Perspectivity
Perspectivity is one of the major achievements of Renaissance period to understand the art of drawing. The artists of that time used principle of perspectivity to produce realistic images of buildings, towns and other sceneries. The main idea behind is
to produce two-dimensional image of a three-dimensional object.
Doing this, lines that are parallel in the original scene may not remain parallel in the drawing. The artists of that time used these principles. These principles still serve as foundation
knowledge for computer in the study of photo/images.
The basic idea behind such principal is as follows:
- Fix the position of object, canvas and viewer’s eye as u.
- For each point p on the object, consider a line from the viewer’s eye u through p and plot a dot p’ on canvas as the ray up meets.
- For any line L in the object, consider a plane Г spanned by L and the viewer’s eye u. By this procedure, the line L will be connected to a line L' in canvas by a view-plane Г.
This principle of mapping where points are mapped to points, lines are mapped to lines, and incidences do preserve, is called perspectivity. In this mapping, parallelism, orthogonality, distances and angles may NOT preserved.
A first systematic treatment
of perspectivity was developed by French architect and engineer Girard Desargues (1591–1661) and later expanded by his student Blaise Pascal (1623–1662). They laid the foundation for “projective geometry”. Almost by 150 years later, a French-men Gaspard
Monge (1746 – 1818) wrote a book descriptive geometry dealing with the problem of making two-dimensional sketches of three-dimensional objects in 1790. Monge also made an exciting observation concerning the relation between geometric object and their
perspective drawing in a theorem. This theorem was later discovered Desargues’s triangle theorem almost after 200 years. Later Monge’s student Jean-Victor Poncelet (1788-1867) investigated properties which remain invariant under projection such as
cross-ratio, perspectivity, involution and the points at infinity in 1822.
Definition
Perspectivity is a method of graphically depicting three-dimensional objects and spatial relationships into a two-dimensional plane.
Let L and L' are two ranges in projective plane π and u be a point neither on L nor on L', then a perspectivity from
L to L' with center u is denoted by \( f:L \pi ^u L' \) and defined by
\( f(x)=L' \cap ux \) for all \(x \in L \)
Note
If \( a,b,c, \ldots \in L \) and \( a',b',c', \ldots \in L' \) then \( f:L \pi ^u L' \) is written as
\( f:L(a,b,c...) \pi ^u L'(a',b',c',...) \) where f(a)=a' , f(b)=b' , f(c)=c' ...
In a perspectivity \( f:L \pi ^u L' \) lines aa' , bb' , cc' ... passes through u, (i.e., object, image and center are always
collinear).
Note
A perspectivity from L to L' is a bijection and its inverse is also perceptivity from L' to L with same center.
\( \alpha \lambda \lambda ' +\beta \lambda \mu ' +\gamma \lambda ' \mu +\delta \mu \mu ' =0\).
Proof
Given \(a= [a_1, a_2, a_3], b= [b_1, b_2, b_3] \in L\) in \( \pi_F\) .
Thus, any arbitrary point c on L is
\( c= [\lambda a_1+\mu b_1, \lambda a_2+\mu b_2, \lambda a_3+\mu b_3] \) for \( \lambda ,\mu \in F\)
Since, \( f:L \overset{u}{\pi} L'\) is a perspectivity, we have
\(f(c)=c'\) for \(c'\in L'\)
Also \(a'= [a_1', a_2', a_3'], b'= [b_1', b_2', b_3'] \in L'\), then \(c'\) can be parametrized as
\(c'= [\lambda 'a_1'+\mu 'b_1', \lambda 'a_2'+\mu 'b_2', \lambda 'a_3'+\mu 'b_3']\) for \(\lambda ',\mu '\in F\)
Since, \(u, c,c'\) are collinear.
