Jean DRABBE emeritus professor University of Brussels jdrabbe@skynet.be |
Chantal GABRIEL-RANDOUR mathematics teacher Athénée Gatti de Gamond maître de stage ULB ch.randour@brutele.be |
(A^{v} , A^{h}) represents A |
The following figures represent a cone and a cylinder |
Imagine a right circular cone mirror standing on the ground plane and an
eye point O directly above the tip of the the cone.
Let P be a point in the
ground plane. It is required to construct au point P' in the ground
plane so that reflected in the mirror - seen from O - it appears
to be P. In other words, P' satisfies the following condition : Let OP intersect the cone at U. A ray from P' striking the cone at U is being reflected along UO. |
Let p be the plane through the cone's axis parallel to the vertical
plane.
In case P is in p, the problem is trivial. We are
going to make use of this special case to solve the general problem. Rotate the ground plane around the (vertical) axis of the cone till P coincides with some point Q in p and solve the problem for Q. This gives R. Apply the inverse rotation to R to find P'. |
In 1973 Michel Parré constructs a device drawing mechanically conical anamorphoses
to be seen from infinity. The tool is based on the pantograph of Scheiner (1605) made to
construct dilatations. We show a reproduction of the device by Jean-Jacques Gabriel. |
The secret of Parré's device lies in some simple mathematics.
Let C be the center of the cone's base circle. R marks the point where the line from C to P cuts the circle. When the eye point is at infinity, the ration PR / RP' does not depend on P. Its value only depends on the cone in use. |
Conical anamorphose of a dog (to be seen for infinity) |
An eye point O and a cylindrical mirror standing on the horizontal
(ground) plane are given. Let p be the polar plane of O with respect to the cylinder (i.e. the plane determined by the lines of contact of the tangent planes through O) . Because of the eye does not distinguish between points on the same visual ray, we assume that the light seen from O is coming from points in p. So we need to solve the problem : Let P be a point in p and let C be the point where the segment OP cuts the cylinder. Determine the point P' in the ground plane so that the ray P'C reflects along CO. We let Q be the point of intersection of OP and the ground plane. The laws of reflexion and some elementary geometry show that P'^{h}C^{h} and C^{h}O^{h} make the same angle with the tangent line at C^{h} to the circular base and that C^{h}P'^{h} = C^{h}Q^{h}. This solves the problem. |
An anamorphose constructed by means of descriptive geometry |