To visualise the concept of the WOP, we created a synthetic image consisting of a single feature at an angle of 130° and a wavelength of 260 px (Fig. ). The image was then divided into 3x3 sub-regions of equal size, and we calculated the amplitudes and angles present in each sub-region. We obtained WOP = 1 for the central region due to the entire image being perfectly homogenous, and thus the angles being well aligned in neighbouring regions. We then rotated all nine sections randomly, leading to WOP = 0 when averaging over 100 random distributions (Fig. ). With an orthogonal orientation, we obtained WOP = -1. Since we only have a single angle, and no intensity modulation, these values are identical to the ones obtained using AFT. However, we can vary the intensity of certain squares to see the impact on the WOP value. We took Fig. and lowered the top row's intensity by 75%. Since the angles are identical, this does not affect the WOP. For the random (Fig. ) example, the mean WOP stays the same, but due to variance in intensity, the standard deviation changes slightly. For the orthogonal (Fig. ) example, the WOP stayed identical, just like in the parallel example. We took the reduced intensity version (Fig. ), and rotated the top row by 90°, making it orthogonal to the central angle (Fig. ). Having a section being orthogonal, but low intensity, still results in a high WOP: 0.75. If we then increase the intensity of the orthogonal section (Fig. and) we obtain a lower WOP, since the contribution of the orthogonal section increases, leading to a WOP of 0.74 and 0.25, respectively. The examples all contained a single angle per sub-window, but overlapping angles can be seen as additional sub-areas. The method can take any number of angles and their specific amplitudes into account to calculate a mean WOP.