OrthopticSurfaceSphere::usage="The orthoptic surface of a surface is the geometric locus of the points through which pass 3 planes orthogonal to each other and tangents to the original surface. References: AI; Personal.";
Development of the orthoptic surface:
Obtaining the orthoptic surface of the sphere
2
x
+
2
y
+
2
z
=4
:
Let the tangent plane (where the normal vector is an unitary vector):
Ax+By+Cz==D;
Condition of tangent between a plane and a sphere centered at the origin:
Tangent plane:
x
0
x
2
a
+
y
0
y
2
a
+
z
0
z
2
a
==1;​​
A
D
x+
B
D
y+
C
D
z==1;​​
x
0
2
a
==
A
D
;
x
0
==
A
2
a
D
;​​
y
0
2
a
==
B
D
;
y
0
==
B
2
a
D
;​​
z
0
2
a
==
C
D
;
z
0
==
C
2
a
D
;​​
Sustituting the point {
x
0
,
y
0
,
z
0
}
in the sphere’s equation and simplifying:
1
2
a
2
A
2
a
D
+
1
2
a
2
B
2
a
D
+
1
2
a
2
C
2
a
D
==1;​​
2
D
==
2
A
2
a
+
2
B
2
a
+
2
C
2
a
;
So for the sphere that concerns us:
2
D
==4
2
A
+4
2
B
+4
2
C
;
So for the three planes we have:
2
D
1
==4
2
A
1
+4
2
B
1
+4
2
C
1
;​​
2
D
2
==4
2
A
2
+4
2
B
2
+4
2
C
2
;​​
2
D
3
==4
2
A
3
+4
2
B
3
+4
2
C
3
;
Adding member to member the three equations and considering that the rows of an orthogonal matrix are also unit vectors:
2
D
1
+
2
D
2
+
2
D
3
==4
2
A
1
+
2
A
2
+
2
A
3
+4
2
B
1
+
2
B
2
+
2
B
3
+4
2
C
1
+
2
C
2
+
2
C
3
;​​
2
D
1
+
2
D
2
+
2
D
3
==4(1)+4(1)+4(1)==12;
Since the three normal vectors are mutually orthogonal and unitary, (and by properties of orthogonal matrices) it is simplified to:
In[]:=
2
D
1
+
2
D
2
+
2
D
3
==Expand[
2
(
A
1
x+
B
1
y+
C
1
z)
+
2
(
A
2
x+
B
2
y+
C
2
z)
+
2
(
A
3
x+
B
3
y+
C
3
z)
]
2
D
1
+
2
D
2
+
2
D
3

2
x
2
A
1
+
2
x
2
A
2
+
2
x
2
A
3
+2xy
A
1
B
1
+
2
y
2
B
1
+2xy
A
2
B
2
+
2
y
2
B
2
+2xy
A
3
B
3
+
2
y
2
B
3
+2xz
A
1
C
1
+2yz
B
1
C
1
+
2
z
2
C
1
+2xz
A
2
C
2
+2yz
B
2
C
2
+
2
z
2
C
2
+2xz
A
3
C
3
+2yz
B
3
C
3
+
2
z
2
C
3
;​​
2
D
1
+
2
D
2
+
2
D
3
==
2
A
1
+
2
A
2
+
2
A
3

2
x
+
2
B
1
+
2
B
2
+
2
B
3

2
y
+
2
C
1
+
2
C
2
+
2
C
3

2
z
+(
A
1
B
1
+
A
2
B
2
+
A
3
B
3
)2xy+(
A
1
C
1
+
A
2
C
2
+
A
3
C
3
)2xz+(
B
1
C
1
+
B
2
C
2
+
B
3
C
3
)2yz;​​
2
D
1
+
2
D
2
+
2
D
3
==
2
x
+
2
y
+
2
z
;
So it turns out how we expected Monge’s sphere:
2
x
+
2
y
+
2
z
==
3
2
a
which is the orthoptic surface of the sphere. There are many similarities in 2D with the circumference and its orthoptic curve (below), the circumference of Monge. Finally, the equation sought is:
2
x
+
2
y
+
2
z
==12;
Animation of the orthoptic surface:
In[]:=
Manipulatea2=4;b2=4;c2=4;​​u1[α_,β_]:={Cos[α]Cos[β],Sin[α]Cos[β],Sin[β]};​​u2[α_]:={-Sin[α],Cos[α],0};​​u3[α_,β_]:={-Cos[α]Sin[β],-Sin[α]Sin[β],Cos[β]};​​d1[α_,β_]:=
a2
2
u1[α,β][[1]]
+b2
2
u1[α,β][[2]]
+c2
2
u1[α,β][[3]]
;​​d2[α_]:=
a2
2
u2[α][[1]]
+b2
2
u2[α][[2]]
+c2
2
u2[α][[3]]
;​​d3[α_,β_]:=
a2
2
u3[α,β][[1]]
+b2
2
u3[α,β][[2]]
+c2
2
u3[α,β][[3]]
;​​Q[α_,β_]:=d1[α,β]u1[α,β]+d2[α]u2[α]+d3[α,β]u3[α,β];​​​​​​Show[​​​​ParametricPlot3D[{2Cos[u]Sin[v],2Sin[u]Sin[v],2Cos[v]},{u,0,2π},{v,0,π},ColorFunction->"SolarColors",MeshNone,MaxRecursion->3],​​ParametricPlot3D[
3a2
Cos[u]Sin[v],
3a2
Sin[u]Sin[v],
3a2
Cos[v],{u,0,2π},{v,0,π},​​ColorFunction->"BlueGreenYellow",PlotStyle->Opacity[0.2],MeshNone,MaxRecursion->3],​​Graphics3D[{Black,Thick,Sphere[Q[α,β],0.1],Line[{Q[α,β],Q[α,β]+u1[α,β]}],Line[{Q[α,β],Q[α,β]+u2[α]}],Line[{Q[α,β],Q[α,β]+u3[α,β]}],{Gray,Opacity[0.4],Hyperplane[u1[α,β],d1[α,β]],Hyperplane[u2[α],d2[α]],Hyperplane[u3[α,β],d3[α,β]]}}]​​,​​BoxRatios->Automatic,​​Axes->True,BoxedFalse,​​AxesOrigin->{0,0,0},​​AxesLabel->{x,y,z},​​PlotRange->6,​​ImageSize->900,ViewPoint{1.3,-2.4,2.}​​],Style["Orthoptic surface: Monge's sphere",Bold,Large]​​,​​{{α,0.00002,"α angle (Turn H)"},0.000001,2π,0.00001},Delimiter,β,
π
4
,"β angle (Turn V)",-
π
2
,
π
2
,0.00001,ControlPlacement->Top,AutorunSequencing->{1,2}​​
Out[]=
​
Orthoptic surface: Monge's sphere
α angle (Turn H)
β angle (Turn V)
Development of the orthoptic curve:
Obtaining the parametric equations from the orthoptic curve to the circumference:
Animation of the orthoptic curve:
Cartesian equation of the orthoptic curve: