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I need to make some pebbles for my scene. How do I get the realistic looking pebble shader in Cycles? This is what I am going for:

enter image description here

And I have got this:

enter image description here

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Here is what I came up with:

It is mostly procedural, I did use an image texture for the grooves but you could take that out and use a voroni texture (I just liked the way the texture looked).

If you are trying to make a ground plane with thousands of pebbles I would NOT go this route. I would use a pebble texture and use displacement. However if you are going for a close up pebble shot this shader should work well.

enter image description here enter image description here

The trick is to build up the layers of detail.

The main left box is the main rock color, It is just broken up with a noise texture and 3 different colors. Use the color ramp to clamp the pattern.

*Notice I put a hue/saturation node (highlighted) I did this to quickly change the color of different rocks (or you could just change the 3 original colors).

The middle and front box are just for the scratches and grooves in the rock. The most important part here is to use the texture as a bump with a different color and material for the chipped grooves.

  • I added in a texture and mapping node to just change the look of the textures across the different rocks (So they don't all look the same).
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  • $\begingroup$ you are too good bro. thanks. Can you help me build a rock texture as well? i am just trying to learn procedural shading? what do you think is the best way to learn it or is there any concept i need to grasp first? $\endgroup$
    – user58005
    Commented Jul 12, 2018 at 7:53
  • $\begingroup$ Honestly its just practice. It seems intimidating at first but you really use the same nodes over and over again so you quickly learn. You can post another question for a certain rock texture here is a salt rock blender.stackexchange.com/questions/95399/… This is kinda advanced but I learned so much recreating it. artstation.com/artwork/KdDxx $\endgroup$
    – icYou520
    Commented Jul 12, 2018 at 14:42

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