nm0925: okay er problem sheet four i wanted to say a word or two about problem 
sheet four which er er i guess you had back the other week er i had vibrations 
from people that er er people were a bit shocked at the marks er and there was 
er a very straightforward reason why the marks on that sheet were somewhat 
lower than the marks on the earlier ones and the reason was that nearly not 
quite but very nearly all of you got in a mess with signs and the result was 
that er er one way or another you nearly all not quite all but nearly all got 
incorrect answers because somewhere along the line you'd got into a mess with 
signs and i'm afraid i did mark it harshly because i think it's quite important 
to make the point that we're now at a stage where signs matter they're not just 
a sort of frill that we add on at the end of a calculation you've actually got 
to keep track of the signs as you go through it otherwise you'll get a silly 
answer so i i want to just remind you of a couple of points which 
will come in again and again in er later problems er to do with getting signs 
right i suspect these calculations involving thermal wind shear are amongst the 
most awkward or have the most potential for getting signs wrong but the same 
goes for other calculations you'll do first thing to remind you is er that er 
in nearly all these problems you do best if you break the wind down into 
components so let me just remind you of the usual convention that we use we 
imagine an X axis which points from the west to the east a Y axis that points 
from the south to the north and if you are told that there is a wind of 
particular magnitude shall we call it U with a particular direction and there's 
various ways of measuring that direction the meteorologists talk about the 
direction from which the wind blows and they measure the angle from true north 
the er mathematicians and this is probably the convention i will stick to will 
measure the angle of the wind vector from the X axis in the anticlockwise sense 
so if we have theta there which describes the er direction of the wind vector 
then 
we have two components we have a component along the X axis which i'll call 
little-U and little-U is big-U times the cosine of that angle theta and then we 
have a component parallel to the Y axis i call that little-V and little-V is 
capital-U times the sine of theta now the way i've drawn it there it's easy 
because er both U and V are positive if the wind vector points into these 
various other quadrants then different signs apply so here we have both U-
positive and V-positive if the wind vector was down in this quadrant then of 
course U would be still positive but in this case V would be directed down the 
negative Y axis so V would be negative similarly if the wind vector turned up 
in this quadrant here we'd have U-negative and V- positive still down here both 
U and V would be negative always worth drawing a sketch like that so that when 
you've done the calculation you can just check that the signs check out against 
that diagram if you're using a calculator to 
calculate the er vectors then strictly you should automatically get the signs 
right if you've got this angle theta and you put in an angle of er two-hundred-
and-eighty degrees or something like that it should automatically take care of 
the signs provided you've got your calculator set up right er but even so it's 
a very good idea when you're doing these calculations not just to blindly trust 
that you've got the signs right but at each stage to check things and a diagram 
is often the er easy way of doing that so that's getting the wind components 
right and many of you fell over at that stage because you just said well you 
just call U and V positive no matter which quadrant the wind vector was 
pointing into the other bit where people went wrong and i think probably more 
people went wrong here er was when we were required to take vertical 
derivatives er we have things like D-U-by-D-P and D-V-by-D-P and what i want to 
do is just to remind you how we would estimate any 
vertical derivative so suppose we have any meteorological quantity i'll just 
call it F for the moment it might be wind or temperature or whatever and we're 
required to estimate D-F-by-D-P the vertical rate of change of F well let's 
draw a little diagram once again to illustrate what we're doing very typically 
we might be given the value of F at say a hundred kilopascals near the surface 
and the value of F at fifty kilopascals halfway up through the atmosphere and 
there's a pressure difference between those two levels delta-P which in this 
case is fifty kilopascals well the way i will estimate derivatives always is by 
what we call finite differences we just take the value at one level minus the 
value at the other level divided by the delta-P and that's an estimate of the 
derivative not necessarily a terribly accurate one but it is an estimate the 
critical bit is getting it the right way round so you get the signs right and 
people do get in a muddle here i have 
to say i get into a muddle often if i'm not thinking the way it works is like 
this we have to take the value of F at the higher pressure a hundred 
kilopascals in this case minus the value of F at the lower pressure fifty 
kilopascals divided by the pressure difference delta-P and this will give us 
