METR 2413 QUIZ 3                                          NAME_______________________________

Answer all questions briefly.    DUE in 15 minutes!

1. You, being the good computer user (with userid somuser), have just issued the ps and ipcs commands and they return the following:
 
 somuser  8013  7974  0 04:19:02 pts/0    0:00 -tcsh
 somuser  8050  8013  0 04:19:07 pts/0    0:02  gplt
 somuser   433   636  0   Feb 18 ?        0:05  ps
 
and
 
Message Queues:
q    1024450   0xbc711e   -Rrw-rw-rw- somuser    ugrad
q    1275751   0x2b904c8  -Rrw-rw-rw- somuser    ugrad
 
In one sentence, explain how you will clean up after yourself before logging off.
 
I would run “kill –9 8050” and “kill –9 433” to remove the two hung processes, and then run
“rm-ipcs somuser” to clear the message queue.  (4 points)

 

2. What directories (folders) would a user of the SoM computing facilities find surface, upper air, and gridded data for use in gempak programs?

/data/gempak/surface/ - Surface data

/data/gempak/upperair/ - Upper air data

/data/gempak/model/ - Gridded data

(4 points)

 

3. Both visible and infrared satellite images show clouds as bright objects and the surface as dark.  Briefly explain how this occurs in terms of visible and infrared radiative transfer.

Visible imagery is derived from reflected solar radiation from the Earth-Atmosphere system.  The intensity of image brightness depends on the albedo of the underlying surface or cloud.  Clouds are more reflective than the surface of the Earth, and will therefore appear brighter. (2 points)

 

IR imagery is derived from terrestrial radiation emitted from the Earth-atmosphere system. Areas with low emitted energy have low brightness values.  However, by convention, emitted energy and brightness values are inverted for thermal IR imagery.  Therefore, the colder clouds will appear brighter on the image, despite emitting less energy than the warmer Earth’s surface. (2 points)

 

 

4. A typical WSR-88D base reflectivity image shows reflectivities ranging from -30dbZ to +30dbZ.  What is the ratio of the maximum reflectivity to the minimum reflectivity in powers of ten that is shown in such an image?

dBZ = 10log10Z or Z=10(dBZ/10)

Therefore, Z+30=1000mm6m-3 and Z-30=0.001mm6m-3 and the ratio of max/min = 1000/.001 = 106 (4 points)

5. Using the following Doppler radial velocity image( copied black and white below, color on the screen), what is the wind direction at the surface? Is the wind veering or backing with height?  What is the approximate wind direction at the height corresponding to the first range ring?  (up is the north direction).


 

http://apollo.lsc.vsc.edu/classes/remote/lecture_notes/radar/doppler/sd_back.html

Surface: S  (1 point)

The wind is backing with height (1 point)

First Range Ring: ESE (SE and E also accepted) (1 point)

And (1 point) just for trying.