METEOROLOGY LAB TUTOR

Lab tutor information from Jeff Haby.

FIRST QUARTER EXERCISE

LAB 1, Pages 1 - 7

Problem #1: The problem uses the assumption that the atmospheric mass will be cut in half with each 5.6 km rise above the surface. Below are how the first 3 bottom rows should look.

```Height (km)     % above
11.2             25%
5.6             50%
sfc            100%
```
Problem #2: The y-axis is the vertical axis on the page while the x-axis is the horizontal axis on the page. Plot increments of 10% along the bottom x-axis (% above value) for each box and plot increments of 2 km along the far left y-axis for each box. The SW corner of diagram is point (0,0). You will then plot 5 dots. The first dot will be at (100%, 0 km) which is the extreme SE corner of diagram, the second dot will be at (50%, 5.6 km), the third dot will be at (25%, 11.2 km) and so forth. After plotting the 5 dots, connect them with a smooth curve. It should look like a slide when you are done.

Problem #3: Change 10% to 100mb, change 20% to 200 mb, change 30% to 300 mb, and so forth

Height/pressure formula: Use a calculator to make sure you can do the example calculation. 0.5^1.58 is 0.5 to the 1.58 power. The order of operation on a scientific calculator is to type in 0.5 then press the ^ key, then press 1.58 then enter. Resulting number is rounded to 0.33

Problem #4: Here is how to find the solution to the first problem referring to the height of a cruising jet.

Method 1: In problem #1 you found that at a height of 11.2 km there is 25% of the atmosphere above you. This will mean also there is 250 mb of air pressure at this level.

Method 2: plug numbers into formula and solve: Pressure = 1000 mb * (0.5)^(11.2/5.6)

Pressure = 1000 * 0.5^2 = 1000 * 0.25 = 250 mb

Problem #5: Convert 21% to a pressure in the same fashion 78% Nitrogen is converted to 780 mb. The total pressure of all gases in the troposphere is about 1000 mb. Since Nitrogen is 780 mb, it is 780/1000 = 78% of the volume.

Problem #6: Example using Mt. Everest. The top of Mt. Everest is a 8.85 km elevation. In the example in the lab book on the left side of page 3 they already solved that the pressure at this elevation is 330 mb. Thus this is 33% of the sea level pressure. Take answer in #5 and multiply it by 0.33 to get the pressure of Oxygen at the top of Mt. Everest.

Problem #7: A lapse rate is referring to how the temperature changes with height. See figure 1-2 in the lab book and notice how the temperature is changing with height when going from the 20 to 30 km level. Is the temperature warming or cooling? This is answered by looking at the temperature lapse rate on the diagram and knowing as stated in the question that ozone is a good absorber of ultraviolet radiation.

Problem #8: The Lapse rate is 6.5 C/km. Since the temperature is 8.5 C at 1000 m then it will be 6.5 C cooler at 2000 m. Temperature at 2000 m = 8.5 - 6.5 = 2 C. Use this same process to fill in the remaining blanks.

Problem #11: An inversion is a temperature increase with height. Near the surface circle the profile that has a temperature increase with height. Inversions are very stable layers.

Problem #12: Notice on the temperature profile in the upper troposphere where the temperature no longer decreases with height.

Problem #13: Thickness means the vertical distance between two pressure levels in the atmosphere. The thickness of the troposphere is the vertical distance from sea level to the tropopause. Thickness is a function of the air temperature. Warmer air is less dense and thus takes up a larger volume. Remember that thickness in tropical regions will be more than in polar areas because warmer air takes up a larger volume.

Problem #14: Pressure at 2 km minus pressure at 4 km

Problem #15: Pressure at 8 km minus pressure at 10 km

Problem #16: Excellent question. Remember the relationship between pressure and height is a curve and not a straight line. See page 23 in hardback textbook. Notice pressure decreases more rapidly with height near the surface as compared to higher aloft.

Problem #17: Pressure at 2 km minus pressure at 4 km

Problem #18: Denser air sinks and less dense air rises. Notice when you boil water the air and vapor from the water rises up to the ceiling and when you open a refrigerator/freezer the cold air sinks down to the floor.

Problem #19: Remember that the atmospheric thickness is less in cold air. Thus the pressure levels are more compacted in cold air. Pressure changes more rapidly with height in cold air as compared to warm air.

See page 4 in lab book for average temperature profile of troposphere. See page 23 in textbook for relationship between pressure and height. The relationship between density and height is a similar relationship to that of pressure and height.

