# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Table 8, Bowditch**

**From:**Lu Abel

**Date:**2009 Jan 12, 23:52 -0800

Ark: For reasons difficult to explain unless you're a graduate electrical engineer or physicist, the lower the frequency of electromagnetic waves, the more they "cling" to the surface of the earth. Despite their very, very high frequencies, radar microwaves are about 1/10,000 the frequency of light waves. And that makes them cling to the surface of the earth (especially conductive surfaces such as seawater) a bit more tightly than light. Or to put it in terms more familiar to us as celestial navigators, they refract more highly. Thus the difference between the multiplicative factors of 1.22 vs 1.15 for distance-to-horizon for radar vs light. An extreme example of this "clinging" occurs at very low radio frequencies, where the radio waves can follow the surface of the earth for hundreds or thousands of miles. Loran capitalizes on this phenomenon -- it runs at just 100KHz (1/100,000 of the frequency of radar) and so-called ground waves are what make the system work. They follow the curvature of the earth for hundreds of miles. Loran compares the time it receives signals vs the time several transmitters transmit their signals and, based on the speed of light, computes the distance of the receiver from the various transmitters. Skywaves -- ones that bounce off the ionosphere -- are useless for Loran distance finding because of the variability of the height of the ionosphere and hence the time it takes for the signal to reach the receiver. Lu Abel inuik@yahoo.com wrote: > Greeting, > > I'm trying to understand why for calculating the approximate radar range at which any given target will return an echo-it is suggested that Bowditch Table 8 can be used with the distances increased by 4-5%% to allow for the slight curvature of radar waves over the geometric horizon but that provision is apparently not used for the visible horizon calc? > > Or the answer is the formulas: for radar 1.22 Sq.Rt of H > for visible horizon 1.15 Sq.Rt of H > > > I think the light refraction is affecting the distance to the visible horizon, so why 4-5% is added to radio waves? > > > Ark > > > > > > > --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@googlegroups.com To unsubscribe, email NavList-unsubscribe@googlegroups.com -~----------~----~----~----~------~----~------~--~---