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.

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.

On 7/25/2010 4:17 AM, Terje Mathisen 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)

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.

Terje


Hehe... it doesn't quite pop out in fact - but who's checking, between
friends?

Brian W

On 7/25/2010 9:54 AM, 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.

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.

Even using alpha math engine, I get 1.07 with those other conversions
that Terje gave. hehe...

Brian W

In article ,
Martin wrote:

On 25/07/10 10:54, Jeff wrote:

Those ranges quoted are for aircraft thousands of feet above sea
level. The jamming range would be much shorter for boats, probably
10 miles or less.


aSSuming it's a linear relationship, I get 202 mile radius...




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)

..and of course that assumes that the emitter is at sea level, so if it
is at any appreciable height (or in another aircraft) then you have to
add the result of the same equation again!!

and ignores the surface effect that allows UK TV & VHF signals to be
picked up as far away as in the Netherlands.

Horizontal Bending doesn't work all that well at 1.6 Ghz, and is
negligible, in its effect. Temperature Inversion Refraction is also
negliable at 1.6 Ghz, and even if you could count on any specific
Inversion System to be present at any one time. Both of the above are
present for frequencies below 1 Ghz, and could contribute a small effect
but above 1 Ghz, it just isn't going to happen....

--
Bruce in alaska
add path after fast to reply

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"

Bruce in alaska wrote:
and ignores the surface effect that allows UK TV& VHF signals to be
picked up as far away as in the Netherlands.

Horizontal Bending doesn't work all that well at 1.6 Ghz, and is
negligible, in its effect. Temperature Inversion Refraction is also
negliable at 1.6 Ghz, and even if you could count on any specific
Inversion System to be present at any one time. Both of the above are
present for frequencies below 1 Ghz, and could contribute a small effect
but above 1 Ghz, it just isn't going to happen....

I'd be willing to trust you one this (I have been a ham since 1977,
la8nw), but we are after all in a sat-nav newsgroup he

Why do we talk about pseudo-range measurements and corrections? Isn't
this due to bending of signals at 1.575 GHz?

OTOH, as long as the effect is in the sub-percent range, it really
doesn't matter when we're discussing 1.2 vs 1.0 rule-of-thumb factors. :-)

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

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 Sun, 25 Jul 2010 23:39:30 +0200, Terje Mathisen "terje.mathisen at
tmsw.no" wrote:

(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?

The CalculatorThatTakesNoPrisoners [HP48] displays 13 plus the
exponent. I think it calculates to 16 places. I suppose there _are_
many cheapies that are not up to it.

Casady

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