cylindrical star track 2-D projection shapes
br77rino
My guess is this: circles, ellipses, elliptical arcs, straight lines, elliptical arcs, ellipses, and circles.
Robert Israel
> When one looks at a time-lapse photograph of the night sky, one sees what
> l=
> ooks like a cylinder, with us inside it. Polaris is the infinitely far
> away=
> end to the North. My question is, what is the transition between conic
> se=
> ction shapes if you would look from one end towards the other? The tracks
> =
> around Polaris are circles, or certainly appear that way, and the tracks
> ri=
> ght above our head are straight lines, aren't they? But shouldn't there be
> =
> transitional shapes? Like ellipses or hyperbolas?=20
>
> My guess is this: circles, ellipses, elliptical arcs, straight lines,
> ellip=
> tical arcs, ellipses, and circles.
Neglecting the earth's motion around the sun, the track of a star on the
celestial sphere is a circle of constant declination (corresponding to
latitude). A photograph is a projective transformation onto a plane: e.g.
if you use a coordinate system such that you are at (0,0,0) and the plane
is z=1, the transformation takes (x,y,z) to (X,Y) = (x/z, y/z). Consider the
celestial sphere to be x^2 + y^2 + z^2 = 1. Consider circle centred at
(c_1,c_2,c_3) on the celestial sphere with angle theta from centre to
circumference. This corresponds to the equations c_1 x + c_2 y + c_3 z = d,
x^2 + y^2 + z^2 = 1, where d = cos(theta), and its projection on the
photograph would have
X^2 + Y^2 + 1 = (x^2 + y^2 + z^2)/z^2 = 1/z^2 with
c_1 X + c_2 Y + c_3 = d/z
and thus d^2 (X^2 + Y^2 + 1) = (c_1 X + c_2 Y + c_3)^2
or
(d^2 - c_1^2) X^2 - 2 c_1 c_2 X Y + (d^2 - c_2^2) Y^2
- 2 c_1 c_3 X - 2 c_2 c_3 Y + d^2 - c_3^2 = 0
For this projection to be a circle, you need c_1 = c_2 = 0 and d <> 0
(i.e. the camera must be pointed directly at the north or south celestial
pole, and the circle is not the celestial equator).
The quadratic part of the equation corresponds to the matrix
[ d^2 - c_1^2 c_1 c_2 ]
[ c_1 c_2 d^2 - c_2^2 ]
which has determinant d^2 (d^2 - c_1^2 - c_2^2). If this is positive,
the projection of the circle is an ellipse. That occurs if the circle
does not intersect the plane z=0 (orthogonal to the direction the camera is
pointed). If it is negative, the projection is a hyperbola. That occurs if
the circle intersects the plane z=0 at two points. If d = 0 (i.e. the circle
is a great circle, namely the celestial equator) the projection is a straight
line c_1 X + c_2 Y + c_3 = 0. If d^2 - c_1^2 - c_2^2 = 0, the circle
intersects the plane z=0 in one point, and the projection is a parabola.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Grover
> When one looks at a time-lapse photograph of the night sky, one sees what looks like a cylinder, with us inside it. Polaris is the infinitely far away end to the North. My question is, what is the transition between conic section shapes if you would look from one end towards the other? The tracks around Polaris are circles, or certainly appear that way, and the tracks right above our head are straight lines, aren't they? But shouldn't there be transitional shapes? Like ellipses or hyperbolas?
>
> My guess is this: circles, ellipses, elliptical arcs, straight lines, elliptical arcs, ellipses, and circles.
A mathematician you may be, but an astronomer you ain't! Polaris is
not at the north celestial pole; it's about a degree away from the
actual pole. Of course, I know what you meant, but you didn't say it
properly. Tsk, tsk.
Regards, Grover Hughes