DEPARTMENT OF GEOSCIENCES NAME_________________________

SAN FRANCISCO STATE UNIV Meteorology 201

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**Synoptic Metr Homework 2 Key
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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 10^{5}m = - 4.0 X 10^{-3}

V_{g} = - [( 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^{-1}m-3.0 X 10^{-1})/3.0 X 10^{5}m = - 6.67 X 10^{-7}

V_{g} = - [( 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^{-1}m-3.0 X 10^{-1})/3.0 X 10^{5}m = - 12.0 X 10^{-7}

V_{g} = - [( 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^{-1}m-3.0 X 10^{-1})/3.0 X 10^{5}m = -1.0 X 10^{-6}

V_{g} = - [( 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^{-1}m-3.0 X 10^{-1})/3.0 X 10^{5}m = - 2.0 X 10^{-6}

V_{g} = - [( 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)