Thus,
\( \begin{vmatrix}u_1& u_2& u_3 \\ \lambda a_1+\mu b_1& \lambda a_2+\mu b_2& \lambda a_3+\mu b_3 \\ \lambda ' a'_1+\mu ' b'_1& \lambda ' a'_2+\mu ' b' _2& \lambda ' a' _3+\mu ' b' _3 \end{vmatrix}=0 \)
or
\( \lambda \lambda ' \begin{vmatrix}u_1& u_2& u_3 \\ a_1& a_2& a_3 \\ a' _1& a' _2& a' _3 \end{vmatrix}+\lambda \mu ' \begin{vmatrix}u_1& u_2& u_3 \\ a_1& a_2& a_3 \\ b' _1& b' _2& b' _3 \end{vmatrix}+
\lambda ' \mu \begin{vmatrix}u_1& u_2& u_3 \\ b_1& b_2& b_3 \\ a' _1& a' _2& a' _3 \end{vmatrix}+\mu \mu ' \begin{vmatrix}u_1& u_2& u_3 \\ bb_1& b_2& b_3 \\ b' _1& b' _2& b' _3 \end{vmatrix}=0 \)
Assuming
\( \alpha =\begin{vmatrix}u_1& u_2& u_3\\ a_1& a_2& a_3\\ a' _1& a' _2& a' _3 \end{vmatrix},\beta =\begin{vmatrix}u_1& u_2& u_3 \\ a_1& a_2& a_3\\ b' _1& b' _2& b' _3 \end{vmatrix},
\gamma =\begin{vmatrix}u_1& u_2& u_3 \\ b_1& b_2& b_3 \\ a' _1& a' _2& a' _3 \end{vmatrix},\delta =\begin{vmatrix}u_1& u_2& u_3 \\ b_1& b_2& b_3\\ b' _1& b' _2& b' _3 \end{vmatrix}\)
We
get
\( \alpha \lambda \lambda ' +\beta \lambda \mu ' +\gamma \lambda ' \mu +\delta \mu \mu ' =0\)
This completes the proof.
Exercise
-
Let a=[1,2,-1], b=[0,1,1] ∈L, and a'=[1,0,1], b'=[0,1,-1] ∈L'. If equation of perspectivity \(f:L\pi^uL'\) is 2λλ'-λµ'+3µµ'=0, then find the center u.
Answer: [3,1,4] - Let a=[1,2,-1], b=[0,1,1] ∈L, and a'=[1,0,1], b'=[0,1,-1] ∈L' are taken as base points and u=[3,1,4]. Then find the equation of perspectivity f:L\pi^uL'.
Answer: 2λλ'-λµ'+3µµ'=0 - Let a=[1,2,-1], b=[0,1,1] ∈L, and a'=[1,0,1], b'=[0,1,-1] ∈L'. Also the equation of perspectivity \(f:L\pi^uL'\) is 2λλ'-λµ'+3µµ'=0. Find the image of a point [1,3,0]
Answer: [1,-1,2] - Find the equation of perspectivity between 〈1,0,0〉 and 〈0,1,0〉 which maps p=[0,1,1], q=[0,2,0], and r=[0,1,3] to the points p'=[1,0,2], q'=[1,0,1], and r'=[2,0,1]. The line 〈1,0,0〉 is parametrized with base points [0,0,1] and [0,1,0]. The line 〈0,1,0〉
is parametrized with base points [1,0,0] and [1,0,1].
Answer: 7λλ'-4λµ'-11 λ'µ +2µµ'=0 - Find the equation of perspectivity between 〈1,0,0〉 and 〈0,1,0〉 which maps p=[0,1,1], q=[0,2,0], and r=[0,1,3] to the points p'=[1,0,2], q'=[1,0,1], and r'=[2,0,1]. The line 〈1,0,0〉 is parametrized with base points [0,1,1] and [0,1,-1]. The line 〈0,1,0〉
is parametrized with base points [1,0,-1] and [0,0,1].
Answer: 3λλ'-λµ'+2`1 λ'µ -12µµ'=0
Proof
Given, equation of perspectivity
\( \alpha\lambda\lambda'+\beta\lambda\mu'+\gamma\lambda'\mu+\delta\mu\mu'=0\)
or
\(\beta\lambda\mu'+\delta\mu\mu'=-\alpha\lambda\lambda'-\gamma\lambda'\mu \)
or
\( \left(\beta\lambda+\delta\mu\right)\mu'=\left(-\alpha\lambda-\gamma\mu\right)\lambda' \)
or
\( \frac{\lambda'}{\mu'}=\frac{\beta\lambda+\delta\mu}{-\alpha\lambda-\gamma\mu} \)
or
\( (\lambda',\mu')=(\beta\lambda+\delta\mu,-\alpha\lambda-\gamma\mu) \)
Since, parameters preserves proportionality, we write
\( \left(\lambda',\mu'\right)=\left(\lambda,\mu\right)\begin{pmatrix}\beta& -\alpha\\\delta& -\gamma\\\end{pmatrix}\) provided \( \alpha\delta-\beta\gamma\neq 0 \)
This completes the proof.
The equation of perspectivity has a form
\( \left(\lambda',\mu'\right)=\left(\lambda,\mu\right)M \) where M is 2x2 nonsingular matrixx.
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