Equation of circle after affine transformation

Question

Let \begin{align*} C = \{(x,y,z) \in \mathbb{R}^3 | x^2+ y^2=1, z=0\} \end{align*} be a unit circle in 3D space and $f$ be a given affine transform $$f(p) = A p^t + b, \ p\in \mathbb{R}^3$$ which can transform $C$ into any ellipse. Moreover we are given a plane by its unit normal vector $n$ then $$P(p) = p - (p n^t) n, \ p\in \mathbb{R}^3 $$ is the orthogonal projection of $p$ onto the plane.

What are the coefficients of the polynomial equation that we get when $C$ is first transformed by $f$ then projected by $P$?

Background: This problem arises in astronomy when we observe an orbit through a camera. Here, $f$ gives the orbit in a coordinate system and the plane is the camera plane.

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Update: As turned out the normal $n$ alone does not provide enough information to deduce the coefficients. If we instead assume we pitch $\theta$ or yaw $\psi$ (but not roll) the ellipse and project the result into the $x,y$ plane we get $P' : \mathbb{R}^3 \longrightarrow \mathbb{R}^2$ $$P' = \left [ \begin{matrix} \cos\theta \cos\psi & - \sin\psi & \sin\theta \cos\psi \\ \cos\theta \sin\psi & \cos\psi & \sin\theta \sin\psi \end{matrix} \right ].$$

Answer 1

The original circle is the intersection of the unit-radius infinite cylinder

$$ x^2 + y^2 = 1 $$

and the $xy$ plane $z = 0$.

If $p = (x, y, z) = \begin{bmatrix} x \\ y \\ z \end{bmatrix} $

And $ Q = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 0 \end{bmatrix}$

Then equivalent to the above, the unit circle in the $xy$ plane is the intersection of the cylinder

$ p^T Q p = 1 $

And the plane

$ e_3^T p = 0 $

where $e_3 = (0,0,1) = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

Now you define an affine transformation on $p$ which produces an image $p'$ corresponding to $p$ as follows

$ p' = f(p) = A p + b $

Further, we project the image $p'$ onto a plane passing through the origin, and having a unit normal vector $n$. Let the projected $p'$ be $p''$ then

$p'' = [I_3 - {nn}^T] p' $

Define matrix $J = I_3 - {nn}^T$ ,then

$p'' = J p' $

Using the expression for $p'$ we get

$ p'' = J ( A p + b ) = (J \ A) \ p + (J \ b) = F p + g$

with $F = J \ A $ and $g = J \ b $.

It is well-known that matrix $J$ is rank-deficient, actually its rank is $2$. So, assuming matrix $A$ is full rank, then $J \ A $ will also be rank deficient, and have a rank of $2$. And because of that, we cannot invert matrix $F$, which we need to solve for $p$ in terms of $p''$.

However we can use vector parametric methods to find what we're looking for.

The original unit circle in the $xy$ plane can be written parametrically as

$ p(\theta) = u_1 \cos(\theta) + u_2 \sin(\theta) $

where $u_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} $ and $u_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} $

Therefore,

$ p'(\theta) = A ( u_1 \cos(\theta) + u_2 \sin(\theta) ) + b $

And

$ p''(\theta) = F p + g = F (u_1 \cos(\theta) + u_2 \sin(\theta) ) + g $

Let's define $w_1 = F \ u_1 $ , $w_2 = F \ u_2 $

$ p''(\theta) = w_1 \cos(\theta) + w_2 \sin(\theta) + g $

Note that we know that $p''$ lies in the plane $ n \cdot (x, y, z) = 0 $

We can attach a reference frame to this plane, with its $x'y'$ plane coinciding with the plane itself, and its $z'$ axis along the normal $n$. If $q = (x', y', 0)$ is the coordinate of point on the projection plane with respect to the arbitrarily selected orientation of the $x'$ and $y'$ axes, then

$ p'' = R q $

Note that the third coordinate of $q$ is zero. The $3 \times 3$ $R$ is a rotation matrix and is given by

$ R = [ \hat{x'}, \hat{y'}, \hat{n} ] $

where $ \hat{x'} $ is the unit vector along the positive $x'$ axis expressed in world coordinates, and $\hat{y'}$ is the unit vector along the positive $y'$ axis expressed in world coordinates.

