# Method for predicting flare

ROB MATSON (ROBERT.D.MATSON@cpmx.saic.com)
20 Apr 1998 16:59:37 -0800

```Hi Sam,

> I understand that there are prediction programs, but I'm interested in =
how
> they work.  I was wondering how these programs predict the flares,
> especially the part on magnitude. Also, do these flares have a really
> narrow ground track?

My program and others work based on "simple" three-dimensional
geometry.  The position of the observer is known, the position of the
satellite at any given time can be calculated, the orientation of the
satellite is maintained relative to its local-vertical and its flight =
direction,
and the location of the three MMA panels and their pointing directions
are also known.  The rest is vector math.

Magnitude estimates are based on two things:  main mission antenna
(MMA) projected area and the specular angle (aka mirror angle).  The
MMA projected area is a function of satellite range and the pointing
direction of the MMA.  This is just the inverse square law combined
with a cosine factor.  The dependency on specular angle has been
estimated from dozens of observations, and approximately follows a
logarithmic curve when magnitude is plotted versus off-axis angle.
This is typical for highly reflective surfaces.  The more mirror-like the
surface, the more steepily the reflected energy drops off with increasing
mirror angle.  (This is a very important quality for reflecting surfaces =
in
an optical instrument).

The MMAs are not perfect mirrors, and of course they weren't meant to be.
However, they are highly reflective and quite flat, so most of the reflect=
ed
energy goes in the specular direction.  A small fraction is reflected in =
a
cone centered on the specular direction, with energy dropping off rapidly
with increasing angle.

With IRIDFLAR, I've actually taken the calculation one step further by
modeling the radiance distribution of the solar disk (the sun is brighter
at the center of the disk than toward the limb).  That's why a flare
with a mirror angle of 0.0 degrees will be predicted to be brighter
than one with an angle of 0.2 degrees.  (Both point to a portion of the
solar disk, but the former will be measurably brighter).

The main source of error in magnitude predictions is the uncertainty in
the satellite orientation.  This uncertainty is quite large for newly
launched satellites, and thus magnitude predictions for these satellites
are pretty much worthless.  Even a small orientation error (e.g. 0.15
degrees) in a cross-track direction will shift the flare centerline more
than 4 miles for a typical geometry.  This would reduce a predicted
centerline flare's brightness by about 3 visual magnitudes.

This leads to your last question -- yes, the brightest part of the flare
track is quite narrow.  For midlatitudes, the instantaneous flare shape
on the ground is elliptical, with the long axis of the ellipse much =
closer
to east-west than north-south. A typical range to Iridium during a
flare is around 580 miles, with a typical elevation of around 55 degrees.
At that range, the 0.5-degree wide solar image is projected onto the
ground as an ellipse about 5 miles wide (roughly north-south) and 9 miles
long (roughly east-west).  This ellipse sweeps over the ground, moving
north-to-south or south-to-north, roughly following a line of longitude.
Thus, anyone within 4-and-a-half miles, east or west of the centerline,
is going to see a direction reflection of a portion of the sun's disk.

Outside the ellipse, the flare brightness drops off rapidly with distance
(i.e. mirror angle) since the intensity is now based on scattered light
from the MMA rather than a direct reflection.

Cheers!

Rob
```