HP 35S: Length of Sunlight During a Day
Source: Total Daily Amount of Solar Radiation - HP 67/97 Energy Conservation Pac, December 1978, Author: Hewlett Packard
(This is a slight variation instead of a direct port)
Input
You are prompted for D and L where:
D = the number of days from March 21, a 365 day year is assumed
L = latitude (North as positive, South as negative), entered as D.MMSS (degrees-minutes-seconds) format
Output
Approximate number of hours of sunlight, in hours, minutes, seconds
Examples
Los Angeles, April 17: latitude of 34°03' N, 27 days after March 21
D = 27, L = 34.03, answer is approximately 12.573501 (12 hours, 57 minutes, 35.01 seconds)
Rome, September 1: latitude 41°51' N, 164 days after March 21
D = 164, L = 41.52, answer is approximately 12.532269 (12 hours, 53 minutes, 22.69 seconds)
Sydney, June 21: latitude 33°51'31" S, 92 days after March 21
D = 92, L = -33.5131, answer is approximately 9.443922 (9 hours, 44 minutes, 39.22 seconds)
Formulas
This version uses the estimate of sun declination:
D = 23.45 sin(d * 0.9856°)
Since 360/365.25 ≈ 0.985626283368
θ = acos(-tan L × tan D)
L = 24 * θ in radians ÷ π
Program
S001 LBL S
S002 DEG
S003 INPUT D
S004 0.9856
S005 ×
S006 SIN
S007 23.45
S008 ×
S009 INPUT L
S010 HMS→
S011 TAN
S012 x<>y
S013 TAN
S014 ×
S015 +/-
S016 ACOS
S017 ->RAD
S018 24
S019 ×
S020 π
S021 ÷
S022 ->HMS
S023 RTN
This blog is property of Edward Shore. 2013
A blog is that is all about mathematics and calculators, two of my passions in life.