Timeline for What does the "area" represent in the luminance of the sky?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Jul 19, 2020 at 8:12 | vote | accept | synecdoche | ||
| Jul 18, 2020 at 14:08 | answer | added | The Photon | timeline score: 1 | |
| Jul 18, 2020 at 14:02 | comment | added | The Photon | You left out "and subtends a given solid angle" in your quote from Wikipedia. It's equivalent to consider the surface area of the source and the solid angle into which it emits, or the surface area of the detector and the solid angle subtended by the source when seen from the detector. In either case, the solid angle term depends on the area of the other surface (detector seen from the source or source seen from the detector) and the distance between source and detector. | |
| Jul 18, 2020 at 12:57 | comment | added | synecdoche | If a detector or surface is being illuminated, it's measured in lux = lumens per square metre, and the area is that denominator. But I'm asking about luminance, not illuminance. Luminance, as that Wikipedia article points out, is a measure of "the amount of light that passes through, is emitted from, or is reflected from a particular area". I'm asking what that area is when a figure is given for the sky as a whole. Any arbitrary cross-section of the sky has an area, but how can luminance be constant if I could choose a cross-section at any distance and thus any area? | |
| Jul 18, 2020 at 5:49 | comment | added | The Photon | But to give a hint: The detector or surface being illuminated by the sky has area. As does a cross section of the sky at any chosen distance above the ground. | |
| Jul 18, 2020 at 5:25 | history | edited | Harish Chandra Rajpoot | CC BY-SA 4.0 |
improved format
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| Jul 18, 2020 at 5:22 | review | First posts | |||
| Jul 18, 2020 at 5:25 | |||||
| Jul 18, 2020 at 5:22 | history | asked | synecdoche | CC BY-SA 4.0 |