Determine cornertype based on xy values

[Solvax]

Hello everyone,

I am trying to use excel for a geometrical automated calculation.
One of the steps that is needed in my calculation is to determine the type of corner. (Inner corner or outer corner)
The X and Y vaues are stored in an chronological list. Each datapoint got a separate column and X and Y values are split up in 2 rows.

I have a dilemma here, and I don't know if excel can handle this in an easy way:
innerouter.png

Let's assume you have three datapoints oriënted in the following way for both an inner (redand an outer corner:
x1 < x2 < x3
y1 > y2 > y3
(see above image)

In that case the location of the points by themselves are not enough to determine the type of corner.
So I assume I need to draw an imaginary line between point 1 and 3 (white line), in other words: is there a formula to make a function outof data point 1 + 3 and tell if datapoint 2 is left or right from this function?

Or maybe you have another solution?

Thank you very much for looking into this!

[Solvax]

I cannot edit my post without rewriting everything..

So therefore just a small correction by reply:
The sentence under the picture should be:
Let's assume you have three datapoints oriënted in the following way for both an inner (red) and an outer corner (blue):

[MrShorty]

I don't know how much of this is "easy".

My first thought, looking at your drawing, is that for an inner corner, the slope [(y2-y1)/(x2-x1)] for the first segment is closer to 0 than the second segment [(y3-y2)/(x3-x2)]. Opposite scenario for an outer corner. If that is true, then a simple sequence of calculations (slope of segment 1, slope of segment 2, compare slopes to see which is closer to 0 and return outer/inner) should be all that is needed.

The approach you are proposing seems like it should be easy enough to program in. Use linear interpolation [(y3-y1)/(x3-x1)=(y2-y1)/(x4-x1)] to solve for x4 (the x coordinate of the point along the white line that is directly across from y2). If x4 is less than x2, then you have an outer corner. If x4 is greater than x2, then you have an inner corner.

I expect that some trig could be brought in to determine the result (is the angle at point 2 a negative/clockwise angle or positive/counter-clockwise).

What difficulties do you run into implementing any of those?

[Solvax]

Thank you very much for your help MrShorty
I managed to solve it also thanks to your approach.

I make a linear function outof points 1 and 2 and also 1 and 3 and by the difference in slope I can then determine where point 2 is located in relation to 1 and 3.