(2) Every talk should be built around a single core argument. What is the take-home message you want your audience to understand? It is good to start a talk with a puzzle/mystery/paradox or some other unexplained phenomenon to provoke the audience interest. What is the purpose of your talk? Why is this topic important? Why should the audience care? A very common beginner mistake is to try to say too much. Only include something if it supports your main argument.
(3) A good rule of thumb is 1 slide/minute, but leave extra time for questions during the talk, after, and for switchover. So for a 20 minute talk, at MOST 15 slides, usually less.
(4) Appeal to the audience on multiple levels - through simple pictures, mathematical reasoning, intuition. Don't just write arguments down and expect people to follow them, or to retain their interest.
(5) Who is your audience? What is their background? What vocabulary do they understand? The purpose of a talk is not to talk about your research, but to build a bridge to where THEY are, so you communicate something.
(6) Giving a scientific talk is mainly a public relations exercise, an advertisement for a paper, where the real content should be. Do not bore the viewer with details of the calculation, but address most important sources of uncertainty.
Quiz 2: October 31, 2011 (Class 22)
Quiz 3: November 30, 2011 (Class 34)
2. An elderly gentlemen is drinking a milkshake through a straw having a length L = 30 cm. The density of the milkshake is 1.2 times the density of water. Atmospheric pressure is 1000 hPa. Calculate the pressure in his lungs at which the milkshake will exit the straw and he will be able to drink. Assume the straw is vertical, and that 30 cm refers to the length of the straw that is above the level of the milkshake in the glass. (Note: I suppose you could do an experiment and estimate the amount by which people could reduce their lung pressure from atmospheric pressure by measuring the height of the tallest straw they could drink from.) Hint: you can first drink from a straw when you are first able to apply a tiny upward acceleration (i.e. assume zero).
3. Viscosity converts parcel kinetic energy to thermal energy (randomized kinetic energy or temperature), via momentum exchange. This is an idealized example to help explain how this occurs. Suppose that a 1 kg air parcel with a velocity of 20 m/s is adjacent to another 1 kg air parcel with a velocity of 10 m/s. After a diffusional exchange of momentum (viscosity), the 1 kg air parcels have velocities of 19 m/s and 11 m/s respectively. Note that this diffusional momentum exchange conserves the momentum of the sum of the two parcels.
(a) How much parcel kinetic energy was lost during this diffusional momentum exchange (in J)?
(b) By conservation of energy, and in the absence of any external energy sources or sinks, this reduction in parcel kinetic energy must be accompanied by an increase in thermal energy (kinetic energy of molecules). Assume that the increase in thermal energy is distributed equally between the two air parcels. Determine the increase in temperature of the two air parcels. (You can assume that this diffusional momentum exchange occurs at constant pressure, and use the specific heat at constant pressure for dry air, usually called cpd.)
2. 1.12 from the text. It is best to assume that the horizontal wind at the service station is zero. The rate of pressure change as measured in the car as it passes the service station is then equal to the local pressure change plus the contribution due to the motion of the car in the pressure field. This contribution is negative since the car is directed (south) to lower pressure. The pressure should therefore be decreasing more rapidly in the car than at the service station. The answer in the text is therefore incorrect: it gives a faster rate of decrease at the service station than in the car. (In the future: give a question where the advective component of the tendency comes from the motion of an air parcel, not a contrived situation like a car.)
3. 2.1 from the text. The answers are in the text. Your grade depends on how clearly you derive the answers. You ignore the effect of the atmosphere: the pressure in the fluid is equal to the hydrostatic pressure. (Hints: for (a) it may be easier to think of solving for the pressure at the bottom of the rotating pan. Assume the pressure in the fluid is given by the hydrostatic approximation, and ignore the pressure of the atmosphere. Under these assumptions, the pressure gradient acceleration is independent of depth. Also note that you are solving for an acceleration not a force, as it suggests in the text. In the (b), ignore the Coriolis Force. The only acceleration is the centripetal acceleration of a rotating fluid parcel.)
Answer for 2.1(c). The best way to calculate the volume of the container as it rotates is to imagine it is made up of rings of fluid, with each ring 2*pi*r in circumference, dr in thickness, and h(r) in height. The volume of the fluid is then V = Int(from 0 to r0) 2*pi*r*h(r)*dr. Use the solution for h(r). Then integrate over dr from 0 to r0. You should get V = pi*r0*r0*h0 + (pi*omega*omega*r0*r0*r0*r0)/4*g. You set this equal to m*r0*r0*z0 to get a relationship for h0 in terms of z0.
