Vray light Lumens

[joconnell]

Ah okay.

So Gijs post is taking into account Vlado's explanation that vray uses watts per square metre to calculate it's intensity so if you're using a different display unit (lets say centimetres), you have to compensate for that with your multiplier values (multiply them bx 10 x 10) so that they become equivalent of metres squared again.

What vlado is saying above is that it doesn't matter what system unit you use once the light is the correct size for the real world too, since vray light intensity is based on multiplier and surface area, you need to make sure that your light / bulb is the correct size for the same bulb in the real world. In your case of lumens, a 100w bulb is giving off about 1700 lumens, so if you are making a vray light you need to make it the same physical size as the equivalent bulb in the real world too, so a bulb with a radius of 5cm in the real world must be either 5cm or 50mm or 0.05 metres.

It also seems that when giving out lumen ratings, the bulb manufacturer are taking into account how physically big the source is too - so the lumen value for the bulb is for a bulb of a very specific size so once you also model a bulb of that specific size, you should get a similar output as it's real world equivalent. I'll do a test one of the evenings with an slr in a room in my house to see if it's in the right ballpark.

[Guest]

so if I'm working in metres 1700 lumens for a 5cm radius bulb is correct. If I am working in millimetres I need to do what with that 1700? times it by 1000 x 1000? that would be 1700000000???

so for a 100W bulb in millimeters, the following is correct?

1000x1000x0.026 = 26000

So if the vray light was a sphere, i would set the vray light U size to about 75mm (roughly the size of the bulb), with a multiplier of 26000?

where did that 0.026 come from?

I'm so confused Do I need to adjust the lumens depending on system unit setup? Or is 1700 lumens correct if my bulb is 5cm or 50mm? I work in mm and everything is to scale. But my results are so dark.

[joconnell]

Yeah It's horrible!

I think all of those nasty multipliers were based around the fact that people were previously using watts which is per square metre so they had to figure out how big their light was and adjust their multiplier accordingly. Lumens seem to be a much kinder value to use since it is actually taking into account the size of the light already. So for example if a manufacturer says their 100w bulb is 1700 lumens, they're saying they measured a 5cm radius 100w bulb and it gave off 1700 lumens of light. If they had a 2.5 cm bulb, it'd have a quarter of the surface area so it'd probably be something like 1700 / 4 lumens - the quoted lumen value is for a specific wattage bulb of a specific size. So in this case, no you don't need to adjust the lumen value, and it also seems that lumens is a total light emitted value, regardless of the size of the light so for the correct shadow size, you'd just have to make your bulb the same size as the light.

I'll try and shoot a few shots at home this evening to see what kind of iso / shutter I get in a room only lit by the ceiling lights!

[Guest]

Thanks for clearing that up There really should be an updated sticky post on this Basically what Vlado posted back in 2005 is now obsolete because Vray lights use lumens now and not watts therefore it doesn't matter what units of measure you use to set up the scene. As long as its the correct scale 1700 lumens is 1700 lumens, no calculation is needed.

I hope that's right

[joconnell]

Exactly - I'd say if you use default multiplier units then it's watts per square metre and all that horrible multiplication. Lumens seem to be an absolute value of light emitted, regardless of the surface area / radius of the light - I did a quick render test of an area light set with lumens and then doubled it's size and it had no effect on the amount of light in the scene - having the right light size is purely for shadow accuracy but 1700 lumens is a total amount, unlike watts which are measured by surface area and thus affected by light size.

[Guest]

but now I am back to my original problem of not being able to get a good amount of light. I have tried various camera settings and its either really dark or really blown out and very noisy. I had to crank up the render settings really high to get rid of splotches and noise. I must be doing something wrong here.

[nils_p]

Take a look at this photo:
http://www.flickr.com/photos/islandgirltalk/5092227212/

The little flame looks very blow out because of the low shutterspeed. You can't even see the flame anymore. Also the room looks very dark. This can't be avoided at all unless you are using (professional) hdri techniques.
This is what vray and all other renderengines do.

