Problem with Two Axis Systems

[hutchay]

I have a problem involving 2 axis systems for the same area.

I have the location of 2 points in both axes and the location of a third point in only 1 axis.

With the objective being to calculate the coordinates for the third point in the other axis.

The diagram (jpg attachment) I put together explains it much more clearly.

Been trying to get starting figuring I need to work out the angle between the axes and the difference in origin location, but it's been so long since I did any of this I'm getting stumped:

 

[hutchay]

because it's not just an excerise it's a solution to a problem I have.

 

[Ishuda]

I have a problem involving 2 axis systems for the same area.

I have the location of 2 points in both axes and the location of a third point in only 1 axis.

With the objective being to calculate the coordinates for the third point in the other axis.

The diagram (jpg attachment) I put together explains it much more clearly.

Been trying to get starting figuring I need to work out the angle between the axes and the difference in origin location, but it's been so long since I did any of this I'm getting stumped:

The standard 'translate and rotate' [or rotate and translate] can be written in matrix form as
[(u-u₀) (v-v₀)] = [(x-x₀) (y-y₀)] A
where A is a two by two matrix with row vectors [cos(\(\displaystyle \theta\)) -sin(\(\displaystyle \theta\))] and [sin(\(\displaystyle \theta\)) cos(\(\displaystyle \theta\))]

From your diagram, we can use
[x₀ y₀] = [A_GX A_GY]
and
[u₀ v₀] = [A_FX A_FY]


For convenience, let
a = u - u₀,
b = v - v₀,
c = x - x₀0,
and
d = y - y₀.
From your diagram we also have that if
[x y] = [B_GX B_GY]
then
[u v] = [B_FX B_FY]
or
a = c cos(\(\displaystyle \theta\)) + d sin(\(\displaystyle \theta\))
b = d cos(\(\displaystyle \theta\)) - c sin(\(\displaystyle \theta\))
I believe this is essentially what you have already done. To finish, multiply the first of these two last equations by c and the second by d, then add to get
a c + b d = (c² + d²) cos(\(\displaystyle \theta\))
or
\(\displaystyle cos(\theta) = \frac{a c + b d}{c^2 + d^2}\)
Simularly
\(\displaystyle sin(\theta) = \frac{a d - b c}{c^2 + d^2}\)

Edit to add: Oh, and now just plug in
[x, y] = [C_GX C_GY]
and compute [u v] to obtain [C_FX C_FY]

 

[hutchay]

To finish, multiply the first of these two last equations by c and the second by d, then add to get
a c + b d = (c² + d²) cos(\(\displaystyle \theta\))

Thank you very much, I'd spent ages trying to find a way to manipulate those formula to give me theta.