Review previous answers and tutor information on how thickness varies.

Compare how much pressure change occurs when going from the surface to 600 km vertical upwards to that of the typical horizontal pressure change at the surface between the center and 600 km from the hurricane center.

LAB 2, Pages 9 - 19

The lab manual donkeyed-up page 11 (Figure 2-4). D should be what C is showing. C should be what D is showing. A should be sun on the horizon. B should be 36.5 degree angle.

Problem #1: June 21. Sun is on left side of Earth. 23.5 degree tilt is directly toward sun. Sun is directly striking at 23.5 degrees N. Noon sun angle is 90 degrees here. For every degree of latitude away from 23.5 N, the sun angle lessens by 1 degree. At 30 N, this is 6.5 degrees away from 23.5 N, thus the noon sun angle at 30 N = 90 - 6.5 = 83.5 degrees.

Problem #2: Add the degrees from equator to 40 N to that of the number of degrees from the equator to 23.5 S.

Problem #3: Figure 2-6 is at the time of the winter solstice. Solar elevation is the same as the noon solar sun angle. Above 66.5 N there is complete darkness. The sun is on the horizon at 66.5 N at noon. Thus, from every degree south of 66.5 you are in Northern Hemisphere then the sun will be that number of degrees above the horizon at solar noon. For example at 60 N, the noon sun angle is 66.5 - 60 = 6.5 degrees above the horizon.

Problem #4: Solar declination is the latitude at which the sun is directly overhead at solar noon. A through D are straight forward since those are on critical dates. March 21 is the equinox thus solar declination is at the equator. The most challenging one will be E. E will not show on the timed test since date is variable.

Here is an example for solving E using the date of August 27. The formula to solve is given on page 14. The number of days to the nearest equinox (September 23) is 27.

Declination = 23.5 * SIN(27) = 10.7 N. At the latitude of 10.7 N the sun is directly overhead at solar noon on August 27.

Problem #5: Here is how the March 21 sun angles are determined. On March 21 the solar declination is at the equator. This means that at the equator the sun angle is 90 degrees (directly overhead) at solar noon. For every degree north of the equator the noon sun angle will be one degree less. Since New Orleans is at 30 N then the noon sun angle is 90 - 30 = 60 degrees. For Helsinki it is 90 - 60 = 30 degrees.

Problem #6: This question will not be on timed test since answer is variable.

Problem #7 - #8: These questions will not be tested on since they are optional questions.

Problem #9: They were nice enough to do half of them for you in the lab book. Go through those examples.

Problem #10: A lower sun angle distributes energy over a larger area and thus the energy is not as concentrated.

Problem #11: Said another way, which latitudes (high or low) has the greatest range of daylight hours during the year?

Problem #12: Graph reading problem. Here is how to complete the answers for the June solstice at 30 N and 60 N. Start at June 21 on graph, go vertically up until you hit the dashed line indicating 30 N. Then read number of daylight hours across horizontal axis. Number of daylight hours is 14 hours at 30 N and a little over 18 hours at 60 N.

Problem #13: This will be the latitude that has the greatest difference of daylight hours between winter and summer.

Problem #14: For this question answer solstice or equinox for each question. Greatest change will be where slope on graph is greatest (slope closest to vertical). Smallest change will be where slope of graph is least (the top of peak and trough area).

Problem #15: Notice the sun takes a more extreme arc shape across the sky at higher latitudes. Notice also day length is more variable during the course of the year at higher latitudes.

Problem #16: Example for March 21. At 30 N the amount of radiation is 1186 W/m^2. At 60 N the amount is 685 W/m^2. This gives a difference of 1186 - 685 = 501 W/m^2.

Problem #17: As the sun angle decreases there is an accelerating loss of energy over a surface.

Problem #18: Same line of logic as #17. For example, when the sun angle changes from 85 to 80 degrees there is not as much of a dramatic change in energy reaching the surface compared to the sun angle changing from 15 to 10 degrees. As the sun becomes closer to the horizon the amount of energy reaching the surface rapidly decreases. In the winter the sun angles are very low in the high latitudes and thus the amount of energy reaching the surface is far less than in lower latitudes.

