平面射影变换计算直线、二次曲线的像

hbghlyj

地理信息系统（GIS）Projective transformation - GIS Wiki The GIS Encyclopedia
射影变换$T$将点$(x,y)$映射到点$(T_x,T_y)$,
\begin{align}T_x&= {\alpha x + \beta y + \gamma \over \delta x + \epsilon y + \zeta}\tag{14}\label1\\
T_y&= {\eta x + \theta y + \kappa \over \delta x + \epsilon y + \zeta}\tag{15}\label2\end{align}

• $T$将直线$y = m x + b$映射到直线$T_y = n T_x + c$, 其中$n,c$为常数
  \[\led
  n&= {m (\epsilon \kappa - \zeta \theta) + b (\delta \theta - \epsilon \eta) + (\delta \kappa - \zeta \eta) \over m (\epsilon \gamma - \zeta \beta) + b (\delta \beta - \epsilon \alpha) + (\delta \gamma - \zeta \alpha)}
  \\c&= {m (\beta \kappa - \gamma \theta) + b (\alpha \theta - \beta \eta) + (\alpha \kappa - \gamma \eta) \over m (\beta \zeta - \gamma \epsilon) + b (\alpha \epsilon - \beta \delta) + (\alpha \zeta - \gamma \delta) }\endled\]比较两式发现$n,c$的分母相同,只差$\times(-1)$.
  通过交换以下系数可交换$c$与$n$
  \[\alpha \leftrightarrow \delta, \ \beta \leftrightarrow \epsilon, \ \gamma \leftrightarrow \zeta . \]
  通过交换以下系数可交换$n$的分子与分母
  \[\alpha \leftrightarrow \eta, \ \beta \leftrightarrow \theta, \ \gamma \leftrightarrow \kappa . \]
  通过交换以下系数可交换$c$的分子与分母
  \[\eta \leftrightarrow \delta, \ \theta \leftrightarrow \epsilon, \ \kappa \leftrightarrow \zeta \]
  所有这些交换对称性相当于交换系数矩阵$     M_T = \begin{bmatrix} \alpha & \beta & \gamma \\ \eta & \theta & \kappa \\ \delta & \epsilon & \zeta \end{bmatrix}$的行。
• $T$将二次曲线$$\tag{17}\label3A x^2 + B y^2 + C x + D y + E x y + F = 0$$映射到二次曲线$$A' T_x^2 + B' T_y^2 + C' T_x + D' T_y + E' T_x T_y + F' = 0\label4\tag{18}$$系数$A',\dots,F'$的表达式见3#.

