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<?xml version="1.0"?>

<!DOCTYPE TEI.2 SYSTEM "base.dtd">




<title>Thermal advection</title></titleStmt>

<publicationStmt><distributor>BASE and Oxford Text Archive</distributor>


<availability><p>The British Academic Spoken English (BASE) corpus was developed at the

Universities of Warwick and Reading, under the directorship of Hilary Nesi

(Centre for English Language Teacher Education, Warwick) and Paul Thompson

(Department of Applied Linguistics, Reading), with funding from BALEAP,

EURALEX, the British Academy and the Arts and Humanities Research Board. The

original recordings are held at the Universities of Warwick and Reading, and

at the Oxford Text Archive and may be consulted by bona fide researchers

upon written application to any of the holding bodies.

The BASE corpus is freely available to researchers who agree to the

following conditions:</p>

<p>1. The recordings and transcriptions should not be modified in any


<p>2. The recordings and transcriptions should be used for research purposes

only; they should not be reproduced in teaching materials</p>

<p>3. The recordings and transcriptions should not be reproduced in full for

a wider audience/readership, although researchers are free to quote short

passages of text (up to 200 running words from any given speech event)</p>

<p>4. The corpus developers should be informed of all presentations or

publications arising from analysis of the corpus</p><p>

Researchers should acknowledge their use of the corpus using the following

form of words:

The recordings and transcriptions used in this study come from the British

Academic Spoken English (BASE) corpus, which was developed at the

Universities of Warwick and Reading under the directorship of Hilary Nesi

(Warwick) and Paul Thompson (Reading). Corpus development was assisted by

funding from the Universities of Warwick and Reading, BALEAP, EURALEX, the

British Academy and the Arts and Humanities Research Board. </p></availability>




<recording dur="00:44:57" n="6048">


<respStmt><name>BASE team</name>



<langUsage><language id="en">English</language>



<person id="nm0925" role="main speaker" n="n" sex="m"><p>nm0925, main speaker, non-student, male</p></person>

<person id="sm0926" role="participant" n="s" sex="m"><p>sm0926, participant, student, male</p></person>

<person id="sf0927" role="participant" n="s" sex="f"><p>sf0927, participant, student, female</p></person>

<person id="sm0928" role="participant" n="s" sex="m"><p>sm0928, participant, student, male</p></person>

<personGrp id="ss" role="audience" size="m"><p>ss, audience, medium group </p></personGrp>

<personGrp id="sl" role="all" size="m"><p>sl, all, medium group</p></personGrp>

<personGrp role="speakers" size="6"><p>number of speakers: 6</p></personGrp>





<item n="speechevent">Lecture</item>

<item n="acaddept">Meteorology</item>

<item n="acaddiv">ps</item>

<item n="partlevel">UG1</item>

<item n="module">Synoptic Meteorology</item>




<u who="nm0925"> okay <pause dur="1.1"/> # <pause dur="1.9"/> problem sheet four i wanted to say a word or two about problem sheet four <pause dur="0.2"/> which # <pause dur="0.3"/> # i guess <pause dur="0.2"/> you had back the other week <pause dur="0.5"/> # <pause dur="0.4"/> i had <pause dur="1.2"/> vibrations from people that # <pause dur="0.5"/> # people were a bit shocked at the marks <pause dur="0.5"/> # <pause dur="0.5"/> and there was # <pause dur="0.2"/> a very straightforward reason why the marks on that sheet <pause dur="0.3"/> were somewhat lower than the marks on the earlier ones <pause dur="0.6"/> and the reason was <pause dur="0.4"/> that nearly not quite but very nearly all of you <pause dur="0.4"/> got in a mess with signs <pause dur="0.7"/> and the result was that # <pause dur="0.3"/> # one way or another you nearly all <pause dur="0.3"/> not quite all but nearly all got <pause dur="0.3"/> incorrect answers because somewhere along the line <pause dur="0.3"/> you'd got into a mess with signs <pause dur="0.8"/> and i'm afraid i did mark it harshly because i think it's quite important to make the point <pause dur="0.5"/> that we're now at a stage where signs matter <pause dur="0.4"/> they're not just a sort of frill that we add on at the end of a calculation <pause dur="0.3"/> you've actually got to keep track of the signs as you go through it <pause dur="0.3"/> otherwise <pause dur="0.2"/> you'll get a silly answer <pause dur="0.7"/> so i i want to just remind you of a couple of points which

will come in again and again in <pause dur="0.4"/> # later problems <pause dur="0.3"/> # to do with getting signs right i suspect these calculations <pause dur="0.3"/> involving thermal wind shear <pause dur="0.5"/> are amongst the most awkward <pause dur="0.6"/> or have the most potential <pause dur="0.4"/> for getting signs wrong <pause dur="0.3"/> but the same goes for other calculations you'll do <pause dur="0.7"/> first thing to remind you is <pause dur="0.4"/> # <pause dur="0.2"/> that # in nearly all these problems <pause dur="0.3"/> you do best if you break the wind down into <pause dur="0.3"/> components <pause dur="0.6"/> so let me just remind you of <kinesic desc="writes on board" iterated="y" dur="2"/> the usual convention that we use <pause dur="0.4"/> we imagine an X axis which points from the west to the east <pause dur="0.7"/> a Y axis that points from the south to the north <pause dur="0.8"/> and if you are told <pause dur="0.7"/> that there is a wind of particular magnitude <pause dur="1.8"/> shall we call it U <pause dur="1.7"/> with a particular direction and there's various ways of measuring that direction <pause dur="0.3"/> the meteorologists <pause dur="0.4"/> talk about the direction from which the wind blows and they measure the angle from true north <pause dur="0.4"/> the <pause dur="0.2"/> # mathematicians and this is probably the convention i will stick to <pause dur="0.3"/> will measure the angle of the wind vector <pause dur="0.2"/> from the X axis in the anticlockwise sense so if we have <pause dur="0.5"/> theta there <pause dur="0.3"/> which describes the <pause dur="0.3"/> # direction of the wind vector <pause dur="0.4"/> then

