How to calculate sun position?

How to calculate sun position? - Printable Version

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How to calculate sun position? - buuuudzik - 13.12.2015

Hello

Do you have some algorithm to calculate a sun position?

RE: How to calculate sun position? - admin - 14.12.2015

This function should work, though it was not tested much:

Code:
-- 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

RE: How to calculate sun position? - buuuudzik - 14.12.2015

(14.12.2015, 08:07)admin Wrote: This function should work, though it was not tested much:

Code:
-- 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

It works. Thank you

I create a library with this function in User library and in resident script I add this script:

Code:
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.

RE: How to calculate sun position? - admin - 14.12.2015

Yes, I think it was ported from JavaScript based on one of online calculators.

RE: How to calculate sun position? - oab_nor - 14.12.2015

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? Cool

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?

RE: How to calculate sun position? - admin - 15.12.2015

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.

RE: How to calculate sun position? - oab_nor - 16.12.2015

(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.

RE: How to calculate sun position? - buuuudzik - 20.12.2015

Can I add this library to LM?

https://github.com/mourner/suncalc

RE: How to calculate sun position? - admin - 21.12.2015

You cannot use this library directly, but rewriting it in Lua should be quite easy.

RE: How to calculate sun position? - Trond Hoyem - 09.05.2017

(14.12.2015, 08:07)admin Wrote: This function should work, though it was not tested much:

Code:
-- 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

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?

RE: How to calculate sun position? - admin - 11.05.2017

Try this one instead:

Code:
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

RE: How to calculate sun position? - Trond Hoyem - 21.05.2017

(11.05.2017, 07:49)admin Wrote: Try this one instead:

Code:
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

Thanks. That one worked.

Trond

RE: How to calculate sun position? - Auzzii - 30.10.2017

(21.05.2017, 20:59)Trond Hoyem Wrote:
(11.05.2017, 07:49)admin Wrote: Try this one instead:

Code:
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

Thanks. That one worked.

Trond

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

RE: How to calculate sun position? - Erwin van der Zwart - 30.10.2017

Hi,

Use function suncalc() like this:

Code:
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)

BR,

Erwin

RE: How to calculate sun position? - Auzzii - 30.10.2017

(30.10.2017, 15:19)Erwin van der Zwart Wrote: Hi,

Use function suncalc() like this:

Code:
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)
BR,

Erwin

Hi,

I get error:

Common functions:212: attempt to index global 'cec' (a nil value)
stack traceback:
Common functions:212: in main chunk

RE: How to calculate sun position? - Erwin van der Zwart - 30.10.2017

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

RE: How to calculate sun position? - Auzzii - 30.10.2017

(30.10.2017, 19:42)Erwin van der Zwart Wrote: 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

After fw upgrade and factory reset all works fine. Big Grin

thank you Erwin

RE: How to calculate sun position? - buuuudzik - 31.10.2017

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:
isdst = os.date('*t').isdst

if isdst then d = d - 3600 end

RE: How to calculate sun position? - Erwin van der Zwart - 31.10.2017

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

RE: How to calculate sun position? - buuuudzik - 31.10.2017

I've checked this and you're right, raw unix time is in UTC. I am using other algorithm and on it I must add the info about time zone and I must consider time shift Wink