Problems #19 - #26: Do not do these problems

LAB 3, Pages 23 - 28

Problem #1a: plug 6,000 K into Stefan-Boltzman equation and then solve

Example Problem: What is E when 5,000 K is plugged into Stefan-Boltzman equation:

E = (5.67 * 10^-8) * (5,000 K)^4 = 35,437,500 W/m^2

Problem #1b: plug 300 K into Stefan-Boltzman equation and then solve

Problem #2a: divide sun temperature by earth temperature from problem #1

Problem #2b: 4th power means to multiply the same number by itself 4 times. i.e. 6^4 = 6 * 6 * 6 * 6 = 1,296

Problem #2c: Divide the answer from 1a by 1b and see if this is equal to the answer from 2b.

Problem #3: Example problem: What is wavelength of maximum emission using a temperature of 1,000 K?

max wavelength of emission = 2898 micrometers*K / 1,000 K = 2.898 micrometers wavelength

Here is a tip you will really like: Problem #1 and #3 are already done for you on page 48 of the textbook

Problem #4: Take wavelength of maximum emission for the sun found in #3. Use table 3-1 to see the type of energy that corresponds to that wavelength

Problem #5: This question will not be tested on since answers are variable

Problem #6: Think of ways the atmospheric can filter out some of the incoming solar radiation before it can make it to the surface.

Problem #7: SW with the arrow pointing up is Shortwave outgoing (radiation from the sun that is reflected off Earth's surface). SW with the arrow pointing down is Shortwave incoming (radiation from the sun that has made it to Earth's surface). 1 pm is hour 13 on the graph. Go vertically up from hour 13 on the graph and read off the SW values. Use formula to determine albedo.

Problem #8: The Earth's emitted radiation is LW (Longwave) outgoing. This value will depend on Earth's surface temperature. Look at graph to see when LW outgoing is highest and then explain why it is highest at that time.

Problem #9: Same line of logic as #8

Problem #10: Explain why the Earth's surface is heated more than air at the 2 km level.

Problem #11: Clouds emit longwave radiation, thus on a cloudy day more longwave radiation will be emitted toward the Earth's surface. On a clear day more shortwave radiation from the sun makes it to the surface and since there are no clouds there is less incoming longwave radiation.

Problem #12: Same line of logic as #8 and #9

Problem #13: Find on graph where longwave incoming decreases most dramatically.

Think of the relationship between cloud cover and the amount of incoming longwave radiation.

Problem #14 - #25: Do not do these problems.

Review Questions (P. 30): They will both vary, however, the changing sun angle during the course of the year and the fact that during part of a 24-hour day there is no solar radiation causes incoming shortwave radiation to vary the most.

On a cloudy night there is more incoming longwave (clouds act like a blanket) and this causes temperature to be warmer than they would if the skies were clear on that night. During the daylight hours, no clouds allows more warming from incoming shortwave radiation.

Do not do lab #4

SECOND QUARTER EXERCISE

LAB 5, Pages 41 - 51

Problem #1 - #9: Do not do these problems

Problem #10: There is a latent heat process that occurs. Give the name of this latent heat process and explain how it accounts for the water remaining at the same temperature although heat continues to be added to the water.

Problem #11: As an example, think of Mississippi and southern Arizona in the summer. Both states are at a similar latitude but Arizona is hotter in the summer. Mississippi though is more humid. Explain through a latent heat process how vegetation and moisture influence the surface air temperature.

Problem #12: First, turn to page 42 and write down the number of calories that are needed for the condensation process. This value is calories per gram although page 42 just says calories. Since the number of grams is given in the problem all that is needed is to multiply the number of grams times the number of calories in the condensation process per gram.

Problem #13: Here is how the second row should look. Water to water vapor is evaporation. Evaporation is a cooling process (heat consumed, or said in another way latent heat is absorbed). An example is evaporative cooling when it starts to rain.

Problem #14: Here is how to do the top row where the temperature is 14 C with a mixing ratio (actual) of 5 g/kg. The saturation mixing ratio graph is on table 5-2 on page 44. When the temperature is 14 C the saturation mixing ratio is 10.140 g/kg by using that table. The relative humidity formula is the actual mixing ratio divided by saturation mixing ratio times 100%. The RH = 5/10.14 * 100% = 49.3%. Problem #15: Hint: Remember that cold air has a low moisture capacity. When that air is heated the air has a capacity to have much more moisture at the same time it has a small amount of moisture. This is the time of the year when you may need chap stick, hand lotion, static guard and a humidifier.

Problem #16: It is the reverse of the previous problem. As air is cooled toward the dewpoint then the relative humidity increases.

