Descriptions and Examples for the POV-Ray Raytracer by Friedrich A. Lohmüller
    Elementary Geometry for Raytracing
Italiano Italiano
Français français
Deutsch Deutsch

Home
- POV-Ray Tutorial

  - Geometrical Basics
    for Raytracing

    Right-angled Triangle
    Pythagorean Theorem
    Trigonometry Basics
    Law of cosines
    Equilateral Triangle
    Regular Polygon
    Polyhedron
      Tetrahedron
      Octahedron
      Cube & Cuboid
      Dodecahedron
      Icosahedron
      Cuboctahedron
      Truncated Octahedron
      Rhombicuboctahedron
      Truncated Icosahedron
    Circles
      Tangent Circles
      Internal Tangents
      External Tangents
     
     
     
     
     
     
     
     
     
   - Geometric 3D Animations

                                               

Tangent circles
In the following we write for the square root of a number the expression "sqrt(ZAHL)"
conforming to the syntax used in POV-Ray.

Note: Here objects in 2D geometry are represented by 3D shapes in the xy-plane. Therefore all coordinates must have the z-components zero! ( <?,?,0>)

Circle 1: center M1 = <0.35,0,0>, radius r1= 0.22.
Circle 2: center on y-axis, tangents Circle 1 from outside in the point S.
Problem 1:
Circle 2 has the radius r2 = 0.30 .
Where on the y-axis (height y2) is the center of the second circle 2 ? (M2 = <0,?,0>)
Problem 2:
Circle 2 has the center M2 = <0,0.40,0>.
How big must be the radius of a second circle 2, if it intersects (touches) the first circle 1 at the nearest point?
Problem 3:
What are the coordinates of the point S ?
Problem 4:
What are the angles at M1 and M2 inside the triangle(0,M1,M2)?
The triangle(O,M1,M2) is rectangular!
Therefore we use the Pythagorean Theorem:

// Problem 1: //--------------------------
We know two sides of the triangle(O,M1,M2):
d(0,M1) = x1 and d(M1,M2) = r1+r2.
   y2 = sqrt( (r1+r2)2 - x12)
   M2 = <0,y2,0>.


// Problem 2: //--------------------------
Because of
   r1+r2 = d(M1,M2) = sqrt(x12 + y22),
we get
   d(M1,M2) = sqrt( x12 - y22 )
   and r2 = d(M1,M2) - r1.

// Problem 3: //--------------------------
There is a simple proportionality:
  xS1/xM1 = r2/ (r1+r2) and yS/yM2 = r1/ (r1+r2),
so: xS = xM1 · r2/ (r1+r2)
      yS = yM2 · r1/ (r1+r2).


// Problem 4: //--------------------------
By the inverse trigonometric function i.e. of tan(x):
   Angle(M1) = atan ( y2/ x1),
   Angle(M2) = 90 - Angle(M1).
Tangent circles rendered with POV-Ray
Note: To avoid any collision with built-in identifiers and reserved words in POV-Ray, it's strongly recommanded to use only words beginning with capital letters for all identifiers of variables declared by the user,
i.e. use "R1" instead of "r1" and use "Y2" instead of "yM2".
#local R1= 0.22;
#local R2= 0.30;
#local M1 = <0.35,0,0>
#local Y2 = sqrt( pow(R1+R2, 2) - pow(M1.x, 2));
#local M2 = <0,Y2,0>; 
Problem 1 in POV-Ray
#local R1= 0.22;
#local M1 = <0.35,0,0>
#local M2 = <0.40,0,0>
#local R2 = sqrt( pow(M1.x,2)-pow(M2.y,2)) - R1;
Problem 2 in POV-Ray
#local XS = M1.x * R2/(R1+R2);
#local YS = M2.y * R1/(R1+R2);
#local S = <XS,YS,0>

#local Angle_M1 = degrees( atan( M2.y / M1.x) ); #local Angle_M2 = 90 - Angle_M1;
Problem 3 + 4 in POV-Ray

For what can we use this geometry?
Here some examples:
We can consider the circle_1 as the cross-section of an horizontal torus and the circle_2 as the cross-section of a bowl:


We can consider the circle_1 and its twin mirrored at the y-axis as the cross-section of two cylinder in z-direction
and the circle_2 as the cross-section of another cylinder to subtract from the yellow rectangle:

Round-bottom Mace
Round-bottom flask
for the labratory
Two cylinders melting together
Chain link for a bike chain
This shape in a macro as a ready made object:
Round_Bottom_Mace_1.
Applied in a macro for the ready made object:
Round_Bottom_Flask_1.
This shape in a macro as a ready made object:
Two cylinders melting together.
Applied in a macro for the ready made object:
Bike_Chain_Link_1.
top
© Friedrich A. Lohmüller, 2010
www.f-lohmueller.de