the right sign notice that very often we're dealing with situations where the 
wind speed increases with height in other words the wind speed decreases with 
pressure so if this was say the U component of wind we might very frequently 
have the situation where U at fifty kilopascals is larger than U at a hundred 
kilopascals and so D-U-by-D-P would be negative and people get in a real muddle 
with this because of course the wind increases with height but that means it 
decreases with pressure so the sign is often perhaps a bit counter-intuitive so 
do remember that again i think the diagram always helps and a little thumbnail 
diagram along with your problems just to check out what 
it is that you're taking differences between to estimate these derivatives will 
not go amiss and again don't just trust that you've done the algebra correctly 
when you work out the signs just check it against the diagram to make sure the 
sign seems to make sense to you and that way er you will get lots of marks and 
feel gratified and everyone'll be happy right so much for the the the the 
problems let me just say as a parting shot the problems really aren't meant to 
be hard if you're finding that you're spending hours and hours on the problems 
i suggest you've rather missed the point most of the problems are a matter of 
either picking formulae or equations out of the lecture notes and then in a 
subsequent problem er rearranging them maybe or putting some numbers in them or 
what have you it really shouldn't take hours and hours and hours to do that so 
er do bear that in mind 
right let's get on to today's topic er today can't find the right page in my 
lecture notes oh here we are today our topic is called thermal advection which 
is a rather grand word for a rather simple concept i'd like to start off by 
thinking about the weather it was a cold morning this morning wasn't it 
ss: mm 
nm0925: it certainly was it was quite a nice afternoon yesterday wasn't it 
ss: yes 
nm0925: yes it was quite warm wasn't it if you went out for a walk in the sun 
right anyone like to give me an explanation of why it was five or ten degrees 
colder this morning than it was say at three o'clock yesterday afternoon yeah 
sm0926: was a clear night last night 
nm0925: yes 
sm0926: a lot of the thermal er a lot of the infrared radiation goes 
nm0925: right good it was a clear night what's more it was dry air as well 
which is important which means there wasn't much water vapour and so the 
infrared radiation from the earth's surface could actually escape rather more 
readily er not everyone knows of course that water vapour is the most important 
greenhouse gas in the atmosphere although it's not the one the 
environmentalists 
make such a fuss about but er er the difference between a a clear night that's 
got dry air and a clear night that's got moist air is actually quite dramatic 
last night it was clear and dry so yeah lots of radiation from the earth's 
surface the earth's surface and the layers of air lying immediately above it 
got pretty cold hence our frost this morning let me take you back to er Friday 
don't know if you can remember Friday i have difficulty remembering events 
before the weekend but if we go back to Friday er the m-, minimum at about six 
o'clock in the morning was a great deal higher it was about four degrees or so 
so it was four or five it was five degrees probably six degrees warmer than it 
was this morning so what was different about Thursday night Friday morning from 
Sunday night Monday morning any ideas why was it fi-, four or six degrees 
warmer the same time on Friday morning as this morning 
sf0927: 
nm0925: a front went past well done you've 
got the right answer straight away i was hoping someone was going to witter on 
about clouds and so on well there are there are two things actually we might 
have said well it was a cloudy night and therefore the radiative effect we've 
just talked about couldn't operate but you're dead right the difference was 
that a front went through we were actually sitting in a different air mass on 
Thursday morning and over the weekend the front came through it went 
dramatically colder it snowed as we shall talk about later on this morning in 
some places anyway and er er so it's altogether colder one mass of air with one 
set of thermodynamic properties has been replaced by another and it's that 
replacement of air with one set of characteristics by air with a different set 
of characteristics which is what's meant by advection and that's what we're 
going to talk about today you see if i sit at a particular met station let's 
just focus our minds on the temperature for the 
moment although the same arguments apply to other meteorological quantities if 
i sit at a meteorological station and i observe a change in the temperature 
then there are two alternative explanations for that change in temperature and 
in many cases of course both of of them operate either the air sitting at that 
station has actually changed its properties in some sense so your explanation 
that there was lots of radiation going on last night lots of infrared escaping 
and therefore the air got colder that could be the case in one set of 
situations and that's what we sometimes call heating or or diabatic processes 