hbghlyj

将$y = m x + b$代入\eqref{1}\eqref{2}
\begin{aligned}T_x &= {\alpha x + \beta (m x + b) + \gamma \over \delta x + \epsilon (m x + b) + \zeta} = {(\alpha + \beta m) x + (\beta b + \gamma) \over (\delta + \epsilon m) x + (\epsilon b + \zeta)}, \\
T_y &= {(\eta + \theta m) x + (\theta b + \kappa) \over (\delta + \epsilon m) x + (\epsilon b + \zeta) }. \end{aligned}If $Ty = n Tx + c$ and $n$ and $c$ are constants, then
\[{\partial T_y \over \partial x} = n {\partial T_x \over \partial x} \]
so that
\[ n = {\partial T_y / \partial x \over \partial T_x / \partial y} \]
Calculation shows that
\[{\partial T_x \over \partial x} = { (\epsilon b + \zeta) (\alpha + \beta m) - (\beta b + \gamma) (\delta + \epsilon m) \over [(\delta + \epsilon m) x + (\epsilon b + \zeta)]^2 } \]
and
\[{\partial T_y \over \partial x} = { (\epsilon b + \zeta) (\eta + \theta m) - (\theta b + \kappa) (\delta + \epsilon m) \over [(\delta + \epsilon m) x + (\epsilon b + \zeta)]^2 } \]
therefore
\[n = {\partial T_y / \partial x \over \partial T_x / \partial y} = { (\epsilon b + \zeta) (\eta + \theta m) - (\theta b + \kappa) (\delta + \epsilon m) \over (\epsilon b + \zeta) (\alpha + \beta m) - (\beta b + \gamma) (\delta + \epsilon m) } \]
We should now obtain $c$ to be
\begin{align*}c &= T_y - n T_x\\
           & = {(\eta + \theta m) x + (\theta b + \kappa) - \left[ { (\epsilon b + \zeta) (\eta + \theta m) - (\theta b + \kappa) (\delta + \epsilon m) \over (\epsilon b + \zeta) (\alpha + \beta m) - (\beta b + \gamma) (\delta + \epsilon m) } \right] \cdot [ (\alpha + \beta m) x + (\beta b + \gamma) ] \over (\delta + \epsilon m) x + (\epsilon b + \zeta) }
\end{align*}
Add the two fractions in the numerator:
\[    c = { \left\{ [(\epsilon b + \zeta) (\alpha + \beta m) - (\beta b + \gamma) (\delta + \epsilon m)] [(\eta + \theta m) x + (\theta b + \kappa)] - [(\epsilon b + \zeta) (\eta + \theta m) - (\theta b + \kappa) (\delta + \epsilon m)] [(\alpha + \beta m) x + (\beta b + \gamma)] \right\} \over [(\delta + \epsilon m) x + (\epsilon b + \zeta)] [(\epsilon b + \zeta) (\alpha + \beta m) - (\beta b + \gamma) (\delta + \epsilon m)] }\]
Distribute binomials in parentheses in the numerator, then cancel out equal and opposite terms:
\[    c = \frac{ - (\beta b + \gamma) (\delta + \epsilon m) (\eta + \theta m) x + (\epsilon b + \zeta) (\alpha + \beta m) (\theta b + \kappa) + (\theta b + \kappa) (\delta + \epsilon m) (\alpha + \beta m) x - (\epsilon b + \zeta) (\eta + \theta m) (\beta b + \gamma) }{ [(\delta + \epsilon m) x + (\epsilon b + \zeta)] [(\epsilon b + \zeta) (\alpha + \beta m) - (\beta b + \gamma) (\delta + \epsilon m)]} \]
Factor the numerator into a pair of terms, only one of them having the numerus cossicus ($x$). There is another numerus cossicus in the denominator. The objective now is to get both of these to cancel out.
\[    c = { \left\{ [(\theta b + \kappa) (\alpha + \beta m) - (\beta b + \gamma) (\eta + \theta m)] (\delta + \epsilon m) x + [(\alpha + \beta m)(\theta b + \kappa) - (\eta + \theta m) (\beta b + \gamma)] (\epsilon b + \zeta) \right\} \over [(\delta + \epsilon m) x + (\epsilon b + \zeta)] [(\epsilon b + \zeta) (\alpha + \beta m) - (\beta b + \gamma) (\delta + \epsilon m)] }\]
Factor the numerator,
\[    c = {[(\theta b + \kappa) (\alpha + \beta m) - (\beta b + \gamma) (\eta + \theta m)] [(\delta + \epsilon m) x + (\epsilon b + \zeta)] \over [(\epsilon b + \zeta) (\alpha + \beta m) - (\beta b + \gamma) (\delta + \epsilon m)] [(\delta + \epsilon m) x + (\epsilon b + \zeta)] }\]
The terms with the numeri cossici cancel out, therefore
\[    c = { (\alpha + \beta m) (\theta b + \kappa) - (\beta b + \gamma) (\eta + \theta m) \over (\alpha + \beta m) (\epsilon b + \zeta) - (\beta b + \gamma) (\delta + \epsilon m) } \]
is a constant. Q.E.D.

