# Luminance and Illuminance

1. Oct 30, 2014

### Howlin

Hi all,
I am not sure this is meant to be a part of the forums but if it is to be changed, please let me know. This is not a homework question but more of a (weird) curiosity question.

I was looking at the different lighting classes that are out there for roads and have noticed that the lighting levels for this vary depending on what type of road and its usage. I was looking through the British standard for public lighting and for Motor traffic zones it shows for a zone it has to have an average road surface Luminance L (bar) of 2.0 cd/m2. I then saw that for a conflict zone the average illuminance E (bar) for a zone must be 30 lux.
I have looked up what luminance and illuminance is: In a nutshell, Luminance is the amount of light given off a surface while illuminance is the amount of light hitting the surface.

I know the equation for illuminance is E=I/D2 or E= I*Cos3θ / D2.

Is there any way to convert illuminance (e.g. 30 lux) to luminance (e.g. 2.0 cd/m2 ) or vice virsa?

I am curious in knowing if there could be much of a difference between the lighting levels for a conflict zone and non conflict zone.

(I know I have a weird curiosity)

If my equations or definitions are wrong, please tell me.

Thank you,
Stephen

2. Oct 30, 2014

### CWatters

Not sure if this helps. Table 2.1 on page 8 suggests conflict areas are divided into 6 classes from ME1 to ME6 with lighting from 2.0 down to 0.3 cd/m^2. Somewhere in the document it may describe what some of the classes are. eg page 7 suggests motorways are ME1 or 2. Apparently it's all in BS5489 but I don't have a copy.

https://www.theilp.org.uk/documents/css-sl1-class-and-quality-of-street-lighting/

3. Oct 31, 2014

### Howlin

I am not asking about the classes. I just mentioned them to give a back round on what I was wanting.

All I want to know is: Is there an easy way of converting luminance to illuminance or illuminance to luminance?

4. Oct 31, 2014

### Andy Resnick

Welcome to the strange world of radiometry/photometry. The best way to proceed is from the units: you have cd/m^2 and lux.

Cd/m^2 is a photometric unit of radiance called 'Luminance', which you already knew. Photometry simply weights the spectrum of light according to human vision. The radiometric unit of radiance is W/(m^2*sr), and so should not be confused with W/m^2 (irradiance). Radiance contains information about the direction that light propagates- an nonuniform source rather than a point source, or a rough reflecting surface rather than a smooth reflecting surface.

Let's back up: the photometric equivalent of W is lm. For example, a 30 lm source outputs a certain number (>30) watts of light, which when spectrally normalized to human vision = 30 lm. 1 lx = 1lm/m^2, so your specification of 30 lx is radiometrically equivalent to irradiance (W/m^2). Thus, the 30lx standard appears to be a statement about the amount light incident on the road (think W/m^2), which you can calculate if you know the distribution of streetlights.

Now, 1 cd = 1 lm/sr, so you can't simply convert the 30lx illuminance into a (scattered) luminance, you need to know how the light is scattered off the road: diffusely? specularly? what is the bi-directional reflectance distribution (BRDF)? Say the road is a Lambertian reflector: perfectly diffuse. Then, the incident illuminance (call it 30 lx) is uniformly scattered into a hemisphere resulting in a luminance of 30/2π cd/m^2, or about 5 cd/m^2 = 5*10^4 stilb. To be totally obnoxious, that's also about 15*10^4 apostilbs = 15*10^7 skots.

Does this help?

5. Oct 31, 2014

### CWatters

Perhaps see page 6 of the doc I posted. As I understand it..

In non conflict area the idea is to reveal objects in silhouette against an lit background which is usually the road surface. So in that case they specify the illluminance (crudely the light reflected off the road surface).

In conflict areas the idea is to illuminate objects directly so they specify Illuminance (light available to be reflected off unknown objects).

You can google for the difference between Illuminance and illluminance.

6. Nov 3, 2014

### Howlin

Thank you, this is what I needed.
Luminance = incident illuminance/2π (assuming it is perfectly diffused)

But if it isn't perfectly diffused
Luminance = Incident illuminance * a diffusion factor / 2π

I have tried to look for the BRDF for a standard (if there is even such a thing) rough diffuse asphalt road but i cannot find it. Do you know of a rough estimate to use or is it better to get a more accurate value?

7. Nov 3, 2014

8. Dec 8, 2014

### John1234

Hi Howlin.

You have mentioned about Lighting classes in according with British Standards. I am street lighting designer and I am struggle to understand conflict (illuminance calculation) areas based on these standards. I'd like to find some examples showing and explaining how to mark / highlight these zones on junctions, roundabout etc. The only site I have found is on design experts forum which comes from www.desexp.com

9. Aug 15, 2016

### technosaurus

Hello,

I know this was a couple of years ago but I hope you can clear up some confusion. You appear to be saying that the luminance emitted by a surface from a perfectly diffuse surface, in relation to the illuminance incident on that surface, can be defined as follows:

luminance = incident illuminance/2π

Which makes sense to me, since a perfectly diffuse surface would reflect evenly over a 180 degree hemisphere, and there are 2π steradians in a hemisphere.

