[meteorite-list] How to compute NEXRAD radar hit times

From: Rob Matson <mojave_meteorites_at_meteoritecentral.com>
Date: Sun, 23 Jan 2011 14:15:26 -0800
Message-ID: <GOEDJOCBMMEHLEFDHGMMAELBEHAA.mojave_meteorites_at_cox.net>

Hi Jim,

> I find the use of Doppler Radar very interesting. I downloaded all
> the NOAA free stuff I could find to view Doppler from my laptop
> using their archived data. However, I can not produce the times
> that are specific to seeing Ash Creek. I was going to use Ash
> Creek as my test to determine if I was doing things right. I am
> not! And, I contacted the Phoenix NWS and they can not help with
> the times. The issue is the time. Only certain specific times are
> archived, so how or where did the data come from for the specific
> time?

You probably DO have the right data -- it's just that you have to
do a bit of extra work to compute the approximate times corresponding
to a particular radar sweep and target location. Strangely, there is
NO information on the web about how to do this -- likely because
meteorologists don't care very much about knowing the scan times
of a particular cloud front.

Here is how I do it. Mind you, I have no confirmation that my
technique is correct, but it is mathematically sensible and is
based on some logical assumtions:

1. The NEXRAD radar sweep rate (in azimuth) is constant.
2. The sweep direction is clockwise as viewed from above.
3. Each elevation scan takes the same amount of time to
    complete.
4. Each full scan starts at the lowest cut angle and ends
    at the highest cut angle, with transitions in cut angle
    occurring when the radar is transmitting due north.
5. That the time-tag of each file corresponds to the start
    time of the lowest cut angle, and that at that time the
    radar is pointing due north.

(I would love to have some confirmation that all of these
assumptions are correct -- Marc Fries can likely confirm or
correct me.)

One remaining uncertainty I have is what physically happens
with the radar when it generates two sweeps at the same cut
angle (e.g. 0.5-degrees and 1.5-degrees are nearly always
swept twice each.) In these cases, does the radar complete
two full revolutions for each cut angle, or are the double
sweeps really just a single sweep, but processed twice (in
software) at different sensitivities? I have a feeling it's
the latter, since I usually see very little movement in
scene features between scans at the same cut angle.

In either case, what you need to do is count up how many
sweeps are done for each full scan -- it's usually at
least 5, and can be quite a bit more when in storm mode.
Divide the time between consecutive full scans (typically
around 10 minutes) by the number of sweeps per scan.

Example:

Image #1 timetag: 02:12:06
Image #2 timetag: 02:21:56
5 sweeps per full scan

(2:21:56 - 2:12:06) = 09:50
09:50 / 5 = 1 minute 58 seconds (1.967 minutes)

This tells you that the radar sweep rate is:

360 degrees / 1.967 minutes = 183.1 degrees/minute, or
3.051 degrees/second.

Now, to determine the time of a specific radar "hit", all
you need is its azimuth (i.e. bearing from the radar),
which is displayed at the bottom of the Toolkit screen
when your cursor is over the hit.

Let's say you see your target in the second sweep out of
five (e.g. the 1.5-degree cut angle) and that it's at
azimuth 227 degrees. Using the above example, the time of
that hit is computed as follows:

02:12:06 + (1 sweep * 1.967 minutes) + (227/360 * 1.967 min)
= 02:12:06 + 1.967 minutes + 1.240 minutes
= 02:15:12

Try this with Ash Creek, and I think you should come very
close to the time that Marc Fries did.

Good luck!
Rob
Received on Sun 23 Jan 2011 05:15:26 PM PST