Problem #17: Here is how to find the saturation mixing ratio and actual mixing ratio of A. The temperature is given as 14 C. Using table 5-2 on page 44, the saturation mixing ratio at 14 C is 10.14 g/kg. Since the relative humidity is 90%, multiply 0.9 times the saturation mixing ratio of 10.14 g/kg to get the actual mixing ratio. 0.9 * 10.14 = 9.126 g/kg actual mixing ratio. Next, determine values for B through E and then rank them as instructed. Ranking will be based on the actual mixing ratio.

Problem #18: Relative humidity is temperature dependent and can be a useless value unless the temperature is also known. The one certain thing about relative humidity is that a value near 100% indicates air near saturation and a low values indicates air with a capacity to evaporate more moisture into the air. However, the relative humidity value by itself does not indicate the actual amount of moisture in the air.

Problem #19: Dry-bulb temperature means the same as temperature. Wet-bulb temperature is new temperature after maximum evaporative cooling. Wet-bulb depression is the temperature difference between dry-bulb and wet-bulb. A depression is NOT a temperature but rather a temperature difference! For example, if the dry-bulb is 35 C and the wet-bulb is 31 C then the wet-bulb depression is 4 units of C. For a depression write it as "units of C" or "C degrees difference" (instead of degrees C) so to not confuse it with a temperature.

To find relative humidity use graph on page 47. For example, if dry bulb is 0 C and wet-bulb depression is 5, then the relative humidity is 11%. The intersection between row and column is the relative humidity value.

Problem #20 - #21: Do not do these problems

Problem #22: Use figure 5-4. Start at a mixing ratio of 17 g/kg on vertical axis. Go across horizontally until graph is intersected. At intersection point draw a line straight down to the bottom. The temperature at the bottom is the dew point.

Problem #23: Saturation mixing ratio can be found by knowing the temperature and using table 5-2 on page 44.

Relative humidity is (mixing ratio / saturation mixing ratio) * 100%. Since relative humidity and the saturation mixing ratio are known, solve for mixing ratio. Mixing ratio = w; saturation mixing ratio is ws; RH = 62%

0.62 = w/ws

w = 0.62 * ws

In formula above, plug ws into formula to solve for w.

Finding dewpoint is same process as in #22. Use mixing ratio (NOT saturation mixing ratio).

Problem #24 - #31: Do not do these problems

Review Question Comments: Define vapor pressure, mixing ratio, dewpoint and relative humidity. Discuss strengths and limitations of each.

Soft drink question: It is a similar process to how dew forms.

Jet question: Similar concept to problem #15 but more extreme.

Sweating question: Hint: latent heat absorption due to evaporative cooling

LAB 6, Pages 53 - 62

Problem #1: For example, when using Table 6-1 on page 54, when the temperature is 22 C then the saturation mixing ratio is 16.963 g/kg

Problem #2: Multiply saturation mixing ratio and decimal form of relative humidity to get actual mixing ratio. Decimal form of 75% is 0.75

Problem #3: Make sure you are plotting actual mixing ratio and not saturation mixing ratio. Both points will be under the curve and this represents unsaturated air.

Problem #4: Find average temperature and average mixing ratio. Plot this point on graph. It is the point plotted by using the answers to #5 and #7

Problem #5: This is the average temperature of the two air samples

Problem #6: This is asking what is the saturation mixing ratio of the answer you got to #5. Use Table 6-1 on page 54.

Problem #7: This is the average of the mixing ratio values you found in #2

Problem #8: This is answer to #7 divided by answer to #6. Value will be greater than 100% (supersaturation). This can be explained by reading the first sentence on question #9.

Problem #9: Hint: relative humidity

Problem #10: Since the dry adiabatic lapse rate (DALR) is 10 C/km, this rate is the same as 1 C / 100 m. For each 100 m the air rises it will cool by 1 degree Celsius. The surface is 35 C. After rises 100 m the temperature cools to 34 C. After rising 200 m the temperature cools to 33 C. etc.

Problem #11: Since the DALR is 10 C/km, the cooling will be 5 C per half kilometer. Table 6-2 is in half kilometer increments. Parcel "A" starts at the surface at 28 C. After rising a half kilometer the temperature will cool to 23 C. The value at 1.5 km is the temperature at the LCL. The temperature at the LCL will be 13 C. Above the LCL the cooling will be wet adiabatic and the problem gives the wet rate as 5 C/ kilometer, which is 2.5 C per half kilometer. Temperature at 2 km level will be 10.5 C. Temperature at 2.5 km level will be 8 C. etc.