on the other hand the temperature at a station may change simply because the 
individual air parcels are are not changing they retain their temperature but 
they move away from your station and they're replaced by another set of air 
parcels that come along with different properties so during the course of 
Friday and Saturday warm moist air moved 
away from the namex area to be replaced by much colder drier air and that's 
what we call advection so the concept is simple enough what we need to do now 
is to er put some er quantitative details on it so i'm going to draw a diagram 
to fix the ideas in our head let's draw a diagram shall we of temperature and 
let's consider one space direction i'll call it X for the moment and suppose i 
have an observing site here i'll call this point A and at the observing site we 
observe a particular temperature there we are that's the temperature at er 
point A but the point is i'm going to suppose that the temperature varies as we 
go along the X axis so if i was to go to another observing site at a different 
position along X i'd observe a different temperature so let's er imagine that 
the temperatures along this X axis they go like this i i'll draw a rather 
simple example where they just go along a an increasing line like that so 
stations to the east of A are 
warm stations to the west of A in this case are cold er let us now suppose that 
a wind is blowing okay and i'm going to keep things dead simple by supposing 
that the wind is blowing parallel to the X axis er in general it won't of 
course in which case we'll have to consider the component of the wind parallel 
to the X axis but let's just suppose there is a wind blowing along the X axis 
there we are wind magnitude U pa-, blowing parallel to the the X-axis and time 
passes so after some time interval delta-T our observing site is no longer 
seeing this parcel of air with a temperature T-A what it will actually be 
sampling is an air parcel that started off some distance upstream so er let's 
suppose we take that point B shall we say that point B is a distance U times 
the time interval delta-T upstream and so after a time delta-T the parcel that 
started off at B will have reached A and so i'll measure a new temperature at 
station A after this time interval er i'll 
measure this temperature i'll call that T-B let's see if i can write down a 
relationship between those two temperatures well i think i can actually i i've 
a feeling i didn't use quite the same notation in the lecture notes if you're 
following in the lecture notes you might want to adjust the notation as as 
we're going along what we're interested in now is the temperature T-B which is 
the temperature that's now reached my observing site and we can see here that 
if this curve here doesn't depart too much from a straight line between 
stations A and B and it won't provided that time interval delta-T is 
sufficiently small er then it's rather easy to do the calculation we could say 
that T- B is going to be T-A minus the distance delta-X to point B multiplied 
by the gradient D-T-by-D-X the rate of change of temperature along the X axis 
so that distance there is delta-X and we can see that that is equal to U times 
delta-T 
so if i rearrange that expression i can get an expression for the rate of 
change of temperature at station A so rate of change of temperature at my 
station A well i'm i'm going to approximate it by T-B minus T-A divided by the 
time interval delta-T and i can write an expression for that from the preceding 
expression by rearranging it if i take the T-A over to that side then i get an 
expression for T-A minus T-B and then i've just to divide it by delta-T and if 
i do that and don't tell me to keep the signs i've got a minus sign here i've 
got a delta-X over a delta-T times D-T-by-D-X it's not hard it's just simple 
algebra here what i'm going to do now is my usual calculus trick i'm going to 
suppose that delta-T and delta-X become very small i'm going to take the limit 
in which delta-X and delta-T tend to zero if i do that then that simply becomes 
the rate of change of temperature with respect to time if you want to 
you can say that's at the point A and that's going to be equal to well in the 
limit del-, of delta-X and delta-T becoming very small delta-X over delta-T 
that's just the rate of change of position of an air parcel along the X axis 
it's the velocity it's the speed U so i'm going to get minus- U times D-T-by-D-
X and so that is an expression which tells us about the rate of change of 
temperature at a fixed point in space a fixed observing site on the assumption 
that the only process that's changing the temperature is this advection effect 
in other words the replacement of one air parcel with an air parcel of 
different properties so we'll put a box round that 'cause it's important so 
it's the rate of change due to advection and the er rate of change due to 
advection is given a name it's sometimes called the Eulerian rate of change 
Euler was a famous French mathematician who had more laws and formulae named 
after him er than most of us have and this 
is one that's named after him put a little a box round that it's an important 
bit of terminology you'll hear referred to the Eulerian rate of change is the 
rate of change at a fixed point in space the rate of change you would estimate 