4. 2.4 from the text. Note that the two weights are for the passenger going in opposite directions in the train but at the same speed. You need the mass of the passenger. You can calculate this from the average (stationary) weight, and a constant value of g. (Note that to calculate the "real" mass of the person, you would have to be given the stationary weight of the person plus the actual "effective" gravity at the location. Remember that "effective" includes the rotational acceleration. You aren't really given this so just use a standard constant value for g.)
2. 2.10. In this case, balanced means a 3 way balance between the friction, pressure gradient, and Coriolis acceleration vectors.
3. 3.5(c) only. You have to assume that, along the axis of the jet, the flow is mostly zonal, and that the acceleration vector is mainly coming from changes in the zonal wind speed.
4. Estimate the Rossby number of the rotational circulation in a cup of tea.
2. 3.9. Don't answer the last part.
3. 3.10. The easiest way to solve this is to enforce conservation of momentum. Changing latitude changes the distance R from the rotation axis. Remember to use the total angular velocity.
4. From Holton: "Isolines of 1000 to 500 hPa thickness are drawn on a weather map using a contour interval of 60 m. What is the corresponding layer mean temperature interval?"
2. 3.1 (a) and (b) from the text. You also have to assume frictionless adiabatic flow. (Notes: the solutions for (a) and (c) given on page 76 appear to be incorrect. Next year: ask to do (c) also.)
3. An air parcel with a temperature of 20 C at 1000 hPa is lifted dry adiabatically. What is its density when it reaches 500 hPa?
4. A high altitude balloon remains at a constant potential temperature as it circles the earth. The balloon is in the lower stratosphere where the temperature is independent of height, and equal to 200 K. If the balloon is displaced a small distance vertically from its equilibrium (netrally buoyant) altitude, what would be the period of the oscillation?
In this question, many of you used vector formulas for relating the temperature gradient vector and the thermal wind. However, you are making life difficult for yourself by introducing extra steps; 99 percent of the time, you are better off using the component forms of the equations for calculations.
(dT/dy) between 900 hPa and 700 hPa: There are a variety of ways of doing this question. One way is to use equations (4.28). At 900 hPa: ug = 0 and vg = 10 m/s. At 700 hPa: ug = 10 m/s and vg = 0. In using (4.28), it is extremely important to note that the derivative is evaluated with p "up" going down. So in going from 700 hPa to 900 hPa, ug decreases by 10 m/s, so dug/dp = (-10 m/s)/(200*100 Pa). For p, you would use the average value of p between 700 hPa and 900 hPa, or 800 hPa. Therefore, dT/dy = [f*(800*100)/R]*[(-10 m/s)/(200*100 Pa)]. R = 287 and f = 9.9E-05. I get dT/dy = -1.38E(-05) K/m.
(dT/dx) between 900 hPa and 700 hPa: Again using (4.28), it is easy to rearrange to get (dT/dx) = -[(f*p)/R]*(dvg/dp). You can use the previous value for f*p/R. The change in vg in going from 700 hPa (0 m/s) to 900 hPa (10 m/s) is positive so dvg/dp is positive. Should again get dT/dx = -1.38E(-05) K/km.
(dT/dy) between 700 hPa and 500 hPa: Same approach as above should get dT/dx = 1.04E-05 K/km. dug/dp is just the negative of that used above for the 900 - 700 hPa layer, and use 600 hPa for the average pressure.
(dT/dx) between 700 hPa and 500 hPa: Same approach as above should get dT/dy = 1.04E-05 K/km.
(ii) Calculate the rate of geostrophic temperature advection in each layer. Do not do the third part of the question.
The mean rate of geostrophic temperature advection in each layer is -(ug,vg)*(dT/dx,dT/dy), where Vg = (ug,vg) refer to the mean layer geostrophic winds. For both layers, (ug,vg) = (5,5), and use you answers for the T gradient vector from above. Note the negative sign.
4.9 Assume the two systems are axisymmetric. By intensity is meant whether the tangential wind increases or decreases with height between the surface and 850 hPa. For each system, draw a cross section of how isobars between the surface and 850 hPa would vary with height. Your horizontal axis should be distance from the center of the low, or high.
