Someone was asking some time ago what are the units of intensity (the “color”) for the VRayLights; finally had some time to figure that out, so here it is.
First, a few assumptions:
(*) VRay always works with Generic units internally. VRayLights also do that and assume that 1 generic unit is equal to 1 meter. If this is not so, the multiplier of the light will need to be adjusted, if you want the light to have specific predefined intensity.
(*) We assume that the light color is pure white.
(*) We assume all other VRayLight options have the default values.
If the “Normalize intensity” option is OFF (the default), then the light multiplier defines the light intensity in watts per square meter (Watt/m^2).
If the “Normalize intensity” option is ON, then the light multiplier defines the light intensity as the total power in watts of the light source.
If the “No decay” option is ON, the light is not a physical light and exact units cannot be defined.
Some other units of measurement in VRay:
(*) The irradiance map stores irradiance as watts per square meter.
(*) Pixel values represent radiance per pixel (the unit for radiance is watt per square meter per hemisphere, where 1 hemisphere = 2*pi steradians), assuming that no color mapping is performed.
explains why I have so much problem working in empirical units
Now a followup question:
If we are working in empirical units (1 unit = 1 foot) how should we be adjusting the multiplier? Should it just be (value X 3.28) ? What about if we check normalize intensity? Same thing? or do we need to square it for area in that case? Or should we be dividing? I’m confused.
If you have the “Normalize intensity” option ON, then you only have to specify the power of the light (in watts) - the VRayLight will do the conversion automatically. If the “Normalize intensity” option is OFF, and you want your light to emit X watts per square foot, then the multiplier should be X/(0.304^2). Again, this assumes that your Generic units are meters, while only the display units are feet. If your Generic units are feet too, then the multiplier is in watts per square foot by default.
what if I set the display units to ‘generic’ and my system units to ‘centimeters’? how do I have to set the vray light multplier if I want get a 60 watt lightsource? to 6000 or …?
When doing research on light, you will come across many different units being used by various light manufacturers. All the units are trying to describe how much light their devices emit. You will see units such as candle power, foot candles, candelas, foot lamberts, lux, lumens and my favorite: watts per steradian. Some units refer to the energy of the light source and others to the power. Many units take only the human eye sensitivity into account. The light units can be even more confusing when you consider that some light sources, such as a common light bulb, launch light in all directions while others, such as a laser, concentrate the light into narrow beams. Rather than confuse you even more by going into a long discussion of what the various units mean, I’m going to try to simplify the problem. Let’s just assume that each light source has a distinctive emission spectrum and a certain emission geometry. You will have to treat each light source differently, according to how it is used with a specific communications system.
In optical communications you only need to consider the light that is sent in the direction of the detector. You also only need to consider the light that falls within the response curve of the detector you use. You should regard all the rest of the light as lost and useless. Since all the light sources discussed in this book rely on electricity to produce light, each source will have an approximate electrical power (watts) to optical power (watts) conversion efficiency, as seen by a silicon detector. You can use the approximate power efficiency and the known geometry of the emitted light to calculate how much light will be emitted, sent in the direction of the light detector and actually collected. Various sections of this book will give you some examples of such calculations.
Do units in the scene affect the falloff of the light?
I am trying to simulate the lights specified by our lighting consultant, and they are constantly telling me that my lights are too “hot” near the light and have too swift of a falloff.
that should be a matter of exposure correction, not falloff.
falloff isn’t unit dependant with the vray light, since there are no start/end settings.
maybe your scenes need more secondary bounces.
What is confusing is that there are two different quantities here - the (electrical) power consumed by the source, and the power of the light that is emitted. These are *different* things. What I wrote for the VRayLight concerns the emitted light power, not the electrical consumption of the source. A 60 watt bulb consumes 60 watts of electrical energy, but it certainly does not emit all of them as light power.
Sorry to bring this up again but I just can’t seem to get this to work the way Vlado describes it. Using a test scene, I placed a 2"x2" vray light underneath the stairs, checked Normalize Intensity, and set the multiplier to 150, assuming it would approximate a 150 watt light source (or at least close to it). The scene was basically black. I kept playing around and got the scene lit OK but needed to set the multipler to 20000! What am I missing?
I’m working in linear space (i.e. gamma 2.2, etc. but I don’t think that would affect it that much, in fact I thought it would help. Using Irr. Map/Light Cache for GI. System units are inches and display units are feet with decimal inches (default inches). Here is a screenshot of the setup…
what it basically means is that Vray is asuming that your lightpatch is not 2" x 2" but 2m x 2m. Therefore Vray will adjust the power as if it was a 2x2 m patch that emits the total amount of light. Since it is not 2x2m, your light will become very dim. 150W has to be multiplied by (1/0.0254)^2. This means that the resulting multiplier has to be about 150 x 1550 which is about 232500. Since a 150W lightbulb produces only about 10% light, this means that 23250 in your case equals a 150W light bulb.
actually i have 15watt bulbs that shine with the intencity of a 60watt bulb so i kinda have to agree that watts isnt really the best term to use is it?
