In Perspective

Geometry Level 4

In perspectival drawings, a 3D situation is projected onto the 2D plane. This involves a projection function ( x , y , z ) ↦ ( ξ , η ) (x,y,z) \mapsto (\xi, \eta) . A simple projection function is of the form ξ = x z ; η = y z . \xi = \frac x z;\ \ \ \ \eta = \frac y z. Now consider in 3D space a circle in a horizontal plane, centered at ( 0 , b , c ) (0, b, c) . Its points satisfy the equations x 2 + ( z − c ) 2 = r 2 ; y = b . x^2 + (z - c)^2 = r^2;\ \ \ y = b. When projected according to the formula above, this circle becomes a curve, which looks like an ellipse. It is obviously symmetric around the ξ \xi -axis.

Is this curve actually an ellipse? I.e. does it satisfy ( ξ ρ ) 2 + ( η − a σ ) 2 = 1 \left(\frac \xi \rho\right)^2 + \left(\frac{\eta-a} \sigma\right)^2 = 1 for some constants ρ , σ , a > 0 \rho, \sigma, a > 0 ?

Yes No

Arjen Vreugdenhil by Arjen Vreugdenhil , 44, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

We have z = b / η z = b/\eta and x = ξ z = b ξ / η x = \xi z = b\xi/\eta , so that the equation for the circle becomes ( b ξ η ) 2 + ( b η − c ) 2 = r 2 . \left(\frac{b\xi}\eta\right)^2 + \left(\frac b\eta - c\right)^2 = r^2. Multiply everything by η 2 \eta^2 and expand the squares: ( ξ b ) 2 + ( b − c η ) 2 = ( r η ) 2 ; ξ 2 b 2 + b 2 − 2 b c η + c 2 η 2 = r 2 η 2 . (\xi b)^2 + (b - c\eta)^2 = (r\eta)^2; \\ \xi^2 b^2 + b^2 - 2bc\eta + c^2\eta^2 = r^2\eta^2. This is a quadratic in η \eta ; we complete the square: ξ 2 b 2 + ( c 2 − r 2 ) η 2 − 2 b c η + b 2 = 0 ; ξ 2 b 2 + ( c 2 − r 2 ) ( η − b c c 2 − r 2 ) 2 = b 2 c 2 c 2 − r 2 − b 2 = b 2 r 2 c 2 − r 2 . \xi^2 b^2 + (c^2 - r^2)\eta^2 - 2bc\eta + b^2 = 0; \\ \xi^2 b^2 + (c^2 - r^2)\left(\eta - \frac{bc}{c^2 - r^2}\right)^2 = \frac{b^2c^2}{c^2-r^2} - b^2 = \frac{b^2 r^2}{c^2-r^2}. We wish the right side of the equation to be 1, so divide by the right-hand term: c 2 − r 2 r 2 ξ 2 + ( c 2 − r 2 ) 2 b 2 r 2 ( η − b c c 2 − r 2 ) 2 = 1. \frac{c^2-r^2}{r^2} \xi^2 + \frac{(c^2 - r^2)^2}{b^2 r^2}\left(\eta - \frac{bc}{c^2-r^2}\right)^2 = 1. Compare this to the ellipse equation 1 ρ 2 ξ 2 + 1 σ 2 ( η − a ) 2 = 1. \frac 1{\rho^2}\xi^2 + \frac 1{\sigma^2}(\eta - a)^2 = 1. The final answer is therefore: Yes \boxed{\text{Yes}} , the projection has the form of an ellipse, with ρ = r c 2 − r 2 ; σ = b r c 2 − r 2 ; a = b c c 2 − r 2 . \rho = \frac r{\sqrt{c^2 - r^2}}; \\ \sigma = \frac {br}{c^2 - r^2}; \\ a = \frac{bc}{c^2 - r^2}.

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Had the exact same result.....great problem, Arjen!

tom engelsman - 5 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...