In message , Timothy Murphy
writes
I'm not really clear on this.
I thought that the whole point of using a parabolic surface
was that the (almost) parallel rays from the satellite
would all be reflected to pass through the focus,
and for that I would have thought that
the directrix of the parabola _would_ have to pass through the satellite.

Yes, but if you move the (let's call them parallel) rays off centre,
then the focal point reflected also moves off centre in the other
direction

--
geoff

On Tuesday, in article
"Timothy Murphy" wrote:

Brian {Hamilton Kelly} wrote:

You are apparently a mathematician; can you not understand the concept of
the dish being a *part* of a parabola, rather than its centre?

For "centre" read "focus" ?

NO! The LNB *is* at the focus of the parabaloid of which the dish you
see is but a part. The diagram of "how an offset antenna works" in the
http://www.lindenreport.com/tech/satworks.html mentioned elsewhere in
the thread shows this quite clearly.

(Although for operation within the UK, I'd have preferred it if the
diagram had been reflected about a line going SW--NE on the page, such
that the "incoming rays" from the satellite were still pointing upwards
and to the right, but at a more realistic elevation angle. That diagram
looks like a set-up for UAE or thereabouts [which ties in with the
article originally having been published in a magazine circulated in that
part of the world].)

--
Brian {Hamilton Kelly}
"I don't think you're in the top class when it comes to thinking
- I suspect I could wade through the depths of your mind and not
wet my ankles." Peter Thomas, in news:uk.telecom 24-Jul-2005

On Tuesday, in article
"Timothy Murphy" wrote:

Brian {Hamilton Kelly} wrote:

Now all these lines are *curved* because they're on the surface of a
[roughly] spherical body: hence the necessity for _spherical_
trigonometry.

If you know where the satellite is, the simplest way would be
to measure the distance on an atlas, and work out the angle subtended
at the earth's centre from that.
This would not require any _spherical_ trigonometry.

Errm, the installer of the dish doesn't care a toss about the distance to
the [projection of] the satellite [onto the equator]. What s/he *does*
need to know is the azimuth of that latter position, relative to the
local *true* south. Spherical trig is ideally suited to this task: it
merely requires a few sin/cos terms of lat & long in various
combinations.

Now those formulae were in our maths textbooks when I did A-Levels in the
early 1960. Of course, back in those days, we studied plane trigonometry
from the age of twelve; AIUI, it's only in the second year of A-Levels
that today's students meet the concept. (At twelve, they *might* have
met fractions, which we did at seven.)

(At one time, about 18 years ago, I derived the requisite formulae myself
from first principles, just for the hell of it; nowadays, I'm quite happy
to use one of the Javascript programs to which you have already been
pointed at various web-sites.)

I know how to calculate the angle,
and I can also see a number of dishes from my window,
so I have a reasonable idea of what the angle is.

My argument was with the original poster (probably you)
who said that you could work out the angle
with "schoolboy trigonometry" (IIRC).

Now you add a knowledge of Newtonian dynamics.

I doubt if many schoolboys could work out
the height of a geostatic satellite.
Could you?

Simple; the satellite's radial velocity needs to match the rotational
speed of the Earth. We did computations like that all the time in the
second-year of VIth Form (which in my case was 1962--63). I don't
believe that we ever did an actual computation for a geostationary
satellite, simply because none existed[1] at that time. Besides, in
those days one didn't perform calculations because they were "useful",
but to demonstrate that one understood the theory, so non-practical
problems might be set.

Anyway, to work (using \TeX notation for the mathematics, which is
portable across plain ASCII and O/S independent):

T^2 = {4\pi^2}\over{GM}r^3

Where T is the orbital period, G the universal graviational constant
(6.672\times10^{-11} Newton-metres-squared per kg squared), M the
mass of the Earth (5.976\times10^{24} kg: the mass of the satellite
itself is, of course, negligible in comparison and omitted) and r the
radius of the orbit relative to the centre of the planet.

The rotational period is 23h 56m 4.09s, or 86164.09 seconds. From the
figures, it is simple to calculate that the radius of the orbit is
approximately 48Mm. Knowing that the nominal equatorial radius of the
Earth is 6.378Mm, then the satellites all sit at about 42,170km above the
Equator.

As I said, we did this sort of computation all the time, *without*
electronic calculators, some 43 years ago. Log tables are quite
adequate.

Incidentally, a number of apparently well-informed posters
have claimed that the parabolic dish may not point to the satellite
because the receiver is not at the focus (what you call the centre).

I'm not really clear on this.
I thought that the whole point of using a parabolic surface
was that the (almost) parallel rays from the satellite
would all be reflected to pass through the focus,
and for that I would have thought that
the directrix of the parabola _would_ have to pass through the satellite.

As has been mentioned in other posts, from myself and others, the dish
you see is only a fraction of a much larger dish, whose *centre* is non-
existent, and offset from the bit you see. The focus of the fragment of
dish naturally corresponds to the focus of the entire imaginary
paraboloid, and the LNB is installed at that spot.

[1] Although Arthur C Clarke's seminal article had appeared in Wireless
World in 1945, by 1962--63 the broadcasters were still playing with non-
geostationary satellites, such as Telstar and Early Bird (and, for that
matter, passive reflectors like Echo I & II balloons).

--
Brian {Hamilton Kelly}
"I don't think you're in the top class when it comes to thinking
- I suspect I could wade through the depths of your mind and not
wet my ankles." Peter Thomas, in news:uk.telecom 24-Jul-2005

On Tue, 9 Aug 2005 12:17:29 UTC, Timothy Murphy
wrote:

Brian {Hamilton Kelly} wrote:

Now all these lines are *curved* because they're on the surface of a
[roughly] spherical body: hence the necessity for _spherical_
trigonometry.

If you know where the satellite is, the simplest way would be
to measure the distance on an atlas, and work out the angle subtended
at the earth's centre from that.
This would not require any _spherical_ trigonometry.

Ah! A Flat-Earther!

Brian {Hamilton Kelly} wrote:

Anyway, to work (using \TeX notation for the mathematics, which is
portable across plain ASCII and O/S independent):

T^2 = {4\pi^2}\over{GM}r^3

Where T is the orbital period, G the universal graviational constant
(6.672\times10^{-11} Newton-metres-squared per kg squared), M the
mass of the Earth (5.976\times10^{24} kg: the mass of the satellite
itself is, of course, negligible in comparison and omitted) and r the
radius of the orbit relative to the centre of the planet.

I agree with your mathematics,
but not with your belief that any schoolboy could come up with it.
I doubt if 0.01% of UK schoolboys (and a lower proportion of schoolgirls)
could follow your argument, let alone repeat it.

Regarding the argument about the angle of the dish,
I misunderstood your comment,
as I have never heard the point where the directrix meets a parabola
referred to as its "centre".

--
Timothy Murphy
e-mail (80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland