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Terje Mathisen Terje Mathisen is offline
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Default May a "landlubber" comment? - was[ Help create better charts]

Ronald Raygun wrote:
Terje Mathisen"terje.mathisen at tmsw.no" wrote:

brian whatcott wrote:
Not linear: for a ground level jammer,
The line of sight estimator for distance versus height above sea level
goes something like this:
distance n.m. = 1.2 sqrt (Height ft MSL)


That calculation follows directly from the Taylor series for Cosine:

1 - x^2/2! + x^4/4! - ...

It means that for very small angles, the height above the sea is

1 - (1 - x^2/2!) = x^2/2! = x^2/2 (when R == 1)


Hmm. Your working suggests that for R=1 the height is equal to 1-cos(x),
but that is not the case, it's actually equal to 1/cos(x)-1. By chance,
for very small angles, these two expressions are approximately equal.


Not "by chance", I (mis-)remembered the result I needed (from doing this
calculation 30+ years ago) and didn't have paper and pen to rederive it
so I picked the first approximation that looked correct. :-(

Anyway, doing a series expansion for your formula leads to the exact
same x^2/2 value for the first term, and since x is very close to zero,
it is the only one we need. :-)

(It is probably_more_ precise than doing regular trig operations on a
calculator, due to the limited precision on said calculator: With a
height of 6 feet we get about 3.5 miles, right?

The ratio of 1.8 m to 6400 km is about 3.5e6, so the second-order term
requires 13 digits while most calculators are happy to show 8 or 10, right?

Insert the radius of the Earth (in nautical miles, 3500 or so) and
multiply the result by the number of feet in a nautical mile (about
6000+) and the 1.2 factor should pop out.


Actually a factor of 1.0 would be a better approximation than 1.2,
since the factor which actually pops out, when I use R=6371km and
conversion factors 1852m/NM and 0.3048m/ft, is 1.064.

OK, that's useful!

Terje

--
- Terje.Mathisen at tmsw.no
"almost all programming can be viewed as an exercise in caching"