, 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,
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.
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 Nachr
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Thank you for reworking your equations for me. I have
om: 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 fo
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
; -- 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
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No rush; thank you. I appreciate you applying your br
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 Goodm
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This is a lot of math for me to digest at once!
One
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
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Thanks for the reference. My ultimate objective is to
Maybe this article will hep.Non-current ephemeris for approximated calculationsacademia.eduAlfonso Pastor Morenogalvag...@gmail.com---
https://lists.uni-koeln.de/mailman/listinfo/sundial
;Bill Gottesman",
Ogg: RE: Azimuth calculation/Wall declination
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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
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 calculati
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
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 no
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 lo
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 Fri
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
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 usi
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 calcula
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