I am trying to convert between two coordinate systems and think I've come up with the answer but would like to make sure my assumptions are correct and to help with some of the math.

The problem is that I have two coordinate systems both Cartesian in nature (one technically isn't but at the scale I'm doing this at the difference is unimportant) and I want to convert a point from one to another. The two coordinate systems will almost certainly not sure share an origin or rotation.

Assumption 1: 3 points with coordinates in both system is the minimum number of points needed to solve this problem when the last point is not collinear too any of the other two points.

Assumption 2: The problem can be solved with a distance conversion using one of the sides of the triangle.

Assumption 3: The problem can be solved by for each pair of points coming up with 4 possible coordinates that the last point can be by using the angle between the line segment and the line segment made by combining one of the endpoints and the last point and the distances. Then finding the only point that is common to all three sets of points.

Have I made any mistakes in my assumptions.

The distance conversion should be simple at distance-2= distance-1 * ( side-1-2/side-1-1) Where distance-2 is the distance to the point I am trying to convert in the coordinate system I am converting to distance-1 being the distance in the system I am converting from(being a simple a^2 = b^2 + c^2) side-1-1 and side-1-2 are an arbitrary side of the triangle in both coordinate systems.

I am not sure what the next step looks like but I assume the final step would include a either a cos or sin of +/- angle.

Thank you for your time.

(Alternatively could the three points be used to make a formula for coordinate conversion)

(Does this become complicated if the x and y coordinates in the second system are not of the same length)


1 Answer 1


After doing so more intense googling I found my answer by the addition of the phrase affine transformation.

The answer mathematically can be found here

Accordingly my first assumption was incorrect, it can be done with only two, therefore all my other assumptions are incorrect.


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