enter image description hereIn the lego movie, the fire, water, and smoke is made of lego bricks. It's not just a simple method of having one type of brick, several different bricks and brick slopes are used to get a really nice effect. Also, the location of bricks and the types of bricks changes over time. I have no idea how to do this in blender. (I want an animated effect, not a still image.) is there an addon, or some other way to do this in blender?

(I'm not expecting someone to have an answer that includes smoke, fire, and water, I just want one of them, I figure that any answer someone gives me can be transferred to another)

  • 1
    $\begingroup$ I would need a gif or something to see some particulars and examples but my initial suggestion would be to look into swapping out transparencies of different pieces. That's the technique that's used for their facial expressions; I imagine they borrowed from the same bag of tricks to perform other magic. $\endgroup$ Oct 8, 2019 at 1:12
  • $\begingroup$ how do I include a gif? I have several images that if played in sequence would make a good gif, but I'm not sure how to include it in the answer. $\endgroup$
    – Millard
    Oct 8, 2019 at 1:22
  • $\begingroup$ I think by editing your initial Ask, you can include gif as a file type. These comments won't take any images. $\endgroup$ Oct 8, 2019 at 1:33
  • $\begingroup$ @Millard you could use a software that will convert it there are many that can do it and some online ones. $\endgroup$
    – Yohello 1
    Oct 8, 2019 at 1:34
  • 2
    $\begingroup$ Really, my bounty isn't really because of a "lack of attention". I just think that it's better to award bounties on such long, difficult questions. $\endgroup$
    – Millard
    Oct 10, 2019 at 19:16

3 Answers 3


The final codes requirements are listed at the end of the post.

The tutorial by Ray Mairlot, which JBlock linked in the comments, is helpful for fluid simulations, or anything with a surface.

I will give insight into a method for volumetrics, specifically smoke, with different pieces. For professional use, you could include pyopenvdb (the install is complicated) functionality.

We can access the density_grid of the smoke with python. From [brickarchitect.com], we gather the dimensions of the smallest brick. (The smallest brick is a third of the heigt in the following image: 3.2mm.)
smallest brick
Click to enlarge

Distinguishing Slopes with Convolution

We will use convolution with the smoke simulations density_grid to decide, if we should use a sloped piece or not.

Im going to use the 2 (5) bricks. brick2 comes in 4 rotation variations, which all have their origin at the bottom left.

smol briggs
Click to enlarge

Understanding convolution is crucial, and I recommend Understanding Convolutions and then Intuitively Understanding Convolutions.

Consider two 2 dimensional 2x2 kernels (red and blue) and a 5x2 image (gray). To do a single convolution step, overlay a kernel over the image, multiply the overlaying number, add them together, then place them on the image at the position of the top-left square of the kernel.


For the red kernel, the convolution of the first pixels, look as follows.

kernel multiplication

For each different shape [5 (rotated) bricks] we will create a kernel and convolve it with the image. Then, we will compare the kernels and choose the maximum value for each pixel. Looking at a single kernel, we compare if its convoluted image has the same value as the maximum image. If yes, we can choose the shape and place it at that pixel.

I noticed, that we will need higher resoluted kernels for the slopes and have chosen a 6x9x1 kernel for a sloped piece.

By changing the weights of each kernel, as well as the maximum image, you can easily adjust the ratio of each piece. In my first render, slopes were quite sparse.

Click on the gif for a higher resolution.

another angle

In practice, we can just construct the kernel as a np.array and divide it. The more we divide it, the less it will contribute.

brick_1.kernel = np.array([[[1, 1, 1]]]) / 8

kernel division


Do a smoke simulation. Import the usual helpful modules for python.

import bpy
import bmesh
import numpy as np
from mathutils import Vector
from math import floor, ceil
import scipy.ndimage

Get the smoke modifier settings off the domain object.
Get the density_grid and calculate the step size. I used a cube, with its bottom-left corner at the world's origin as the domain.

depsgraph = bpy.context.evaluated_depsgraph_get()
smoke_obj = bpy.data.objects['domain'].evaluated_get(depsgraph)
smoke_domain_mod = smoke_obj.modifiers[0]
settings = smoke_domain_mod.domain_settings

grid = settings.density_grid
thres = max(grid)/2
res = settings.resolution_max
max_dim = max(smoke_obj.dimensions)
x_step, y_step, z_step = 1/res, 1/res, 1/res
grid = np.array(grid).reshape((res, res, res))

If we construct a mesh from this data, and add vertices if the density_grids values are over a certain threshold, then we can visualize the data with vertices.

bm = bmesh.new()

for z in range(0, grid.shape[0]):
    for y in range(0, grid.shape[1]):
        for x in range(0, grid.shape[2]):
            if grid[z, y, x] > thres:
                bm.verts.new((x*x_step, y*y_step, z*y_step))


grid cubes

After this prototype, for real now. Create a class to sample an object once and instantiate later in another bmesh for the brick types.

