How did it start? Well, it actually started a long time ago, maybe a year ago, most probably more. I got myself tumbling down the beautiful rabbit hole of Blender based mapping. The idea is very simple: if you have a DEM, you can build a 3D representation of the terrain and use a renderer to build your map. To me the most striking thing about those maps were not their 3D-ness (which to some it's starting to be tiresome , and I agree), but the shadows. I've been pursuing the best shadow for a while, and this seemed like the perfect fit.

So, like I said, one year ago or so I took "the" Blender relief tutorial and run with it. I got to the point where I could reproduce it with a 1x1, 3600x3600px DEM from mapzen, but when I tried to automate it, I found out that Blender has a python console where it prints out the commands that are equivalent to the actions you make in the UI, but the resulting script was too horrible to my eyes and run out of breath (another of those cases of the perfect being the enemy of the good).

Then a few days ago I read that first link and got some steam build up. In fact, it was two passages in it that lit up the fire:

Most of these use a tool called Blender, an extremely powerful open-source tool for all kinds of 3D modeling and rendering. A few cartographers use other tools, such as Aerialod, or R‘s Rayshader plugin.

R! I can easily automate this!

If we stick to a very zoomed-out map, or if we have a really huge computer running Blender, we could try to do a hillshade for the entire world, and then slice that up for our map tiles. But there’s no way we could do this at a high-enough resolution so you could zoom all the way in, as in the sample tiles above.

Challenge accepted! (Spoiler: I'm not there yet).

I tried Rayshader. I wanna be honest: it's easy, quick, but I didn't like the results. It seemed like no matter how high you put the sun, it always drew very long shadows. So despite its pragmaticism, I left it on a side.

So I picked up what I did in the past and tried to apply it to a map. I re-rendered everything and applied it to my style. The outcome was encouraging:

To start with, I probably did that render with not enough render passes, so the output looks grainy. Second, the material color is too bright, so the height tints are washed out. Still, we can see the big shadow cast over the valley some 3200m below the Mont Blanc/Monte Bianco.

This proved to be a good place to test the method, because of the great difference between the valley and the peak casting the shadow over it, and that lead me to think: are there more extreme places in the world? An easy bet is yes, and the place to look for them was the Himalayas. The Aconcagua could have been a good contender, but the valley at its SE is some 4550m already. Surprisingly, the way I searched for a good place was to use existing maps with the usual hill shade technique, looking for big dark spots, specially those wide in the NW-SE direction. I first found my spot in the Tukuche Peak, that looms some 4350m above the Gandaki River, and then the nearby Dhaulagiri, that's even some 1250m higher, clocking at 8167m. Here's how they look (not the big structure in the upper left, but the bigger grey [8000m+] blob to the East of it; the river snakes in the right all the way down):

I had to add 3 more color bands to my style and reshuffle them because I never rendered any 8k'er before, so the colors were haphazardly concocted for the rendering and are not definitive. At least it lets you behold the grandiosity of that rock jutting through thousands and thousands of meters with very steep sides.

Time to get real. I usually render regions were I'll be going, and next time it's the Upper Segre Valley, so I rendered N42E001-2 in one go. That's 7201x4884px after reprojecting to WebMercator (and compensating as described in the second link!), so some 35Mpx. Blender took some 44m+ on a 4.5yo medium-high spec'ed laptop at 200 render samples, which means that I can continue to render small regions this way, but that for the moment I won't be applying this technique to the whole Europe.

Up to here I was just applying the same style in QGIS, which has been an indispensable tool to develop this style. But trying these TIFFs in mapnik for the first time needed an extra step. Well, two, in fact. Blender does not save the TIFFs georeferenced, so you have to copy the data from the original DEM. For that, use -a_srs ... -a_ullr ... with the right ESPG and the data from the output of gdalinfo. Next, for some reson, it always use 16bits integers, even when explicitly saying to use 8. This little snippet takes care of that:

import imageio

image = imageio.imread('pirinoak-blender-10x.tif')
image = image / 256
image = image.astype('uint8')
imageio.imwrite('pirinoak-blender-10x-8bits.tif', image)

Thank $DEITY (and developers!) for good libraries.