we have two components <pause dur="0.6"/> we have a component along the X axis which i'll call <pause dur="0.3"/> little-U <pause dur="0.6"/> and little-U is big-U <pause dur="0.4"/> times <pause dur="0.4"/> the cosine <pause dur="0.2"/> of that angle theta <pause dur="1.1"/> and then we have <pause dur="0.3"/> a component <pause dur="0.3"/> parallel to the <pause dur="0.3"/> Y axis i call that little-V <pause dur="0.4"/> and little-V is capital-U <pause dur="0.2"/> times <pause dur="0.3"/> the <pause dur="0.2"/> sine <pause dur="0.2"/> of theta <pause dur="1.2"/> now <pause dur="1.0"/> the way i've drawn it there it's easy because <pause dur="0.3"/> # <pause dur="0.3"/> both U and V are positive <pause dur="0.9"/> if the wind vector points into these various other quadrants then different signs apply <pause dur="0.5"/> so here <pause dur="0.2"/> we have both U-positive <pause dur="0.8"/> and V-positive <pause dur="1.9"/> if the wind vector was down in this quadrant <pause dur="1.2"/> then of course U <pause dur="0.3"/> would be still positive but in this case V would be directed down the negative Y axis so <pause dur="0.4"/> V would be negative <pause dur="1.3"/> similarly if the wind vector turned up in this <pause dur="0.2"/> quadrant here <pause dur="0.4"/> we'd have U-negative <pause dur="0.4"/> and V-<pause dur="0.2"/>positive still <pause dur="0.3"/> down here both U and V would be negative <pause dur="1.0"/> always worth drawing a sketch like that <pause dur="0.3"/> so that when you've done the calculation you can just check that the signs <pause dur="0.5"/> check out against that diagram <pause dur="1.4"/> if you're using a calculator to

calculate the # vectors <pause dur="0.8"/> then <pause dur="0.4"/> strictly you should automatically get the signs right if you've got this angle theta and you put in an angle of <pause dur="0.3"/> # <pause dur="0.3"/> two-hundred-and-eighty degrees or something like that it should automatically <pause dur="0.3"/> take care of the signs <pause dur="0.4"/> provided you've got your calculator set up right <pause dur="0.5"/> # <pause dur="0.2"/> 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 <pause dur="0.4"/> but at each stage <pause dur="0.3"/> to check things and a diagram is often the # easy way of doing that <pause dur="2.4"/> 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 <pause dur="1.2"/> the other <pause dur="0.7"/> bit where <pause dur="0.3"/> people went wrong and i think probably more people went wrong here <pause dur="0.6"/> # was when <pause dur="0.2"/> we were required to take vertical derivatives <pause dur="1.1"/> # <pause dur="0.2"/> we have things like D-U-by-D-P <pause dur="0.2"/> and D-V-by-D-P <pause dur="1.5"/> and what i want to do <pause dur="0.2"/> is just to remind you how we would estimate <pause dur="0.3"/> any

vertical derivative <pause dur="0.4"/> so suppose we have any <pause dur="0.2"/> meteorological quantity i'll just call it F for the moment <pause dur="0.4"/> it might be wind or temperature or whatever <pause dur="1.5"/> and we're required to estimate D-F-by-D-P <pause dur="0.3"/> the vertical rate of change of F <pause dur="1.3"/> well let's draw a little diagram once again to illustrate what we're doing <pause dur="0.3"/> very typically we might be <pause dur="0.3"/> given the value of F at say a hundred kilopascals near the surface <pause dur="0.4"/> and the value of F <pause dur="0.4"/> at fifty kilopascals halfway up through the atmosphere <pause dur="1.4"/> and <pause dur="0.2"/> there's a pressure difference between those two levels <pause dur="1.0"/> delta-P which in this case is fifty kilopascals <pause dur="1.3"/> well <pause dur="0.2"/> the way i will estimate derivatives always <pause dur="0.4"/> is <pause dur="0.3"/> by what we call finite differences we just take the value at one level <pause dur="0.3"/> minus the value at the other level <pause dur="0.3"/> divided by the delta-P <pause dur="0.7"/> and that's an estimate of the derivative <pause dur="0.6"/> not <pause dur="0.4"/> necessarily a terribly accurate one but it is an estimate <pause dur="1.9"/> the critical bit is getting it the right way round so you get the signs right <pause dur="0.2"/> and people do get in a muddle here i have

to say i get into a muddle often if i'm not thinking <pause dur="2.5"/> the way it works is like this <pause dur="0.4"/> we have to take the value of F at the higher pressure <pause dur="0.2"/> a hundred kilopascals in this case <pause dur="0.9"/> minus <pause dur="0.5"/> the value of F at the lower pressure <pause dur="1.2"/> fifty kilopascals <pause dur="1.1"/> divided by the pressure difference <pause dur="1.0"/> delta-P <pause dur="2.6"/> and <pause dur="1.4"/> this will give us the right sign <pause dur="0.8"/> notice that very often <pause dur="0.8"/> we're dealing with situations where the wind speed increases with height <pause dur="1.0"/> in other words the wind speed decreases with pressure <pause dur="0.3"/> so if this was say the U component of wind we might very frequently have the situation <pause dur="0.4"/> where U at fifty kilopascals is larger <pause dur="0.4"/> than U at a hundred kilopascals and so <pause dur="0.4"/> D-U-by-D-P would be negative <pause dur="1.1"/> and people get in a real muddle with this because of course the wind increases with height <pause dur="0.6"/> but that means it decreases with pressure <pause dur="0.4"/> so the sign is often <pause dur="0.2"/> perhaps a bit counter-intuitive <pause dur="0.4"/> so do remember that <pause dur="0.3"/> again <pause dur="0.2"/> i think the diagram always helps and a little thumbnail diagram <pause dur="0.3"/> along with your problems <pause dur="0.3"/> just to check out what

it is that you're taking differences between to estimate <pause dur="0.4"/> these derivatives will not go amiss and again <pause dur="0.5"/> don't just trust that you've done the algebra correctly when you work out the signs <pause dur="0.4"/> just check it against the diagram to make sure the sign seems to make sense to you <pause dur="0.4"/> and that way <pause dur="0.3"/> # <pause dur="0.2"/> you will get lots of marks and feel gratified and everyone'll be happy <pause dur="1.3"/> right <pause dur="1.5"/> so much for the the the the problems <pause dur="0.2"/> let me just say as a <pause dur="0.2"/> parting shot the problems really aren't meant to be hard if you're finding <pause dur="0.4"/> that you're spending hours and hours on the problems <pause dur="0.4"/> i suggest you've rather missed the point <pause dur="0.4"/> most of the problems are a matter of either picking <pause dur="0.5"/> formulae or equations out of the lecture notes <pause dur="0.4"/> and then <pause dur="0.2"/> in a subsequent problem <pause dur="0.3"/> # rearranging them maybe <pause dur="0.6"/> or putting some numbers in them <pause dur="0.2"/> or what have you it really shouldn't take hours and hours and hours to do that <pause dur="0.6"/> so <pause dur="0.2"/> # <pause dur="2.9"/> do bear that in mind <pause dur="1.4"/>

