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CX2SA > SATDIG 02.10.12 21:03l 560 Lines 20434 Bytes #999 (0) @ WW
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To : SATDIG@WW
Today's Topics:
1. Re: maximum AO-7 distance (Thomas Doyle)
2. Re: ISS CubeSat deployment Oct 4 (Trevor .)
3. First time ever satellite related net on VHF
(Raidel Abreu Espinet)
4. Re: maximum AO-7 distance (Dave Guimont)
5. Re: maximum AO-7 distance (Gus 8P6SM)
6. Re: maximum AO-7 distance (Stephen Melachrinos)
7. Re: maximum AO-7 distance (i8cvs)
8. Re: maximum AO-7 distance (Gus 8P6SM)
9. SatPC32 & Flex5000A (Richard Lawn)
10. Re: ISS CubeSat deployment Oct 4 (Mineo Wakita)
11. Time Is Running Out (Martha)
12. Japan AO-7A (Bob- W7LRD)
----------------------------------------------------------------------
Message: 1
Date: Mon, 1 Oct 2012 08:18:49 -0500
From: Thomas Doyle <tomdoyle1948@xxxxx.xxx>
To: Bob- W7LRD <w7lrd@xxxxxxx.xxx>
Cc: amsat-bb@xxxxx.xxx
Subject: [amsat-bb] Re: maximum AO-7 distance
Message-ID:
<CAHnRQRJwJLc9r16QCrGpjuu0cVURko21yULijyeCSB09VqP+2g@xxxx.xxxxx.xxx>
Content-Type: text/plain; charset="iso-8859-1"
Bob,
Since the height above ground varies due to the elliptical nature of
orbits the max range also varies. Where the two stations are located
relative to the apogee of the orbit can also come into play. If you
are on the wrong side of the earth when the satellite is at its apogee
that could be a problem.
Here is my best shot at the range of max distances for AO-07 - compare
them to numbers you get from other folks and let me know what you
find. My numbers could be and are probably wrong.
73 W9KE tom...
On Mon, Oct 1, 2012 at 1:47 AM, Bob- W7LRD <w7lrd@xxxxxxx.xxx> wrote:
> to my AO-7 afictionados...what is the maximum distance one can work edge
to edge, and how did you figure that out?
> 73 Bob W7LRD
> _______________________________________________
> Sent via AMSAT-BB@xxxxx.xxx. Opinions expressed are those of the author.
> Not an AMSAT-NA member? Join now to support the amateur satellite program!
> Subscription settings: http://amsat.org/mailman/listinfo/amsat-bb
--
Sent from my computer.
tom ...
------------------------------
Message: 2
Date: Mon, 1 Oct 2012 20:12:14 +0100 (BST)
From: "Trevor ." <m5aka@xxxxx.xx.xx>
To: amsat-bb@xxxxx.xxx
Subject: [amsat-bb] Re: ISS CubeSat deployment Oct 4
Message-ID:
<1349118734.35565.YahooMailClassic@xxxxxxxxx.xxxx.xxx.xxxxx.xxx>
Content-Type: text/plain; charset=utf-8
Deployment times being quoted as 18.35 MSK for first pod and the second at
19.35 MSK, see http://tinyurl.com/ISS-CubeSat-Announcement
I make that 1435 and 1535 UT.
73 Trevor M5AKA
------------------------------
Message: 3
Date: Mon, 1 Oct 2012 16:54:32 +0200 (CEST)
From: Raidel Abreu Espinet <cl2esp@xxxxxx.xx.xx>
To: amsat-bb@xxxxx.xxx
Subject: [amsat-bb] First time ever satellite related net on VHF
Message-ID: <9654127.256.1349103272911.JavaMail.root@xxxx>
Content-Type: text/plain; charset=utf-8
Hello all,
As a part of our effort to continue promoting the proper use of the amateur
radio satellites by the Cuban ham community and added to the several
radio-conferences that Hector (CO6CBF), Pavel (CO7WT) had performed in the
past months.
It's a big pleasure to announce that we start this past Friday a VHF FM Net
devoted entirely to the Amateur Radio Satellites on the 2m band on a simplex
frequency for the Havana Province (capital of Cuba), with few coverage also
to the surrounding provinces of Mayabeque and Artemisa. The net will be on
the air every Friday from 19:00 to 20:00 local time on 145.550 MHz, CM2ESP
is the Net Control Station.