Problem #12: Same process as #11 EXCEPT the wet rate is 7 C/km which is 3.5 C per half kilometer.

Problem #13: Hint: it deals with a latent heat process

Problem #14: Here is a critical concept that may not seem intuitive. Even with the same amount of moisture in a parcel of air, the dewpoint decreases as air rises (and increases as air sinks). The reason deals with thermodynamic properties of air and how they change when pressure changes. In unsaturated air the dewpoint decreases at a rate of 2 C/km. For this problem you label how the temperature and dewpoint decrease as unsaturated air rises in 0.5 km increments. The surface temperature is 30 C with a dewpoint of 14 C. The temperature cools at the DALR. After rising 0.5 km the temperature cools to 25 C. Since the dewpoint lapse rate is 2 C/km, the lapse rate is 1 C per half kilometer. After the parcel rises 500 m the temperature is 25 C with a dewpoint of 13 C. Use this same process to fill in the rest of the temperatures and dewpoint for the remaining heights.

Problem #15: Here is how to do example A. Surface temperature is 32 C with a dewpoint of 20 C. After this unsaturated air rises 1 km the temperature will have cooled to 22 C and the dewpoint will have decreased to 18 C. After rising another half kilometer the temperature will have cooled to 17 C and the dewpoint will decrease to 17 C. This works out nice. LCL is exactly 1.5 km. Use the same type of process to find LCL for example B.

Problem #16: The LCL is the cloud height. Note which example has a higher cloud height and why that is the case.

Problem #17: Do not do this problem.

Problem #18: Keep the following in mind: In unsaturated air rising air cools at DALR. DALR is 10 C/km which is 5 C per half kilometer (500 m). In unsaturated air rising air changes dewpoint at a rate of 1 C DECREASE per half kilometer. The LCL is the height where the temperature and dewpoint first become equal as air rises. Above the LCL the temperature and dewpoint change at the same rate which is the wet adiabatic lapse rate. The wet rate is given as 5 C per kilometer in this problem (2.5 C per half kilometer). When air sinks it warms at the dry adiabatic rate and the dewpoint INCREASES at a rate of 2 C per kilometer.

Problem #19: This is the same height as the LCL

Problem #20: Notice the surface temperature and dewpoint are different on the windward and leeward sides. The temperature is warmer on leeward side due to the latent heat of condensation release that occurred on windward side above the LCL. Dew point is less on leeward side since some moisture was lost to condensation on windward side.

Problem #21: The side with condensation is the cloudy side.

Problem #22-#23: Fill in table 6-3. Height is in 0.5 km increments. DALR is 5 C per half kilometer. Dewpoint lapse rate is decrease of 1 C per half kilometer in unsaturated air. Wet adiabatic rate is given as 5 C/kilometer which is 2.5 C per half kilometer. LCL is height at which rising unsaturated air first becomes saturated. Dewpoint lapse rate and wet adiabatic lapse rate are the same in saturated air. Notice the values on table are filled in at 4 km level and above. You will know if your values you filled in make sense if your values merge correctly which the values given at 4.0 km.

Problem #24-#25: Plot the two graphs. Stable layer is where parcel is colder than environmental temperature. Unstable layer is where parcel is warmer than environmental temperature. Convective cloud development is possible within unstable layer (assume convective clouds will develop in unstable layer for this problem).

Review Questions: Rising air cools to dewpoint and then condenses. In an unstable atmosphere the theoretical parcel temperature is warmer than the actual (environmental) temperatures. Basically, when warm air is placed within cold air the warm air will rise.

LAB 7, Pages 63 - 68

Problem #1: RH = (vapor pressure / saturation vapor pressure over water) * 100%

Problem #2: Hint: phase change

Problem #3: RH = (1.72 mb / saturation vapor pressure over ice) * 100%

Answer will be greater than 100%, representing supersaturation relative to ice.

Problem #4: Hint: Bergeron Process

Problem #5: Take the 0.025 mm radius and cube it (which means multiply that number by itself 3 times). For example, 5 cubed is equal to 5 * 5 * 5 = 125.

Cubing 0.025 mm will produce a small number. To get final answer take this small number and multiply by PI then multiply by 4/3. PI = 3.14156. Final units are volume units of mm^3 (millimeters cubed).

