Search in
Sort by:

Question Status:

Search help

  • Simple searches use one or more words. Separate the words with spaces (cat dog) to search cat,dog or both. Separate the words with plus signs (cat +dog) to search for items that may contain cat but must contain dog.
  • You can further refine your search on the search results page, where you can search by keywords, author, topic. These can be combined with each other. Examples
    • cat dog --matches anything with cat,dog or both
    • cat +dog --searches for cat +dog where dog is a mandatory term
    • cat -dog -- searches for cat excluding any result containing dog
    • [cats] —will restrict your search to results with topic named "cats"
    • [cats] [dogs] —will restrict your search to results with both topics, "cats", and "dogs"

Cast a shadow with the object's collision shape and not the object itself?

I have a large collection of trees that will be viewed from far away, and I'd like to cast shadows with the greatly-simplified collision shape I generated in the static mesh editor instead of the object itself. Is there a way to do this without adding a shadow-blocking volume to each individual tree (because the collision shape is still a lot more complex than a cube)?

Product Version: Not Selected
more ▼

asked Nov 02 '14 at 04:27 PM in Rendering

avatar image

1 2 4

(comments are locked)
10|2000 characters needed characters left
Viewable by all users

1 answer: sort voted first

Hey heyx3 -

No shadowing is limited to the rendered objects and not just the calculated geometry. I would suggest looking into the Distance Field Ray Trace Shadowing though it might help you get a much more performant lighting environment for your trees.

Thank You

Eric Ketchum

more ▼

answered Nov 18 '14 at 10:35 PM

avatar image

Lovecraft_K ♦♦ STAFF
36.7k 702 260 736

(comments are locked)
10|2000 characters needed characters left
Viewable by all users
Your answer
toggle preview:

Up to 5 attachments (including images) can be used with a maximum of 5.2 MB each and 5.2 MB total.

Follow this question

Once you sign in you will be able to subscribe for any updates here

Answers to this question