From this, it follows that

$ q(\theta) = R^{-1} p''(\theta) = R^{T} ( w_1 \cos(\theta) + w_2 \sin(\theta) + g ) $

Define $ y_1 = R^{T} w_1$ and $y_2 = R^T w_2 $ and $q_0 = R^T g $ , then

$ q(\theta) - q_0 = \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} y_1 && y_2 \end{bmatrix} \begin{bmatrix} \cos(\theta) \\ \sin(\theta) \end{bmatrix} $

Define

$Y = \begin{bmatrix} y_1 && y_2 \end{bmatrix} $ and $ u = \begin{bmatrix} \cos(\theta) \\ \sin(\theta) \end{bmatrix} $

then

$ q - q_0 = Y u \hspace{50pt}(1)$

Note that $q$ and $q_0$ have their third component identically zero.

Also, note that $Y$ is a $3 \times 2$ matrix, and $u$ is a $2 \times 1$ unit vector.

Pre-multiply both sides by $Y^T$, you get

$ Y^T q = (Y^T Y) u \hspace{50pt}(2)$

The matrix $(Y^T Y)$ is $2 \times 2$. Solving for $u$ we get

$ u = (Y^T Y)^{-1} Y^T (q - q_0) $

We know that $u$ is a unit vector, i.e. $u^T u = 1$ , therefore,

$ (q - q_0)^T \ Y (Y^T Y)^{-T} (Y^T Y)^{-1} Y^T \ (q - q_0) = 1 \hspace{50pt}{(3)} $

Recall that $q = \begin{bmatrix} x' \\ y' \\ 0 \end{bmatrix} $

Equation $(3)$ will give you the coefficients of $x'^2, y'^2, x'y', x', y'$ plus the constant term.

We can slightly modify $(3)$, by noting that matrix $Y$ can be written as

$ Y = R^T W $

where $W = \begin{bmatrix} w_1 && w_2 \end{bmatrix} $

Then $ Y^T Y = W^T R R^T W = W^T W $

Therefore, equation $(3)$ becomes

$ (q - q_0)^T \ R^T W (W^T W)^{-T} (W^T W)^{-1} W^T R \ (q - q_0) = 1 \hspace{30pt} (4) $

And recall that $q_0 = R^T g $.

A substantially different approach, starts with the equation

$ p''(\theta) = w_1 \cos(\theta) + w_2 \sin(\theta) + g \hspace{50pt}(5) $

which I copied from above.

I can now replace $ \theta $ with $(t - t_0)$ which $t$ is arbitrary and $t_0$ is a fixed value that is to be determined. With this, equation $(5)$ becomes

$ p''(t) = w_1 \cos( t - t_0) + w_2 \sin(t - t_0) + g \hspace{50pt}(6)$

Expand the trigonometric terms, to get

$ p''(t) = \cos(t) [ w_1 \cos(t_0) - w_2 \sin(t_0) ] + \sin(t) [ w_1 \sin(t_0) + w_2 \cos(t_0) ] + g \hspace{50pt} (7) $

By properly choosing $t_0$ we can make the vector $[ w_1 \cos(t_0) - w_2 \sin(t_0) ]$ orthogonal to the vector $[ w_1 \sin(t_0) + w_2 \cos(t_0) ]$, this requires that

$[ w_1 \cos(t_0) - w_2 \sin(t_0) ]\cdot [ w_1 \sin(t_0) + w_2 \cos(t_0) ] = 0 \hspace{50pt}(8)$

Expanding, gives us

$ (w_1 \cdot w_1 - w_2 \cdot w_2) ( \sin(t_0) \cos(t_0) ) + (w_1 \cdot w_2 ) ( \cos^2(t_0) - \sin^2(t_0) ) = 0 \hspace{50pt}(9) $

Using trigonometric identities, this in turn implies,

$ \dfrac{1}{2} \sin(2 t_0) (w_1 \cdot w_1 - w_2 \cdot w_2) + \cos(2 t_0) (w_1 \cdot w_2 ) = 0 \hspace{50pt}(10)$

And this finally gives us

$ t_0 = \dfrac{1}{2} \tan^{-1} \dfrac{ 2 (w_1 \cdot w_2 ) }{ w_2 \cdot w_2 - w_1 \cdot w_1 ) } \hspace{50pt}(11)$

This means that at this calculated $t_0$ the vectors

$ v_1 = w_1 \cos(t_0) - w_2 \sin(t_0) $

and

$ v_2 = w_1 \sin(t_0) + w_2 \cos(t_0) $

represent the major and minor semi-axes of the ellipse that is a result of transformation and projection.

