Dear Pat
Mrs Naughtin put's it very well when she says the BBC are "on the horns of
cleft stick".
I notice that even within the article Dr Whitehouse can't make up his mind
which should be the primary unit for the measurement data:
... 9km (29,000ft) ... 45-52,000ft (13.7-15.9km). ...
Phil Hall
----- Original Message -----
From: "Pat Naughtin" <[EMAIL PROTECTED]>
To: "U.S. Metric Association" <[email protected]>
Sent: Saturday, December 24, 2005 5:54 AM
Subject: [USMA:35456] Re: Lightning and thunder
Dear Jim,
And all I was looking for was a quick rule of thumb to estimate, in
kilometres, the distance between me and a storm.
I'll stick to my 1 - 3 - 5 -7 idea but < with my thanks to you < I will be
much more conscious of the edge effects at the horizon.
By the way, in 1998, Hurricane OBonnie¹ was estimated at 18 000 metres high.
This report from the BBC at http://news.bbc.co.uk/1/hi/sci/tech/163582.stm
gives details and an interesting computer image but, as usual, the BBC
remains in the never never land of conversion confusion. As my wife, Wendy,
humorously puts it, with respect to the BBC metrication policy, they remain
Oon the horns of a cleft stick¹.
Cheers,
Pat Naughtin
PO Box 305 Belmont 3216
Geelong, Australia
61 3 5241 2008
[EMAIL PROTECTED]
http://www.metricationmatters.com
On 23/12/05 3:50 PM, "James R. Frysinger" <[EMAIL PROTECTED]> wrote:
Martin,
Euclid was pretty good with that geometry stuff but he was weak on
refraction.
The rule of thumb I stated takes refraction into account, which is
important
to observers near the Earth's surface. Hence the factor of 3.84 rather
than
the value of 3.56 that Euclid would have predicted. For radar ranges to
the
horizon, add another 2 % to 3 %, especially for the lower frequency
radars.
This is of course of great interest to celestial navigators, who can even
fudge it a bit more based on extreme weather conditions. There are look-up
tables available to add an additional correction to sextant readings that
are
based on barometric pressure and temperature.
Of course, navigators add the range to the horizon based on their height
of
eye as well when calculating visibility ranges to lights. The two tangents
to
Earth's surface are co-linear. If one is calculating the expected sighting
of
a low light both distances to the horizon are important.
In the case of lightning, where the upper portion is at least a few
kilometers
above the Earth's surface, one may omit the few kilometers the observer's
horizon is from him or her. In fact, at that great distance above the
Earth's
surface, the geometry and the allowance for refraction break down a bit.
The
rule of thumb is just for estimating, of course, which is why I didn't
mind
rounding the factor off to 4. How well can the observer determine the
height
of the bolt?
I suggest Dutton's _Navigation_and_Piloting_ and Bowditch's
_American_Practical_Navigator_ as references. A more terse explanation can
be
found in the _Nautical_Almanac_ each year, which also has the corrections
to
sextant readings for extreme weather conditions. Bowditch can now be
downloaded for free online; at least it was available at one time and I
suppose it still is.
As one who has done and who has taught celestial navigation I find it
quite
interesting. A really nifty technique is that of shooting an amplitude of
the
Sun and that depends on the refraction correction. It's a great way to do
a
quick calculation for a compass check, though! And of course, refraction
lets
us see the Sun before it actually rises above the geometric horizon and
for a
few moments after it has set below the geometric horizon.
I fondly recall using metric navigation charts in the '70s, by the way.
So,
there. We're still on topic.
Jim
On Thursday 22 December 2005 16:20, Martin Vlietstra wrote:
Just to be pedantic, I calculated the rule of thumb as being
3.56*sqrt(h).
It is quite easy to demonstrate using Euclidean Geometry:
Consider a circle and a point X that lies outside the circle.
Construct a line XAB such that AB is the diameter of the circle.
Construct a line XC such that XC is a tangent to the circle.
There is a theorem in Euclidean Geometry that states:
XC^2 = XA*XB
If XA is small compared to AB, then we can replace XB by AB.
Set XB to the Earth's diameter (40000/pi)
Set XA to the height of the observation station (or object) above the
Earth's surface
Solving for XC gives
XC = sqrt((XA/1000) * 40000/pi) (the factor of 1000 is to convert metres
to km)
Simplifying gives XC=3.56*sqrt(XA)
It should be noted of course that the Earth's Circumference is as close
as
makes no difference 40,000km as everybody who knows anything about the
metric system should know..
----- Original Message -----
From: "James R. Frysinger" <[EMAIL PROTECTED]>
To: "U.S. Metric Association" <[email protected]>
Sent: Thursday, December 22, 2005 5:05 PM
Subject: [USMA:35440] Re: Lightning and thunder
On Wednesday 21 December 2005 15:42, Pat Naughtin wrote:
Dear All,
....
From what distance can lightning be seen?
A rough rule of thumb for estimating distance to the horizon for visible
light is based on the height of the source. If that is given in meters
as
h,
then the distance d is given in kilometers by d=3.84sqrt(h).
