ATM OCN (Meteorology) 100

Answers for Homework 1

Summer 2004

Date Due: Thursday, 26 June 2004

The total maximum points were 60. Point distribution for each question noted below.


1a. The barometric pressure associated with one standard atmosphere at mean sea level is
           [you may round to nearest whole number]:

(5 pts - 1 each)

1 atmosphere of pressure is equivalent to: 

29.92 (or 30) inches of mercury

76.0 centimeters of mercury

14.7 (or 15) pounds per square inch (psi)

1013.25 (1000) millibar (mb)

34 feet of water
(note this answer is equivalent to approximately 10 meters of water, but feet were asked.)

1b. The lowest recorded sea level corrected pressure in the world was ________. [Please include units!]

Lowest pressure: 870 mb = 25.68 inches of mercury

    The highest recorded sea level corrected pressure in the world was ________. [Please include units!]

Highest pressure: 1083.8 mb = 32.01 inches of mercury

    The range between the record lowest and highest sea level corrected pressure (above)
     is approximately ________.

The range: Range = (High - Low) 
= (1083.8 - 870) mb = 214 mb
=(32.01 - 25.68) in = 6.33 in of mercury

(4 pts - 1,1,2)

1c. What is the weight exerted by the atmosphere upon the flat, horizontal roof of a 25-foot by 50-foot building? [Assume standard sea level conditions; English units may be used here]. Clearly show your work for partial credit!

From Pressure =  Weight / area, we can determine that Weight = Pressure x area 

Pressure = 15 pounds per square inch (approx.) 

Area = 25 ft x 50 ft = 1250 sq. ft. Since 1 sq. ft. = 144 sq. in. (count them - since 12 inches on each side of the square), then 1250 sq. ft. = 1500 x 144 = 180,000 sq. inches. 


Weight = 15 lb per sq. in x 180,000 sq. in. = 2,700,000 lb or 1350 ton

(If 14.7 psi were used, the weight would be 2,646,000 lb or 1323 ton) 

Note that units check too! 

While this answer may sound large, the roof does not collapse from the weight exerted by the atmosphere since the air pressure is pushing on the roof in all directions.

(5 pts)

1d. A football fan brought an aneroid barometer to Mile High Stadium in Denver (elevation of 1 mile) and made a reading of 835 mb. What would be the approximate sea level corrected pressure if we assumed that the pressure decreases at approximately 1 mb per 10 meters ascent through the atmosphere?

Mile High Stadium in Denver is 5280 ft above mean sea level (MSL), or 1600 m MSL. 

Since the air pressure is assumed to decrease at a rate of 1 mb per 10 m, the pressure at the stadium should be 160 mb less than at the mean sea level directly below the stadium. 

Because the observed station (or in this case, stadium) pressure was 835 mb, by descending to sea level, the pressure would increase, or at the stadium would be [840 + 160] mb or 995 mb

(4 pts)

How does this sea level pressure that you calculated compare with the standard sea level pressure?

The sea level pressure below Denver on this particular day (995 mb) is slightly less than a typical value of sea level pressure (1000 mb) and it is 18 mb less than standard sea level pressure (1013 mb)

(3 pts)

2. Current Weather on the Web (6 pts.)
This portion of the homework was designed to have you access current weather and climate information from a local National Weather Service Office on the Internet.  Any "reasonable answer" that fell within the range of values for the past week's weather in Madison was accepted.

3. Convert the following temperature readings:

41ºF = 5ºC = 278 K

-40ºC = -40ºF = 233 K 

258 K = -15ºC = 5ºF

Note: Be careful of signs! If the negative sign does not appear in your answer where appropriate, the answer is not correct.

(6 pts)

4a. The record highest temperature for Madison, WI was 107ºF (41.7ºC) on 14 July 1936, while the record low was -37ºF (-38.3ºC) on 30 Jan 1951. What is the range of Madison's extreme temperatures?

Range = (High - Low) = 107ºF - (-37)ºF = 144 Fahrenheit degrees

(1 pt)

4b. Compare these record temperatures and range with those of the United States and the world.
[Please include units!]

The following values were obtained from the links off the Lecture #3 (Temperature) page:
     For the United States:     "All-time temperature extremes by state "
and for the world: "observed extremes in temperature by continent (from NCDC)".


Record High 

Record Low 

Range = (High - Low)

United States 

134ºF or

-79.8ºF or
(includes Alaska) 

-69.7ºF or
(for lower 48 states) 

213.8ºF or
(includes Alaska) 

203.7ºF or
112.7ºC or
(for lower 48 states)


136ºF or

-129ºF or 

265ºF or 

(6 pts)

5. The National Weather Service at Madison reported the following information for individual days during this past January. The "normal" high and low temperatures for these days are also included and represent the 30-year averages for the 1971-2000 climatological interval.


Average Temperature

Average Temperature*




26 Jan 2004

[22 + 15]/2 = 19ºF

[25 + 9]/2 = 17ºF

28 Jan 2004

[10 + (-6)]/2 = F

[26 + 9]/2 = 18ºF

(12 pts.)

i.) Actual Heating Degree Day Units:

HDDU = [65ºF  - Average daily temperature]

26 Jan 2004: 65ºF  - 19ºF = 46 HDDU

28 Jan 2004: 65ºF  -  2ºF = 63 HDDU

ii.) Normal HDDU

26 Jan: 65ºF  - 17ºF  = 48 HDDU

28 Jan: 65ºF  - 18ºF  = 47 HDDU

iii.) How would the amount of energy required for space heating on each of those dates compare with that of the climatological (or "normal") average for the corresponding dates? Explain your reasoning.

The second day, 28 January 2004 would require more energy for heat.
Reason: Since 28 January 2004 was a day that was cold (daily average was +2ºF) or "below normal" temperature-wise (with normal at 17ºF), 63 HDDU were accumulated as compared to the "normal" of 47 HDDU.  Thus, more energy than normal would have to be consumed to heat your house. However, two days earlier (25 January 2003), the daily average of 19ºF  was slightly above "normal" (17ºF), meaning a "normal" amount of heat would be expected (other factors such as wind and sunlight that are not included in this computation would also affect the fuel consumption), since 46 HDDU were accumulated as compared with the typical 48 HDDU.

(4 pts.)

Latest revision: 27 June 2004 (2100 UTC)

Produced by Edward J. Hopkins, Ph.D.
Department of Atmospheric and Oceanic Sciences
University of Wisconsin-Madison, Madison, WI 53706

URL Address: aos100/homework/s04hmk01a.html

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