Lab 4 Atmospheric Pressure and winds

Imagine yourself standing at several elevations (refer to column 1 ) when you fill in your answers in Table 1 below. Determine the atmospheric mass above and below for each elevation above sea-level in Table 4.1. The first two lines (responses in purple) have been completed for you. So, when you look at the percent atmospheric mass above in column 2 at being 100, this means as you stand at 0.0 km (surface of the Earth), 100 percent of the atmospheric mass is above you and 0% is below you (refer to column 2 ). To get values in column 3 , you will use the equation (100 – mass above). The atmospheric pressure on the surface is 1013.2 mb ( column 4 ). When you are standing at 5.6 km in elevation, you already know that half the atmospheric mass is above you and half is below you. So, to get the column 2 (percent atmospheric mass above), you divide by half, so 100 divided by 2 equals 50. You will divide by two all the way up in column 2 . So, when you stand at 5.6 km above the earth’s surface, 50 % of the atmosphere is above you. How much is below you now? If 50% is above you, 50% is below (this totals to 100%). The formula for the third column is 100 – mass above (100 - 50 = 50). So, at 5.6 km, 50% of the atmospheric mass is below you. What about atmospheric pressure? You would divide each value up by two in column 4 . So, 1013.2 mb divided by 2 equals 506.6 mb. Fill in the table below. You will notice that the blank cells have numbers encased in parentheses. The numbers will correspond to the lab assessment (and questions below) when you are ready to submit your answers online. Use proper units in (% or mb) for the questions below. Your responses should go to one decimal point, if needed. (refer to column 4 answers below). You can round your answers if, for example, you have an answer of 15.65, you can round to 15.7. If you have 15.43, you can submit your answer as 15.4. Note: You will have decimal points for columns 2 and 3 and 4 as you work through the table. Column 1 Column 2 Column 3 Column 4 Table 1. Atmospheric Mass and Pressure Height (km) Percent Atmospheric Mass Above Percent Atmospheric Mass Below Atmospheric Pressure (mb) 33.6 (13) (14) (15) 28.0 (10) (11) (12) 22.4 (7) (8) (9) 16.8 (4) (5) (6) 11.2 (1) (2) (3) 5.6 50% 50% 506.6 mb 0.0 100% 0% 1013.3 mb

Coriolis Effect As stated in the previous section, the pressure gradient is the driving force of winds. The winds are directed perpendicular to isobars away from high pressure to low pressure. The wind always blow from high to low. However, the winds do not move perpendicular to isobars due to the Coriolis effect. Once air has been set in motion by the pressure gradient force (PGF), it undergoes an apparent deflection from its path, as seen by an observer on Earth. The Coriolis effect defined is an apparent deflection of moving bodies, such as winds, caused by the Earth’s rotation. The winds in the Northern Hemisphere tend to be deflected to the right, and deflected to the left in the Southern Hemisphere. The air movement at the surface of the Earth is at an angle across isobars toward low pressure (more on this below). On a weather map, isobars are closed circles that illustrate areas of high and low pressure. A low pressure cell (cyclone) illustrates where air is rising (ascending). High pressure cells, or an anticyclone illustrates where air is sinking (descending). As winds move toward the center of low pressure (cyclone) in the northern hemisphere, they circulate counter-clockwise. Winds circulate clockwise in the northern hemisphere around high pressure cells (anticyclone). Conversely, in the southern hemisphere, the coriolis effect deflects moving bodies such as winds to the left, so circulation around high pressure (anticyclone) is counter-clockwise and clockwise around low pressure (cyclone). An air parcel will move from high pressure (anticyclonic) to low pressure (cyclonic) because of the PGF, and is deflected by the Coriolis force (effect) which is to the right in the northern hemisphere (to the left on the southern hemisphere). As the wind gains speed, the deflection increases until the Coriolis force equals the pressure gradient force. At this point, the wind will be blowing parallel to the isobars. So, when this happens, the wind is referred to as a geostrophic wind, an upper level wind in which friction is not an issue ( Figure 2) below. Figure 2 Image Source: http://www.goes-r.gov/users/comet/tropical/textbook_2nd_edition/media/graphics/forces_winds.jpg

25. Using Figure 4, which illustration with its corresponding letter illustrates cyclonic flow on the Earth’s surface in the northern hemisphere? (0.5 point) A) Letter A B) Letter B C) Letter C D) Letter D

Map Interpretation The map below is an example of barometric pressures (mb). The lines on the map are isobars indicating lines of equal barometric pressure in millibars (mb) for the eastern half of the United States and parts of Canada and Mexico. Each line (isobar) represents a specific barometric pressure at a particular time/day. For example, in the map below, the barometric pressure in San Antonio, Texas is approximately 1032 mb and Oklahoma City, Oklahoma has a barometric pressure reading of 1036 mb. The pressure is increasing as you drive north from San Antonio to Oklahoma City. Image Source: http://mathsheet.ifcpnice.com/isobars-and-air-pressure-worksheet-answers/