What value does slope represent, how is it calculated from neighboring cells of a raster DEM?

Question

In QGIS, I have a Digital Terrain Model. Using Slope (Menu Raster > Analysis > Slope..., in fact using GDAL Slope algorithm), I get an output raster, either in percents or degrees, the same principle as described here for ArcGIS. So far, so clear.

The question I have is (I guess) not so software specific, but more a general, conceptual question: how is the slope of a cell calculated, based on a raster DEM?

I can't make any sense of the values of my slope raster. I tried to manually calculate the slope value and I'm not sure how it is calculated: I suppose that eight neighboring cells are used for calculation. So the slope is calculated separately for each of these 8 neighbours? And then some kind of mean value is calculated?

Based on the data I got, I was not able to come even close to the slope raster's values by calculating this value manually. See screenshot for example data with a point grid, labeled with the elevation and slope values (0.5*0.5 m, corresponding to resolution and CRS of elevation and slope rasters). So how can you calcaluate the slope of 41.1% the cell indicated in yellow from the 8 neighbours (point 2)? Horizontal distance is 0.5, slope is in percent:

1. Elevation current cell: 504.58
2. Elevation 8 neighbours, red values (clockwise, from upper left): 504.75, 504.64, 504.51, 504.35, 504.15, 504.33, 504.3, 504.56
3. Slope current cell, yellow value: 41.1

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Edit:

Based on the graphic linked by @FelixIP (see below), I manually calculated the angles (slope) of each cell with this forumula (pseudocode): degrees (arctan ((elevation(a)-elevation(e))/distance)) = slope-angle, where distance is 0.5 for cells b,f,h, and d (straight line from centroid to centroid) and 0.707=sqrt(0.5^2+0.5^2) for a,c,i,g (diagonal distance from centroid to centroid).

The resulting angles in degree looks like. So the question, more specifically, is: how do these values relate to the slope in degrees calculated for cell e as 41.1 degrees? None of these values seems to be close to the result:

Manually calculated slope angle from each cell to cell e:

a   13.51828463
b   6.842773413
c   -5.653572474
f   -24.70243023
i   -31.30427926
h   -26.56505118
g   -21.60256512
d   -2.290610043

Answer 1

Using ArcGIS documentation and cell naming (see equations at the bottom of help page):

and your data. This script:

a,b,c,f,i,h,g,d,e = 4.75,4.64,4.51,4.35,4.15,4.33,4.3,4.56,4.58

import math
dzBydx = ((c + 2*f + i)*4/4 - (a + 2*d + g)*4/4) / (8 * 0.5)
dzBydy = ((g + 2*h + i)*4/4 - (a + 2*b + c)*4/4) / (8 * 0.5)
rise_run = math.hypot(dzBydx, dzBydy)
print 'Planar method {:6.1f}'.format(rise_run*100)

import numpy as np
X = [-0.5,0,0.5,0.5,0.5,0,-0.5,-0.5,0.0]
Y = [0.5,0.5,0.5,0.0,-0.5,-0.5,-0.5,0.0,0.0]
mX = zip(X,Y)
dzBydx,dzBydy = np.linalg.lstsq(mX, [a,b,c,f,i,h,g,d,e])[0]
rise_run = math.hypot(dzBydx, dzBydy)
print 'Geodesic method {:6.1f}'.format(rise_run*100)

outputs:

UPDATE: GIS does NOT compute 8 slopes. It approximates terrain around cell in the middle by plane using least square technique:

So it replaces your actual 'terrain':

by plane:

Plane equation:

Z = AX + BY + C

in you case:

due to:

Your slope equals 42.4 percent, not degrees.