Re: Longest day calculations

[John Goodman]

Thanks to all for the formulas. For me, the format that's most convenient is 
the one sent by Kevin Karney:

[acos(-tan{latitude} x tan{declination})] / 7.5

where declination = 23.5 at midsummer and -23.5 at midwinter. 

If you flip the signs on the declination angles, so that midsummer = -23.5 and 
midwinter = 23.5, you can get rid of the minus sign in the equation, making it:

[acos(tan{latitude} x tan{declination})] / 7.5

The divide by 7.5 works as long as you're calculating in degrees. Computers 
like to work in radians, so I divide by pi, which gives the answer as a 
fraction of a whole day. I can multiply by 24 if I want the result to be in 
hours.

Before consulting the list, I had found a website with a much more confusing 
explanation. In case anyone here is interested and hasn't seen it, the address 
is - http://herbert.gandraxa.com/length_of_day.aspx  I'm much happier with the 
advice given here.

Thanks again for sharing your knowledge,
John


On Jul 1, 2010, at 1:57 PM, Karney wrote:

> [acos(-tan{latitude} x tan{declination})] / 7.5
> 
> Declination = 23.5 at  midsummer , 0 at equinox -23.5 at midwinter
> 
> Kevin Karney
> Freedom Cottage, Llandogo, Monmouth, NP25 4TP
> Phone 01594 539 595. Mobile 07595 024 960
> 
> On 1 Jul 2010, at 18:12, John Goodman <johngood...@mac.com> wrote:
> 
>> Hello all, 
>> 
>> I usually shudder when I see equations in messages posted to the list, but 
>> now I have a problem than needs a mathematical solution.
>> 
>> 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. 
>> 
>> Thanks very much,
>> John
>> 
>> ---------------------------------------------------
>> https://lists.uni-koeln.de/mailman/listinfo/sundial
>> 

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