Since $v_1$ is orthogonal to $v_2$ we can take these two directions to be the $x'$ and $y'$ axes direction.

So the equation of the ellipse now is

$ p''(t) = g + \cos(t) v_1 + \sin(t) v_2 \hspace{50pt}(12) $

where $v_1 \perp v_2$, therefore, taking the $x'$ coordinate to be along the $v_1$ direction and $y'$ to be along the $v_2$ direction, and further, if we let the origin of the coordinate system be at $g$ (which is known), then the quadratic polynomial describing the ellipse is

$ \dfrac{x'^2}{\| v_1 \|^2 } + \dfrac{ y'^2}{\| v_2 \|^2} = 1 \hspace{50pt}(13)$

To see this, look at equation $(12)$, we have

$ p''(t) - g = \cos(t) v_1 + \sin(t) v_2 \hspace{50pt}(12) $

Define the matrix

$ R = \begin{bmatrix} \dfrac{v_1}{\| v_1 \|} && \dfrac{v_2}{\| v_2 \| } && n \end{bmatrix} $

Now define the local coordinates vector $r = R^T (p'' - g) $

Then

$ r = \begin{bmatrix} x' \\ y' \\ 0 \end{bmatrix} = \begin{bmatrix} \| v_1 \| && 0 \\ 0 && \| v_2 \| \\ 0 && 0 \end{bmatrix} \begin{bmatrix} \cos(t) \\ \sin(t) \end{bmatrix} \hspace{50pt} (14) $

Disregarding the third row that is zero on both sides of the equation, and inverting, we get

$ \begin{bmatrix} \cos(t) \\ \sin(t) \end{bmatrix} = \begin{bmatrix} \dfrac{1}{\| v_1 \|} && 0 \\ 0 && \dfrac{1}{\| v_2 \|} \end{bmatrix} \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} \dfrac{x'}{\| v_1 \|} \\ \dfrac{y'}{\| v_2 \|} \end{bmatrix} $

Since $\cos^2(t) + \sin^2(t) = 1 $ , then

$ \dfrac{x'^2}{\| v_1 \|^2 } + \dfrac{ y'^2}{\| v_2 \|^2} = 1$

as expressed in equation $(13)$.

Answer 2

I will follow the usual convention in linear algebra to express a vector in $\mathbb R^3$ as a column vector. Hence $C$ is the set of all vectors $p$ in the form of $(\cos\theta,\sin\theta,0)^T$, $f(p)=Ap+b$ and $P(p)=(I-nn^T)p$. It follows that $$ q=P(f(p))=(I-nn^T)Ap+(I-nn^T)b=(I-nn^T)A\pmatrix{I_2\\ 0}\pmatrix{\cos\theta\\ \sin\theta}+(I-nn^T)b. $$ Let $M=(I-nn^T)A\pmatrix{I_2\\ 0}\in\mathbb R^{3\times2}$ and $c=(I-nn^T)b\in\mathbb R^3$. Then $q=M\pmatrix{\cos\theta\\ \sin\theta}+c$. The rank of $M$ is at most two, but $M$ can be rank-one even if $A$ is invertible; if $A$ is singular, $M$ may even be zero.

• If $M$ is rank-two, then $\pmatrix{\cos\theta\\ \sin\theta}=M^+(q-c)=(M^TM)^{-1}M^T(q-c)$. Hence the locus of $q$ can be expressed as a single polynomial equation $\left\|(M^TM)^{-1}M^T(q-c)\right\|^{\,2}=1$. It is an ellipse.
• If $M$ is rank-one, we may write it as $uv^T$ where $u$ is a nonzero vector in $\mathbb R^3$ and $v$ is a unit vector in $\mathbb R^2$. Hence the locus of $q=uv^T\pmatrix{\cos\theta\\ \sin\theta}+c$ is the line segment joining $c-u$ and $c+u$.
• If $M=0$, then the locus of $q$ is the single point $c$. If you insist on writing it as a single polynomial equation, you may write $(q_1-c_1)^2+(q_2-c_2)^2+(q_3-c_3)^2=0$.