Write the final answer in scientific notation. Below is scientific notation tutor information:

Scientific notation is used as a simpler way to write very big and very small numbers
The number is written using a coefficient and a base 10 exponent
The exponent determines the number of spaces the decimal point is moved while the sign of the exponent determines the direction the decimal point is moved
Exponent sign is positive if number is bigger than 1 and negative if number is smaller than 1
A number in scientific notation will have one number to the left of the decimal point. This number is called the coefficient and it will have a value between 1 and 10
678,500,000,000,000.0
In scientific notation is written as 6.785 * 10^14
Decimal has been moved to the left by 14 spaces
6.785 is called the coefficient. 10^14 is called a base 10 with an exponent of 14
0.00000000000045
In scientific notation is written as 4.5 * 10^-13 (note the negative sign indicating a small number)
Decimal has been moved to the right by 13 spaces

Problem #6: First, find volume of a 2.5 mm radius drop. Second, take this answer and divide by the answer to problem #5 to get the "how many times greater" answer. It will be a large number.

Problem #7: Area = PI*(radius^2). The radius^2 is the radius squared. Squaring is taking a number and multiplying it by itself. For example, with a radius of 5, radius^2 = 5 * 5 = 25. The units will be millimeter squared (mm^2) for this problem.

Problem #8: First, solve for area using 2.5 mm radius. Next, take that answer and divide by the answer to problem #7 to get the "how many times greater" answer.

Problem #9: Basically the question is asking which grows faster: cubing a number as radius grows or squaring a number as radius grows.

Problem #10: Use table 7-1. Assume drop does not evaporate. Distance is known as 2,000 m. The fall rate of a 50-um drop is on table 7-1. Formula: distance = rate * time. Solve for time. Express final answer in hours/minutes/seconds.

Problem #11: Same process as problem #10 except drop size is different. Express final answer in seconds.

Problem #12: Use a combination of table 7-1 and table 7-2 to determine answer.

Problem #13: Use figure 7-4. Cloud thickness is labeled on bottom. Draw a vertical line from a cloud thickness of 1. A low cloud base is 0 - 700 m. Where the vertical line intersects the low cloud base, read off the probability of drizzle and then rain on vertical axis of graph.

Problem #14: Same process as problem #13 but using a different thickness

Problem #15: More time for droplets to grow, more moisture to fall through

Problem #16: a. Same process as problem #13 but using a different thickness

b. Same process but use the higher cloud base

Problem #17: Hint: see table 7-2 on page 66

Review Questions: See table 7-1, 7-2 and 7-4 when formulating answer.

THIRD QUARTER EXERCISE

Lab 8, pages 69 - 88

Problem #1a: It will be where isobars are packed closest together

b: Pressure Gradient Force points directly from higher toward lower pressure

c: It will be where isobars are the most spread out

Problem #2: Do not do this problem

Problem #3-#6: Your line should deflect and curve to the right of the path of motion for Northern Hemisphere and deflect to the left for Southern Hemisphere.

Problem #7: Your rocket should deflect to the right of the path of motion. The rocket starts at the North Pole and moving on the 120 degree longitude but 4 hours later it ends up on the 60 degree longitude line.

Problem #8: Rocket deflects toward left

Problem #9: Rocket deflects toward right and travels a greater distance than it did in problem #7

Problem #10: Compare problem #7 to problem #9 to develop conclusions

Problem #11: It is confusing how this problem is set up so go ahead and skip it. However, what you need to know is that the magnitude of the Coriolis is greater at higher latitudes.

Problem #12: To help answer this look at figure 1-23 on page 23 of the hardback textbook. Notice the relationship between pressure and altitude is not linear.

Problem #13: This question and the problems related to it are very important concepts in meteorology. This is the basis for how the pressure-height charts are developed that you see on the forecast models. On figure 8-11 draw the 500 mb and 300 mb lines using the height data from figure 8-10 just as was done for the 850 mb line.

Problem #14: Colder air is denser than warmer air. Thus, in colder air the pressure surfaces will be closer together. You will be able to see this relationship by looking at the data you plotted on figure 8-11. Also see the data in figure 8-10 and notice the temperatures at both locations at the various elevations. The pressure is going to decrease more rapidly with height at a location where the pressure surfaces are closer together. Colder air = denser air = height surfaces closer together = thickness is less. Notice the vertical distance from the 850 mb level to the 300 mb level is less in a colder location.

Problem #15: Notice the thickness between height surfaces is greater in the warmer location.

Problem #16: Horizontal pressure gradient force points from higher toward lower pressure.