right let's get on to today's topic <pause dur="2.0"/> # today <pause dur="0.9"/> can't find the right page in my lecture notes <pause dur="8.3"/> oh here we are <pause dur="5.6"/> today our topic is called <pause dur="0.2"/> thermal advection <pause dur="6.0"/> which is a rather grand word for a rather simple concept <pause dur="2.5"/> i'd like to start off by thinking about the weather <pause dur="0.9"/> it was a cold morning this morning wasn't it </u><u who="ss" trans="latching"> mm </u><pause dur="0.7"/> <u who="nm0925" trans="pause"> it certainly was <pause dur="0.4"/> it was quite a nice afternoon yesterday wasn't it </u><pause dur="0.3"/> <u who="ss" trans="pause"> yes </u><u who="nm0925" trans="latching"> yes <pause dur="0.6"/> it was quite warm wasn't it if you went out for a walk in the sun <pause dur="0.8"/> right anyone like to give me an explanation of why <pause dur="0.2"/> it was <pause dur="0.5"/> five or ten degrees colder <pause dur="0.3"/> this morning <pause dur="0.3"/> than it was say at three o'clock yesterday afternoon <pause dur="1.1"/> yeah </u><pause dur="0.2"/> <u who="sm0926" trans="pause"> was a <pause dur="0.4"/> clear night last night </u><u who="nm0925" trans="latching"> yes </u><u who="sm0926" trans="latching"> a lot of the thermal # a lot of the infrared radiation <unclear>goes</unclear> <pause dur="0.8"/> <gap reason="inaudible" extent="2 secs"/></u><u who="nm0925" trans="overlap"> right <pause dur="0.6"/> good <pause dur="0.9"/> 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 <pause dur="0.3"/> and so the infrared radiation from the earth's surface could actually escape rather more readily <pause dur="0.6"/> # not everyone knows of course that water vapour is the most important greenhouse gas in the atmosphere <pause dur="0.3"/> although it's not the one the environmentalists

make such a fuss about <pause dur="0.4"/> but # <pause dur="0.3"/> # <pause dur="0.6"/> the difference between a <pause dur="0.6"/> a clear night that's got dry air and a <pause dur="0.2"/> clear night that's got moist air is actually quite dramatic <pause dur="0.3"/> last night it was <pause dur="0.3"/> clear and dry <pause dur="0.6"/> so yeah <pause dur="1.1"/> lots of radiation from the earth's surface the earth's surface and the layers of air lying immediately above it got pretty cold <pause dur="0.5"/> hence our frost this morning <pause dur="0.8"/> let me take you back <pause dur="0.8"/> to # <pause dur="0.5"/> Friday <pause dur="1.9"/> don't know if you can remember Friday i have difficulty remembering events before the weekend <pause dur="0.4"/> but if we go back to <shift feature="voice" new="laugh"/>Friday <shift feature="voice" new="normal"/><pause dur="1.0"/> # the <pause dur="0.4"/> <trunc>m</trunc> minimum <pause dur="0.3"/> at about six o'clock in the morning <pause dur="0.5"/> was a great deal higher it was about four degrees or so <pause dur="1.0"/> so it was four or five it was five degrees probably six degrees warmer than it was this morning <pause dur="0.7"/> so what was different <pause dur="0.9"/> about <pause dur="0.3"/> Thursday night Friday morning from Sunday night <pause dur="0.4"/> Monday morning <pause dur="1.3"/> any ideas <pause dur="2.8"/> why was it <pause dur="0.8"/> <trunc>fi</trunc> four or six degrees warmer <pause dur="0.5"/> the same time on Friday morning <pause dur="0.3"/> as this morning </u><pause dur="0.2"/> <u who="sf0927" trans="pause"> <gap reason="inaudible" extent="1 sec"/></u><pause dur="1.2"/> <u who="nm0925" trans="pause"> a front went past well done <pause dur="0.3"/> you've

got the right answer straight away i was hoping someone was going to witter on about clouds and so on <pause dur="0.5"/> well <pause dur="0.2"/> there are there are two things actually <pause dur="0.3"/> we might have said well it was a cloudy night and therefore <pause dur="0.2"/> the radiative effect we've just talked about couldn't operate <pause dur="2.3"/> but you're dead right <pause dur="0.8"/> the difference was <pause dur="0.4"/> that a front went through <pause dur="0.3"/> we were actually sitting in a different air mass <pause dur="0.3"/> on <pause dur="0.2"/> Thursday morning <pause dur="0.5"/> and over the weekend <pause dur="0.3"/> the front came through it went dramatically colder it snowed <pause dur="0.8"/> as we shall talk about later on this morning <pause dur="0.5"/> in some places anyway <pause dur="0.4"/> and # <pause dur="0.4"/> # so it's altogether colder one mass of air <pause dur="0.2"/> with one set of thermodynamic properties <pause dur="0.5"/> has been replaced by another <pause dur="0.6"/> and it's that replacement of air with one set of characteristics by air with a different set of characteristics <pause dur="0.6"/> which is what's meant by advection <pause dur="0.2"/> and that's what we're going to talk about today <pause dur="1.3"/> 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 <pause dur="0.6"/> if i sit at a meteorological station <pause dur="0.7"/> and i observe a change in the temperature <pause dur="1.5"/> then there are two <pause dur="0.3"/> alternative explanations for that change in temperature <pause dur="0.3"/> and in many cases of course <pause dur="0.2"/> both of of them operate <pause dur="1.0"/> either <pause dur="1.0"/> the air sitting at that station has actually changed its properties in some sense <pause dur="0.5"/> so your explanation that <pause dur="0.2"/> there was lots of radiation going on last night lots of infrared <pause dur="0.3"/> escaping <pause dur="0.2"/> and therefore the air got colder <pause dur="0.7"/> that could be the case in one set of situations <pause dur="2.3"/> and that's what we sometimes call heating or or diabatic processes <pause dur="2.2"/> on the other hand <pause dur="0.2"/> the temperature at a station <pause dur="0.3"/> may change <pause dur="0.8"/> simply because <pause dur="0.5"/> the individual air parcels are are not changing they retain their temperature but they move away from your station <pause dur="0.4"/> and they're replaced by another set of air parcels that come along with different properties <pause dur="0.9"/> so <pause dur="0.3"/> during the course of Friday and Saturday <pause dur="0.5"/> warm moist air <pause dur="0.3"/> moved

away from the <gap reason="name" extent="1 word"/> area <pause dur="0.3"/> to be replaced by much colder drier air <pause dur="1.5"/> and that's what we call advection <pause dur="1.0"/> so the concept is simple enough what we need to do now <pause dur="0.3"/> is to <pause dur="0.2"/> # put some # quantitative details on it <pause dur="1.3"/> so i'm going to draw a diagram to fix the ideas in our head <pause dur="0.3"/> let's draw a diagram shall we <pause dur="0.3"/> of <pause dur="0.3"/> temperature <pause dur="0.9"/> and <pause dur="0.5"/> let's consider one space direction i'll call it X for the moment <pause dur="0.7"/> and suppose i have an observing site here <pause dur="1.0"/> i'll call this <pause dur="0.2"/> point A <pause dur="0.9"/> and at the observing site we observe a particular temperature <pause dur="0.4"/> there we are that's the temperature <pause dur="0.5"/> at <pause dur="0.5"/> # point A <pause dur="5.3"/> 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 <pause dur="0.3"/> i'd observe a different temperature <pause dur="0.6"/> so <pause dur="0.3"/> let's # imagine that the temperatures along this X axis <pause dur="0.5"/> they go like this i i'll draw a rather simple example where they just go along a <pause dur="0.2"/> an increasing line like that <pause dur="0.9"/> so stations to the east of A <pause dur="0.2"/> are

