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How to calculate sun position?
#1
Hello

Do you have some algorithm to calculate a sun position?
Reply
#2
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
Reply
#3
(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.
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#4
Yes, I think it was ported from JavaScript based on one of online calculators.
Reply
#5
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?
Reply
#6
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.
Reply
#7
(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.
Reply
#8
Can I add this library to LM?

https://github.com/mourner/suncalc
Reply
#9
You cannot use this library directly, but rewriting it in Lua should be quite easy.
Reply
#10
(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?
There are 10 kinds of people in the world; those who can read binary and those who don't  Cool
Reply
#11
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
Reply
#12
(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
There are 10 kinds of people in the world; those who can read binary and those who don't  Cool
Reply
#13
(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

Attached Files Image(s)
   
Reply
#14
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
Reply
#15
(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
Reply
#16
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
Reply
#17
(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
Reply
#18
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
Reply
#19
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
Reply
#20
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 shiftWink
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