[USMA:35456] Re: Lightning and thunder

[Pat Naughtin]

Title: Re: [USMA:35446] 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 ‘Bonnie’ 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 ‘on the horns of a cleft stick’.

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" <usma@colostate.edu>
>> 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).