To avoid the noisy render, try increasing the hsph. subdivs and interp. samples.

[Guest]

Is this correct:

Vray always works with generic units and assumes the 1 unit is equal to 1 metre. The default light multiplier defines the light intensity in watts per square meter. So 1 unit (watt) would cover 1 metre squared (Watt/m^2). If a Vray plane light has a half length and width of 5 metres, the actual light size is 10 metres x 10 metres. This means that the total emitted light is 100 watts.

That was worked out from a post by Vlado some 5 years a go. Is it still the same? http://www.chaosgroup.com/forums/vbu...ght+efficiency

How does it work with a sphere light?

If you change the multiplier to luminous power then the size of the light does not effect the intensity only the value entered defines intensity.

[Guest]

Or does default (image) units have no connection to wattage or lumen's its just an intensity value dependant on the lights size. This would be useful for scenes that haven't been modelled to real world scale.

[add101]

That was what i was wondering. Is there a method to get close to the human eye, rather than a photo?

I think if a camera was closer to the human eye it would only be in-focus in the middle part, and it would have different exposures for different parts of the image depending on how bright they are + almost 180 degree peripheral vision with no 'noticeable' perspective distortion (and a couple of other amazing features).

The way that photographers deal with this (for example- interior with one bright light source, or exterior in daytime with bright sun outside) is either to add extra lighting of their own to the shot, or to take multiple exposures and composite them together as HDR composite images.
Here is an example http://farbspiel.wordpress.com/2010/...hdr-vertorama/

I came across similar issues when I wanted an exterior to be visible through the window of an interior scene. Instead of adding extra lights inside i 'faked it' by having an environment override for illumination, and lowering the visible environment sky.

Maybe try doing some tests with a camera at different brightness/exposure to see how the light is captured photographically. One bright light illuminating a space (and directly visible to camera/viewer) will be a lot more blown out than lots of smaller low intensity lights, or light from a big source or projected through a large diffuser.
From there you can use some 'artisitic license' to adjust the render closer to the way you perceive it.

[Guest]

Vlado can you give an updated explanation on how the lights calculate referring to my posts above?

[Guest]

Is this correct:

Vray always works with generic units and assumes the 1 unit is equal to 1 metre. The default light multiplier defines the light intensity in watts per square meter. So 1 unit (watt) would cover 1 metre squared (Watt/m^2). If a Vray plane light has a half length and width of 5 metres, the actual light size is 10 metres x 10 metres. This means that the total emitted light is 100 watts.

That was worked out from a post by Vlado some 5 years a go. Is it still the same? http://www.chaosgroup.com/forums/vbu...ght+efficiency

How does it work with a sphere light?

If you change the multiplier to luminous power then the size of the light does not effect the intensity only the value entered defines intensity.

Can anyone confirm the above? Or give an up to date explanation of how the lights calculate. The help help still doesn't explain how default (image) calculates. Is it in watts?

[Kr0no]

Just came across this post while searching info about the system unit and the precision of real world lighting, I have some work where I have a lot of info about the lighting and must be as precise as possible, and Im wondering if the system unit= 1 generics unit= 1 meters but the info is 5 years old, is it still like this? I found different information from here and there in different post and Im not quite sure which way to go?? Can Vlado can confirm or shed some light on this?

Would be much appreciated


This is what as been posted but in 2004!?

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.

Best regards,
Vlado

Is this still right, I can t see the option of normalize intensity, what did changed since that time and what is the correct way to go?

[Kr0no]

Vlado, can you confirm the info above?

Any input would be much appreciated!

thx in advance.

[winberg]

I have started to dabble with real world light intensitys too, what confuses me is the internal coordinates of 3dsmax and the "working coordinates". If vray always asumes that 1 unit is one meter, should the internal coordinates always be meters? Does it matter what the working coordinates is then?

-Tom