warm <pause dur="0.6"/> stations to the west of A in this case are cold <pause dur="2.6"/> # let us now suppose that a wind is blowing <pause dur="0.2"/> okay and i'm going to keep things dead simple <pause dur="0.4"/> by supposing that the wind is blowing parallel to the X axis <pause dur="0.6"/> # 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 <pause dur="0.4"/> but let's just suppose there is a wind <pause dur="0.4"/> blowing along the X axis <pause dur="1.0"/> there we are wind magnitude U <pause dur="0.4"/> <trunc>pa</trunc> blowing parallel to the the X-axis <pause dur="2.4"/> and <pause dur="0.3"/> time passes <pause dur="2.0"/> so <pause dur="0.8"/> after <pause dur="0.2"/> some time interval delta-T <pause dur="1.2"/> our observing site is no longer seeing this parcel of air with a temperature T-A <pause dur="0.7"/> what it will actually be sampling is an air parcel that started off <pause dur="0.5"/> some distance upstream <pause dur="0.7"/> so <pause dur="0.2"/> # <pause dur="1.8"/> let's suppose we take that point <pause dur="0.4"/> B shall we say <pause dur="0.9"/> that point B is a distance U <pause dur="0.7"/> times the time interval delta-T <pause dur="0.2"/> upstream <pause dur="2.7"/> and so after a time delta-T <pause dur="0.4"/> the parcel that started off at B <pause dur="0.9"/> will have reached <pause dur="0.3"/> A <pause dur="0.9"/> and so i'll measure a new temperature at station A after this time interval <pause dur="0.5"/> # i'll

measure this temperature i'll call that T-B <pause dur="10.3"/> let's see if i can write down <pause dur="1.3"/> a relationship <pause dur="1.0"/> between those two temperatures <pause dur="2.9"/> 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 <pause dur="0.4"/> you might want to adjust the notation as as we're going along <pause dur="1.7"/> what we're interested in now is the temperature T-B <pause dur="0.8"/> which is the temperature that's now reached my observing site <pause dur="2.1"/> and we can see here <pause dur="0.2"/> that if this <pause dur="0.6"/> curve here <pause dur="0.7"/> doesn't depart too much from a straight line between stations A and B <pause dur="0.3"/> and it won't provided that time interval delta-T is sufficiently small <pause dur="1.0"/> # <pause dur="0.2"/> then it's rather easy to do the calculation we could say that T-<pause dur="0.3"/>B is going to be T-A <pause dur="0.8"/> minus <pause dur="1.5"/> the distance delta-X <pause dur="0.4"/> to point B <pause dur="1.0"/> multiplied by the gradient <pause dur="2.7"/> D-T-by-D-X the rate of change of temperature along the X axis <pause dur="0.5"/> so that distance there is delta-X and we can see that that is equal to <pause dur="0.4"/> U times delta-T <pause dur="4.3"/>

so <pause dur="0.8"/> if i rearrange that expression <pause dur="0.4"/> i can <pause dur="0.2"/> get an expression for the <pause dur="0.2"/> rate of change <pause dur="0.3"/> of temperature <pause dur="0.7"/> at <pause dur="0.3"/> station <pause dur="0.6"/> A <pause dur="6.8"/> so rate of change of temperature at <pause dur="0.3"/> my station A <pause dur="0.5"/> well <pause dur="0.2"/> i'm i'm going to approximate it by T-B <pause dur="0.4"/> minus T-A <pause dur="0.6"/> divided by the time interval <pause dur="0.4"/> delta-T <pause dur="3.1"/> 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 <pause dur="0.6"/> then i get an expression for T-A <pause dur="0.4"/> minus T-B <pause dur="0.2"/> and then i've just to divide it by delta-T and if i do that <pause dur="0.2"/> and don't tell me to keep the signs i've got a minus sign here <pause dur="0.7"/> i've got a delta-X over a delta-T <pause dur="1.8"/> times <pause dur="1.0"/> D-T-by-D-X <pause dur="6.4"/> it's not hard it's just simple algebra here <pause dur="1.3"/> what i'm going to do now <pause dur="0.4"/> is my usual calculus trick <pause dur="0.7"/> i'm going to suppose <pause dur="0.4"/> that delta-T and delta-X become very small <pause dur="0.2"/> i'm going to take the limit <pause dur="0.3"/> in which delta-X and delta-T tend to zero <pause dur="0.7"/> if i do that <pause dur="0.6"/> then that simply becomes the rate of change <pause dur="0.7"/> of temperature with respect to time <pause dur="0.7"/> if you want to

you can say that's at the point A <pause dur="2.1"/> and that's going to be equal to <pause dur="0.6"/> well in the limit <trunc>del</trunc> of delta-X and delta-T becoming very small <pause dur="0.7"/> delta-X over delta-T that's just the rate of change of position <pause dur="0.4"/> of an air parcel <pause dur="0.4"/> along the X axis <pause dur="0.5"/> it's the <pause dur="0.2"/> velocity it's the speed U <pause dur="1.3"/> so i'm going to get minus-<pause dur="0.2"/>U <pause dur="0.7"/> times <pause dur="0.5"/> D-T-by-D-X <pause dur="3.5"/> and so <pause dur="1.5"/> that is an expression which tells us about the rate of change <pause dur="0.4"/> of temperature <pause dur="0.5"/> at <pause dur="0.5"/> a fixed point <pause dur="0.3"/> in space a fixed observing site <pause dur="0.9"/> on the assumption <pause dur="0.6"/> that the only process that's changing the temperature <pause dur="0.3"/> is this advection effect in other words <pause dur="0.3"/> the replacement of one air parcel <pause dur="0.4"/> with an air parcel of different properties <pause dur="4.5"/> so we'll put a box round that 'cause it's important <pause dur="1.3"/> so it's the rate of change <pause dur="3.2"/> due to <pause dur="1.1"/> advection <pause dur="4.2"/> and the # <pause dur="3.2"/> rate of change due to advection is given a name <pause dur="1.5"/> it's sometimes called <pause dur="0.4"/> the Eulerian <pause dur="0.3"/> rate of change <pause dur="2.0"/> Euler was a famous <pause dur="0.8"/> French mathematician <pause dur="0.9"/> who had more laws and formulae named after him <pause dur="0.6"/> # than most of us have <pause dur="0.6"/> and this

