# Calculating photographic exposure - confused about luminance, illuminance, luminosity etc

On first glance this sounds like a photography SE question, but it's more concerned with physical measurement of luminosity, luminance etc - the photographic application of the value calculated is almost incidental.

I'm working on a calculator for astrophotography, so as to get a ballpark figure for what my shutter speed ought to be for an object of a given size (i.e a planet, not an effective point source like a star) and luminosity (calculated from apparent magnitude (i.e, within the visual spectrum) and in lumens). Wiki gives the equation for shutter speed as:

$$\frac{N^2}{t}=\frac{LS}{K}$$

with N as aperture, S as ISO (both constant in my case), K as a constant (subjective, but usually about 12), and L as average scene luminance, from which one can find t as shutter speed.

it also gives:

$$\frac{N^2}{t}=\frac{ES}{C}$$

with N as aperture, S as ISO (both constant in my case), C as a constant (subjective, but usually about 300), E as illuminance, and t as shutter speed.

The equation itself is of little relevence, but I'm having trouble working out how to connect my luminosity value to a measure of brightness used in either. I'm in over my head with all these different measures and units of intensity, flux, luminosity, luminance ad infinitum. From here, I also run into the problem of lumens vs candelas. I know my object's luminance in lumens - I can't really comprehend the difference to candelas, but I believe it's something to do with the amount of light being emitted in total, while candelas refer to that from some specific solid angle.

What I mean to ask is: how can I get from the luminosity of an object of a known size (or, to put it another way, a known percentage of an otherwise totally black image) to getting a value for t out of the aforementioned equation?

• Using wikipedia I see sunlight energy is 42 % in visible and solar irradiance is 1400 watts/ m2, so 588 w/m2 visible. Using "apparent maginitude" you could convert sunlight to starlight to w/m2. Just a guess for now. Then for w/m2 you can get lumens and then lux which is illuminance. Jan 26, 2022 at 0:10

## 1 Answer

Using wikipedia I see sunlight energy is 42 % in visible and solar irradiance is 1400 watts/ m2, so 588 w/m2 visible. Using "apparent maginitude" you could convert sunlight to starlight to w/m2. Just a guess for now. Then for w/m2 you can get lumens and then lux which is illuminance.