Relating a point to the shifted axes.

Question

In the figure below, point O is the origin with lines marked with yellow, point O' is the shifted origin with lines marked with white color. The point (P,Q) is at a distance PO from the v axis and QO form the normal (green line) to the v axis. With regards to O', point (P,Q) is at a distance PO' from v' and QO' from u'. Point O is shifted to O' by x and y distance (along the horizontal and vertical of the picture). The angle between u' and the line normal to v is $\theta$.

How to relate the point (P,Q) with O, O', angle $\theta$, and the shifts x and y?

Answer 1

Until I have time to write a more detailed answer, you can still look through this page on wikipedia: rotation of axes.

Edit

I'm going to make a pretty important assumption in the following, which is that everything has an algebraic value and not a geometric one. To put that into perspective, and in case I made a bad translation from my mother tongue, algebraic distances may have negative values whereas geometric distances are always non-negative. In a nutshell, geometric = $\lvert$algebraic$\rvert$. As is, in your figure, $y<0$ (according to my notations). So you'll have to adapt depending on what you actually have.

Notations
There are three coordinates systems naturally attached to your problem:

• $uO\bar u$, where $O\bar u$ denotes the axis orthogonal to $Ou$ (and $uO\bar u$ is a right-handed). This is the natural system to express the shift from $O$ to $O'$ (the values of $x$ and $y$).
• $vO\bar v$, same as above but with axis $Ov$. In this system, the point you called "$(P,Q)$" has abscissae $QO$ and ordinate $PO$.
• $v'O'\bar{v'}$, same logic as above but the origin is $O'$. In a sense "$\bar{v'}=-u'$" (if this even makes sense). I'll still favour the notation $\bar{v'}$ because right-handed coordinates system are more natural for me ($v'O'u'$ is left-handed if I interpret the figure properly). In this system, point $(P,Q)$ has abscissae $QO'$ and ordinate $PO'$.

Denote by $\varphi$ the oriented angle between the axes $Ou$ and $Ov$. Likewise $\theta$ is the oriented angle between axes $Ov$ and $Ov'$. We're going to express everything in the coordinates system $vO\bar v$.

Working the relations between the various coordinates
First up is the vector $OO'$ which has coordinates $(x,y)$ in $uO\bar u$. this same vector has coordinates $(X,Y)$ in $vO\bar v$ where \begin{align*} X&=x\cos\varphi +y\sin\varphi\\ Y&=-x\sin\varphi +y\cos\varphi \end{align*}

Next, consider the auxiliary coordinate system $v''O'\bar{v''}$ with origin $O'$ and exactly the same orientation as $vO\bar{v}$; in other words the oriented angle between axes $Ov''$ and $Ov$ is $-\theta$. To go from system $v'O'\bar{v'}$ to system $vO\bar v$, you can first go from $v'O'\bar{v'}$ to $v''O'\bar{v''}$ with a simple rotation of angle $-\theta$, then go from $v''O'\bar{v''}$ to $vO\bar v$ with a translation of vector $-OO'$. Assume you have some point with coordinates $(a,b)$ in $vO\bar v$, $(a',b')$ in $v'O'\bar{v'}$, and $(a'',b'')$ in $v''O'\bar{v''}$. Then \begin{align} a'' &=& a'\cos(-\theta) + b'\sin(-\theta) &=& a'\cos\theta -b'\sin\theta \\ b'' &=& -a'\sin(-\theta) + b'\cos(-\theta) &=& a'\sin\theta +b'\cos\theta \\ a'' &=& a-X\\ b'' &=& b-Y \end{align}

Putting everything together \begin{align*} a-x\cos\varphi-y\sin\varphi &= a'\cos\theta -b'\sin\theta \\ b+x\sin\varphi-y\cos\varphi &= a'\sin\theta +b'\cos\theta \end{align*} In particular for your point "$(P,Q)$" \begin{align*} QO-x\cos\varphi-y\sin\varphi &= QO'\cos\theta -PO'\sin\theta &(1)\\ PO+x\sin\varphi-y\cos\varphi &= QO'\sin\theta +PO'\cos\theta &(2) \end{align*}

Figuring out $x$, $y$ and $\theta$?
From your answer to my comment it seems that you want some way to compute/figure out $x$, $y$ and $\theta$ given the measurements of $PO$, $PO'$, $QO$, $QO'$ and maybe $\varphi$. This is impossible as is. Although the equations are not linear with respect to $\theta$, you only have $2$ equations for $3$ unknowns, and in this particular case you have an infinity of possible values for $x$, $y$ and $\theta$. You need the measurements for another point to get unique values for those parameters. There is however a "nice" relationship between $x$, $y$ and $\theta$.

Notice that $O'$ lies on the circle centered at $(P,Q)$ with radius the distance between $O'$ and $(P,Q)$ (this can be proved from equations $(1)$ and $(2)$): $$ (x-x_0)^2 + (y-y_0)^2 = (d')^2 $$ where $(d')^2=(PO')^2+(QO')^2$, and $x_0=QO\cos\varphi -PO\sin\varphi$, $y_0=QO\sin\varphi +PO\cos\varphi$ are the coordinates of $(P,Q)$ in $uO\bar u$. [Note that $y_0$ should be null.] The possible values of $(x,y)$ are such that there exists some real number $t$, $x=x_0+d'\cos t$ and $y=y_0+d'\sin t$.

Let $\psi$ such that $\cos\psi=\frac{QO'}{d'}$ and $\sin\psi=\frac{PO'}{d'}$. If you re-inject the values $x(t)$ and $y(t)$ into equations $(1)$ and $(2)$ you eventually find (after an annoying development) that $\theta+\varphi+\psi = t+\pi$ (if you use radians), or $\theta+\varphi+\psi = t+180$ (with °).

So to sum this up, the possible triplets are $ \left\{ \Big( x(t),\ y(t),\ \theta(t) \Big) \mid t\in\mathbb R \right\} $ where \begin{align*} x(t) &= x_0 +d'\cos t \\ y(t) &= y_0 +d'\sin t \\ \theta(t) &= t+180-\varphi-\psi \end{align*}

Edit 2

Corrected abscissae and ordinate in the above Also note that you have $\cos\varphi = -\frac{QO}{d}$ and $\sin\varphi = \frac{PO}{d}$, where $d^2=QO^2+PO^2$. So you technically don't have to measure $\varphi$ if you already have $PO$ and $QO$. This also implies that $y_0=0$ and $x_0=-d$.

Also noticed I messed up part of my development for $\theta(t)$ so here are a little more details. When you check out $QO\times(1)+PO\times(2)$ you end up with $$ d'd\cos t = -d'd\cos(\varphi+\psi+\theta) $$ whereas $PO\times(1)-QO\times(2)$ gives $$ -d'd\sin t = d'd\sin(\varphi+\psi+\theta) $$