We are talking about the time angle in this case. It is often referred to
as "t", "HA" or "LHA". We solve for the angle and then divide by 15 for
the sunrise or sunset time in hours.
This is derived from the altitude equation in celestial navigation.
Sin Hc = Sin D * Sin L + Cos D * Cos L * Cos HA
For sunrise and sunset, the altitude Hc = 0 so Sin Hc = 0. Rearrange this
equation by moving half the stuff to the other side, dividing through by
Cos L and Cos D and substituting Tan for Sin / Cos. The simple, elegant
equation for the approximate time of sunrise and sunset results.
Cos HA = - Tan D * Tan L
Note this convention gives the negative sign. Also note that this is an
approximation. Corrections for semi-diameter and refraction have not been
included but for sundial design, this equation is great.
Roger Bailey
On Sun, 29 Dec 1996 [EMAIL PROTECTED] wrote:
>
> At 11:26 PM 12/28/96 +-100, you wrote:
> >
> >Dear Mac,
> >
> >It depends on what point of zero-reference for the hour angle HA is taken:
> >
> >1. Rohr, Mayall:
> >
> > Hour angle HA1 is measured from the north (zero) to the east
> (positive).
> >
> > Resulting formulae: cosHA1 = +tanL.tanD
> >
> >2. Otherwise
> >
> > Hour angle HA2 is measured from the south (zero)
> >
> > Resulting formulae: cosHA2 = -tanL.tanD
> >
> >
> >(Note: cos(a+180)=cos(a-180)=-cos(a), giving the same result)
> >
> >
> >Regards,
> >
> >
> >
> >Hans Sassenburg
> >
> >
> >
> Hans,
>
> Please forgive this humble and ignorant diallist if I am wrong, but I get
> the impression you are talking about the sun's AZIMUTH when you refer to
> measuring from either the north or the south. I think of the sun's hour
> angle as being an arc on the celestial sphere measured east and west of the
> local meridian with the measurement being centered on the Earth's axis and
> the arc parallel to the celestial equator.
>
> Charles Gann
>
>
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