Answer: I will draw the diagrams in class. However, one good way to think is that the thermal wind is parallel to isopleths of thickness with the warm air on the right. Therefore in both cases the thermal wind is counter clockwise (CCW). The surface wind is CCW for the low and CW for the high. Therefore the thermal wind adds to the surface wind for the low (more instense with height), and cancels the surface wind for the high (less intense with height).
4.12. Do all parts. This question seems to require that you make many assumptions.
(a) You have to work back from knowledge of the geostrophic wind at
850. What is the CA? What is the PGA? What is the density at 850 hPa?
What is the PG at 850 hPa? What is the PG at 1005 hPa? What is the PGA
at 1005 Pa? What is the Vg at 1005 hPa? Assume that the 1005 - 850 hPa
layer temperature
Most of you got 30 m/s, which is correct. The book has 26.2 m/s. This
seems to be obtained by assuming that the mean layer
(b) Eq (3.51) is useful. By kinetic energy generation is meant the rate
of kinetic energy increase due to acceleration by the PGA. It also looks
like you have to assume the actual wind speed V is approximately equal to
the geostrophic wind speed Vg (but the directions of the velocity vectors
must be different). In principle, the actual wind speed should be used in
the expression for KE generation by the PGA, but I don't see how to solve
this. If using natural coordinates would need R, which is not given. Note
that the cross-isobar angle means the angle between the pressure isobars
and the wind vector, not the angle between the velocity vector and the
PGA which you initially solve for.
Most of you got an angle close to 13 degrees, whereas the book has 14.81
degrees, which is obtained by using their slightly incorrect value for
the geostrophic wind speed at the surface from (a). Basically use dKE/dt
from pressure gradient work equals (-1/rho)*V*dot*PG =
(-1/rho)*Vg*PG*cos(theta), where theta is the angle between the velocity
vector and the pressure gradient vector, and assuming the actual speed is
close to the geostrophic wind speed. Use your answer for the pressure
gradient from above, and substract theta from 90 degrees to get 13
degrees. The book is not asking for the angle between the pressure
gradient vector and the velocity vector (theta = 77 degrees), but for the
angle the wind vector makes with the isobars.
(c) You can assume the rate of change in geopotential height is zero, and
that there exists a steady state balance between KE generation by the PGA
and KE loss due to friction.
Should get about -6.5E-04 N.
(d) answer: about 276 km.
4.13.
(i) Note that the T advection must be negative to cancel the positive heating.
Answer: The best approach is to use the natural coordinate
expression (4.38a). Draw a diagram
in which you have geopotential contours aligned in a north-south direction with
high geopotential (warm air) to the east.
The wind vector is pointing northeast, so cutting
diagonally uphill across geopotential contours. You would therefore expect the
parcel to be experiencing a deceleration. From the geostrophic wind,
dphi/dy = 0, and dphi/dx = f*vg. You want to find dphi/ds, where ds
is a segment alnog the parcel trajectory. from your diagram, you should
show dx = ds*sin(30). So dV/dt = -sin(30)*dphi/dx = -0.001 m/s2.
Some of you tried to solve for the ageostrophic wind vector and then use Eq
(3.41). However, you are not given the actual wind speed (only the direction),
so it is impossible to solve for the ageostrophic wind.
It is therefore also impossible to solve for the
acceleration vector. So if you did solve
for the acceleration vector, you must have made some additional assumption.
2. (3.4 Holton) (a) What is the geostrophic wind speed (m/s) on an isobaric
surface for a geopotential height gradient of 100 m per 1000 km?
Answer: Straightforward application of $4.39b).
Almost all got 9.8 m/s. Note that speed is always positive.
Also note you must convert 100 m of geopotential height to geopotential
by multiplying by g.
In all
cases below, assume the absolute value of R is 1000 km and f = 1.0E-04.
(b) What is the gradient wind speed for a regular low?
9 m/s.
(c) What is the gradient wind speed for an anomalous low?
108 m/s.
(d) What is the gradient wind speed for a regular high?
11 m/s.
(e) What is the gradient wind speed for an anomalous high?
89 m/s.
