Darrin Rasberry
It's the curse of every math teacher to have to answer the question "but where do we use this?"
This time around through Calculus II, though, I want to give my students an example of where the mathematics I'm teaching them is used.
Since we have both calculus in polar coordinates and area between two curves on the syllabus, I'm thinking about creating a (very) basic SkewT/Hodograph question. The students will work on this question in groups for a "take home" assignment.
However, being unfamiliar with the exactness of terminology and presentation, I would like to request suggestions and corrections from anyone here at STORMTRACK. Although I've taken liberties in this problem, I would still like to report accurate information to my students.
Thank you all in advance for your time. I did not put this in Education since I'm not really asking about chasing in general; if this needs to be moved there, mods, please do so (or to the hidden scientific forum if need be, but I'll need to be PMd the posts
)
-----------------------
Question:
When assessing thunderstorm and tornado risk, weather forecasters look at many factors. Two important measurements they consider are the Skew-T Diagram and the Hodograph.
Consider a small "piece" of moist air from the surface. When conditions are right, these "parcels" will lift from the surface, or become "buoyant." When air reaches its dewpoint, it condenses - like the dew you see on the grass some mornings, except in most cases in thunderstorm prone weather the air will condense higher up in the atmosphere than the ground.
When an unstable atmosphere exists, the air packet will cool when lifted higher, but since it is lifting so quickly it will not have time to adjust to the rapidly changing temperature around it. Thus air lifted from the ground will continue to rise through the atmosphere (since warm air rises above cooler air) until the parcel from the ground finally cools off to the temperature of the atmosphere. The instability this "parcel" experiences is the difference in the parcel's temperature (measured in a special way) vs. the temperature of the cooler surrounding atmosphere.
Instability is realized for the most part when the rising surface air is already "condensed" (for the purposes of this problem, it is "condensed" from the beginning, although this is not always so in reality).
A Skew-T Diagram examines the temperature of various situations and compares them to each other to determine many aspects of the atmosphere on a given day. Once such factor is the aforementioned instability, which is known to meteorologists as CAPE (Convective Available Potential Energy).
A Skew-T Diagram plots temperatures of various parts of the atmosphere (x-axis) with the height the temperature occurs (y-axis). Although a Skew-T diagram is more complicated than the Cartesian Coordinate System, our problem will utilize the Cartesian Coordinate System as we have done all semester, for simplicity.
Once again, since warm air rises, it is logical to examine the places where a parcel of air rising from the surface is cooler than its surrounding atmosphere. Observe the temperature of a surface packet versus the temperature of the surrounding air on the following Skew-T diagram.
[NOTE: Axes have been realigned for convenience. Height (usually measured in terms of pressure, since pressure decreases as height increases) has been restructured to fit the problem for our purposes as well. The piece you view is but a mere part of a typical Skew-T diagram!]
(Fig 1 with diagram graphing the functions below)
The temperature of the surrounding air in this case may be mathematically modeled by the function
x = -10 * y + 10
The temperature of the rising air packet from the surface can be mathematically modeled by the function
x = -y^2 + 100
a) Between what two coordinates does the rising air's temperature equal or exceed the environmental temperature?
Ans: between (0,10) and (100,0)
b) Cape = g*integral [(temp of parcel - temp of air)/(temp of air)]*dy
Constant g is gravity. The limits of your integration should yield only positive area. Let g = 10 m/s^2.
Suppose the forecaster says that your model requires the above answer to be scaled (multiplied) by a factor of 100 to be accurate. What is the CAPE (in Joules per Kilogram) on this day?
Ans: 5000 J/kg.
c) A hodograph measures the flow of fluid through height. Since air is a fluid, hodographs may be used to determine how wind changes in both speed and direction with height.
A plot of wind speed on a hodograph is given by its magnitude and direction. Thus it is a vector, and we may best represent a hodograph with the polar coordinate system.