is one that's named after him <pause dur="6.9"/> put a little a box round that it's an important bit of terminology you'll hear referred to <pause dur="0.4"/> the Eulerian rate of change <pause dur="0.4"/> is the rate of change <pause dur="0.2"/> at a fixed point in space <pause dur="0.9"/> the rate of change you would estimate for example <pause dur="0.5"/> if you're a meteorologist sitting <pause dur="0.3"/> at a fixed observing site on the earth's surface <pause dur="0.4"/> watching the winds blow past you <pause dur="2.3"/> course it's not the only rate of change <pause dur="0.3"/> that we might consider <pause dur="0.8"/> suppose we didn't sit at a fixed observing site <pause dur="0.4"/> suppose we attached ourselves to <pause dur="0.2"/> some sort of balloon <pause dur="0.5"/> which was wafted around by the winds <pause dur="1.4"/> in that case <pause dur="0.2"/> if we had a thermometer attached to that balloon <pause dur="0.8"/> # we wouldn't see the Eulerian rate of change at all <pause dur="1.1"/> in fact if we designed the balloon very carefully <pause dur="0.5"/> we would actually see the rate of change of individual air parcels because our balloon <pause dur="0.4"/> would effectively be always embedded in the same lump of air as it moved around <pause dur="0.6"/> actually that's a pretty bad example 'cause it's very hard <pause dur="0.4"/> to design a

balloon that does that <pause dur="0.3"/> the problem is that the <pause dur="0.3"/> balloon usually floats at a constant height <pause dur="0.3"/> so it follows the winds at that height but it doesn't rise and sink with the air parcels <pause dur="0.6"/> # anyway that's # <pause dur="0.2"/> that's a that's a technical difficulty let's suppose we could do that <pause dur="2.3"/> so we have this other rate of change <pause dur="0.7"/> the rate of change following an individual fluid element <pause dur="1.4"/> and that in fact is identical to the rate of change that we mentioned earlier on if if an air parcel cools <pause dur="0.4"/> because of radiation <pause dur="0.4"/> we're actually talking there about this rate of change following the fluid element we're thinking of this air parcel <pause dur="0.3"/> that's sitting there <pause dur="0.3"/> radiating infrared radiation to space and thereby changing its temperature <pause dur="1.2"/> so we need another notation for that <pause dur="0.6"/> 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 <pause dur="0.5"/> with notation on occasions <pause dur="0.2"/> so <pause dur="0.8"/> # <pause dur="1.7"/> i'm now going to consider the rate of change <pause dur="4.5"/><kinesic desc="writes on board" iterated="y" dur="16"/> following <pause dur="5.2"/> an

individual <pause dur="3.6"/> fluid parcel <pause dur="4.9"/> so what we would do here <pause dur="0.3"/> is to measure the temperature of our air parcel at some time T <pause dur="0.5"/> we would then measure the <pause dur="0.3"/> # temperature of our air parcel at some later time T-plus-delta-T shall we say <pause dur="0.4"/> we'd subtract the two divide by the time interval <pause dur="0.4"/> take the limit as delta-T tends to zero <pause dur="0.3"/> and we'd have a rate of change of temperature <pause dur="0.5"/> following the fluid parcel <pause dur="0.7"/> and we'll use a special notation for that <pause dur="1.2"/> we will define that <kinesic desc="writes on board" iterated="y" dur="11"/> as capital-D <pause dur="2.3"/> D-T <pause dur="0.3"/> i'm using too many Ts here but still <pause dur="2.1"/> by D-little-T <pause dur="2.0"/> and that's the notation we will use <pause dur="0.7"/> this is sometimes called <pause dur="2.0"/> it's the rate of change for an individual fluid parcel <pause dur="0.3"/> and this is sometimes called the Lagrangian <pause dur="4.4"/><kinesic desc="writes on board" iterated="y" dur="11"/> rate of change <pause dur="5.7"/> Lagrange was also <pause dur="0.3"/> an eighteenth century French mathematician <pause dur="0.4"/> who has <pause dur="0.3"/> # a vast number of equations and formulae named after him and this is just one of them <pause dur="1.3"/> so they they they were good at getting their names on equations in the eighteenth century French republic anyway <pause dur="0.3"/>

# <pause dur="0.6"/> rates of change <pause dur="0.3"/> so <pause dur="0.5"/> we've got these two rates of change <pause dur="0.4"/> and very important they are too <pause dur="0.5"/> the rate of change at a fixed point in space <pause dur="0.7"/> now that's the sort of thing you might say meteorologists are interested in 'cause we have <pause dur="0.4"/> fixed observing sites <pause dur="0.4"/> we're required to produce weather forecasts for fixed points on the earth's surface <pause dur="0.3"/> the forecast for <gap reason="name" extent="1 word"/> or whatever <pause dur="1.5"/> and in other words <pause dur="0.4"/> the meteorologist is interested in the <pause dur="0.5"/> Eulerian <pause dur="0.2"/> rate of change of air properties <pause dur="2.8"/> but in a sense <pause dur="0.2"/> more physically fundamental is the <pause dur="0.2"/> Lagrangian rate of change if i follow an individual air parcel what happens to it <pause dur="0.6"/> does heat enter or leave it <pause dur="0.9"/> does it rise or sink in the atmosphere changes pressure <pause dur="0.6"/> whatever <pause dur="1.8"/> and if you cast your mind back to <pause dur="0.8"/> your work last term <pause dur="0.3"/> and our work this term <pause dur="0.9"/> most of the physical laws that we've written down <pause dur="0.2"/> that apply to air <pause dur="0.4"/> for example <pause dur="0.3"/> the first law of thermodynamics <pause dur="0.6"/> for example Newton's laws of motion <pause dur="0.9"/> they all apply to air parcels <pause dur="0.8"/> we ask how does an air parcel

accelerate <pause dur="0.3"/> how does its temperature change when you pump heat into it <pause dur="0.2"/> and so on <pause dur="1.1"/> most of the physical laws <pause dur="0.3"/> that govern the atmosphere <pause dur="0.4"/> are expressed in terms of Lagrangian <pause dur="0.2"/> rates of change <pause dur="0.8"/> so we have this problem <pause dur="0.7"/> what we measure and what we want to predict are Eulerian rates of change <pause dur="1.3"/> what our physics tells us about are Lagrangian <pause dur="0.2"/> rates of change <pause dur="0.6"/>