The net is divided in three sections: News; A Technical Topic; and Question
and Answer Session.
This first net was very appreciated by many hams, and a lot more were not
able to check-in due to the severe thunderstorm that affected Havana right
close to the starting time.
The Cuban Satellite Group named Grupo de Radioaficionados para Operaciones
Satelitales (GROS) will continue promoting and creating new activities to
increase the presence of Cuban Amateur Radio Stations on the satellites and
the activation of new grids.
Several demonstrations and workshops are currently being scheduled.
Thanks to all for the support and good wishes we had receive since we start
this idea on November 2011 which its greatest moment was the official
constitution of the Group on March this year.
73,
Raydel Abreu Espinet, CM2ESP
GROS Coordinator.
PS: Sorry for bad english....
-------
Este mensaje ha sido enviado mediante el servicio de correo electronico que
ofrece la Federacion de Radioaficionados de Cuba a sus miembros para
respaldar el cumplimiento de los objetivos de la organizacion y su politica
informativa. La persona que envia este correo asume el compromiso de usar el
servicio a tales fines y cumplir con las regulaciones establecidas.
------------------------------
Message: 4
Date: Mon, 01 Oct 2012 15:06:11 -0700
From: Dave Guimont <dguimon1@xxx.xx.xxx>
To: amsat-bb@xxxxx.xxx
Subject: [amsat-bb] Re: maximum AO-7 distance
Message-ID: <506A13D3.3070307@xxx.xx.xxx>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
Hi Bob,
Draw the ground circle on a lambert conformal chart, and you can measure
the distance directly...From your location use the distance from you to
the west end of Florida, according to my ANCIENT OSCAR LOCATOR....That
will probably warm the cockles of some old timer's hearts...Who is still
around??
Best 73 to all Dave WB6LLO
On 9/30/2012 11:47 PM, Bob- W7LRD wrote:
to my AO-7 afictionados...what is the maximum distance one can work edge
to edge, and how did you figure that out? 73 Bob W7LRD
_______________________________________________ Sent via
AMSAT-BB@xxxxx.xxx. Opinions expressed are those of the author. Not an
AMSAT-NA member? Join now to support the amateur satellite program!
Subscription settings: http://amsat.org/mailman/listinfo/amsat-bb
------------------------------
Message: 5
Date: Mon, 01 Oct 2012 21:43:27 -0400
From: Gus 8P6SM <8p6sm@xxxx.xxx>
To: Bob- W7LRD <w7lrd@xxxxxxx.xxx>
Cc: amsat-bb@xxxxx.xxx
Subject: [amsat-bb] Re: maximum AO-7 distance
Message-ID: <506A46BF.9060109@xxxx.xxx>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
On 10/01/2012 02:47 AM, Bob- W7LRD wrote:
> to my AO-7 afictionados...what is the maximum distance one can
> work edge to edge, and how did you figure that out?
> 73 Bob W7LRD
Maximum communications distance:
MCD = 2 R arccos [ R / (R + h) ]
where
R = Radius of earth (Spherical earth: 6371 Km)
h = Height at apogee
(Eq 12.4, Sat. Experimenters Handbook 2nd Ed; Davidoff)
All you need is the apogee height for the satellite in question, and a
calculator. For AO-7 the apogee height USED to be 1460 Km, and probably
hasn't altered much over the years. Therefore:
MCD = 2 * 6371 * arccos [ 6371 / (6371 + 1460) ]
= 12756 * arccos [ 6371 / 7831 ]
= 12756 * arccos [ 0.8135614864 ]
= 12756 * 0.620545318575
= 7915.7 Km (4918.59 miles)
What is the "official" distance?
With a good station, the non-spherical earth working for you at your QTH
and hoping for some sub-horizon love, who knows?
--
73, de Gus 8P6SM
The Easternmost Isle
------------------------------
Message: 6
Date: Mon, 01 Oct 2012 20:48:56 -0500 (CDT)
From: Stephen Melachrinos <melachri@xxxxxxx.xxx>
To: w7lrd@xxxxxxx.xxxx amsat-bb@xxxxx.xxx
Subject: [amsat-bb] Re: maximum AO-7 distance
Message-ID: <9877125.1549663.1349142537019.JavaMail.root@xxxxxxxxxxx>
Content-Type: text/plain; charset=UTF-8
On 10/01/12, Bob- W7LRD<w7lrd@xxxxxxx.xxx> wrote:
> to my AO-7 afictionados...what is the maximum distance one can work edge
to edge, and how did you figure that out?