Problem #17: Notice the slope of each of the three lines on figure 8-11. The line with the most slope represents the height with the pressure horizontal pressure gradient force. This question also answers why the jet stream is located in the upper troposphere.

Problem #18: Winds are stronger where the pressure gradient force is strongest.

Problem #19: PGF points from higher heights toward lower heights (for same reason PGF points from higher toward lower pressure). Coriolis is to the left of the path of motion in S. Hemisphere. Wind direction (geostrophic wind vector) is determined by knowing air initially heads toward lower heights but is then deflected to the left. A geostrophic wind is parallel to the height contours.

Problem #20: PGF points toward lower pressure. With the low, the length of the Coriolis vector is shorter than the PGF vector since both the centrifugal and Coriolis are adding together to balance the PGF. With the high, the length of the Coriolis vector is longer than the PGF vector since the Coriolis is balanced by both the PGF and centrifugal vectors. Centrifugal is the apparent force you can feel when you turn a corner in a car and feel your body being pushed away from the center of rotation. An important point to make is that when wind speed changes and the pressure gradient remains the same, then the PGF can not change but the Coriolis can change. This is the basis of the question.

See the following website more great help on this topic: http://weather.ou.edu/~metr4424/Files/CurvedFlow.pdf

Problem #21: The wind speed is a function of the length of the Coriolis vector.

Problem #22: The website given above in #20 will answer and explain this question.

Problem #23: The BIG difference between what was done in problems #20-#22 compared to figure 8-14 is that in figure 8-14 the contour spacing is NOT equal. In figure 8-14 the wind speed will be highest where the height contours are closer together.

Problem #24: The idealized problem has an equal PGF, but in reality the contour spacing varies and will be more closely spaced in a trough.

Problem #25: Friction changes the magnitude of the Coriolis force

Problem #26: PGF points directly from higher toward lower heights. Coriolis acts to the right of the path of motion (directly to right of wind vector drawn on figure). Friction force acts in opposite direction of wind vector drawn on figure.

Problem #27: Winds flow counterclockwise around low pressure in Northern Hemisphere and clockwise in Southern Hemisphere. Winds flow clockwise around high pressure in Northern Hemisphere and counterclockwise in Southern Hemisphere. Winds converge toward low pressure and diverge away from high pressure.

Problem #28: This is at the surface since wind vectors are crossing isobars. Go by the same process as problem #26.

Problem #29: See figure 8-21 and explain why and how sinking air produces fair weather and rising air produces clouds and precipitation.

Review Questions: The PGF initiates the wind

See figure 8-17 for circulation around high and low pressure systems at surface. The process is similar aloft except the flow is more parallel to contours.

See figure 8-21 to answer the vertical motion question

Lab 9, pages 89 - 104

Problem #1: Study figure 9-1 on page 91. Keep the following in mind:

a. Make sure to label units on temperature to distinguish between a Fahrenheit and Celsius temperature

b. The average surface barometric pressure is 1013 mb. When pressure is coded the leading 10 or 9 is left off. Your job is to add the leading 10 or 9, whichever is most realistic. In other words, either add 10 or 9, whichever will make it closest to the average of 1013 mb. The decoded pressure is in 10th of millibars. For example, for the first problem in the upper left a value of 024 is given. The pressure is either 1002.4 or 902.4. 902.4 is unrealistic since that would be the surface pressure in an extremely powerful hurricane. The realistic answer is 1002.4 mb.

Problem #2: Connect the points that have equal temperature and label that contour.

Problem #3: Before doing this problem take a pencil and trace over the dashed lines on figure 9-3 on page 93. Understand why the dashed line is going where it is going. Make sure to fully understand interpolation and how it is done on a chart. Carefully read the information given on pages 92 and 93.

P. 94: The temperature is the value in the upper left corner and the dewpoint is below that value. It is a good idea for the first contour to go through the center of the data. Using an isotherm of 50 F would be a good starting contour. Once you have the first contour drawn in makes drawing the rest of them easier.

Problem #4: Same process as #3 but make the dewpoint contours dashed

Problem #5: The arrow over Kansas is an example. Arrow is showing the direction in which the wind is coming from (from the NW in that example). Draw several arrows on the diagram to note the wind direction over the entire area.

Problem #6: See page 235 in the hardback textbook

Problem #7: Follow stated directions. Keep the following in mind:

a. Front is drawn at the leading edge of air mass change

b. Review previous questions in exercise for drawing and labeling isobars

c. Low is located where barometric pressure is lowest. Review rules for decoding pressure value

d. If circle is completely darkened then that location has greater than 75% cloud coverage

Problem #8: The heights are in meters. Write down the highest and lowest height values that are on the chart.