for example if you're a meteorologist sitting at a fixed observing site on the 
earth's surface watching the winds blow past you course it's not the only rate 
of change that we might consider suppose we didn't sit at a fixed observing 
site suppose we attached ourselves to some sort of balloon which was wafted 
around by the winds in that case if we had a thermometer attached to that 
balloon er we wouldn't see the Eulerian rate of change at all in fact if we 
designed the balloon very carefully we would actually see the rate of change of 
individual air parcels because our balloon would effectively be always embedded 
in the same lump of air as it moved around actually that's a pretty bad example 
'cause it's very hard to design a 
balloon that does that the problem is that the balloon usually floats at a 
constant height so it follows the winds at that height but it doesn't rise and 
sink with the air parcels er anyway that's er that's a that's a technical 
difficulty let's suppose we could do that so we have this other rate of change 
the rate of change following an individual fluid element and that in fact is 
identical to the rate of change that we mentioned earlier on if if an air 
parcel cools because of radiation we're actually talking there about this rate 
of change following the fluid element we're thinking of this air parcel that's 
sitting there radiating infrared radiation to space and thereby changing its 
temperature so we need another notation for that and so i'm going to introduce 
a notation and this is where some people get a bit confused 'cause they're a 
bit sloppy with notation on occasions so er i'm now going to consider the rate 
of change following an 
individual fluid parcel so what we would do here is to measure the temperature 
of our air parcel at some time T we would then measure the er temperature of 
our air parcel at some later time T-plus-delta-T shall we say we'd subtract the 
two divide by the time interval take the limit as delta-T tends to zero and 
we'd have a rate of change of temperature following the fluid parcel and we'll 
use a special notation for that we will define that as capital-D D-T i'm using 
too many Ts here but still by D-little-T and that's the notation we will use 
this is sometimes called it's the rate of change for an individual fluid parcel 
and this is sometimes called the Lagrangian rate of change Lagrange was also an 
eighteenth century French mathematician who has er a vast number of equations 
and formulae named after him and this is just one of them so they they they 
were good at getting their names on equations in the eighteenth century French 
republic anyway 
er rates of change so we've got these two rates of change and very important 
they are too the rate of change at a fixed point in space now that's the sort 
of thing you might say meteorologists are interested in 'cause we have fixed 
observing sites we're required to produce weather forecasts for fixed points on 
the earth's surface the forecast for namex or whatever and in other words the 
meteorologist is interested in the Eulerian rate of change of air properties 
but in a sense more physically fundamental is the Lagrangian rate of change if 
i follow an individual air parcel what happens to it does heat enter or leave 
it does it rise or sink in the atmosphere changes pressure whatever and if you 
cast your mind back to your work last term and our work this term most of the 
physical laws that we've written down that apply to air for example the first 
law of thermodynamics for example Newton's laws of motion they all apply to air 
parcels we ask how does an air parcel 
accelerate how does its temperature change when you pump heat into it and so on 
most of the physical laws that govern the atmosphere are expressed in terms of 
Lagrangian rates of change so we have this problem what we measure and what we 
want to predict are Eulerian rates of change what our physics tells us about 
are Lagrangian rates of change 
so what's going to be the relationship between the Eulerian and the Lagrangian 
rate of change well it's fairly straightforward that of course is the rate of 
change of temperature the Eulerian rate of change of temperature if i assume 
air parcels are not changing it's the rate of change in other words if capital- 
D-T-by-D-T is zero all that's happening is that one air parcel is moving away 
and another one with different properties is taking its place but the 
individual air parcels aren't changing their properties if the air parcels are 
actually changing their properties at the same time so suppose that i had this 
curve that was advecting along but at the same time all the air parcels were 
getting colder with time so that curve was dropping down as as it was moving 
along then i'd have to add into that formula the Lagrangian rate of change so 
what i will write is that er in general when there are both Lagrangian and 
Eulerian changes then we can say the rate of change at a particular location D-
U-by-D-T is going to be the Lagrangian rate of change capital- D-by-D-T minus 
this advective rate of change minus-U times D-T-by-D-X and that's really the 
central result of today's lecture it gives us the relationship between rates of 
change at a position and the rate of change for an individual parcel and 