> 73 Bob W7LRD
Bob -
Here's my take on your question. In addition to the orbital variations,
there are at least two other considerations: the altitude of the stations
and the elevation angle of their horizons (whether they can work down to
zero degrees elevation angle or even lower). There are many variations on
this, but let's look at a simple case where the stations are each at zero
altitude (relative to the earth's radius) and both can work down to exactly
zero degrees elevation angle.
At any given moment, the in-view area of a satellite will be almost exactly
a circle drawn on the earth's surface. (If the earth were perfectly round,
it WOULD be a circle, but variations in surface altitude create minor
distortions. So let's assume the earth is perfectly round, at least within
the in-view region.) The maximum distance question you ask becomes "what is
the diameter of this circle?" The diameter of the circle (call it D) is a
function of both the altitude of the satellite (call it A) and the radius of
the earth (call it R). For two points on the diameter of the circle (and the
satellite directly above the center of the circle), this three-dimensional
question degenerates into a two-dimensional problem.
[Okay, here I'm going to describe a geometry problem and use trigonometry to
solve it. After all, Bob asked "how did you figure it out?" and this is the
way I solve these problems at work. If you want to skip the derivation and
get to the answer, jump past the end of the brackets. This would be easier
if I could draw a picture, but I'll try to explain in words. Also, see the
PPS below.
For the case where the ground station elevation angle is zero, the 2D view
becomes two right triangles. For each triangle, the vertices are at the
satellite, the center of the earth, and the ground station. The zero degree
elevation angle results in the right angle being at the ground station. So
the two legs of the right triangle are the slant range to the satellite and
the radius of the earth. The hypotenuse is the radius of the satellite's
orbit, equal to its altitude plus the radius of the earth. The two triangles
are positioned hypotenuse to hypotenuse.
What we need to determine from each triangle is the included angle between
the hypotenuse and the leg that goes from the earth center to the ground
station. Trig tells us this angle (call it B) is represented by the
following equation:
B = Arccos( (earth radius) / (satellite radius) )
Substitute R for earth radius and A for satellite altitude, and this
equation becomes
B = Arccos ( R / (R + A) )
Once we have this angle, we double it to determine the earth-center angle
between the two ground stations. Divide this angle by 360 degrees to
determine what fraction of a circle this represents, then multiply that
fraction by the circumference of a perfectly circular/spherical earth to
determine the distance between the two points along the surface of that
perfectly spherical earth.
Thus the distance between the two points (call that D) is represented by the
following equation:
D = (2 * B / 360) * (2 * pi * R)
Replace B with the equation above to get:
D = (2 * Arccos( R / (R + A) ) / 360) * 2 * pi * R
which simplifies to: ] <<<<<----- End of derivation!!!!
D = (pi * R / 90) * Arccos( R / (R + A) )
I put this equation in an Excel spreadsheet, so let's plug some numbers in.
AO-7 is listed on the AMSAT web page with an apogee of 1459 km and a perigee
of 1440. Wikipedia says the earth's radius varies between 6353 km and 6384
km, but most models come up with a mean radius of 6371. So plugging in R =
6371 and A = 1459 results in a D of 7904.7 km. At perigee (A = 1440), this
drops to 7861.2.
In reality, you have to subtract a little bit from these distances; since
the satellite is moving, you need to be able to complete a QSO while both
stations are in coverage, so it's not quite as wide as the full diameter of
the coverage circle. But as I said earlier, there are additional variations.
Propagation might let your signal work below your actual horizon. The
distance from your station to the center of the earth may/will vary from the
average of 6371 km. The big kicker is if your antenna is at the top of a
mountain so you can see below zero degrees elevation angle; then the
earth-center angle widens and the distance increases. But the geometry
becomes more complex. And the trick is to find mountains of just the right
height, just the right distance apart, with a satellite at apogee exactly
between them.
