Specular cylinders (e.g. HST, Centaurs)
Matson, Robert (ROBERT.D.MATSON@cpmx.saic.com)
Thu, 6 May 1999 20:27:32 -0700
Hi Mike,
Responding to an older message of yours regarding solar
specular reflections from cylinders, you wrote:
> My opinion is that you can get a specular reflection along the length
> of the cylinder and that this explains the 4 or 5 magnitudes brighter
> flashing of Centaurs and SL-8s.
You are absolutely right, Mike. I hadn't given this enough thought,
for it is a very interesting calculation to compare lambertian
cylinders, specular cylinders and specular planar surfaces. An
example illustrates this.
Assume that a cylinder is 1 meter in diameter and 3 meters
long, with the long axis perpendicular to the sun's direction (i.e.
the "end caps" are not illuminated). If this cylinder is lambertian,
then 1400 W/m^2 x 3 m^2 = 4200 W is collected, and then
redirected primarily into the hemisphere on the sunward side of
the cylinder. (Obviously some sunlight is reflected into the rear
hemisphere since any point in space that has a line of sight to
the illuminated portion of the cylinder will receive some energy.)
Due to projected area effects, the amount of reflected energy
will also drop off as you approach the poles above and below
the end caps.
So the overall shape of the reflected energy is going to be a
"hump", the broad peak of which will be back in the direction
of the sun for the geometry we've selected. By the time you
are 60 degrees off-axis from the sun-cylinder line, the radiance
will have dropped by a factor of two. More than 85% (the area
under the cosine curve from 0 to 60 degrees) of the 4200 W
(3570 W) will thus be within 60 degrees of the "specular return
direction". This is pi steradians -- half of a hemisphere by
solid angle.
Now, what happens if the hemisphere is specular instead of
Lambertian? Well, to oversimplify, you lose a dimension.
Instead of the energy going into a cone with a 60-degree
half-angle, it goes into a long, narrow strip 1/2-degree wide
by 120 degrees long (again assuming that ~85% of the energy
is within +/- 60 degrees of the specular direction). The solid
angle is roughly (2/3) pi * pi/360 steradians = .0183 steradians.
This is about 1/170th the area of the lambertian case, so the
average brightness has gone up by a factor of 170 -- about
5 1/2 visual magnitudes! Gee, what a coincidence!
So what about an end-cap (planar) specular reflection? In
our example, the maximum collected energy would be pi/4
square meters * 1400 W/m^2 = 1100 W. But this 1100 W
is primarily reflected into a cone with a half-angle of only
0.25 degrees -- that's only about 0.00006 steradians. So
that's a solid angle reduction of a factor of 300 over the
specular cylinder. If you scale that by the reduced power
fraction (1100W / (0.85*4200W) = 0.3, the end-cap reflected
energy density is more than 90 times brighter than that
from the curved part of the cylinder -- another 4.9 visual
magnitudes.
So tumbling specular cylinders make for very interesting
brightness curves. You can get a fairly low signature if
you don't happen to be within the 120-degree by 0.5-deg
strip; but twice per tumble that strip must sweep over
your location, causing a five-plus magnitude increase
in the signature. And if the geometry is right during some
part of the pass, you could catch an end-cap specular
reflection which would be 10 or more visual magnitudes
above the "background" level.
One interesting side effect of highly specular cylinders is
that they are much, much dimmer than flat-painted
cylinders when the specular strip isn't sweeping over you.
Basically an optical analog to low-RCS aircraft. This
explains why HST can be such a "dud" sometimes
during a pass, and why Centaurs can be quite dim
between flashes. --Rob