DEPARTMENT OF GEOSCIENCES                  NAME_________________________

SAN FRANCISCO STATE UNIV              Meteorology 201

 

Synoptic Metr Homework 3 Key

                          

You are provided the 500 mb analysis (without data plots) for 12 UTC 9 February 2005.  Note the station in southeastern Oklahoma (indicated by the square).  The line segment shown is a portion of the "n" axis (normal to the flow) at that location, with the segment length shown there (delta n) = 300 km.

 

1.  The geostrophic wind speed can be calculated from the expression

 

 

 

 

Part 1:

1. (30 points)

g = 9.8 m s-2

f= 10 X 10-5 s-1

∆z/∆n= -(5580m-5700m)/3.0 X 105m = - 4.0 X 10-3

Vg = - [( 9.8 m s-2)/10 X 10-5 s-1] [- 4.0 X 10-3] = 39.1 m s-1=141.1 km h-1

(39.1 m s-1) X 100 cm/m X 1 inch/2.54 cm X 1 foot/12 in X 1 nautical mile/6040 ft X 3600 s/h = 77.1 knots

 

2.  With a long arrow/streamline drawn right on the chart, locate the position of the fastest current (the polar jet stream). (10 points)

 

Part II:

1.*Note--Error in delta n...it was closer to 200 km.

Eastern Pacific (20 points)

g = 9.8 m s-2

f= 10 X 10-5 s-1

delta z/delta n= -(1.0 X 10-1m-3.0 X 10-1)/3.0 X 105m = - 6.67 X 10-7

Vg = - [( 9.8 m s-2)/10 X 10-5 s-1] [- 6.67 X 10-7 ] = 0.065 m s-1= .24 km h-1

(0.065 m s-1) X 100 cm/m X 1 inch/2.54 cm X 1 foot/12 in X 1 nautical mile/6040 ft X 3600 s/hr = .13 knots

Western Pacific (20 points)

g = 9.8 m s-2

f= 10 X 10-5 s-1

delta z/delta n= -(-1.0 X 10-1m-3.0 X 10-1)/3.0 X 105m = - 12.0 X 10-7

Vg = - [( 9.8 m s-2)/10 X 10-5 s-1] [- 12.0 X 10-7 ] = .13 m s-1= .48 km h-1

(.13 m s-1) X 100 cm/m X 1 inch/2.54 cm X 1 foot/12 in X 1 nautical mile/6040 ft X 3600 s/hr = .26 knots

For corrected delta n

Eastern Pacific (20 points)

g = 9.8 m s-2

f= 10 X 10-5 s-1

delta z/delta n= -(1.0 X 10-1m-3.0 X 10-1)/3.0 X 105m = -1.0 X 10-6

Vg = - [( 9.8 m s-2)/10 X 10-5 s-1] [-1.0 X 10-6 ] = 0.098 m s-1= .35 km h-1

(0.098 m s-1) X 100 cm/m X 1 inch/2.54 cm X 1 foot/12 in X 1 nautical mile/6040 ft X 3600 s/hr = .19 knots

Western Pacific (20 points)

g = 9.8 m s-2

f= 10 X 10-5 s-1

delta z/delta n= -(-1.0 X 10-1m-3.0 X 10-1)/3.0 X 105m = - 2.0 X 10-6

Vg = - [( 9.8 m s-2)/10 X 10-5 s-1] [-2.0 X 10-6 ] = .20 m s-1= .71 km h-1

(.13 m s-1) X 100 cm/m X 1 inch/2.54 cm X 1 foot/12 in X 1 nautical mile/6040 ft X 3600 s/hr = .39 knots

(10 points)