In this class

  • store a brick bmesh
  • store a kernel
  • add a method to instantiate the bmesh in another bmesh, we can do this by looping of the vertices and faces in the original mesh and adding them to the new mesh
  • add a method to instantiate multiple copies. It takes a grid with True and False values as the input and operates in steps of X = 0.008 | Y = 0.008 | Z = 0.0096, the values we read from the original brick image.
  • as a bonus I pass along a "color_grid", which we can use to give certain faces a different material index. Currently, faces get assigned to two material slots, so make sure that the target object, which will hold the lego geometry has in fact two materials on slot 0 and 1.
class Brick(object):
    def __init__(self, name):
        self.bm = bmesh.new()
        self.kernel = None
    def instantiate(self, bm_target, at = Vector((0, 0, 0))):
        verts = [bm_target.verts.new(v.co + at) for v in self.bm.verts]
        faces = []
        for f in self.bm.faces:
            face_verts = [verts[v.index] for v in f.verts]
        return faces
    def place(self, bm_target, grid, color_grid):
        sx, sy, sz = 0.008, 0.008, 0.0096
        for z, y_ in enumerate(grid):
            for y, x_ in enumerate(y_):
                for x, value in enumerate(x_):
                    if value:
                        faces = self.instantiate(bm_target, Vector((sx*x, sy*y, sz*z)))
                        mat = 1 if color_grid[z, y, x] > 40 else 0
                        for f in faces:
                            f.material_index = mat

brick_1 = Brick('brick_1')
brick_1.instantiate(bm, Vector((0.2, 0.1, 0)))

Since the evenly spaced voxels are not the appropriate size for the bricks (which have different x, y, z dimensions), we have to resample the grid with scipys ndimage.zoom function. Note, that the original grid will have stored the values in z, y, x order, which we will work with.

  1. Get the dimensions of the smoke_obj in the 3D space and reverse them.

  2. Calculate the amount of sample which the smoke simulation has, by using the smoke_domains settings' resolution_max.

  3. Calculate how many lego samples we require. I used the original bricks dimension and multiplied it with 3. Then we end up with a 6x9

    dimensions = np.flip(np.array(smoke_obj.dimensions)) samples_smoke = dimensions / np.max(dimensions) * domain_settings.resolution_max samples_smoke = np.floor(samples_smoke) samples_lego = dimensions / np.array([0.0032, 0.008, 0.008]) samples_lego = np.floor(samples_lego)

    zoom_factor = samples_lego / samples_smoke * 3

    grid = np.array(grid).reshape(samples_smoke.astype(int)) grid = scipy.ndimage.zoom(grid, zoom_factor, order=1)

Example: If the z, y, x dimension of the domain cube are 0.51, 0.32, 0.32, and the resolution_max is 64, then the voxels are resoluted as 64, 40, 40. The amount of small (3.2mm x 8mm x 8mm) lego pieces in that area would be 159, 40, 40, which is stored in samples_lego. This results in a zoom_factor of 7.45, 3, 3 and a resampled grid the size of 477, 120, 120. We have to use ndimage.zoom with and order=1 interpolation, because with higher order interpolation, we could receive negative numbers.

I will construct the kernels for these bricks with numpy.

brick_1.kernel = np.array([[[1, 1, 1]]]) / 8
brick_1.kernel = np.repeat(np.repeat(brick_1.kernel, 9, axis = 0), 3, axis=1)
brick1_grid = scipy.ndimage.convolve(grid, brick_1.kernel)[::9,::3,::3]

brick2a.kernel = np.array([ 
                            [[-6, -6, -1, 0, 1, 1]],
                            [[-6, -5, 0, 1, 1, 1]],
                            [[-5, -3, 0, 1, 1, 1]],
                            [[-4, -1, 0, 1, 1, 1]],
                            [[-3, 1, 0, 1, 1, 1]],
                            [[-2, 0, 1, 1, 1, 1]],
                            [[-1, 0, 1, 1, 1, 1]],
                            [[1, 3, 1, 1, 1, 1]],
                            [[3, 3, 1, 1, 1, 1]]
                            ]) / 84
brick2a.kernel = np.repeat(brick2a.kernel, 3, axis=1)

The first kernel has dimensions of 9, 3, 3 and the second kernel has dimensions of 9, 3, 6, since it is a 2 slot brick.