[More geometry: I typically solve this by adding more right triangles to
make the trig easy. For example, you define a grazing point where the signal
"hits" the surface of the perfectly spherical earth. Then you have one right
triangle from the ground station to the grazing point, and a second from the
grazing point to the satellite. I'm sure there's a way to do this with
non-right triangles, but the trig is easier for me to remember this way.]
Steve Melachrinos
W3HF
P.S. Nothing in this derivation is satellite-specific, so it works for
anything with similar geometry. Plug in AO-27's altitude to see its max
range (6076.2 km, subject to the same limitations and assumptions). Or
geostationary altitude. Or the height of your repeater's tower to determine
its coverage area. Or put in the nominal six-foot height of a human to show
that the distance to the visible horizon is about 3 miles.
P.P.S. So I just Googled "human visible horizon" to verify my recollection
that the visible horizon for a six-foot-tall person is about 3 miles away.
And I came upon a web page, linked below, that does a similar calculation.
His objective was to determine the distance to the horizon from the
observer's eyes, which in our model is the satellite slant range. But he has
a picture that is similar enough to what I was trying to describe that it
should help. Just imagine the head of the stick figure is really AO-7.
http://blogs.discovermagazine.com/badastronomy/2009/01/15/how-far-away-is-the-
horizon/
------------------------------
Message: 7
Date: Tue, 2 Oct 2012 04:09:47 +0200
From: "i8cvs" <domenico.i8cvs@xxx.xx>
To: "Dave Guimont" <dguimon1@xxx.xx.xxx>, "Amsat - BBs"
<amsat-bb@xxxxx.xxx>
Subject: [amsat-bb] Re: maximum AO-7 distance
Message-ID: <000001cda043$47858c80$0401a8c0@xxxxxx>
Content-Type: text/plain; charset="iso-8859-1"
Hi Dave, WB6LLO
I am also still around with my old OSCAR LOCATOR with aquisition
circle and orbital track for OSCAR-7 ,it warm the cockles of my old
timer heart !
I developed an OSCAR LOCATOR for the orbit of the first P3-A HEO
satellite wich failed to go into orbit due to a malfunction of second
stage engine on ARIANE-5 flight nr?6
In the time been, between P3-A and P3-B i.e. OSCAR-10, a software
made for HEO satellites was developed by Tom Clark W3IWI and
it was publisched into the ORBIT magazine and it's math is the
fundamental base for all satellite traking programs to day.
Nice to remember the old days of the OSCAR LOCATOR my friend.
73" de
i8CVS Domenico
----- Original Message -----
From: "Dave Guimont" <dguimon1@xxx.xx.xxx>
To: <amsat-bb@xxxxx.xxx>
Sent: Tuesday, October 02, 2012 12:06 AM
Subject: [amsat-bb] Re: maximum AO-7 distance
> Hi Bob,
>
> Draw the ground circle on a lambert conformal chart, and you can measure
> the distance directly...From your location use the distance from you to
> the west end of Florida, according to my ANCIENT OSCAR LOCATOR....That
> will probably warm the cockles of some old timer's hearts...Who is still
> around??
>
> Best 73 to all Dave WB6LLO
>
> On 9/30/2012 11:47 PM, Bob- W7LRD wrote:
>
> to my AO-7 afictionados...what is the maximum distance one can work edge
> to edge, and how did you figure that out? 73 Bob W7LRD
> _______________________________________________ Sent via
> AMSAT-BB@xxxxx.xxx. Opinions expressed are those of the author. Not an
> AMSAT-NA member? Join now to support the amateur satellite program!
> Subscription settings: http://amsat.org/mailman/listinfo/amsat-bb
>
> _______________________________________________
> Sent via AMSAT-BB@xxxxx.xxx. Opinions expressed are those of the author.
> Not an AMSAT-NA member? Join now to support the amateur satellite program!
> Subscription settings: http://amsat.org/mailman/listinfo/amsat-bb
------------------------------
Message: 8
Date: Mon, 01 Oct 2012 22:12:35 -0400
From: Gus 8P6SM <8p6sm@xxxx.xxx>
To: Bob- W7LRD <w7lrd@xxxxxxx.xxx>
Cc: amsat-bb@xxxxx.xxx
Subject: [amsat-bb] Re: maximum AO-7 distance
Message-ID: <506A4D93.5020803@xxxx.xxx>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
On 10/01/2012 02:47 AM, Bob- W7LRD wrote:
> to my AO-7 afictionados...what is the maximum distance one can work
> edge to edge, and how did you figure that out?