For example, suppose a reading at 3 km height is 35 miles per hour (hodographs in practice use a nearly identical wind unit called knots). The direction in which the wind is blowing is from the south. The vector will thus be plotted:
(Fig 2 easy vector plot)
Suppose a weather forecaster has sent up a weather balloon measuring wind speeds at different heights in the atmosphere. The forecaster returns with the data and asks you to plot the following four vectors using the Polar Coordinate System. Note in practice the angles are measured from a different standard, but for our purposes we will use the Chapter 10 standard.
0 km height (surface): 5 mph from the south/southeast (theta = 7pi/12 radians)
1 km height: 10 mph from the south (theta = pi/2 radians)
2 km height: 20 mph from the south/southwest (theta = pi/3 radians)
3 km height: 27 mph from the west/southwest (theta = pi/6 radians)
(Figure 3 blank hodograph [or polar coordinate system] plot.)
d) The Helicity is a measure of "turning" a lifted parcel experiences with height. It is calculated from the hodograph. Winds that blow stronger and turn more with height will be more favorable for "supercell" (dangerous thunderstorms) and tornado development. The helicity between two given heights is the polar coordinate area between those heights.
Find the 0-1km helicity and the 0-3 km helicity for part (c) if the curve is most closely modeled by the limacon
r = 10 + 20*cos(theta)
from theta = pi/6 to theta = 7pi/12.
Answer: 0-1km: 7.5
0-3km: 221
e) A cold front is moving through your area. This means storms may develop! Look at your calculated data over parts a-d.
The Energy Helicity Index is calculated
(CAPE*Helicity) / 160,000
If using the 0-1km Helicity, values greater than .25 indicate tornado possibility, while values greater than two indicate strong tornado possibility.
If using the 0-3km Helicity, values greater than 1 indicate that supercells (strong thunderstorms) are possible, values greater than three are associated with strong tornadoes.
Your forecaster that you're working for makes an evaluation "supercells are likely, with a significant possibility of strong and damaging tornadoes." Given the other parameters for storms that we haven't discussed are in proper place, is your forecaster correct with his bold assertion? What would you use in your findings to both object to and to support his forecast?
Ans: 0-1km EHI: .23
0-3km EHI: 6.9
High EHI may support his assertion, but the student may want to look further with the forecaster into the low value for the 0-1km EHI and see what this means.
This time around through Calculus II, though, I want to give my students an example of where the mathematics I'm teaching them is used.
Since we have both calculus in polar coordinates and area between two curves on the syllabus, I'm thinking about creating a (very) basic SkewT/Hodograph question. The students will work on this question in groups for a "take home" assignment.
However, being unfamiliar with the exactness of terminology and presentation, I would like to request suggestions and corrections from anyone here at STORMTRACK. Although I've taken liberties in this problem, I would still like to report accurate information to my students.
Thank you all in advance for your time. I did not put this in Education since I'm not really asking about chasing in general; if this needs to be moved there, mods, please do so (or to the hidden scientific forum if need be, but I'll need to be PMd the posts
) -----------------------
Question:
When assessing thunderstorm and tornado risk, weather forecasters look at many factors. Two important measurements they consider are the Skew-T Diagram and the Hodograph.
Consider a small "piece" of moist air from the surface. When conditions are right, these "parcels" will lift from the surface, or become "buoyant." When air reaches its dewpoint, it condenses - like the dew you see on the grass some mornings, except in most cases in thunderstorm prone weather the air will condense higher up in the atmosphere than the ground.
When an unstable atmosphere exists, the air packet will cool when lifted higher, but since it is lifting so quickly it will not have time to adjust to the rapidly changing temperature around it. Thus air lifted from the ground will continue to rise through the atmosphere (since warm air rises above cooler air) until the parcel from the ground finally cools off to the temperature of the atmosphere. The instability this "parcel" experiences is the difference in the parcel's temperature (measured in a special way) vs. the temperature of the cooler surrounding atmosphere.
Instability is realized for the most part when the rising surface air is already "condensed" (for the purposes of this problem, it is "condensed" from the beginning, although this is not always so in reality).
A Skew-T Diagram examines the temperature of various situations and compares them to each other to determine many aspects of the atmosphere on a given day. Once such factor is the aforementioned instability, which is known to meteorologists as CAPE (Convective Available Potential Energy).