so <pause dur="0.4"/> what's going to be the relationship <pause dur="0.6"/> between <pause dur="0.5"/> the <pause dur="0.2"/> Eulerian and the Lagrangian rate of change <pause dur="0.6"/> well it's fairly straightforward <pause dur="0.4"/> that of course is the rate of change of temperature the Eulerian rate of change of temperature <pause dur="0.4"/> if i assume air parcels are not changing <pause dur="0.7"/> it's the rate of change in other words if capital-<pause dur="0.3"/>D-T-by-D-T <pause dur="0.3"/> is zero <pause dur="0.9"/> all that's happening is that one air parcel is moving away <pause dur="0.3"/> and another one with different properties is taking its place but the individual air parcels aren't changing their properties <pause dur="2.7"/> if <pause dur="0.4"/> the air parcels are actually <pause dur="0.4"/> 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 <pause dur="0.4"/> with time so that curve was dropping down as <pause dur="0.2"/> as it was moving along <pause dur="1.0"/> then <pause dur="0.3"/> i'd have to add into that formula <pause dur="0.5"/> the Lagrangian rate of change <pause dur="0.6"/> so <pause dur="0.4"/> what i will write is that # <pause dur="0.6"/> in general <pause dur="7.0"/><kinesic desc="writes on board" iterated="y" dur="47"/> when <pause dur="2.4"/> there are both <pause dur="2.7"/> Lagrangian <pause dur="4.1"/> and Eulerian <pause dur="0.3"/> changes <pause dur="4.8"/> then we can say <pause dur="0.2"/> the rate of change at a particular <pause dur="0.5"/> location <pause dur="1.7"/> D-U-by-D-T <pause dur="0.4"/> is going to be the Lagrangian rate of change <pause dur="2.4"/> capital-<pause dur="0.5"/>D-by-D-T <pause dur="0.3"/> minus <pause dur="0.2"/> this advective rate of change <pause dur="0.4"/> minus-U times D-T-by-D-X <pause dur="2.3"/> and that's really the central result <pause dur="0.2"/> of today's lecture <pause dur="4.3"/> it gives us the relationship between rates of change at a position <pause dur="1.3"/> and <pause dur="0.2"/> the rate of change for an individual parcel <pause dur="1.2"/> and they're related <pause dur="0.6"/> via <pause dur="0.2"/> the velocity <pause dur="0.3"/> and the temperature gradient <pause dur="5.0"/> 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 <pause dur="0.6"/> it's dead easy to generalize this <pause dur="0.3"/> if i have a general wind <pause dur="0.4"/> then that will have components parallel to all three axes X Y and Z <pause dur="0.9"/> and of course the temperature may vary in all three directions X Y and Z as in general it does <pause dur="0.6"/> so i can generalize this <pause dur="0.5"/> for # three-dimensional <pause dur="0.2"/> motion <pause dur="3.0"/><kinesic desc="writes on board" iterated="y" dur="52"/> and i won't go through the arguments of course they just work exactly the same way <pause dur="0.9"/> but just write a result down <pause dur="1.1"/> in three dimensions then we can say the rate of change at a fixed position the Eulerian rate of change <pause dur="0.4"/> is equal to the Lagrangian rate of change <pause dur="1.0"/> the rate of change <pause dur="0.4"/> for an individual air parcel <pause dur="0.4"/> minus well the same term <pause dur="0.8"/> U times D-T-by-D-X U remember now is the component of wind parallel to the X axis <pause dur="1.2"/> minus <pause dur="0.2"/> V <pause dur="0.2"/> times D-T-by-D-Y <pause dur="0.7"/> V is the component of wind parallel to the Y-axis <pause dur="0.5"/> D-T-D-Y <pause dur="0.3"/> is the <pause dur="0.2"/> rate of change of temperature along the Y-axis <pause dur="0.4"/> and then of course we've got the vertical one minus-W <pause dur="0.4"/> times D-T-by-D-Z <pause dur="0.6"/> it looks a bit long <pause dur="0.6"/> but that's just because <pause dur="0.7"/> we've written

out the same terms three times yes </u><pause dur="0.4"/> <u who="sm0928" trans="pause"> # <pause dur="0.4"/> # i might be <pause dur="1.0"/> going along the totally wrong lines but </u><u who="nm0925" trans="overlap"> mm </u><u who="sm0928" trans="overlap"> can you not use the <gap reason="inaudible" extent="1 sec"/><pause dur="0.3"/> the <gap reason="inaudible" extent="1 sec"/><pause dur="0.2"/> # <gap reason="inaudible" extent="1 sec"/></u><u who="nm0925" trans="overlap"> we could <pause dur="0.4"/> # <pause dur="0.2"/> i'm not going to do that 'cause people'll find it so hard their brains will seize up <pause dur="0.5"/> but <trunc>w</trunc> we could <pause dur="0.6"/> and # i'll # thank you for asking that this has # this has # brought a warm <pause dur="0.3"/> glow to an old man's heart <pause dur="0.4"/><kinesic desc="writes on board" iterated="y" dur="11"/> # <pause dur="0.5"/> we could write this much more neatly <pause dur="0.3"/> as D-by-D-T <pause dur="0.2"/> minus the vector wind <pause dur="0.5"/> U <pause dur="0.5"/> dot <pause dur="0.3"/> grad <pause dur="0.5"/> temperature <pause dur="1.1"/> okay </u><pause dur="0.3"/> <u who="sm0928" trans="pause"> yeah </u><pause dur="0.2"/> <u who="nm0925" trans="pause"> happy <pause dur="0.2"/> right <pause dur="0.2"/> write that down <pause dur="1.1"/> # <pause dur="0.5"/> this is notation that we will come to <pause dur="0.2"/> much more compact notation <pause dur="0.4"/> which we will come to <pause dur="0.3"/> # <pause dur="0.4"/> next year <pause dur="0.2"/> next autumn's geophysical fluid dynamics course we'll redo this argument <pause dur="0.5"/> # using vector notation and it all comes out a lot more compact <pause dur="0.4"/> i'm not going to use that for this course <pause dur="0.4"/> because not all of you have met this notation yet <pause dur="0.4"/> and # <pause dur="0.3"/> # people find it hard enough in the second year when they do so <pause dur="1.1"/> we'll we'll save that for next year <pause dur="2.9"/> okay <pause dur="2.0"/> well <pause dur="1.5"/> 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 <pause dur="0.5"/> due to <pause dur="0.2"/> advection <pause dur="0.7"/> replacement of one air parcel by another <pause dur="0.8"/> and i <pause dur="0.4"/> i'm going to make an additional assumption <pause dur="2.0"/> suppose we have a situation where the wind is essentially <pause dur="0.3"/> geostrophic <pause dur="1.1"/> that's the situation that's we've been # <pause dur="0.4"/> # assuming in most of our analysis of synoptic weather charts <pause dur="2.5"/> then we get <pause dur="0.2"/> a very <pause dur="0.2"/> useful <pause dur="0.5"/> # <pause dur="0.4"/> # <pause dur="0.9"/> set of relationships <pause dur="1.4"/> and the reason that we get them you see is that # <pause dur="0.5"/> thermal wind relationship <pause dur="0.8"/> if you # recall <pause dur="0.4"/> 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 <pause dur="0.7"/> and the change of the wind in the vertical <pause dur="0.3"/> D-P-by-D-Z <pause dur="0.2"/> or sorry D # sorry <pause dur="0.3"/> D-U-by-D-P <pause dur="0.4"/> or #<pause dur="0.2"/> D-V-by-D-P <pause dur="1.4"/> so <pause dur="0.2"/> there's an intimate relationship <pause dur="0.2"/> between advection <pause dur="0.6"/> and the wind changing with height <pause dur="1.3"/> and # <pause dur="0.3"/> # <pause dur="0.4"/> i want to <pause dur="0.5"/> # <pause dur="0.9"/> take a look at that now <pause dur="1.8"/> you'll remember that at the end of the # lecture two lectures ago when we talked about thermal wind balance <pause dur="0.4"/> i introduced this concept of the thermal wind <pause dur="3.4"/><kinesic desc="writes on board" iterated="y" dur="6"/> as a