hbghlyj

$T$ 将二次曲线 \eqref{3} 上的每个点 $(x,y)$ 映射到 \eqref{1}\eqref{2} 给出的 $(T_x,T_y)$.
逆映射 $T'$ 将 $(T_x,T_y)$ 映射到 $(x,y)$. 它的系数矩阵为$M_T$的逆矩阵
$$ M_{T'} = \begin{bmatrix} \alpha' & \beta' & \gamma' \\ \eta' & \theta' & \kappa' \\ \delta' & \epsilon' & \zeta' \end{bmatrix}. $$
在 \eqref{3} 中以 $(T_x,T_y)$ 代 $(x,y)$
\begin{multline*}A \left( {\alpha' T_x + \beta' T_y + \gamma' \over \delta' T_x + \epsilon' T_y + \zeta'} \right)^2 + B \left( {\eta' T_x + \theta' T_y + \kappa' \over \delta' T_x + \epsilon' T_y + \zeta'} \right)^2\\+ C \left( {\alpha' T_x + \beta' T_y + \gamma' \over \delta' T_x + \epsilon' T_y + \zeta'} \right) + D \left( {\eta' T_x + \theta' T_y + \kappa' \over \delta' T_x + \epsilon' T_y + \zeta'} \right)\\
+ E \left( {\alpha' T_x + \beta' T_y + \gamma' \over \delta' T_x + \epsilon' T_y + \zeta'} \right) \left( {\eta' T_x + \theta' T_y + \kappa' \over \delta' T_x + \epsilon' T_y + \zeta'} \right) + F = 0\end{multline*}乘以分母$(\delta' T_x + \epsilon' T_y + \zeta')^2$
\begin{multline*}A (\alpha' T_x + \beta' T_y + \gamma')^2 + B (\eta' T_x + \theta' T_y + \kappa')^2 \\+ C (\alpha' T_x + \beta' T_y + \gamma') (\delta' T_x + \epsilon' T_y + \zeta') + D (\eta' T_x + \theta' T_y + \kappa') (\delta' T_x + \epsilon' T_y + \zeta')\\ + E (\alpha' T_x + \beta' T_y + \gamma') (\eta' T_x + \theta' T_y + \kappa') + F (\delta' T_x + \epsilon' T_y + \zeta')^2 = 0\, \end{multline*}整理为$T_x$和$T_y$的多项式
$$ \begin{matrix} (A \alpha'^2 + B \eta'^2 + C \alpha' \delta' + D \eta' \delta' + E \alpha' \eta' + F \delta'^2) T_x^2 \\ + (A \beta'^2 + B \theta'^2 + C \beta' \epsilon' + D \theta' \epsilon' + E \beta' \theta' + F \epsilon'^2) T_y^2 \\ + (2 A \alpha' \gamma' + 2 B \eta' \kappa' + C (\alpha' \zeta' + \gamma' \delta') + D (\eta' \zeta' + \kappa' \delta') + E (\alpha' \kappa' + \gamma' \eta') + 2 F \delta' \zeta') T_x \\ + (2 A \beta' \gamma' + 2 B \theta' \kappa' + C (\beta' \zeta' + \gamma' \epsilon') + D (\theta' \zeta' + \kappa' \epsilon') + E (\beta' \kappa' + \gamma' \theta') + 2 F \epsilon' \zeta') T_y \\ + (2 A \alpha' \beta' + 2 B \eta' \theta' + C (\alpha' \epsilon' + \beta' \delta') + D (\eta' \epsilon' + \theta' \delta') + E (\alpha' \theta' + \beta' \eta') + 2 F \delta' \epsilon') T_x T_y \\ + (A \gamma'^2 + B \kappa'^2 + C \gamma' \zeta' + D \kappa' \zeta' + E \gamma' \kappa' + F \zeta'^2) = 0. \end{matrix} \tag{19}\label{5}$$
\eqref{5} 与 \eqref{4} 具有相同的形式.
现在只需将 $M_T$ 的逆矩阵算出来, 在 \eqref{5} 中以 $\alpha,\dots,\zeta$ 代 $\alpha',\dots,\zeta'$
为计算 $M_T$ 的逆矩阵, 对 $M_T$ 用 Cramer's rule
$$ M_{T'} = {1 \over \Delta} \begin{bmatrix} \left| \begin{matrix} \theta &\kappa \\ \epsilon & \zeta \end{matrix} \right| & \left| \begin{matrix} \epsilon & \zeta \\ \beta & \gamma \end{matrix} \right| & \left| \begin{matrix} \beta & \gamma \\ \theta & \kappa \end{matrix} \right| \\ \quad & \quad & \quad \\ \left| \begin{matrix} \kappa & \eta \\ \zeta & \delta \end{matrix} \right| & \left| \begin{matrix} \zeta & \delta \\ \gamma & \alpha \end{matrix} \right| & \left| \begin{matrix} \gamma & \alpha \\ \kappa & \eta \end{matrix} \right| \\ \quad & \quad & \quad \\ \left| \begin{matrix} \eta & \theta \\ \delta & \epsilon \end{matrix} \right| & \left| \begin{matrix} \delta &\epsilon \\ \alpha & \beta \end{matrix} \right| & \left| \begin{matrix} \alpha & \beta \\ \eta & \theta \end{matrix} \right| \end{bmatrix} \tag{20}$$
where $Δ$ is the determinant of the unprimed coefficient matrix.

