Stanislav Darula Institute of Construction and Architecture SAS Bratislava, Slovakia

1. Principles of daylighting calculations Stanislav Darula Institute of Construction and Architecture SAS Bratislava, Slovakia

2. Daylighting in exterior Daylighting (sunlight and skylight) Pv – parallelbeam illuminance Dv – diffuse illuminance Gv – global illuminance Gv = Dv + Pv Luminous solar constant 133.8 klx Top of the atmosphere Sky radiation Pv, Dv, Gv Ground level Brno 24.07.07_3

3. Luminance distribution under real skies Nikon Coolpix 990 Nikon Fisheye Converter FC-E8 Program LumDISP Brno 3.12.07_11

4. Interior illuminance Daylight Factor, DF e = eo + ei + ee eo - sky komponent ei - interreflextion component ee - exterior reflextion component 3 CIE overcast sky ( Moon-Spencer, 1942) Lz E L g eo Interior daylighting ei Ei e 1 ee

5. 1.2 2.8 1.6 21.1° 0.9 0.85 Section Plan 2 wR 2 lR Example of Daylight Factor caclulation Daylight Factor izolines Obstruction Room 5 m x 5 m

6. Exterior sky illuminance Lv- luminance of the sky element in cd.m2

7. Calculation of the standard luminance distribution

8. Example Luminancecalculation of sky element L Inputs: For Sky IV.4 is (a, b, c, d, e parametes, taken from ISO 15469:2004) Position of Sun: s = 38.02, As= 147.67 Pozítion of sky element:  = 10, A = 130 Calculation procedure: Sun zenith angle: Zs = 90 - s =90 - 38.02 = 51.98 = 0.907 rad Azimuth of sky element measured from sun meridian Az = s - = 147.67 -130= 17.67 Angular distance between position of Sun and sky element  cos  = cos Zs cos Z + sin Zs sin Z cos Az cos  = cos 51.98 cos 80 + sin 51.98 sin 80 cos 17.67 = 0.8462  = 32.20  = 0.562 rad

9. Input data in radians Example Calculation of luminance of sky element L Calculation of gradation and indicatrix functions: Gradation function for the sky element: (Z) = 1 + a exp (b/cos Z) (Z) = 1 - 1 exp(-0.55 / cos 80) = 0.9579 Gradation function for zenith: (0) = 1 + a exp (b/cos 0) (0) = 1 + a exp b = 1 - 1 exp(-0.55) = 0.4231 Indikatrix function for the sky element : f () = 1 + c[exp(d) - exp(d/2)] + e cos2 f () = 1 + 10[exp(-3 * 0.562) - exp(-3 /2)] + 0.45*0.84622 = 3.0847 Indikatrix function for zenith: f (Zs) = 1 + c[exp(dZs) - exp(d/2)] + e cos2Zs f (Zs) = 1 + 10 [exp(-3*0.907) - exp(-3 /2)]+ 0.45 cos2 51.98 = 1.7385

10. Example Calculation of luminance of sky element L Relative luminance predetermination L / Lz =  (Z) f () / (((0) f (Zs)) L / Lz = 0.9579* 3.0847/(0.4231* 1.7385) = 4.0175 Relative sky luminace distribution for IV.4 standard when solar altitudes = 38,02 ifthe measuredLvz = 4404 cd/m2 then absolute luminance is L=Lvz (L / Lz)=4404* 4.0175=17693 cd/m2 DARULA, S. - KITTLER, R. CIE General Sky standard defining luminance distributions. The bi-annual conference of IBPSA Canada eSim 2002. http://www.esim.ca/2002/documents/Proceedings/other2.pdf

11. P P Overast sky conditions

12. P P Clear sky conditions

13. Conclusions • For precise calculation we need a computer and information: • plan and section drawings of the building design • be careful to check the inputs: • - - room and window dimensions, glass transmittance • - - interior reflectances • - - exterior obstructions and their reflectances • apply an appropriate algorithms • - - check the results Today the laptop will do ...

14. What was the vision of scientists in 1954 how will look like a „home computer“ after 50 years later, i.e. in 2004?

15. Thanks for your attention