I have a scene with objects organized hierarchically like this:

enter image description here

I would like to get the overall bounding box of the 00000_361.L3B EMPTY object - however, since this is an empty it has no dimensions, and its bounding box also has no dimensions.

I am thinking of recursing through the hierarchy like explained here but I wonder if there is a better/easier way to do that at the scene level.

  • $\begingroup$ The bounding box of all mesh objects that are children of the empty? $\endgroup$
    – batFINGER
    May 27, 2021 at 8:57
  • $\begingroup$ @batFINGER - yes, correct. Or even better - if there were more empties, of all of them together. $\endgroup$
    – simone
    May 27, 2021 at 10:14

1 Answer 1


Bounding box coords of all meshes in scene.

Pretty sure this has been answered before, but finding it is another matter.

Here is some numpy code I use to create a bounding box for all mesh objects in a scene

Basically, list coordinates of all bounding box coordinates of all mesh objects in the scene.

Use the minima / maxima of each axis produce an axis aligned bounding box encompassing all others.

import bpy
from mathutils import Vector
from bpy import context
import numpy as np
import itertools

# multiply 3d coord list by matrix
def np_matmul_coords(coords, matrix, space=None):
    M = (space @ matrix @ space.inverted()
         if space else matrix).transposed()
    ones = np.ones((coords.shape[0], 1))
    coords4d = np.hstack((coords, ones))
    return np.dot(coords4d, M)[:,:-1]
    return coords4d[:,:-1]

# get the global coordinates of all object bounding box corners    
coords = np.vstack(
    tuple(np_matmul_coords(np.array(o.bound_box), o.matrix_world.copy())
         for o in  
            if o.type == 'MESH'
print("#" * 72)
# bottom front left (all the mins)
bfl = coords.min(axis=0)
# top back right
tbr = coords.max(axis=0)
G  = np.array((bfl, tbr)).T
# bound box coords ie the 8 combinations of bfl tbr.
bbc = [i for i in itertools.product(*G)]
  • $\begingroup$ thanks - it looks like it works. If it's not too much trouble for you, would you add a couple comments in the code? It's a bit more idiomatic than the Python I can handle. But I can understand if you don't have the time $\endgroup$
    – simone
    May 27, 2021 at 13:37
  • $\begingroup$ what does space @ matrix @ space.inverted() if space else matrix do? I read it as matrix multiplication of space by matrix - but I am stuck with the multiply by space.inverted if something. Sorry for the basic question - it looks like numpy - but it's hard to google for "if". BTW - the comments help a lot $\endgroup$
    – simone
    May 27, 2021 at 14:58
  • $\begingroup$ ...and whats's the second return statement in np_matmul_coords? $\endgroup$
    – simone
    May 27, 2021 at 15:26
  • 1
    $\begingroup$ To put the bounding box in the local space of another object if need be.From a personal addon to scale a lattice to encompass objects and make an oblique projection eg blender.stackexchange.com/a/195675/15543 The second return statement will never be run and can be ignored or removed. Be a remnant from a copy paste job from a method to convert 3d coords to 4d $\endgroup$
    – batFINGER
    May 27, 2021 at 15:33

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