The Rytz’s axis construction is a basic method of descriptive geometry to find the axes, the semi-major axis and semi-minor axis and the vertices of an ellipse, starting from two conjugated half-diameters. If the center and the semi axis of an ellipse are determined the ellipse can be drawn using an ellipsograph or by hand (see ellipse).
Rytz’s construction is a classical construction of Euclidean geometry, in which only compass and ruler are allowed as aids. The design is named after its inventor David Rytz of Brugg (1801–1868).
Conjugate diameters appear always if a circle or an ellipse is projected parallelly (the rays are parallel) as images of orthogonal diameters of a circle (see second diagram) or as images of the axes of an ellipse. An essential property of two conjugate diameters
d1, d2
The parallel projection (skew or orthographic) of a circle that is in general an ellipse (the special case of a line segment as image is omitted). A fundamental task in descriptive geometry is to draw such an image of a circle. The diagram shows a military projection of a cube with 3 circles on 3 faces of the cube. The image plane for a military projection is horizontal. That means the circle on the top appears in its true shape (as circle). The images of the circles at the other two faces are obviously ellipses with unknown axes. But one recognizes in any case the images of two orthogonal diameters of the circles. These diameters of the ellipses are no more orthogonal but as images of orthogonal diameters of the circle they are conjugate (the tangents at the end points of one diameter are parallel to the other diameter !). This is a standard situation in descriptive geometry:
C
P,Q
(1) rotate point
P
C
D
\overline{P'Q}
P'Q
D
C
A,B
CA
CB
\overline{AB}
a+b
Q
a=|AQ|
b=|BQ|
a>b
a
If one performs a left turn of point
P
a=|AQ|
b=|BQ|
The standard proof is performed geometrically.[1] An alternative proof uses analytic geometry:
The proof is done, if one is able to show that
U,V
P'Q
C
D
U=A
V=B
|UQ|=a, |VQ|=b .
(1): Any ellipse can be represented in a suitable coordinate system parametrically by
\vecp(t)=(a\cost, b\sint)T
Two points
\vecp(t1), \vecp(t2)
t2-t1=\pm\tfrac\pi2 .
Q=(a\cost, b\sint)
P=(a\cos(t+\tfrac\pi2), b\sin(t+\tfrac\pi2))=(-a\sint,b\cost)
two points on conjugate diameters.
Then
P'=(b\cost,a\sint)
\overline{P'Q}
D=\left(\tfrac{a+b}{2}\cost,\tfrac{a+b}{2}\sint\right)
P'Q
x\sint+y\cost=(a+b) \cost\sint .
The intersection points of this line with the axes of the ellipse are
U=\left(0, (a+b)\sint\right) , V=\left((a+b)\cost, 0\right) .
|UD|=|VD|=|CD|
U,V,C
D
|CD| .
Hence
A=U, B=V .
|UQ|=a, |VQ|=b .
The proof uses a right turn of point
P
If one performs a left turn of point
P
P=(a\cos(t{\color{red}-}\tfrac\pi2), b\sin(t{\color{red}-}\tfrac\pi2))=(a\sint,-b\cost)
To find the vertices of the ellipse with help of a computer,
C,P,Q
\vecx=\vecp(t)=\vecf0+\vecf1\cost+\vecf2\sint .
\vecf0=\vec{OC}
\vecf1=\vec{CP}, \vecf2=\vec{CQ}
If necessary: With
\cot(2t0)=\tfrac{\vec
2 | |
f | |
1 |
-\vec
2 | |
f | |
2 |
\vecp(t0), \vec
p(t | ||||
|
), \vecp(t0+\pi) .