I'm wondering if someone could confirm my interpretation of how to do this.
First of all are they surface winds calculated [less local issues friction
etc]?

On the EGRR charts - top left on every one is a graph for determining the
geostrophic wind speed.

So for the S of the UK - taking the 50N line - one looks at the isobar
spacing on the chart across London say and compare it with the distance from
the left hand edge of the graph.

I find that by doing this - the wind speeds come out quite high i.e today
60KTs plus..

Phil

--

http://homepage.ntlworld.com/phil.layton/meteo.htm

In message , Phil Layton
writes

So for the S of the UK - taking the 50N line - one looks at the isobar
spacing on the chart across London say and compare it with the distance from
the left hand edge of the graph.

I find that by doing this - the wind speeds come out quite high i.e today
60KTs plus..

That's how I teach my students Phil, but don't forget that this doesn't
equate with surface winds because the geostrophic gradient refers to
speeds without impairment from friction induced by hills, trees,
buildings etc.

Well that's my understanding anyway!
--
Steve Jackson,
Bablake Weather Station,
Coventry, UK
http://www.netlink.co.uk/users/bws

"Steve Jackson" wrote in message
...
In message , Phil Layton
writes

So for the S of the UK - taking the 50N line - one looks at the
isobar
spacing on the chart across London say and compare it with the
distance from
the left hand edge of the graph.

I find that by doing this - the wind speeds come out quite high i.e
today
60KTs plus..

That's how I teach my students Phil, but don't forget that this
doesn't
equate with surface winds because the geostrophic gradient refers to
speeds without impairment from friction induced by hills, trees,
buildings etc.

... and don't forget to allow for curvature ...

There is more on all this in the FAQ ... see
Q/A 2A.21
and
Q/A 2A.30

Martin.


--
FAQ & Glossary for uk.sci.weather at:-
http://homepage.ntlworld.com/booty.weather/uswfaqfr.htm

On Sun, 8 Feb 2004 18:34:05 +0000, Steve Jackson wrote in


In message , Phil Layton
writes

So for the S of the UK - taking the 50N line - one looks at the isobar
spacing on the chart across London say and compare it with the distance from
the left hand edge of the graph.

I find that by doing this - the wind speeds come out quite high i.e today
60KTs plus..

That's how I teach my students Phil, but don't forget that this doesn't
equate with surface winds because the geostrophic gradient refers to
speeds without impairment from friction induced by hills, trees,
buildings etc.

Well that's my understanding anyway!

Yes that's true. You have to allow for two factors 1) friction, as Steve
says and 2) any curvature of the isobars. If that curvature is cyclonic,
the wind (gradient speed VG) will be less than the geostrophic speed Vg.
There is a formula for calculating VG in relation to Vg if you know
curvature - say as a function of distance from centre of rotation.

--
Mike 55.13°N 6.69°W Coleraine posted to uk.sci.weather 08/02/2004 18:57:24 UTC

"Mike Tullett" wrote in message
...
On Sun, 8 Feb 2004 18:34:05 +0000, Steve Jackson wrote in


Yes that's true. You have to allow for two factors 1) friction, as Steve
says and 2) any curvature of the isobars. If that curvature is cyclonic,
the wind (gradient speed VG) will be less than the geostrophic speed Vg.
There is a formula for calculating VG in relation to Vg if you know
curvature - say as a function of distance from centre of rotation.

And yesterday's small but intense depression was a case
in point ... it had an exceptionally steep pressure gradient, but
in an area where the cyclonic curvature was very tight indeed.
That's why Stornoway didn't blow away.

pe

"Steve Jackson" wrote in message
...
In message , Phil Layton
writes

So for the S of the UK - taking the 50N line - one looks at the isobar
spacing on the chart across London say and compare it with the distance
from
the left hand edge of the graph.

I find that by doing this - the wind speeds come out quite high i.e today
60KTs plus..

That's how I teach my students Phil, but don't forget that this doesn't
equate with surface winds because the geostrophic gradient refers to
speeds without impairment from friction induced by hills, trees,
buildings etc.

Well that's my understanding anyway!
--
Steve Jackson,
Bablake Weather Station,
Coventry, UK
http://www.netlink.co.uk/users/bws

Not to mention other ageostrophic effects, especially curvature of the flow.
In fact, the geostrophic scale applies strictly to flow which is straight
and through which the air experiences no acceleration. Having measured the
speed from the geostrophic scale, it then needs to be corrected for
curvature to obtain the gradient wind, which then has to be corrected for
friction to obtain the surface or 10m wind. The gradient wind is closely
related to the flow at 700 m to 1000 m above the surface.
For cyclonic flow, for example, of radius of curvature 90 nautical miles, it
is possible to measure a geostrophic wind of 160 knots, and have a gradient
wind of just 60 knots. Conversely, for anticyclonic flow of radius of
curvature 400 nm, a geostrophic flow of 40 knots will give a gradient wind
of 80 knots. (These figures apply strictly to 55N ).

--
Bernard Burton
Wokingham, Berkshire, UK.


Satellite images at:
www.btinternet.com/~wokingham.weather/wwp.html

On Sun, 8 Feb 2004 19:06:23 -0000, Philip Eden wrote in
k

"Mike Tullett" wrote in message
...
On Sun, 8 Feb 2004 18:34:05 +0000, Steve Jackson wrote in


Yes that's true. You have to allow for two factors 1) friction, as Steve
says and 2) any curvature of the isobars. If that curvature is cyclonic,
the wind (gradient speed VG) will be less than the geostrophic speed Vg.
There is a formula for calculating VG in relation to Vg if you know
curvature - say as a function of distance from centre of rotation.

And yesterday's small but intense depression was a case
in point ... it had an exceptionally steep pressure gradient, but
in an area where the cyclonic curvature was very tight indeed.
That's why Stornoway didn't blow away.

Indeed... and another ageostrophic factor is the speed at which the low is
moving. A fast moving low heading east will cause the trajectory to the
south to approach a straight line, and thus VG will not be far off Vg. To
the north, the opposite effect will occur and VG will be a lot less than
Vg.[1]

To give Phil an idea of how the two are related, this is the standard
formula relating gradient with geostrophic speeds in a low.

VG = Vg - [(VG)squared]/rf

where r is the radius of curvature
and f is the coriolis parameter - 2 omega sine phi
where omega is a measure of the earth's angular velocity and phi is
latitude. That means "f" is fixed for a particular latitude and can be
inserted into that formula as just a number.
As "r" gets bigger, the straighter the isobars are, the smaller the last
term becomes, and the nearer VG gets to Vg.

The solution of that uses the technique for solving a quadratic equation.

[1] That's the reason whenever I see a deepening [2]low heading for
Ireland, I hope it will travel to my south.

[2] A developing low introduces yet another ageostrophic factor, as the
pressure gradient is changing with time.

--
Mike 55.13°N 6.69°W Coleraine posted to uk.sci.weather 08/02/2004 19:48:13 UTC

And thanks all on this thread. I think the message is that quite a bit of
interpolation is required rather than cutting the grid out and using as a
ruler :-)

Phil