My problem is that every other source I go to (including a student handout from Cornell University) says this:

luminance = incident illuminance/π

I usually concede that when so many sources agree on something like this they must be right, but I cannot see how. I have tried to think of what I might have missed for days now. So either you and I are correct (because your formula seems to me to make perfect sense) or I have indeed missed something vital. I'm hoping you can throw some light on this or maybe point me in the direction of a reliable source document.

10. Aug 15, 2016

### Tom.G

11. Aug 15, 2016

### technosaurus

Thanks Tom,
My interpretation of this definition is that luminance is the luminous intensity from an infinitessimally small area (i.e. a point) in candelas divided by the orthagonal projection of this area. Since for a Lambertian (perfectly diffuse) surface the light is reflected through all angles up to 180 degrees about the normal to the surface, I would interpret this as saying that the luminous intensity acting in the direction of the orthagonal projection would be the same fraction of the total luminous intensity of the point as the orthagonal projection area is of the surface of a hemisphere.

Maybe I should simplify, or at least clarify, what I am getting at. Here are the terms of reference as I understand them:

1) The illuminance incident on a surface is measured in lumens per square metre (i.e. lux).
(Illuminance = Luminous power incident on a surface in lumens per square metre).

2) Luminance is measured in candelas per square metre, so we are talking about lumens per steradian per square metre, since candela are measured in lumens per steradian.
(Luminance = Luminous power per unit solid angle per unit projected source area).

3) The light output from a point (in lumens) due to luminance is due to the illuminance falling on that point, and can at most equal the light input (in lumens) due to the illuminance.

4) I am assuming a lambertian surface, so light is reflected in all directions around the normal to the surface in a 180 degree arc (in other words, it affects a solid angle of a hemisphere which is 2π steradians. I am assuming also a reflectance factor of 1 (i.e. no absorbtion of light enegy by the surface; it is all reflected in some direction or other).

Now, the way I see it, from any point on the surface, an equal number of lumens must be going out as are coming in, but they are scattered over a solid angle of 2π steradians. The luminence (remembering that this is measured in candelas, which are lumens per steradian per square metre) must then surely be the illuminance (which is measured in lux, which are lumens per square metre) divided by the number of steradians. And the number of steradians is 2π, and not π.

Therefore there are a lot of sources out there, including some published by top universites, that are incorrect.

If I have missed something incredibly obvious, I apologise, but at the moment I can't see it.

Last edited: Aug 15, 2016
12. Aug 15, 2016

### technosaurus

Hello,

(I think I put this in the wrong place before)

I know this was a couple of years ago but I hope you can clear up some confusion. You appear to be saying that the luminance emitted by a surface from a perfectly diffuse surface, in relation to the illuminance incident on that surface, can be defined as follows:

luminance = incident illuminance/2π

Which makes sense to me, since a perfectly diffuse surface would reflect evenly over a 180 degree hemisphere, and there are 2π steradians in a hemisphere.

My problem is that every other source I go to (including a student handout from Cornell University) says this:

luminance = incident illuminance/π

I usually concede that when so many sources agree on something like this they must be right, but I cannot see how. I have tried to think of what I might have missed for days now. So either you and I are correct (because your formula seems to me to make perfect sense) or I have indeed missed something vital. I'm hoping you can throw some light on this or maybe point me in the direction of a reliable source document.

13. Aug 15, 2016

### technosaurus

Can't believe I didn't realise this before. I forgot to take Lambert's cosine law into account - the size of the area seen by an observer tends to zero as the obeserver's position approaches 90 degrees from the normal (cosine 90 = 0). Doh! So yes, it's pi and not 2 x pi.

14. Aug 15, 2016

### Andy Resnick

Wolfe's book has a paragraph on exactly this point (page 19 for those reading along at home). Considering a Lambertian emitter (emitted radiance is independent of the angle), the radiant excitance is π times the radiance. The erroneous argument goes something like you state- there are 2π radians in a half-sphere, so the exitance must be 2π*radiance. However, this reasoning omits the *projected* area of the emitter. Averaged over a hemisphere, the projected area is half of the actual area, so the exitance is only π*radiance.

Note- your equation is using photometric units rather than radiometric units, but geometry is retina-independent so the relationship is the same.

ah- while I was typing this you realized your error. Hope this is helpful, anyway.

15. Aug 15, 2016

### technosaurus

Yes, thanks for taking the trouble. If I hadn't had my gestalt moment I would still be going round in circles, so your answer would have put me out of my misery :-)