they're related via the velocity and the temperature gradient i've deliberately 
kept things terribly simple in this discussion because i supposed that 
temperature was only varying in the X direction and the wind was only blowing 
parallel to the X 
axis it's dead easy to generalize this if i have a general wind then that will 
have components parallel to all three axes X Y and Z and of course the 
temperature may vary in all three directions X Y and Z as in general it does so 
i can generalize this for er three-dimensional motion and i won't go through 
the arguments of course they just work exactly the same way but just write a 
result down in three dimensions then we can say the rate of change at a fixed 
position the Eulerian rate of change is equal to the Lagrangian rate of change 
the rate of change for an individual air parcel minus well the same term U 
times D-T-by-D-X U remember now is the component of wind parallel to the X axis 
minus V times D-T-by-D-Y V is the component of wind parallel to the Y-axis D-T-
D-Y is the rate of change of temperature along the Y-axis and then of course 
we've got the vertical one minus-W times D-T-by-D-Z it looks a bit long but 
that's just because we've written 
out the same terms three times yes 
sm0928: er er i might be going along the totally wrong lines but 
nm0925: mm 
sm0928: can you not use the the er 
nm0925: we could er i'm not going to do that 'cause people'll find it so hard 
their brains will seize up but w-, we could and er i'll er thank you for asking 
that this has er this has er brought a warm glow to an old man's heart er we 
could write this much more neatly as D-by-D-T minus the vector wind U dot grad 
temperature okay 
sm0928: yeah 
nm0925: happy right write that down er this is notation that we will come to 
much more compact notation which we will come to er next year next autumn's 
geophysical fluid dynamics course we'll redo this argument er using vector 
notation and it all comes out a lot more compact i'm not going to use that for 
this course because not all of you have met this notation yet and er er people 
find it hard enough in the second year when they do so we'll we'll save that 
for next year okay well what i want to do now is to take these ideas these 
ideas of thermal advection of temperature 
changing at a particular place in the atmosphere due to advection replacement 
of one air parcel by another and i i'm going to make an additional assumption 
suppose we have a situation where the wind is essentially geostrophic that's 
the situation that's we've been er er assuming in most of our analysis of 
synoptic weather charts then we get a very useful er er set of relationships 
and the reason that we get them you see is that er thermal wind relationship if 
you er recall tells us the relationship between the change of temperature in 
the horizontal things like D-T-by-D-X and D-T-by-D-Y and the change of the wind 
in the vertical D-P-by-D-Z or sorry D er sorry D-U-by-D-P or er D-V-by-D-P so 
there's an intimate relationship between advection and the wind changing with 
height and er er i want to er take a look at that now you'll remember that at 
the end of the er lecture two lectures ago when we talked about thermal wind 
balance i introduced this concept of the thermal wind as a 
reminder the thermal wind which er i'll just write its magnitude down for the 
moment i called it V-subscript-T for thermal and it's equal to G-over-F times 
the gradient of the thickness so it's D-by-D-Y of Z at some high level minus-Z-
one at some lower level typically Z-two would be the fifty kilopascal er 
pressures at the height of the fifty kilopascal surface Z-one would be the 
height of the hundred kilopascal pressure surface so we could plot it on a 
chart er it bears just the same relationship to lines of thickness as the 
geostrophic wind vector bears to lines of constant pressure or geopotential 
height and we can use these ideas now to discuss a number of typical cases of 
the wind varying with height and i'm going to give you three examples and 
you'll be able to look later on this morning at your charts and spot what's 
happening let's take the simplest case first of all where the wind speed 
changes with height but the wind direction does not so we have a 
geostrophic wind near the surface which is in a particular direction the wind 
at say five-hundred millibars is in the same direction but stronger not an 
uncommon situation let's do a little drawing of what's going on er let me draw 
thickness contours in red and let me draw er s-, the er er height contours at 
the lower level and i'll draw them looking like that so that's shall we say the 
hundred kilopascal Z and the red contours show the thickness lines so that's 
the hundred to fifty kilopascal thickness if you look on the charts in the 
corridor you'll see that's what we actually plot people tend to put the surface 
pressure say but with lines of constant thickness on as well now in this case 
the surface geostrophic wind looks like that er i'm going to call that V-one 
and the thermal wind well remember that's parallel to lines of constant 
thickness and inversely proportional to their spacing so there's going to be V-
two in the same direction and if i add the 
two together i'll get the wind at the second height at er the five-hundred 