Problem #9: Read the following webpage for information on interpreting thermal advection:

http://www.theweatherprediction.com/habyhints/254/

Problem #10: Trace over the 0 C isotherm

Problem #11: The heights are in meters. Write down the highest and lowest height values that are on the chart.

Problem #12: Take two adjacent contours. Subtract the smaller value from the larger one to get the contour interval. In other words, how many meters does it change when going from one contour to the next.

Problem #13: The 500 mb winds are the "steering wind" for low pressure systems.

Problem #14: Same type of process as problem #12

Problem #15: A triangle is 50 knots. 60 knots is a triangle with a full line next to it. 70 knots is a triangle with two full lines next to it. Shade areas with winds of 70 knots or greater.

Problem #16: Same as problem #15, circle the two shaded areas

Review Questions: Show position of fronts, structure of low pressure systems, air mass contrasts, and wind patterns

FOURTH QUARTER EXERCISE

Lab 10, pages 105 - 118

Problem #1: mT (maritime tropical) is warm and moist air; cP (continental polar) is cool and lower dewpoint air

Problem #2: Winds will be anywhere from northwesterly to southwesterly behind the cool front. Winds will be from the south to southwest in the warm sector. North of the warm front winds will tend to be from an easterly type direction.

Problem #3: See how the wind direction changes when going from Monday to Tuesday to Wednesday and pick the location that is most realistic to wind direction change. The four blanks below question 3 will be filled in with either location A, B, C or D

Problem #4: The pressure on Nov. 9 12Z minus the pressure on Nov. 10 0Z

Problem #5: When the isobars are closer together then the pressure gradient is stronger and the wind is stronger

Problem #6: Compare the cloud and precipitation pattern to that predicted by cyclone model

Problem #7: Give temperatures and dewpoints in the warm sector

Problem #8: Wind direction is the direction winds are coming from

Problem #9: For example, compare temperature in Jackson, MS to temperatures in eastern Oklahoma

Problem #10: Look at wind barbs on each side of cold front

Problem #11: There are a total of 4 cities. Look at the meteogram for each of the 4 cities on pages 111-112 and find which one has lowest pressure

Problem #12: Find difference between dewpoint at 10/0900 and dewpoint at 9/0300

Problem #13: When pressure minimizes, that will be the time the front passes. Also look for Changes in wind direction, temperature and dewpoint.

Problem #14: Give wind direction before and after each frontal passage.

Problem #15-#16: Look for time of lowest pressure. Also look for wind shift and temperature/dewpoint changes

Problem #17: Oh yes, you have found these already when answering the previous questions. "Surface pressure typically drops as a front approaches and rises after it passes", thus front passes when pressure is the lowest.

Problem #18: Warm Air Advection - warmer air being blown toward the forecast area. As a warm front approaches there will be warm air advection.

Cold Air Advection - colder air being blown toward the forecast area. After a cold front passes there will be cold air advection.

Problem #19: Describe where rising OR sinking air is in relations to the fronts (i.e. behind cold front, ahead of warm front)

Problem #20: Widespread precipitation tends to occur where mT air is lifted over a warm front. This same region is experiencing warm air advection and lifting.

Problem #21: Draw over 0 C isotherm at 850 mb with a colored pencil. Next. see figure 10-18 on page 117 to see if snow is dominate where precipitation is occurring where it is colder than 0 C at 850 mb and rain is dominate where precipitation is occurring where it is warmer than 0 C at 850 mb.

Problem #22: Upper level divergence tends to occur on the right side of a trough. Divergence is the air spreading apart as it flows forward. This contributes to rising air on the synoptic scale.

Problem #23: Speed is distance divided by time (miles per hour in this case).

Problem #24: Use figure 10-14 on page 115 as a guide

Problem #25: Warm conveyor belt is in vicinity of where warm front will be. Often a broad region of clouds will be north of warm front due to the warm and moist air lifting over the warm front.

Review Questions: Use what you have learned in this lab to answer the questions.

Lab 11, pages 119 - 130

Questions 1 through 22 and review questions: E-mail instructor or post on bulletin board your questions you need help with on this lab.

Lab 12, pages 131-142

Questions 1 through 17 and review questions: E-mail instructor or post on bulletin board your questions you need help with on this lab.

END OF LAB TUTOR