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

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,

Well, that's reassuring, because it's what we should expect. I'm not able
in the small amount of time I'm prepared to devote to this, to derive the
series expansion for 1/cos(x).

and since x is very close to zero, it is the only one we need. :-)

Indeed.

(It is probably_more_ precise than doing regular trig operations on a
calculator, due to the limited precision on said calculator:

Indeed, you don't want to use a calculator to work out the cosine of
very small angles (or the arccos of numbers very close to 1). The
purpose of the trig based formulae is merely to act as the basis from
which to derive the rule of thumb formula.

Another poster mentioned Pythagoras, which is probably an easier basis
for deriving the same rule of thumb. The difference is that your
method aims to calculate the distance along the arc of the Earth's
surface, from horizon to observer's foot, whereas the Pythagoras approach
calculates the straight-line distance from horizon to observer's eye:

d^2 = (R+h)^2 - R^2 = 2Rh + h^2 = 2Rh(1 + h/R)

where for small angles (hR) the h/R term can be dropped and we get
d = sqrt(2Rh). Under small angle conditions the straight-line eye
distance and curved foot-distance can be treated as equal.

With a height of 6 feet we get about 3.5 miles, right?

Wrong :-)

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?

Quite.

Ronald Raygun wrote:
Terje Mathisen"terje.mathisen at tmsw.no" wrote:
d^2 = (R+h)^2 - R^2 = 2Rh + h^2 = 2Rh(1 + h/R)

Actually 2Rh(1 + h/2R) but that's OK.

where for small angles (hR) the h/R term can be dropped and we get

I believe that h/2R term is significantly smaller than the normal
atmospheric effects, i.e. today I can see features on the other side of
the Oslo fjord that I know should be below the horizon. :-)

d = sqrt(2Rh). Under small angle conditions the straight-line eye
distance and curved foot-distance can be treated as equal.

With a height of 6 feet we get about 3.5 miles, right?

Wrong :-)

Oops!

Doing sqrt(3) ~= 1.7321 or so, from memory, and then multiplying by 2
instead of sqrt(2) ~= 1.4142 gave a pretty bad end result.

sqrt(6) should be close to 2.45, since 25^2 is 625 and 24^2 is 576.

Thanks for catching my error.

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?

Quite.



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

On 7/26/2010 1:12 PM, Terje Mathisen wrote:

I believe that h/2R term is significantly smaller than the normal
atmospheric effects, i.e. today I can see features on the other side of
the Oslo fjord that I know should be below the horizon. :-)


You betcha! It's usual to take the notional radius of the Earth as 20%
bigger than the real radius (which of course itself varies because of
the oblateness of the spheroid) to account for refraction.

As usual, there is a wiki on the topic.

Brian W

brian whatcott wrote:

On 7/26/2010 1:12 PM, Terje Mathisen wrote:

I believe that h/2R term is significantly smaller than the normal
atmospheric effects, i.e. today I can see features on the other side of
the Oslo fjord that I know should be below the horizon. :-)


You betcha! It's usual to take the notional radius of the Earth as 20%
bigger than the real radius .. to account for refraction.

Ah, so that's why the fiddle factor in the "standard" rule of thumb
is 1.2 instead of 1.0 (or 1.064)! Or is it because the rule is for
non-nautical miles?

(which of course itself varies because of
the oblateness of the spheroid)

But only by a negligible amount (by comparison to the refraction effect),
the equatorial radius exceeding the polar radius by only about one third
of a percent.

One counterintuitive aspect worth mentioning is that while of course the
polar radius is smaller than the equatorial radius, the "radius" we must
use for horizon distance purposes is biggest at the poles and smallest
at the equator. That's when the horizon is N or S of the observer. An
equatorial observer needs to use a different radius when looking E or W
than when looking N or S.