Equation (20) allows primed coefficients to be expressed in terms of unprimed coefficients. But performing these substitutions on the primed coefficients of equation (19) it can be noticed that the determinant Δ cancels itself out, so that it can be ignored altogether. Therefore

    $ A' = A (\theta \zeta - \kappa \epsilon)^2 + B (\kappa \delta - \eta \zeta)^2 + C (\theta \zeta - \kappa \epsilon) (\eta \epsilon - \theta \delta) + D (\kappa \delta - \eta \zeta) (\eta \epsilon - \theta \delta) + E (\theta \zeta - \kappa \epsilon) (\kappa \delta - \eta \zeta) + F (\eta \epsilon - \theta \delta)^2 \,$

    $ B' = A (\epsilon \gamma - \zeta \beta)^2 + B (\zeta \alpha - \delta \gamma)^2 + C (\epsilon \gamma - \zeta \beta) (\delta \beta - \epsilon \alpha) + D (\zeta \alpha - \delta \gamma) (\delta \beta - \epsilon \alpha) + E (\epsilon \gamma - \zeta \beta) (\zeta \alpha - \delta \gamma) + F (\delta \beta - \epsilon \alpha)^2 \,$

    $ C' = 2 A (\theta \zeta - \kappa \epsilon) (\beta \kappa - \gamma \theta) + 2 B (\kappa \delta - \eta \zeta) (\gamma \eta - \alpha \kappa) + C [ (\theta \zeta - \kappa \epsilon) (\alpha \theta - \beta \eta) + (\beta \kappa - \gamma \theta) (\eta \epsilon - \theta \delta)]\,$

        $+ D [ (\kappa \delta - \eta \zeta) (\alpha \theta - \beta \eta) + (\gamma \eta - \alpha \kappa) (\eta \epsilon - \theta \delta) ] + E [ (\theta \zeta - \kappa \epsilon) (\gamma \eta - \alpha \kappa) + (\beta \kappa - \gamma \theta) (\kappa \delta - \eta \zeta) ] + 2 F (\eta \epsilon - \theta \delta) (\alpha \theta - \beta \eta) \,$

    $ D' = 2 A (\epsilon \gamma - \zeta \beta) (\beta \kappa - \gamma \theta) + 2 B (\zeta \alpha - \delta \gamma) (\gamma \eta - \alpha \kappa) + C [ (\epsilon \gamma - \zeta \beta) (\alpha \theta - \beta \eta) + (\beta \kappa - \gamma \theta) (\delta \beta - \epsilon \alpha) ]\,$

        $+ D [ (\zeta \alpha - \delta \gamma) (\alpha \theta - \beta \eta) + (\gamma \eta - \alpha \kappa) (\delta \beta - \epsilon \alpha) ] + E [ (\epsilon \gamma - \zeta \beta) (\gamma \eta - \alpha \kappa) + (\beta \kappa - \gamma \theta) (\zeta \alpha - \delta \gamma) ] + 2 F (\delta \beta - \epsilon \alpha) (\alpha \theta - \beta \eta) \,$

    $ E' = 2 A (\theta \zeta - \kappa \epsilon) (\epsilon \gamma - \zeta \beta) + 2 B (\kappa \delta - \eta \zeta) (\zeta \alpha - \delta \gamma) + C [(\theta \zeta - \kappa \epsilon) (\delta \beta - \epsilon \alpha) + (\epsilon \gamma - \zeta \beta) (\eta \epsilon - \theta \delta)]\,$

        $+ D [ (\kappa \delta - \eta \zeta) (\delta \beta - \epsilon \alpha) + (\zeta \alpha - \delta \gamma) (\eta \epsilon - \theta \delta)] + E [ (\theta \zeta - \kappa \epsilon) (\zeta \alpha - \delta \gamma) + (\epsilon \gamma - \zeta \beta) (\kappa \delta - \eta \zeta)] + 2 F (\eta \epsilon - \theta \delta) (\delta \beta - \epsilon \alpha) \,$

    $ F' = A (\beta \kappa - \gamma \theta)^2 + B (\gamma \eta - \alpha \kappa)^2 + C (\beta \kappa - \gamma \theta) (\alpha \theta - \beta \eta) + D (\gamma \eta - \alpha \kappa) (\alpha \theta - \beta \eta) + E (\beta \kappa - \gamma \theta) (\gamma \eta - \alpha \kappa) + F (\alpha \theta - \beta \eta)^2 \,$

The coefficients of the transformed conic have been expressed in terms of the coefficients of the original conic and the coefficients of the planar transformation $T$. Q.E.D.