3. (3.10 Holton) The mean temperature in the 750 - 500 hPa layer decreases
eastward by 3 C per 100 km. The 750 hPa geostrophic wind is from the
southeast at 20 m/s.
(a) What is the geostrophic wind speed and direction at 500
hPa? Let f = 1.0E-04.
Answer: It helps to draw a diagram. Use (4.28) with dT/dy = 0, so only
vg changes between 750 hPa and 500 hPa. Since dT/dx < 0, dvg/dp > 0,
so vg should decrease in going from 750 hPa to 500 hPa. The decrease
is -34.9 m/s, so vg goes from 14.14 m/s at 750 hPa to -20.76 m/s
at 500 hPa. Since ug is unchanged, at 500 hPa (ug,vg) = (-14.14,-20.76) m/s.
The speed is 25.1 m/s, and points 55.7 degrees south of westward.
(b) What is the mean rate of geostrophic
temperature advection in the 750 - 500 hPa layer? Express in K/day.
Note: To find the average geostrophic temperature advection of the layer, you
must use the average geostrophic wind 0.5*(Vg(750) + Vg(500)).
Should get -1.5 K/hour, or -37 K/day.
In Future replace by:
(3.10 Holton) The mean temperature in a 750 - 500 hPa layer decreases eastward
by 3 C per 100 km. The 750 hPa geostrophic wind is from the southeast at 20 m/s.
(i) What are the (u,v) components of the geostrophic wind at 500 hPa?
Assume that formulas like (4.28) in the text can be applied to finite
pressure layers.
(ii) What are the (u,v) of the thermal wind of the 750 - 500 hPa layer?
(iii) What is the sign of the geostrophic temperature advection in the 750-500
hPa layer? (Don't have to do the calculation; can show graphically) Would
expect veering or backing?
4. 4.17(a) from the text. The angle convention is CCW from north. For
example, 0 degrees is pointing south, 90 degrees is pointing west,
180 degrees is pointing north, and 270
degrees is pointing east. Please convert your divergence
value to /day (easier to understand than /sec).
2. (4.1 Holton) (i) What is the circulation about a square of 1000 km on a side
for an easterly (i.e. westward flowing) wind that decreases in magnitude toward
the north at a rate of 10 m/s per 500 km?
(ii) What is the mean relative
vorticity in the square?
3. The surface pressure over both the land and ocean is 1000 hPa. The average
1000-900 hPa temperature over the ocean is 280 K. The average 1000-900
hPa temperature over the land is 290 K. You can assume f = 0.
(i) Estimate the rate of change of absolute circulation dC/dt of a loop
where the two ends are 40 km apart.
(ii) Estimate an average geopotential height of the loop.
(iii) Let < v > refer to the mean tangential velocity along a rectangle
where one end goes from 1000-900 hPa over the ocean, while the other
end goes from 1000-900 hPa over the land.
Suppose < v > = 0 at t = 0. Estimate < v > after 10
minutes.
4. 5.6 from the text (a) and (b).
Assignment 7. Due: Wednesday November 16, 5:00.
1. (3.2 Holton) The actual wind is directed 30 degrees to the right of
the geostrophic wind. If the geostrophic wind speed is 20 m/s, what is
the rate of change of wind speed (i.e.solve for dV/dt). Let f =
1.0E-04. Assume the geostrophic wind is pointing north.
Assignment 8. Due: Wednesday November 23, 5:00.
1. (3.22 of Holton) The divergence of the horizontal wind at various
pressure layers above a given station is as follows (units of divergence
in E-05 per inverse s): (i) 0.6 in the 1000-850 hPa layer,
(ii) 0.45 in the 850-700 hPa layer, (iii) 0.15 in the 700-500 hPa layer,
(iv) -0.3 in the 500-300 hPa layer, (v) and -0.8 in the 300-100 hPa layer.
Assume that the vertical velocity w = 0 at 1000 hPa, and that the
atmosphere is isothermal with a constant temperature of 260 K. Find the
vertical velocity w and the pressure velocity omega at the six levels:
(i) 1000 hPa, (ii) 850 hPa, (iii) 700 hPa, (iv) 500 hPa, (v) 300 hPa,
(vi) 100 hPa. The easiest way to solve this question is to "bootstrap" (or iterate)
yourself up from the surface using the surface boundary condition.
Assignment 9. Due: Wednesday December 7, 5:00.