hi guys, as the title says, i need a node recipe for thickish clear glass( old style, for a kerosene lantern)

TIA!  :))

thanks guys!!!

Anthanasius shows us the wonderfully simple way using P9/PP2012.

Note, however, that the effective IOR of thin glass is higher due to multiple internal reflections. To match what I showed in the first shader, use IOR of 1.99 in the Fresnel Blend node.

cool!  thankees!

Hum ... 1.99 ?

Something wrong with the fresnel blender node ?

Ok i use it a lot for my cornae shader ... Cornea IOR is close to 1.37 ... If i understand what you say Bagginsbill, i need to add some value.

For the glass, a classic glass is 1.5 IOR, i need to put 1.99 ... For cornea ? 1.37+?

I'm a bit lost !

Cornea does not show multiple internal reflections added together. Thin glass does.

Thin glass looks brighter than thick glass.

Light passes through the front and some of that also gets reflected from the back side and comes out the front.

Everybody who wants to understand Fresnel effect must read this until it is memorized.

http://en.wikipedia.org/wiki/Fresnel_equations

Of particular note about this latest point:

Quote - For common glass, the reflection coefficient is about 4%. Note that reflection by a window is from the front side as well as the back side, and that some of the light bounces back and forth a number of times between the two sides. The combined reflection coefficient for this case is 2R/(1 + R), when interference can be neglected

If anybody is carefully doing math, you may notice that I have been suggesting 11% reflection, whereas if we assume that the IOR is 1.5, it would be 4% for single reflection, and about 7.7% for multiple (thin glass) reflection.

However, many forms of glass are IOR 1.6 or even 1.65. In this range you can get 6% single reflection, and about 11% for multiple (thin glass) reflection.

Since people like shiny things, and since glass often really is that shiny, I give the 11% figure in my shader.

The equivalent multiple-bounce reflectivity, R, as an IOR, n, is given by the equation:

n = (1+sqrt(R))/(1 - sqrt(R))

For 11% we use

n = (1 + sqrt(.11))/(1 - sqrt(.11) = 1.99 .... voila

That worked quite well.

Bagginsbill, how did you make the chest board tiles in the above sample image? I tried tiles with clouds, but I'm not even getting close to that.

No offense meant, GeneralNutt ...but he is Bagginsbill!

...a name, a warranty in what he does and what he says.

I'm preparing tile shaders for the market, and that one is rather complicated.

However, the basic idea is not. You use a second tile node with a "softened" mortar to darken the edges of the white tile.

Check the screen shot - a highly simplified version of the shader.

Thanks. After you post it, it looks so simple (to get close), but when doing it myself, I just can't think it though like that.

"I'm preparing tile shaders for the market, and that one is rather complicated."

Another shader set that sounds worth the investment. Any news on the metal shader set?