Dear Martin and Jim,
Thanks for your thoughts on lightning and thunder. I have interspersed some
remarks.
On 23/12/05 8:20 AM, "Martin Vlietstra" <[EMAIL PROTECTED]> wrote:
> Just to be pedantic, I calculated the rule of thumb as being 3.56*sqrt(h).
To be even more pedantic the distance to the horizon from a height of 1
metre is nearer to 3.57 kilometres (rounded from 3.568 248).
I am not being pedantic for the sake of point scoring but because I find
this relationship 1 metre and 3.57 kilometres easy to remember given that
they contain the odd number series 1 - 3 - 5 - 7
Using this I can easily estimate that from (say) 4 metres I can see to the
horizon that is a little over 7 kilometres away.
> 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..
I used a similar method, but I based it on the Earth's radius. My steps
were:
1 Draw a circle to represent the Earth.
2 Draw a radius 'r' and extend this a little above the surface 'h'. This
line is the length of a radius plus the length of your point of observation
above the surface. Call this 'r + h'
3 Draw a tangent from your observation height to the surface of the Earth.
Call the length of this line 'd' for the distance to the horizon.
4 Draw a line to join the centre of the Earth to where the tangent meets
the Earth's surface and not that this line and the tangent meet at right
angles.
5 Pythagoras' theorem can now be used to write the formula:
r^2 + d^2 = (r + h)^2
6 Transpose this as follows:
r^2 + d^2 = (r + h)^2
r^2 + d^2 = r^2 + 2rh + h^2
d^2 = r^2 - r^2 + 2rh + h^2
d^2 = 2rh + h^2
7 Now note that since the size of h (a few metres) is usually small
relative to r (40 000 000 ÷ ¼ metres) the expression h^2 can be ignored in
most practical calculations. This leaves:
d^2 = 2rh
7 Substitution gives:
d^2 = 2rh
= (40 000/1000 ÷ ¼) * h
= 12.732 395 45 * h
or:
d = 3.568 248 232 * sqrt (h)
Say:
d = 3.57 * sqrt (h)
Sorry for the elaborate detail; I am not aiming to be pedantic but rather
pedagogic!
Cheers,
Pat Naughtin
Geelong, Australia
> ----- 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).
>