Since we are using a grid, with a higher resolution than our brick size, we can adjust the step size through that grid to get to the discrete steps the size of our smallest brick.

brick2a_grid = scipy.ndimage.convolve(grid, brick2a.kernel)[::9,::3,::3]
brick2b_grid = scipy.ndimage.convolve(grid, brick2b.kernel)[::9,::3,::3]

To check which if the convolved grids, has the highest value, construct a max grid with a small initial value (0.065). Where there is no smoke, the other grids will have values of 0. When comparing them to our small number in each cell of the max_grid, the max grid will be larger. This results in no bricks filling those spots. No smoke -> no bricks.
For each brick, we check if it has larger values in its convoluted grid.

max_grid = np.ones(grid.shape, dtype=np.float)[::9,::3,::3] * 0.065

max_grid = np.maximum(max_grid, brick1_grid)
max_grid = np.maximum(max_grid, brick2a_grid)

If we compare a grid of a brick with the max_grid (brick1_grid == max_grid), then we get a binary array, which holds True for the positions, where we have to place the brick kind. For that, I use the place function which we wrote earlier.

bm = bmesh.new()
brick_1.place(bm, brick1_grid == max_grid, density_grid)
brick2a.place(bm, brick2a_grid == max_grid, density_grid)

For the rotated brick2's (brick2b, brick2c, brick2d), we can simply switch flip the kernel of brick2a and/or swap its axes.

Download the script here

The final code (github: smoke_to_lego.py) assumes

  • a baked smoke domain object named 'domain'
  • the brick objects brick_1, brick_2a, brick_2b, brick_2c, brick_2d
  • a target mesh object named result
  • an installed scipy module. You can install a module for your local python distribution, but I choose the unclean, easy way and simply did a pip install (python -m pip install scipy) in Blender 2.8 python executable directory.

After downloading the file, bake the smoke simulation. Then execute the script on a frame with smoke. The script takes approx 5-20s in the given resolution during which Blender is unresponsive.

To render all of the frames, simply call the wrapped operation while looping through all frames.

# call the method to generate the geometry for this current frame
if True:

# loop through all the frame to render all frames
# this will be slow and won't show visual feedback
if False:
    scn = bpy.context.scene
    for frame in range(scn.frame_start, scn.frame_end + 1):
        scn.render.filepath = "//output/" + str(frame).zfill(4) + ".png"
        bpy.ops.render.render(animation=False, write_still=True)

I won't update the script in the .blend. The github link is up to date.

  • $\begingroup$ will I have to run the script for every frame or will it run for an animation? $\endgroup$
    – Millard
    Oct 18, 2019 at 12:52
  • $\begingroup$ how did you do the materials? did you just have a different material on each sample brick? $\endgroup$
    – Millard
    Oct 18, 2019 at 13:00
  • $\begingroup$ Yes, correct, youd would have to run the script on every frame. I will extend the script to automate that, which is what I used to render the preview. (I can do that later probably). If you look in the place method for the bricks, you will see how a material index is assigned for every brick. This requires two materials on the target object result. I just realized, that this would be easier to process with vertex colors. I will try to update that asap as well. Thanks for your patience. $\endgroup$
    – Leander
    Oct 18, 2019 at 13:19
  • $\begingroup$ Thank you for the script! i look forward to future updates! $\endgroup$
    – Millard
    Oct 18, 2019 at 13:31
  • 1
    $\begingroup$ @Robin If you want to use your own inside/outside of a mesh you should use pyopenvdb, which will provide many functions by it's own. The better remesher here is an example, but as you can see from the threads posts compiling openvdb for your system sounds tedious. You're right, in theory there are many convoluted squares which aren't necessary. Since we only use a single position of all summed values, you probably wouldn't even need convolution but some multiplication and a reduce operation, but convolution is fast and I'm lazy. $\endgroup$
    – Leander
    Oct 26, 2019 at 10:05

There is also an addon made specifically for this in Blender; may be worth looking into as it streamlines the whole process: https://blendermarket.com/products/bricker

enter image description here

enter image description here


Use geometry node - position to determine node set responsible for coloring legos in effect. You can also create separate legos, create geometry and add particle system that will emit lego bricks at this geometry(faces/verts). If you choose to emit random lego cube from set of cubes you made earlier, choose their lifetime very short, and emit them often you should get nice effect :) Try looking for addon for generating effect geometry. Im not sure if shape keys would work in this case.lego So I created geometry from landscape addon I set large emmit number. Lifetime 1. The problem is that despite I can control water movement on the time of creation of landscape by just changing its offset value, I cannot animate it after :( Photo above shows slopes. If you create grid-like mesh steps can be exactly 1 lego block (tested). You will than need to use grid mesh with geometry boolean modifier.

Update: You can actually use shape keys, but not in direct meaning of this. You create multiple meshes from landscape modifier, that differ a bit using offset x or y. This will make them look like next frames of water. Then make shape keys on those objects and then you can hit space "transfer shape keys" into mesh that creates particle simulation. And make transition between tham making key frames on their proportion.

  • $\begingroup$ it's a good idea, so i upvoted it, but It wouldn't work for slopes, as the particle system doesn't know where to put slopes and where not to. $\endgroup$
    – Millard
    Oct 12, 2019 at 20:45
  • $\begingroup$ (there is some slopes in the the gif i showed, but they are the 1x1 brick types, and this simulation isn't quite as detailed as for instance, the ocean scene in lego movie) $\endgroup$
    – Millard
    Oct 12, 2019 at 20:47

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