3

I'm creating a poster in Inkscape using various complex objects containing a lot of nodes. Inkscape is lagging quite a lot lately and so I thought I could edit the separate objects in separate files first and combine them later. This does have many downsides (like not being able to compare the multiple objects to one another) and therefore I have the following

Question: can I tell Inkscape not to draw certain objects, such that it wouldn't lag when working with the current objects?

Improve this question
4
  • You can hide objects yes. (not drawing means its not shown)
    – joojaa
    Commented May 3, 2022 at 14:44
  • Have you tried working with clones? For example instead of duplicating objects, clone them instead. That way, you are not creating more objects (with more and more nodes), merely using a clone.
    – Billy Kerr
    Commented May 3, 2022 at 15:25
  • 1
    You need to provide more details: Inkscape version, Operating system, and more importantly, how many objects you're dealing with and what dialogs are open.
    – s.ouchene
    Commented May 3, 2022 at 18:07
  • I've found Inkscape on Macs to be much slower than on Linux machines. If you have Linux available it might be worth a try for this use case particularly.
    – chicks
    Commented May 3, 2022 at 19:11

1 Answer 1

Reset to default
4

Close the non-active items in the objects panel. You can make it fast if items which belong together are a group. Rename the group to be able to find it easily in the Objects panel.

As well a bunch of items can be in its own layer. A layer can be closed and opened with a single click in the Objects panel or in the Layers panel.

A suggestion: Have a rasterized copy or a simple outline of a complex object (lock it to be sure it's not moved) to see where the invisible complex object is.

Improve this answer

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.