A Skew-T Diagram plots temperatures of various parts of the atmosphere (x-axis) with the height the temperature occurs (y-axis). Although a Skew-T diagram is more complicated than the Cartesian Coordinate System, our problem will utilize the Cartesian Coordinate System as we have done all semester, for simplicity.
Once again, since warm air rises, it is logical to examine the places where a parcel of air rising from the surface is cooler than its surrounding atmosphere. Observe the temperature of a surface packet versus the temperature of the surrounding air on the following Skew-T diagram.
[NOTE: Axes have been realigned for convenience. Height (usually measured in terms of pressure, since pressure decreases as height increases) has been restructured to fit the problem for our purposes as well. The piece you view is but a mere part of a typical Skew-T diagram!]
(Fig 1 with diagram graphing the functions below)
The temperature of the surrounding air in this case may be mathematically modeled by the function
x = -10 * y + 10
The temperature of the rising air packet from the surface can be mathematically modeled by the function
x = -y^2 + 100
a) Between what two coordinates does the rising air's temperature equal or exceed the environmental temperature?
Ans: between (0,10) and (100,0)
b) Cape = g*integral [(temp of parcel - temp of air)/(temp of air)]*dy
Constant g is gravity. The limits of your integration should yield only positive area. Let g = 10 m/s^2.
Suppose the forecaster says that your model requires the above answer to be scaled (multiplied) by a factor of 100 to be accurate. What is the CAPE (in Joules per Kilogram) on this day?
Ans: 5000 J/kg.
c) A hodograph measures the flow of fluid through height. Since air is a fluid, hodographs may be used to determine how wind changes in both speed and direction with height.
A plot of wind speed on a hodograph is given by its magnitude and direction. Thus it is a vector, and we may best represent a hodograph with the polar coordinate system.
For example, suppose a reading at 3 km height is 35 miles per hour (hodographs in practice use a nearly identical wind unit called knots). The direction in which the wind is blowing is from the south. The vector will thus be plotted:
(Fig 2 easy vector plot)
Suppose a weather forecaster has sent up a weather balloon measuring wind speeds at different heights in the atmosphere. The forecaster returns with the data and asks you to plot the following four vectors using the Polar Coordinate System. Note in practice the angles are measured from a different standard, but for our purposes we will use the Chapter 10 standard.
0 km height (surface): 5 mph from the south/southeast (theta = 7pi/12 radians)
1 km height: 10 mph from the south (theta = pi/2 radians)
2 km height: 20 mph from the south/southwest (theta = pi/3 radians)
3 km height: 27 mph from the west/southwest (theta = pi/6 radians)
(Figure 3 blank hodograph [or polar coordinate system] plot.)
d) The Helicity is a measure of "turning" a lifted parcel experiences with height. It is calculated from the hodograph. Winds that blow stronger and turn more with height will be more favorable for "supercell" (dangerous thunderstorms) and tornado development. The helicity between two given heights is the polar coordinate area between those heights.
Find the 0-1km helicity and the 0-3 km helicity for part (c) if the curve is most closely modeled by the limacon
r = 10 + 20*cos(theta)
from theta = pi/6 to theta = 7pi/12.
Answer: 0-1km: 7.5
0-3km: 221
e) A cold front is moving through your area. This means storms may develop! Look at your calculated data over parts a-d.
The Energy Helicity Index is calculated
(CAPE*Helicity) / 160,000
If using the 0-1km Helicity, values greater than .25 indicate tornado possibility, while values greater than two indicate strong tornado possibility.
If using the 0-3km Helicity, values greater than 1 indicate that supercells (strong thunderstorms) are possible, values greater than three are associated with strong tornadoes.
Your forecaster that you're working for makes an evaluation "supercells are likely, with a significant possibility of strong and damaging tornadoes." Given the other parameters for storms that we haven't discussed are in proper place, is your forecaster correct with his bold assertion? What would you use in your findings to both object to and to support his forecast?
Ans: 0-1km EHI: .23
0-3km EHI: 6.9
High EHI may support his assertion, but the student may want to look further with the forecaster into the low value for the 0-1km EHI and see what this means.