reminder the thermal wind <pause dur="1.7"/> which <pause dur="0.2"/> # i'll just write its magnitude down for the moment i called it V-subscript-T for thermal <pause dur="1.1"/> and <pause dur="0.2"/> it's equal to G-over-F <pause dur="1.1"/> times the gradient of the thickness so it's D-by-D-Y <pause dur="1.0"/> of <pause dur="0.5"/> Z at some high level minus-Z-one at some lower level <pause dur="1.3"/> typically Z-two would be the fifty kilopascal <pause dur="0.7"/> # <pause dur="0.4"/> pressures at the height of the <pause dur="0.3"/> fifty kilopascal surface <pause dur="0.3"/> Z-one would be the height of the hundred kilopascal <pause dur="0.5"/> pressure surface <pause dur="2.8"/> so <pause dur="0.3"/> we could plot it on a chart <pause dur="0.5"/> # it bears just the same relationship to lines of thickness as the geostrophic wind vector bears to <pause dur="0.3"/> lines of constant pressure or geopotential height <pause dur="4.6"/> and <pause dur="0.5"/> we can use these ideas now to <pause dur="0.2"/> discuss a number of <pause dur="0.7"/> typical cases of the wind varying with height <pause dur="0.6"/> 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 <pause dur="0.5"/> what's happening <pause dur="2.2"/> let's take the simplest case first of all <pause dur="0.4"/> where the wind speed <pause dur="0.2"/> changes with height but the wind direction does not <pause dur="0.8"/> so we have a

geostrophic wind near the surface which is in a particular direction <pause dur="0.5"/> the wind at say five-hundred millibars is in the same direction but stronger <pause dur="1.0"/> not an uncommon situation <pause dur="1.0"/> let's do a little drawing of what's going on <pause dur="2.6"/> # let me draw thickness contours <pause dur="0.5"/> in <pause dur="1.9"/><kinesic desc="writes on board" iterated="y" dur="35"/> red <pause dur="1.3"/> and let me <pause dur="0.4"/> draw # <trunc>s</trunc> the # <pause dur="0.4"/> # height contours at the lower level <pause dur="2.2"/> and <pause dur="0.7"/> i'll draw them looking like that so that's shall we say the hundred kilopascal <pause dur="2.1"/> Z <pause dur="1.4"/> and the red contours <pause dur="0.3"/> show the thickness lines <pause dur="1.5"/> so that's the <pause dur="0.5"/> hundred to fifty kilopascal <pause dur="1.4"/> thickness <pause dur="1.4"/> if you look on the charts in the corridor you'll see that's what we actually plot <pause dur="0.4"/> people <pause dur="0.3"/> tend to <pause dur="0.2"/> put the surface pressure say <pause dur="0.3"/> but with lines of constant thickness on as well <pause dur="0.8"/> now in this case the surface geostrophic wind <pause dur="1.4"/><kinesic desc="writes on board" iterated="y" dur="20"/> looks like that <pause dur="0.6"/> # i'm going to call that <pause dur="0.3"/> V-one <pause dur="3.8"/> and the thermal wind well remember that's parallel to lines of constant thickness <pause dur="2.3"/> and inversely proportional to their spacing so <pause dur="0.9"/> there's going to be <pause dur="0.3"/> V-two <pause dur="0.2"/> in the same direction <pause dur="1.8"/> and if i add the

two together <pause dur="0.3"/> i'll get the wind at the <pause dur="0.3"/> second height at # the five-hundred millibar level <pause dur="0.5"/> so the wind at the five-hundred millibar level <pause dur="0.3"/> is just going to be the sum of those two <pause dur="0.5"/> it's going to look like that <pause dur="3.9"/> notice <pause dur="0.8"/> that both those wind vectors V-one and V-two are parallel <pause dur="0.9"/> to the thickness contours <pause dur="0.8"/> thickness remember is proportional to temperature <pause dur="0.4"/> that means <pause dur="0.3"/> that there are no variations of temperature <pause dur="0.3"/> in the direction the wind is blowing <pause dur="1.3"/> 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 <pause dur="0.4"/> with one of the same temperature <pause dur="0.7"/> in other words in this situation <pause dur="0.6"/> there is no <pause dur="0.2"/> thermal advection <pause dur="0.6"/> the

temperature <pause dur="0.2"/> change <pause dur="0.4"/> <trunc>i</trunc> <pause dur="0.2"/> is not due to replacing air parcels 'cause we're just replacing one air parcel with another one <pause dur="0.2"/> of the same temperature <pause dur="1.3"/> so we have <kinesic desc="writes on board" iterated="y" dur="8"/> no <pause dur="1.1"/> thermal <pause dur="0.3"/> advection <pause dur="6.3"/> i have to say that's not an uncommon situation but it's a <pause dur="0.2"/> boring situation <pause dur="0.7"/> let's <pause dur="0.2"/> consider a much

more exciting and interesting situation <pause dur="4.7"/> what we've had over this weekend is a situation which is called cold advection <pause dur="1.0"/> that is to say warm air <pause dur="0.3"/> has been displaced by cold air blowing down from the north <pause dur="0.9"/> let's have a look at what that implies in terms of <pause dur="0.3"/> the height and thickness <pause dur="2.1"/> let me again draw the thousand millibar height <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="10"/> contours looking like that <pause dur="0.3"/> 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 <trunc>w</trunc> i ought to stress that i'm talking about vectors here <pause dur="0.4"/> the direction is important as well as the magnitude <pause dur="0.6"/> so <pause dur="0.9"/> there's V-one <pause dur="0.2"/> that's the surface geostrophic wind but now this time <pause dur="0.5"/> i'm going to suppose <pause dur="0.6"/> that the <pause dur="0.2"/> # height contours <pause dur="0.4"/> are no longer <pause dur="0.2"/> is the sorry the thickness contours are no longer parallel <pause dur="0.2"/> to these height contours <pause dur="0.4"/> so i'm going to draw my <pause dur="0.3"/><kinesic desc="writes on board" iterated="y" dur="3"/> thickness contours <pause dur="0.7"/> looking like that <pause dur="1.4"/> and i'm going to suppose it's colder <pause dur="0.3"/> here <pause dur="0.5"/> so that's low values of thickness <pause dur="0.4"/> and it's warmer here <pause dur="0.3"/> so we have <pause dur="0.3"/> large values of

