sundial  

Re: Longest day calculations

Willy Leenders
Thu, 01 Jul 2010 14:02:46 -0700

The formula of John Schilke is a good one when you divide the result by 15

Willy LEENDERS
Hasselt in Flanders (Belgium)

Visit my website on the sundials in the province of Limburg in Flanders 
(Belgium) and on worthwhile facts about sundials
www.wijzerweb.be



Op 1-jul-2010, om 22:58 heeft Willy Leenders het volgende geschreven:

> Using the formula of John Schilke I become 244,74 hours (above the horizon), 
> for latitude 51° and declination 23,44°
> Using my formula (see picture) I become 16,32 hours.
> 
> 
> Willy LEENDERS
> Hasselt in Flanders (Belgium)
> 
> Visit my website on the sundials in the province of Limburg in Flanders 
> (Belgium) and on worthwhile facts about sundials
> www.wijzerweb.be
> 
> <formula1.jpg>
> 
> Op 1-jul-2010, om 22:39 heeft John Schilke het volgende geschreven:
> 
>> 
>> On 2010 July 1, at 10:12, John Goodman wrote:
>>      I'm looking for a simple formula which calculates the number of  
>> hours that the sun will be above the horizon on the summer solstice  
>> for any given latitude. I don't need to worry about refraction, or  
>> take into account the sun's diameter. I'm just interested in the  
>> simple geometric case, using the center of the sun's disk for  
>> determining the start and end times.
>>      For my purposes, these simple relationships will be true: If the  
>> answer for summer solstice hours is SSH, then winter solstice hours  
>> will be 24 - SSH, and equinox hours will be 12, regardless of the  
>> latitude.
>> 
>> John,
>> 
>> It would seem simplest to use the old formula
>> 
>>      cos T = – tan φ tan δ
>>              (latitude and declination, respectively)
>> 
>> then, H (hours above the horizon) = 2 arc cos T, approximately.
>> 
>> That is, in fact, how I usually do it, quickly and easily.
>> 
>> Your other statements are correct.
>> 
>> Best wishes,
>> 
>> John 
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