The first thing I noticed was that we have been lied by maps (again!) for a long time. Most hill shading algos use a 45° high sun (the direction does not matter much). But if you think about it, how many mountains have sides 45°+ steep? According to a (real, not like me!) cartographer friend, for continental Argentina it's less than 1% at 30arcsecs of resolution (note that SRTM is 1arcsec). Still, some shadows are there, and they help us (and we get used to that) to recognize slope direction. And now we're asking a raytracing program to calculate real shadows? The result I initially got was underwhelming, even when I was already asking Blender to exaggerate height by 5x!:

So, I bit the bullet and went all in with 10x:

Much better, but not definitive. I still have to render Dhaulagiri again, and at least some region I already know well by having being there a lot.

Now some notes about "the" Blender relief tutorial. I followed it to the letter, but with my experience I had to make some changes. One you already know, using a displacement scale of 10x instead of 0.3. I have no exact idea why his initial rendering were so spiky, but I suspect that the DEM grid unit was not meters.

Second, since that first Mount Blanc/Monte Bianco render, we know the color is too bright. I lowered it to 0.6 (and later I found that that's what he actually suggests at the end of the plane section) and then compared the grey in a plain (#b5b5b5) to what GDAL outputs and compensated using a simple coefficient. The final value is 0.402.

Third, I was having issues rendering: I was getting a lot of terracing. After a long chat with Viktor_smg from they figured out that the sRGB color space in the Image Texture is broken and that I should use XYZ instead. This meant installing Blender by hand instead of relying on the one in Debian Unstable because it's too old and does not have it. They also gave me pointers about how to automate it.

Last, you can't apply this technique DEM by DEM because you want the shadows from the neighbouring tiles to spill over the current one. That link shows how to render the tile and its 8 neighbouring ones, but I think that you can optimize it in two ways: First, since shadows come from the NW, just add the tiles that lie in that general direction. Second, no shadow would cast over 10s of kilometers. You could even get away with just adding a smaller band around the tile.

That's it for now. The next step is to automate this an publish that. $DEITY knows when that will happen.

python openstreetmap gdal elevation hillshading imageio gis dem

Posted dom 23 oct 2022 00:43:19 Tags: dem

It all started with the following statement: "my renders are too slow". There were three issues as the root case: osm-carto had become more complex (nothing I can do there), my compute resources were too nimble (but upgrading them would cost money I don't want to spend in just one hobby) and the terrain rasters were being reprojected all the time.

My style uses three raster layers to provide a sense of relief: one providing height tints, one enhancing that color on slopes and desaturating it on valleys, and a last one providing hill shading. All three were derived from the same source DEM, which is projected in WGS 84 a.k.a. EPSG 4326. The style is of course in Pseudo/Web Mercator, a.k.a. EPSG 3857; hence the constant reprojection.

So my first reaction was: reproject first, then derive the raster layers. This single step made it all hell break loose.

I usually render Europe. It's the region where I live and where I most travel to. I'm using relief layers because I like mountains, so I'm constantly checking how they look. And with this change, the mountains up North (initially was around North Scandinavia) looked all washed out:

The reason behind this is that GDAL's hill and slope shading algorithms don't take in account the projection, but just use pixel values as present in the dataset and the declared pixel size. The DEMs I'm using are of a resolution of 1 arc second per pixel. WGS84 assumes a ellipsoid of 6_378_137m of radius (plus a flattening parameter which we're going to ignore), which means a circumference at the Equator of:

In [1]: import math
In [2]: radius = 6_378_137  # in meters
In [3]: circunf = radius * 2 * math.pi  # 2πr
In [4]: circunf
Out[4]: 40_075_016.68557849  # in meters

One arc second is hence this 'wide':