thickness here <pause dur="1.5"/> in this case <pause dur="0.7"/> the thermal wind vector <pause dur="0.7"/> is going to be <pause dur="0.6"/> parallel to the thickness contours <pause dur="1.1"/> and with <pause dur="0.2"/> cold air on the left so the thermal wind vector <pause dur="0.4"/> is going to <pause dur="0.2"/> look like that <pause dur="2.3"/> okay <pause dur="1.8"/> now <pause dur="0.2"/> what about the wind at five-hundred millibars <pause dur="1.0"/> well it's simply going to be the vector sum <pause dur="0.4"/> of the low level wind <pause dur="0.5"/> plus the thermal wind <pause dur="0.7"/> so if i # <pause dur="0.5"/> construct that using my usual <pause dur="0.4"/> # <pause dur="0.4"/> triangle <pause dur="0.8"/> construction <pause dur="3.8"/> i will find that V-two will look like that <pause dur="2.4"/> very interesting isn't it <pause dur="0.4"/> what you will see is there <pause dur="0.3"/> that as you go up from the surface <pause dur="0.4"/> to the upper layer <pause dur="1.5"/> the wind vector <pause dur="0.4"/> turns <pause dur="2.5"/> it turns in an anticlockwise sense <pause dur="0.7"/> and we describe this <pause dur="0.4"/> as the wind <pause dur="1.7"/> backing with height <pause dur="4.7"/> and this is a situation where we have cold advection <pause dur="7.7"/> that's the sort of situation we might look out for this weekend and if i can get hold of some charts <pause dur="0.5"/> for the weekend <pause dur="0.2"/> later on we'll we'll we'll take a look at that <pause dur="4.0"/> let me finish <pause dur="0.6"/> by <pause dur="0.2"/> taking the opposite case i hope you can guess what's going to happen now <pause dur="0.3"/> we're now going

to consider warm advection <pause dur="8.1"/> once again i'll put my low level height contours in the same direction <pause dur="0.4"/> my V-one <pause dur="0.3"/> same direction and same strength <pause dur="1.5"/> that's my surface geostrophic wind and my near-surface geostrophic wind <pause dur="0.6"/> this time <pause dur="0.3"/> i'm going to tip <pause dur="0.3"/> the <pause dur="0.4"/> <trunc>hei</trunc> thickness contours in the opposite <pause dur="0.3"/> sense <pause dur="0.4"/> so this time i'm going to draw thickness contours that look like this with the warm air down here <pause dur="1.3"/><kinesic desc="writes on board" iterated="y" dur="4"/> cold air <pause dur="1.3"/> here <pause dur="1.5"/> and <pause dur="0.7"/> again the <pause dur="0.3"/> thermal wind vector is parallel to the thickness contours <pause dur="0.4"/> cold air on the left <pause dur="0.7"/><kinesic desc="writes on board" iterated="y" dur="3"/> so <pause dur="0.2"/> it's going to look like that <pause dur="0.9"/> that's my thermal wind vector <pause dur="1.7"/> let's get the winds now at the upper level by doing the vector sum <pause dur="0.2"/> of the low level wind and the thermal wind <pause dur="1.0"/> usual parallelogram construction <pause dur="1.5"/><kinesic desc="writes on board" iterated="y" dur="5"/> like so <pause dur="1.4"/> that's V-two <pause dur="4.0"/> this time <pause dur="0.2"/> you will see <pause dur="0.2"/> that the wind <pause dur="0.2"/> has turned in a clockwise sense <pause dur="0.3"/> with height <pause dur="2.2"/> and so warm advection <pause dur="0.4"/> is associated with the wind <pause dur="1.5"/><kinesic desc="writes on board" iterated="y" dur="10"/> veering <pause dur="3.8"/> with height <pause dur="4.4"/> i find this very fascinating <pause dur="0.4"/> it means that if i have a single radiosonde ascent that returns the the the winds <pause dur="0.5"/>

by looking at the way the wind vector is changing <pause dur="0.4"/> as the # balloon ascends i can actually say something about the spatial distribution of the temperature <pause dur="0.6"/> can work out what direction the thickness contours <pause dur="0.3"/> must have <pause dur="0.5"/> at that station <pause dur="7.8"/> gosh i've drawn all over my lecture notes in my excitement <pause dur="1.0"/> right <pause dur="1.3"/> # 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 <pause dur="0.4"/> the thickness <pause dur="0.4"/> and if you all say i don't know how to analyse thickness well you've got a treat in store because <pause dur="0.4"/> # <pause dur="0.3"/> # <gap reason="name" extent="1 word"/>'s going to tell you all about that in in a couple of minutes <pause dur="0.6"/> # <pause dur="0.2"/> so <pause dur="0.9"/> that's the next stage in our practicals <pause dur="0.2"/> i'm going to hand over # to <gap reason="name" extent="1 word"/> in a moment <pause dur="0.4"/> # i just want to make one announcement the # <pause dur="0.2"/> # the lads down in the lab have fixed the radiosonde system <pause dur="0.3"/> they have soldered the wire <pause dur="0.3"/> back onto its connector <vocal desc="laughter" n="ss" iterated="y" dur="1"/> inside the # receiver <pause dur="0.4"/> and i'm assured it works <pause dur="0.3"/> so <pause dur="0.2"/> we are

hoping <pause dur="0.6"/> when <gap reason="name" extent="1 word"/>'s finished <pause dur="0.3"/> that we will have a a second launch today <pause dur="0.3"/> 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 <pause dur="0.4"/> and # we'll try and # plot an ascent <pause dur="0.4"/> during the course of the morning <pause dur="0.3"/> so <pause dur="0.2"/> # if the people who wanted to volunteer to <pause dur="0.3"/> write down some numbers can remember who they were <pause dur="0.3"/> arms jerking there <pause dur="0.3"/> # <pause dur="0.2"/> then # <pause dur="0.2"/> we'll we'll arrange a rota for you to go down <pause dur="0.2"/> i i'll let <gap reason="name" extent="1 word"/> finish and then we'll sort out the timing <pause dur="0.6"/> okay <pause dur="3.2"/> <gap reason="name" extent="1 word"/> you've got to be wired for sound today