VRayLight units

[Gijs]

Maybe it is a .03 bug, I don't know. Have you assigned a Vray material to that box? Otherwise, post the scene and I'll give it a go

[TRGraphics]

One thing to note when working with a vray light is that size does matter when normalize intestity is checked. The size of the light directly effects shadows.

So, if you use a vray plane light to approximate a standard light bulb it should be set to the general size of the light source for correct shadows. Same goes for a sphere light. If you don't, the shadows will be diffused compared to what they should be. This is old news, but I thought it needed mentioning when using normalize intensity.

Tony

[_jaro]

TRGraphics,
thanx for usefull tip, concerning light size and shadows.

Gijs,
I'll post the scene during weekend.

[_jaro]

Gijs,

here is my test scene, version is 3ds max 7.

http://www.drevco.sk/misc.%20picture...20settings.max

[dlparisi]

I experimented a little with your file (nothing wrong with the file AFAIK but i had it open so I kept going ). The screenshot below shows my trials. If I checked double sided for the plane light I needed to cut the multiplier in half (i.e. 1.3) to get an equivalent render. This makes sense because I now have two radiating surfaces. But when I switch to a sphere I need to divide the original 2.6 by Pi (3.14) to get a multiplier of .828 to get an equivalent render.

I'm not sure how this math works exactly since the surface area of a sphere is 4*Pi*r^2 and the area of a plane rectangle would be just r^2 (or x*y). Is it just coincidence that a multiplier of 3.14 is close to the other renders or is there really something to it?

[_jaro]

dlparisi,

I'm not sure how this math works exactly since the surface area of a sphere is 4*Pi*r^2 and the area of a plane rectangle would be just r^2 (or x*y). Is it just coincidence that a multiplier of 3.14 is close to the other renders or is there really something to it?

I think we can call Vlado to make it clear ( definitely )

[WWX]

Great thread

have a couple of questions:

1 - does these calculation only relate to VRay lights or can it also be applied to normal Max spot &omni lights? if not is there a route to follow in order to get the right amount of light out put from them?

2 - same ? for photometric lights

3 - this image show the result of a 5 X 5 m space being lit by a single Max spot. the light settings can be seen in the image. a vray material has been applied to all elements with. why is it that the camera viewport appears totall dark, but the image render "correctly



4 - the same scene as above, but with the Max spot being replaced by a Vray light. This time the camera viewport lights up and are totally washed out, but when rendered it comes out black?



p.s. my system and display units are set to mm

Thanks

[WWX]

Anybody?

[zoranm]

Let me give it a try:

1. No. With stanard lights you manually define the cone of light and that in turn defines the light contribution to scene. For example, if you have a standard light with cone hotspot/falloff=30/30 and multiplier=1 and than you just change the falloff velue to 150 everything'll overburn because you're letting much more ligt out. Therefore, listen to Vlado's advice and use Vlights as much as possible.

2.Photometric lights have their own distribution web predefined and you, theoretically, shouldn't change them except through colour mapping.Although, I, personally, haven't yet use a photometric light without changing (read "adjusting") its multiplier value.

3. Check help file for viewport preferences/default lighting.

4. Read Gijs post from before. Since you are working in mm and Vray internally works in m, you need to multiply your values by 1000000 to get it right and than diveide by 10 to accomodate light loss. So if you want a 40w=4000000, 50w=5000000 etc (with Normalize intensity=ON)!
As for viewport problem, see 3.

Hope this might help

Zoran

[RErender]

I think I have finaly seen the light

I found this at http://www2.sjsu.edu/faculty/watkins/eye.htm

Here is an illustration. Suppose a viewer is one meter from a 60W light bulb. The area of a sphere r meters in diameter is equal to 4πr2. Thus a one meter sphere has an area of 12.566 square meters and hence the intensity of the light from a 60W bulb is 60/12.566 or 4.775 Watts per square meter. The photometric intensity in lumens per square meter depends upon the wavelength of the light. If the bulb radiated pure yellow green light of wavelength 0.555 μm then the luminous intensity would be L = 685*1*4.775 = 3271 lumens per square meter. The same power bulb radiating pure green light at 0.510 μm where the efficiency of the eye is 50 percent would have a luminous intensity of L = 685*(0.5)*4.775 = 1635 lumens per square meter. For a bulb radiating red light at 0.48 μm, where the eye has an efficiency of about 10 percent the luminous intensity would only be 327.1 lumens per square meter.

So according to that if your generic units are set to meters and you want a 60 watt bulb the multiplier should be 4.775 or 60/12.566 with normalize intensity on . If you generic units are set to centimeters it would be 47500. It is interesting that the human eye is more or less efficient at different wavelengths, I have read that white light can vary from .4 to .7 so there is quite a range to adjust the multiplier to a more percieved output.

Also lux/685=multiplier

Some more good info http://www.technika.com/navpage/light.htm

[Guest]

I have started looking at using correct light values in my scenes instead of placing random sized plane lights with off the bat multipliers. This is the first thread that I have come across in regards to calculating the correct amount of light for a typical bulb. But it is 5/6 years old, can some one give an updated explanation of how this all works? I notice that "normalize intensity" is no longer in the Vraylight setup.

I typically work in mm with LWF. I would like to know how to light a room with a Vray sphere light that is the size of a typical 60watt bulb but emits the right amount of light.

[Morne]

I typically work in mm with LWF. I would like to know how to light a room with a Vray sphere light that is the size of a typical 60watt bulb but emits the right amount of light.

Did you eventually figure this one out James? Could you share your findings please?

[vlado]

Simply dial in the right lumen value. F.e. this page:
http://lumennow.org/lumens-vs-watts/

mentions that:
40-watt incandescent bulb = 450 lumens
60-watt incandescent bulb = 800 lumens
100-watt incandescent bulb = 1600 lumens

However it is best to check the specification of the particular light bulb because the lumens might vary - 60 Watt bulbs from different manufacturers might produce different illumination. The lumens are specified on the packaging.

Best regards,
Vlado