Texture image resolution - how does Revit choose?

[bill_gilliss]

As far as I can tell, the image textures in C:\Program Files (x86)\Common Files\Autodesk Shared\Materials\Textures\1\Mats are identical to the ones in ...\2\Mats and ...\3\Mats except for resolution: folder 1 is low-res, 2 is medium-res, 3 is high-res. The default ones in the architectural template all seem to be from the low-res folder.

How does Revit decide when to use the others? Does Revit automatically switch to higher-res images when a rendering is large?

Lastly, when creating new materials using these textures, which folder should I be using? And why?

Thanks!

[barthbradley]

They are not all the same. They are different resolutions. 

Example: "american_cherry.png"  

Default Resolution=256x256

Base Resolution=512x512 

Medium Resoultion=1024x1024 (for close-ups and larger scale renders)

[bill_gilliss]

I understand that they are different resolutions. My question was, when does Revit use the higher-resolution images instead of the low-resolution ones in the ...\1\Mats folder that are used to define each material?

Does it automatically switch to using a higher resolution texture image if the rendering is large? What triggers the switch?

[RDAOU]

@bill_gilliss 

When you are using the out of the box material appearance assets, Revit automatically uses the Highest resolution in the Render Quality Settings.

Your next question would be what are the low and medium for or when are they used; Revit uses the low and medium resolution for balanced performance when using View specific setting such as in Realistic View and/or when using Simplified accuracy in the Render Quality Settings.

[ysautodesklover]

Hi @RDAOU 

Let me ask further question about this topic.

What if I put my original texture in each folder, will Revit recognize each texture as you explained? 

[JM_K]

Just to clarify, do you want to add your own custom Maps into the Revit folders or do you want to copy over the lower resolution images with the high resolution images (or vice versa)?