Question re. thermal winds..MET

[Fastair345]

Have gone 50% of the way through the Oxford notes and find them pretty clear however I am struggling with thermal winds (no puns please !). I am hitting a brick wall :ugh:with the understanding and would appreciate some clarity on this question...

In the north'n hemispere the wind at 2000 ft is 210/60. the thermal wind is 20 kts with the lowest mean temp in the northwest

the upper wind is ;
a) 210/40
b) 210/80
c) 300/40
d) 030/80

Has someone found a set of notes that explains thermal winds more clearly?

cheers Fastair

[paco]

The thermal wind will be 235/20 - Buys Ballot applies with the colder air to the left and your back to the wind. NW - 90 degs = SW. As westerly winds will increase their speed but tend to keep their direction, I would go for B.

Phil

[Obs cop]

Fastair,

This is a vector resolution question requiring the CRP5.

First you need to grasp Buys Ballots Law fully. It applies equally to thermal winds and geostrophic winds.

In essence with the wind to your back in the Northern hemisphere the low temperature (for a thermal wind) or the low pressure (for a geostrophic wind) will be to your left.

The question states that the low temperature is to the Northwest, so placing that out to your left would mean the wind hitting your back is coming from the Southwest. Hence the thermal wind is 225/20.

However the geostrophic wind and the thermal wind will interact with each other.

Putting the geostrophic wind on the CRP5 (use the high speed side of the slide and mark the wind down from the centre point) as 210/60. You then turn the wheel to the thermal wind direction of 225 degrees. Move the slide so that your wind mark falls on the very top line of the chequerboard in the bottom section of the slide. This will now allow you to put a second wind mark directly below your first and 2 divisions below it corresponding to a 20 kt wind.

This second mark is now your resolved wind and you should find that if you turn the wheel to put it directly beneath the centre point you should now be able to measure of the combined wind of 213/80.

Hence the answer is B) 210/80

I'm pretty confident that this is right as I only sat my met today and it's still pretty fresh in my mind. I don't believe there are any questions on the origins of thermal winds, but a good grasp of Buys Ballots Law for geostrophic and thermal winds is essential for the exam (quite a few questions on it today).

Finally, I found the Oxford Met CD excellent in understanding Met as moving pictures make the subject much easier to recall, and I would highly recommend the Bristol Online Question database as a revision tool.

Hope this helps,

Obs cop

[Fastair345]

Cheers paco and obs cop.:D

Will get back soon when I have studied your explanations.

Fastair345

[vkchopra]

Thanks for explaining how to find thermal winds.
Can some one explain me answer to the following one.

winds 090/20 at ground level.
winds 270/40 at high level

what is thermal wind at high level.??/

[bfisk]

The "thermal wind" does in fact not exist at high level, so that last question is nonsensical.

You could ask either what is the thermal wind, or what is the wind at high level.

Remember that the thermal wind is not in fact a measurable, actual wind, it is merely the term used to describe the change in wind, usually as a result of changes in air temperature; which cause the isobaric surfaces to slope, thus creating a pressure gradient that changes the wind.

In your example;
09020KT at SFC and 27040KT at say, 12000 feet.
The change in wind is 27060KT (you need to draw this up with vectors), but it might look something like

<-----27040---------o--09020-->
<-----------27060---------------<

So, 27060KT in 12000 ft is 27005KT pr 1000 feet.

Ie the thermal wind is westerly at 5 kt pr 1000 feet -- so you would derive the following:
SFC -- 09020KT
1000' -- 09015KT
2000' -- 09010KT
3000' -- 09005KT
4000' -- CALM
5000' -- 27005KT
6000' -- 27010KT
etc.

The easterly flow at the surface could possibly be due to a low pressure to the south in the northern hemisphere. So as we see -- the thermal wind is affecting the wind to the extent of making it 5 knots more westerly for every thousand feet -- this would normally be caused by cold air to the north.

Of course wind patterns could be, and often are, way more complex. I'm sure you've read about high and low pressures, sea and land breezes, and local phenomenons and terrain affecting the wind direction and speed. Such variances are usually stronger (more affecting) than the thermal component (thermal wind), however the thermal wind is a factor all the way to the tropopause, and will therefore be more significant at high altitude.

When solving questions involving pressures, temperatures and winds, I like to picture the following:
The atmosphere is an accordion (the musical instrument), pulled out so that the folds are parallell to the surface. It rests on a table. Where the pressure is low, that end rests below the table. Where pressure is high, that end of the accordion floats above the table. Where the air is cold, the accordion is slightly compressed, where it is warm, the accordion is pulled out (expanded).
Little marbles are free to roll on the planes that the folds make up, they roll downwards a bit before turning 90 degrees right (northern hemisphere).

http://www.clipartheaven.com/clipart/holidays/st-patricks-day/leprechaun-with-accordion-clipart.gif
So this fellow right here is telling me that:
on the left side of the bottom of the instrument (say Ireland!) the pressure is high (QNH 1020), while on the right side (say England), the pressure is low (QNH 995). The air above Ireland is cold and compact, while that over England is somewhat warmer. The surface wind will go from left to right, add 90 degrees right turn, and you'll know the wind will be northerly. At altitude it will progressively reduce, before becoming southerly way up on the top of the instrument. So the thermal wind is simply the change between each fold.


Good luck :)