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).