In [5]: circunf / 360 / 60 / 60
Out[5]: 30.922080775909325  # in meters

In fact the DEMs advertises exactly that:

mdione@diablo:~/src/projects/osm/data/height/mapzen$ gdalinfo <span class="createlink">N79E014</span>.hgt
Size is 3601, 3601
Coordinate System is:
    DATUM["World Geodetic System 1984",
        ELLIPSOID["WGS 84",6378137,298.257223563,
        AXIS["geodetic latitude (Lat)",north,
        AXIS["geodetic longitude (Lon)",east,
Pixel Size = (0.000277777777778,-0.000277777777778)

Well, not exactly that, but we can derive it 'easily'. Deep down there, the declared unit is degree (ignore the metre in the datum; that's the unit for the numbers in the ELLIPSOID) and its size in radians (just in case you aren't satisfied with the amount of conversions involved, I guess). The projection declares degrees as the unit of the projection for both axis, and the pixel size is declared in the units of the projection, and 1 arc second is:

In [6]: 1 / 60 / 60  # or 1/3600
Out[6]: 0.0002777777777777778  # in degrees

which is exactly the value declared in the pixel size[1][2]. So one pixel is one arc second and one arc second is around 30m at the Equator.

The 'flatness' problem arises because all these projections assume all latitudes are of the same width; for them, a degree at the Equator is as 'wide' as one at 60°N, which is not; in reality, it's twice as wide, because the width of a degree decreases with the cosine of the latitude and math.cos(math.radians(60)) == 0.5. The mountains we saw in the first image are close to 68N; the cosine up there is:

In [7]: math.cos(math.radians(68))
Out[7]: 0.37460659341591196

So pixels up there are no longer around 30m wide, but only about 30 * 0.37 ≃ 11 # in meters!

How does this relate to the flatness? Well, like I said, GDAL's algorithms use the declared pixel size everywhere, so at f.i. 68°N it thinks the pixel is around 30m wide when it's only around 11m wide. They're miscalculating by a factor of almost 3! Also, the Mercator projection is famous for stretching not only in the East-West direction (longitude) but also in latitude, at the same pace, based on the inverse of the cosine of the latitude:

stretch_factor = 1 / math.cos(math.radians(lat))

So all the mountains up there 'look' to the algorithms as 3 times wider and 'taller' ('tall' here means in the latitude, N-S direction, not height AMSL), hence the washing out.

In that image we have an original mountain 10 units wide and 5 units tall; but gdal thinks it's 20 units wide, so it looks flat.

Then it hit me: it doesn't matter what projection you are using, unless you calculate slopes and hill shading on the sphere (or spheroid, but we all want to keep it simple, right? Otherwise we wouldn't be using WebMerc all over :), the numbers are going to be wrong.

The Wandering Cartographer mentions this and goes around the issue by doing two things: Always using DEMs where the units are meters, and always using a local projection that doesn't deform as much. He can do that because he produces small maps. He also concedes he is using the 111_120 constant when the unit is degrees. I'm not really sure about how that number is calculated, but I suspect that it has something to do with the flattening parameter, and also some level of rounding. It seems to have originated from gdaldem's manpage (search for the -s option). Let's try something:

In [8]: m_per_degree = circunf / 360
In [9]: m_per_degree
Out[9]: 111_319.49079327358  # in meters

That's how wide is a degree is at the Equator. Close enough, I guess.

But for us world-wide, or at least continent-wide webmappers like me, those suggestions are not enough; we can't choose a projection that doesn't deform because at those scales they all forcibly do, and we can't use a single scaling factor either.

gdal's algorithm uses the 8 neighboring pixels of a pixel to determine slope and aspect. The slope is calculated on a sum of those pixels' values divided by the cell size. We can't change the cell size that gdal uses, but we can change the values :) That's the third triangle up there: the algorithm assumes a size for the base that it's X times bigger than the real one, and we just make the triangle as many times taller. X is just 1 / math.cos(math.radians(lat)).

If we apply this technique at each DEM file individually, we have to chose a latitude for it. One option is to take the latitude at the center of the file. It would make your code from more complicated, specially if you're using bash or, worse, make because you have to pass a different value for -s, but you can always use a little bit of Python or any other high level, interpreted language.

But what happens at the seams where one tile meets another? Well, since each file has it's own latitude, two neighbouring values, one in one DEM file and the other in another, will have different compensation values. So, when you jump from a ~0.39 factor to a ~0.37 at the 68° of latitude, you start to notice it, and these are not the mountains more to the North that there are.

