I'm trying to find how to generalise the calculation of the Gaussian-weighted area to triangles for convolution purposes. Let's start with how that works when there's only one line. If there's a line going through the Euclidean plane, we can define the origin-centred Gaussian-weighted area of either side of the line by calculating the distance $d$ between the line and the origin and use the raised error function to find the area, $f(x) = \frac{1 + \operatorname{erf}(d)}{2}$ for the area of one side of the line and $f(-x)$ or $1 - f(x)$ for the other side, with $0$ meaning an empty area and $1$ for a full area, and $0.5$ if the line goes through the origin, the weight applied being $e^{-x^2+y^2}$.

[![Triangle multiplied by Gaussian weighting][1]][1]

<sup><i>Here's a way to visualise the problem, the area inside the triangle is multiplied by Gaussian weights from the origin, the Gaussian-weighted area is approximately proportional to the sum of these pixel values.</i></sup>

So for $d = 0.5$, meaning a line that comes to within 0.5 units of the origin we get a Gaussian-weighted area of $f(0.5) \approx 0.76$ for the side that the origin is on, meaning that about 76% of the Gaussian weights are on that side of the line.

The problems start when we try to find the Gaussian-weighted area for the union of the given sides of two lines. If the two lines are parallel and the sides measured overlap then the Gaussian-weighted area of the overlap is given by $f(d_1) + f(d_2) - 1$, $d_1$ and $d_2$ being the closest distance of each line to the origin, and can be positive or negative values depending on the sides chosen. If the two lines are perpendicular we instead use $f(d_1) \times f(d_2)$. But what can we do about other angles? I feel like there's something that should be possible but can't see what. With addition we would count the weighted area of the same cones twice.

I was able to figure out that for non-parallel lines the weighted area on the line that bisects the angle is given by $e^{-t^2} * [\operatorname{erf}(t \times tan\frac{\theta}{2}) \times H(t)]$ if $t$ is the axis represented by that bisecting line, $*$ being the convolution operator, $H$ being the Heaviside step function and $\theta$ being the angle formed by the two lines. In other words it's the Gaussian function convolved by the positive half of the error function scaled in $x$ by the slope of the lines wrt the bisecting line. That's as far as I was able to get and I was unable to find any literature on the subject.


  [1]: https://i.sstatic.net/tnIDx.png