 2000, Fernando Cabral wrote:
>>
>> >Hello Friends
>> >
>> >It's been a long, long time since I last disturbed you with my novice
>> >questions.
>> >I was just acculating credits do be entitled to ask the following
>> >question
>> >that has more to do with navigation than any other thing:
>> >
>> >a) If I am using UTM coordenates, what is the easiest way to calculate
>> >the bearing from point A to point B in the chart;
>> >b) Same questions if  am using latitude and longitude
>> >
>> >When using UTM I have come accross a solution that works but I must
>> >confess I hate it because I don't think it is elegant and it takes a lot
>> >of
>> >time so I am sure there must be a better solution.
>> >
>> >For the UTM it is much simpler because I can always create a Pythagorean
>> >
>> >triangule whose sides are the difference of Northing and Easting of the
>> >points,
>> >so I have three sides and an angle. Now, if I make the origin point the
>> >origen
>> >of a Cartesian system I can find the Azimuth adding together the angle I
>> >found
>> >plus 0, 90, 180 or 270 if the destination point is on the first, second,
>> >third or
>> >fourth quadrant.
>> >
>> >I does work, but there must be a simpler way to do it.
>>
>> Nothing realy wrong with your approach but if you use the ATAN2 Fortran
>> function (or the equivalent in other programming languages or the Cartesian 
>> to
>> polar coordinate function on many calculators), you can specify the sign of
>> delta x and delta y and the azimuth will then be automatically reported in 
>> the
>> correct quadrant.
>>
>> >If what I have are the geographic coordinates it takes me much more time
>> >
>> >with the spherical triangles and I am never sure I've done the right
>> >thing.
>> >
>> >I've read some books on celestial navigation and position astronomy. I
>> >can see
>> >the solution is there, but it does not seem I have the expertise to
>> >convert all those
>> >useful information in a simple formula for this specific calculation.
>> >
>> >And I feel very unhappy when I have to fill out a couple of pages with
>> >ugly
>> >calculations just to find out an azimuth I can easily find with a GPS or
>> >with
>> >a protactor and a map!
>>
>> The use of plane trigonometry to find distances and angles is an important
>> feature of UTM.  But for highest accuracies one has to be careful about the
>> differences between geodetic and grid azimuth.  See my article in the 
>> February
>> 1998 issue of GPS World magazine for more details.
>>
>> -- Richard Langley
>>Professor of Geodesy and Precision Navigation
>>
>> >Best regards
>> >
>> >- fernando
>> >
>> >
>> >
>> >
>> >
>> >--
>> >Fernando Cabral Padrao iX Sistemas Abertos
>> >mailto:[EMAIL PROTECTED]  http://www.pix.com.br
>> >Fone Direto: +55 61 329-0206mailto:[EMAIL PROTECTED]
>> >PABX: +55 61 329-0202   Fax: +55 61 326-3082
>> >15? 45' 04.9" S 47? 49' 58.6" W
>> >19? 37' 57.0" S 45? 17' 13.6" W
>> >
>> >
>> >
>>
>>
>> ===
>>  Richard B. LangleyE-mail: [EMAIL PROTECTED]
>>  Geodetic Research Laboratory  Web: http://www.unb.ca/GGE/
>>  Dept. of Geodesy and Geomatics EngineeringPhone:+1 506 453-5142
>>  University of New Brunswick   Fax:  +1 506 453-4943
>>  Fredericton, N.B., Canada  E3B 5A

Re: Azimuth calculation

[Richard Langley]

On Thu, 16 Nov 2000, Fernando Cabral wrote:

>Hello Friends
>
>It's been a long, long time since I last disturbed you with my novice
>questions.
>I was just acculating credits do be entitled to ask the following
>question
>that has more to do with navigation than any other thing:
>
>a) If I am using UTM coordenates, what is the easiest way to calculate
>the bearing from point A to point B in the chart;
>b) Same questions if  am using latitude and longitude
>
>When using UTM I have come accross a solution that works but I must
>confess I hate it because I don't think it is elegant and it takes a lot
>of
>time so I am sure there must be a better solution.
>
>For the UTM it is much simpler because I can always create a Pythagorean
>
>triangule whose sides are the difference of Northing and Easting of the
>points,
>so I have three sides and an angle. Now, if I make the origin point the
>origen
>of a Cartesian system I can find the Azimuth adding together the angle I
>found
>plus 0, 90, 180 or 270 if the destination point is on the first, second,
>third or
>fourth quadrant.
>
>I does work, but there must be a simpler way to do it.

Nothing realy wrong with your approach but if you use the ATAN2 Fortran
function (or the equivalent in other programming languages or the Cartesian to
polar coordinate function on many calculators), you can specify the sign of
delta x and delta y and the azimuth will then be automatically reported in the
correct quadrant. 

>If what I have are the geographic coordinates it takes me much more time
>
>with the spherical triangles and I am never sure I've done the right
>thing.
>
>I've read some books on celestial navigation and position astronomy. I
>can see
>the solution is there, but it does not seem I have the expertise to
>convert all those
>useful information in a simple formula for this specific calculation.
>
>And I feel very unhappy when I have to fill out a couple of pages with
>ugly
>calculations just to find out an azimuth I can easily find with a GPS or
>with
>a protactor and a map!

The use of plane trigonometry to find distances and angles is an important
feature of UTM.  But for highest accuracies one has to be careful about the
differences between geodetic and grid azimuth.  See my article in the February
1998 issue of GPS World magazine for more details.

-- Richard Langley
   Professor of Geodesy and Precision Navigation


>Best regards
>
>- fernando
>
>
>
>
>
>--
>Fernando Cabral Padrao iX Sistemas Abertos
>mailto:[EMAIL PROTECTED]  http://www.pix.com.br
>Fone Direto: +55 61 329-0206mailto:[EMAIL PROTECTED]
>PABX: +55 61 329-0202   Fax: +55 61 326-3082
>15? 45' 04.9" S 47? 49' 58.6" W
>19? 37' 57.0" S 45? 17' 13.6" W
>
>
>


=== 
 Richard B. LangleyE-mail: [EMAIL PROTECTED]  
 Geodetic Research Laboratory  Web: http://www.unb.ca/GGE/
 Dept. of Geodesy and Geomatics EngineeringPhone:+1 506 453-5142  
 University of New Brunswick   Fax:  +1 506 453-4943  
 Fredericton, N.B., Canada  E3B 5A3
 Fredericton?  Where's that?  See: http://www.city.fredericton.nb.ca/
===