millibar level so the wind at the five-hundred millibar level is just going to 
be the sum of those two it's going to look like that notice that both those 
wind vectors V-one and V-two are parallel to the thickness contours thickness 
remember is proportional to temperature that means that there are no variations 
of temperature in the direction the wind is blowing so although the air 
parcel's moving past we're not replacing it by an air parcel with a different 
temperature we're replacing the air parcel with one of the same temperature in 
other words in this situation there is no thermal advection the 
temperature change i-, is not due to replacing air parcels 'cause we're just 
replacing one air parcel with another one of the same temperature so we have no 
thermal advection i have to say that's not an uncommon situation but it's a 
boring situation let's consider a much 
more exciting and interesting situation what we've had over this weekend is a 
situation which is called cold advection that is to say warm air has been 
displaced by cold air blowing down from the north let's have a look at what 
that implies in terms of the height and thickness let me again draw the 
thousand millibar height contours looking like that and V-one is going to be 
the same as in the previous diagram i really ought to underline these Vs 'cause 
i i i w-, i ought to stress that i'm talking about vectors here the direction 
is important as well as the magnitude so there's V-one that's the surface 
geostrophic wind but now this time i'm going to suppose that the er height 
contours are no longer is the sorry the thickness contours are no longer 
parallel to these height contours so i'm going to draw my thickness contours 
looking like that and i'm going to suppose it's colder here so that's low 
values of thickness and it's warmer here so we have large values of 
thickness here in this case the thermal wind vector is going to be parallel to 
the thickness contours and with cold air on the left so the thermal wind vector 
is going to look like that okay now what about the wind at five-hundred 
millibars well it's simply going to be the vector sum of the low level wind 
plus the thermal wind so if i er construct that using my usual er triangle 
construction i will find that V-two will look like that very interesting isn't 
it what you will see is there that as you go up from the surface to the upper 
layer the wind vector turns it turns in an anticlockwise sense and we describe 
this as the wind backing with height and this is a situation where we have cold 
advection that's the sort of situation we might look out for this weekend and 
if i can get hold of some charts for the weekend later on we'll we'll we'll 
take a look at that let me finish by taking the opposite case i hope you can 
guess what's going to happen now we're now going 
to consider warm advection once again i'll put my low level height contours in 
the same direction my V-one same direction and same strength that's my surface 
geostrophic wind and my near-surface geostrophic wind this time i'm going to 
tip the hei-, thickness contours in the opposite sense so this time i'm going 
to draw thickness contours that look like this with the warm air down here cold 
air here and again the thermal wind vector is parallel to the thickness 
contours cold air on the left so it's going to look like that that's my thermal 
wind vector let's get the winds now at the upper level by doing the vector sum 
of the low level wind and the thermal wind usual parallelogram construction 
like so that's V-two this time you will see that the wind has turned in a 
clockwise sense with height and so warm advection is associated with the wind 
veering with height i find this very fascinating it means that if i have a 
single radiosonde ascent that returns the the the winds 
by looking at the way the wind vector is changing as the er balloon ascends i 
can actually say something about the spatial distribution of the temperature 
can work out what direction the thickness contours must have at that station 
gosh i've drawn all over my lecture notes in my excitement right er you'll have 
a chance to see these ideas in practice as we go through the present case study 
we'll be looking at the winds at different levels we'll be analysing the 
thickness and if you all say i don't know how to analyse thickness well you've 
got a treat in store because er er namex's going to tell you all about that in 
in a couple of minutes er so that's the next stage in our practicals i'm going 
to hand over er to namex in a moment er i just want to make one announcement 
the er er the lads down in the lab have fixed the radiosonde system they have 
soldered the wire back onto its connector [laughter] inside the er receiver and 
i'm assured it works so we are 
hoping when namex's finished that we will have a a second launch today i don't 
think we all need to troop down to see the equipment 'cause you saw it last 
time but we will do a launch and er we'll try and er plot an ascent during the 
course of the morning so er if the people who wanted to volunteer to write down 
some numbers can remember who they were arms jerking there er then er we'll 
we'll arrange a rota for you to go down i i'll let namex finish and then we'll 
sort out the timing okay namex you've got to be wired for sound today