The next step would be to start cutting DEM tiles into smaller ones, so the 'jumps' in between are less noticeable... but can we make this more continuous-like? Yes we can! We simply compensate each pixel based on its latitude.

This image cheats a little; this one is based on EU-DEM and not mapzen like the original. You can see where the DEM file ends because the slope and hillshading effects stop at the bottom left. I still have to decide which dataset I will use; at some point I decided that EU-DEM is not global enough for me, and that it's probably better to use a single dataset (f.i. mapzen) than a more accurate but partial one, but since I usually render the regions I need, I might as well change my workflow to easily change the DEM dataset.

This approach works fine as shown, but has at least one limitation. The data type used in those TIFFs is Int16. This means signed ints 16 bits wide, which means values can go between -32768..32767. We could only compensate Everest up to less than 4 times, but luckily we don't have to; it would have to be at around 75° of latitude before the compensated value was bigger than the maximum the type can represent. But at around 84° we get a compensation of 10x and it only gets quickly worse from there:

75:  3.863
76:  4.133
77:  4.445
78:  4.809
79:  5.240
80:  5.758
81:  6.392
82:  7.185
83:  8.205
84:  9.566
85: 11.474
86: 14.336
87: 19.107
88: 28.654
89: 57.299  # we don't draw this far, the cutout value is around 85.5 IIRC
90: ∞       # :)

The solution could be to use a floating point type, but that means all calculations would be more expensive, but definitely not as much as reprojecting on the fly. Or maybe we can use a bigger integer type. Both solutions would also use more disk space and require more memory. gdal currently supports band types of Byte, UInt16, Int16, UInt32, Int32, Float32, Float64, CInt16, CInt32, CFloat32 and CFloat64 for reading and writing.

Another thing is that it works fine and quick for datasets like SRTM and mapzen because I can compensate whole raster lines in one go as all its pixels have the same latitude, but for EU-DEM I have to compensate every pixel and it becomes slow. I could try to parallelize it (I bought a new computer which has 4x the CPUs I had before, and 4x the RAM too :), but I first have to figure out how to slowly write TIFFS; currently I have to hold the whole output TIFF in memory before dumping it into a file.

Here's the code for pixel per pixel processing; converting it to do row by row for f.i. mapzen makes you actually remove some code :)

#! /usr/bin/env python3

import sys
import math

import rasterio
import pyproj
import numpy

in_file =[1])
band =  # yes, we assume only one band

out_file =[2], 'w', driver='GTiff',
                         height=in_file.height, width=in_file.width,
                         count=1, dtype=in_file.dtypes[0],
               , transform=in_file.transform)
out_data = []

transformer = pyproj.Transformer.from_crs(, 'epsg:4326')

# scan every line in the input
for row in range(in_file.height):
    if (row % 10) == 0:

    line = band[row]

    # EU-DEM does not provide 'nice' 1x1 rectangular tiles,
    # they use a specific projection that become 'arcs' in anything 'rectangular'

    # so, pixel by pixel
    for col in range(in_file.width):
        # rasterio does not say where do the coords returned fall in the pixel
        # but at 30m tops, we don't care
        y, x = in_file.xy(row, col)

        # convert back to latlon
        lat, _ = transformer.transform(y, x)

        # calculate a compensation value based on lat
        # real widths are pixel_size * cos(lat), we compensate by the inverse of that
        coef = 1 / math.cos(math.radians(lat))

        line[col] *= coef


# save in a new file.
out_file.write(numpy.asarray(out_data, in_file.dtypes[0]), 1)

That's it!

python openstreetmap gdal elevation hillshading rasterio pyproj gis dem crs

[1] Except for the negative value. That has to do with the fact that pixels count from top to bottom, but degrees from South (negative) to North.

[2] The other number, 0.0174532925199433, is how much a degree is in radians.

[3] All that extra math is converting degrees into radians so cos() is happy.

Posted sáb 26 mar 2022 09:52:37 Tags: dem