13.12.2015, 16:23
Hello
Do you have some algorithm to calculate a sun position?
Do you have some algorithm to calculate a sun position?
How to calculate sun position?
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13.12.2015, 16:23
Hello
Do you have some algorithm to calculate a sun position?
14.12.2015, 08:07
This function should work, though it was not tested much:
Code: 12345678910111213141516171819202122232425262728 -- solar altitude, azimuth (degrees)
function sunposition(latitude, longitude, time)
time = time or os.time()
if type(time) == 'table' then
time = os.time(time)
end
local date = os.date('*t', time)
local timezone = (os.time(date) - os.time(os.date('!*t', time))) / 3600
if date.isdst then
timezone = timezone + 1
end
local utcdate = os.date('*t', time - timezone * 3600)
local latrad = math.rad(latitude)
local fd = (utcdate.hour + utcdate.min / 60 + utcdate.sec / 3600) / 24
local g = (2 * math.pi / 365.25) * (utcdate.yday + fd)
local d = math.rad(0.396372 - 22.91327 * math.cos(g) + 4.02543 * math.sin(g) - 0.387205 * math.cos(2 * g)
+ 0.051967 * math.sin(2 * g) - 0.154527 * math.cos(3 * g) + 0.084798 * math.sin(3 * g))
local t = math.rad(0.004297 + 0.107029 * math.cos(g) - 1.837877 * math.sin(g)
- 0.837378 * math.cos(2 * g) - 2.340475 * math.sin(2 * g))
local sha = 2 * math.pi * (fd - 0.5) + t + math.rad(longitude)
local sza = math.acos(math.sin(latrad) * math.sin(d) + math.cos(latrad) * math.cos(d) * math.cos(sha))
local saa = math.acos((math.sin(d) - math.sin(latrad) * math.cos(sza)) / (math.cos(latrad) * math.sin(sza)))
return 90 - math.deg(sza), math.deg(saa)
end
14.12.2015, 09:50
(14.12.2015, 08:07)admin Wrote: This function should work, though it was not tested much: It works. Thank you I create a library with this function in User library and in resident script I add this script: Code: 1234567 latitude = 50.990
longtitude = -23.179
time = nil
altitude, azimuth = sunposition(latitude, longitude, time)
log(altitude, azimuth)Result is very close to the results from online calculators.
14.12.2015, 10:20
Yes, I think it was ported from JavaScript based on one of online calculators.
14.12.2015, 14:18
Hi, while we’re on this topic.
Has anyone tried to work out an algorithm to calculate slate position of venetian blinds? Given the orientation of a window/blind, and the solar altitude and azimuth (given in post above). What is the optimal slate position to avoid direct sunlight? ![]() I’we been playing with some calculations. Even tried to “reverse engineer” observations from a major brand “KNX Wetterstation”, but never really got hold of it. Anyone with good experiences on this? Any suggestions on how to approach this is welcome. E.g. Is there one formula? Should i use a lookup table with different altitude- intervals for different azimuth “ranges”? Should i use a combination?
15.12.2015, 13:24
For starters, you can try with just using altitude since blinds only move on one axis. Blinds should be perpendicular to current sun position. Then you can tweak it by checking azimuth so you don't adjust blinds when there's no direct sunlight.
16.12.2015, 08:47
(15.12.2015, 13:24)admin Wrote: For starters, you can try with just using altitude since blinds only move on one axis. Blinds should be perpendicular to current sun position. Then you can tweak it by checking azimuth so you don't adjust blinds when there's no direct sunlight. Tanks, maybe i'm just making it too complex. There should, of course, be a 1:1 relation between sun altitude and slate position (e.g. 0-100%) – regardless of azimuth and orientation of the blinds/window.
20.12.2015, 15:03
21.12.2015, 07:08
You cannot use this library directly, but rewriting it in Lua should be quite easy.
09.05.2017, 12:22
(14.12.2015, 08:07)admin Wrote: This function should work, though it was not tested much:Hi I used this today. What I see is that when the azimuth get to 180, it starts counting down rather than going on to the full 360... Any ideas why?
There are 10 kinds of people in the world; those who can read binary and those who don't
![]()
11.05.2017, 07:49
Try this one instead:
Code: 12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061626364656667686970717273747576 function suncalc(lat, lng, time)
-- sun calculations are based on http://aa.quae.nl/en/reken/zonpositie.html formulas
local PI = math.pi
local sin = math.sin
local cos = math.cos
local tan = math.tan
local asin = math.asin
local atan = math.atan2
local acos = math.acos
local deg = math.deg
local rad = PI / 180
local e = rad * 23.4397 -- obliquity of the Earth
local daysec = 60 * 60 * 24
local J1970 = 2440588
local J2000 = 2451545
local function toDays(time)
return time / daysec - 0.5 + J1970 - J2000
end
local function rightAscension(l, b)
return atan(sin(l) * cos(e) - tan(b) * sin(e), cos(l))
end
local function declination(l, b)
return asin(sin(b) * cos(e) + cos(b) * sin(e) * sin(l))
end
local function azimuth(H, phi, dec)
return atan(sin(H), cos(H) * sin(phi) - tan(dec) * cos(phi))
end
local function altitude(H, phi, dec)
return asin(sin(phi) * sin(dec) + cos(phi) * cos(dec) * cos(H))
end
local function siderealTime(d, lw)
return rad * (280.16 + 360.9856235 * d) - lw
end
local function astroRefraction(h)
if h < 0 then -- the following formula works for positive altitudes only.
h = 0 -- if h = -0.08901179 a div/0 would occur.
end
-- formula 16.4 of "Astronomical Algorithms" 2nd edition by Jean Meeus (Willmann-Bell, Richmond) 1998.
-- 1.02 / tan(h + 10.26 / (h + 5.10)) h in degrees, result in arc minutes -> converted to rad:
return 0.0002967 / math.tan(h + 0.00312536 / (h + 0.08901179))
end
-- general sun calculations
local function solarMeanAnomaly(d)
return rad * (357.5291 + 0.98560028 * d)
end
local function eclipticLongitude(M)
local C = rad * (1.9148 * sin(M) + 0.02 * sin(2 * M) + 0.0003 * sin(3 * M)) -- equation of center
local P = rad * 102.9372 -- perihelion of the Earth
return M + C + P + PI
end
local function sunCoords(d)
local M = solarMeanAnomaly(d)
local L = eclipticLongitude(M)
return declination(L, 0), rightAscension(L, 0)
end
local lw = rad * -lng
local phi = rad * lat
local d = toDays(time or os.time())
local dec, ra = sunCoords(d)
local H = siderealTime(d, lw) - ra
local alt, az = altitude(H, phi, dec), azimuth(H, phi, dec)
return deg(alt), 180 + deg(az)
end
21.05.2017, 20:59
(11.05.2017, 07:49)admin Wrote: Try this one instead: Thanks. That one worked. Trond
There are 10 kinds of people in the world; those who can read binary and those who don't
![]()
30.10.2017, 13:58
(21.05.2017, 20:59)Trond Hoyem Wrote:(11.05.2017, 07:49)admin Wrote: Try this one instead: Hi, Maybe someone have example code? I need to get sun position every 5 min. 1) Elevation degree (0-100%) 2) Azimuth (0-360*) thx
30.10.2017, 15:19
(This post was last modified: 30.10.2017, 15:20 by Erwin van der Zwart.)
Hi,
Use function suncalc() like this: Code: 1234567 lat = 52.5167747 -- Zwolle - Netherlands
lng = 6.0830219 -- Zwolle - Netherlands
time = os.time()
alt, az = suncalc(lat, lng, time)
grp.update('33/1/1', alt)
grp.update('33/1/2', az)Erwin
30.10.2017, 18:43
(30.10.2017, 15:19)Erwin van der Zwart Wrote: Hi,Hi, I get error: Common functions:212: attempt to index global 'cec' (a nil value) stack traceback: Common functions:212: in main chunk
30.10.2017, 19:42
(This post was last modified: 30.10.2017, 19:44 by Erwin van der Zwart.)
Hi,
Did you placed the function inside common functions? Do you do the call from a resident script? I runned the function and the call from the same resident script and that worked perfect.. BR, Erwin (30.10.2017, 19:42)Erwin van der Zwart Wrote: Hi, After fw upgrade and factory reset all works fine. ![]() thank you Erwin
31.10.2017, 06:46
What about time zone and time shift? Please be carefull especially with the time shift because after it your calculations can be offset by 1 hour.
Code: 123 isdst = os.date('*t').isdst
if isdst then d = d - 3600 end
31.10.2017, 08:09
(This post was last modified: 31.10.2017, 08:10 by Erwin van der Zwart.)
Hi,
I don't think that is needed, i send the raw unix timestamp and time extracted from the raw stamp already includes dst.. BR, Erwin |
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