> 73 Bob W7LRD
Did you spot the error in my math, deliberately made to test you? :)
No? Well Steve did. Hard to believe I actually used a calculator and
STILL got that multiplication wrong. Thanks, Steve!
The correct calculations are:
MCD = 2 * 6371 * arccos [ 6371 / (6371 + 1460) ]
= 12742 * arccos [ 6371 / 7831 ]
= 12742 * arccos [ 0.8135614864 ]
= 12742 * 0.620545318575
= 7,906.98 Km (4913.16 miles)
--
73, de Gus 8P6SM
The Easternmost Isle
------------------------------
Message: 9
Date: Mon, 1 Oct 2012 23:07:04 -0400
From: Richard Lawn <rjlawn@xxxxx.xxx>
To: Amsat BB <amsat-bb@xxxxx.xxx>
Subject: [amsat-bb] SatPC32 & Flex5000A
Message-ID:
<CADQmrTEE=m3aU_zqoSo0768p7FJ7uyHT3YGw1hKGeeVD_27Mvw@xxxx.xxxxx.xxx>
Content-Type: text/plain; charset=ISO-8859-1
Ive started experimenting with converters and/or transverters on my
Flex5KA, using it as my receiver and an FT-847 as the uplink. I use SatPC32
to control both radios. I've got the Flex setup as rdaio 1 using the
Kenwood TS-2000 configuration and radio 2 is set to the FT-847. It seemed
to work for a short time. but I made some changes to the CAT Delay and the
status of the RTS +12 check box and now things are getting strange. When I
start SatPC32 the Flex is suddenly set to 65 mhz. Anyone have any
suggestions particularly about the CAT Delay setting and RTS +12 status?
TNX
73 de W2JAZ
Rick
------------------------------
Message: 10
Date: Tue, 2 Oct 2012 19:35:46 +0900
From: "Mineo Wakita" <ei7m-wkt@xxxxxxxxx.xx.xx>
To: <amsat-bb@xxxxx.xxx>
Subject: [amsat-bb] Re: ISS CubeSat deployment Oct 4
Message-ID: <D18962BAD20847A38038E419640B13CD@xxxxxxxxxxxxxx>
Content-Type: text/plain; format=flowed; charset="iso-2022-jp";
reply-type=original
ISS_KIBO CubeSats deployment Update
The deployment of five CubeSats from the International Space Station (ISS)
should take place on Thursday, October 4 commencing with the first pod at
1430 UT and the second at 1535 UT. It is planned to broadcast the deployment
live on the web. October 4 is also the 55th anniversary of the launch of
the first satellite Sputnik-1.
4th October 2012,
First, 14:30-14:40 UTC: Order of WE-WISH, RAIKO by Astronaut Akihiko Hoshide
Second, 15:35-15:45 UTC: Order of TechEdSat, NanoRack/F-1, FITSAT-1 by JAXA GS
http://iss.jaxa.jp/iss/jaxa_exp/hoshide/library/live/
http://www.amsat-uk.org/?p=10119
http://www.ne.jp/asahi/hamradio/je9pel/sputnik1.htm
JE9PEL, Mineo Wakita
------------------------------
Message: 11
Date: Tue, 2 Oct 2012 11:36:12 -0400
From: Martha <martha@xxxxx.xxx>
To: AMSAT BB <amsat-bb@xxxxx.xxx>
Subject: [amsat-bb] Time Is Running Out
Message-ID:
<CAPk0USyvVZHk75epAHnikT_wkY5BSDANxno6pG0J+kMftziUPA@xxxx.xxxxx.xxx>
Content-Type: text/plain; charset=ISO-8859-1
*Time is running out to register for the 2012 AMSAT Space Symposium and
Annual Meeting*. This year the meeting will be held in Orlando FL,
October 26-28th. You can register on-line by using the link on the main
web page.
--
73- Martha
------------------------------
Message: 12
Date: Tue, 2 Oct 2012 18:18:16 +0000 (UTC)
From: Bob- W7LRD <w7lrd@xxxxxxx.xxx>
To: amsat-bb@xxxxx.xxx
Cc: jamsat-bb@xxxxxx.xx.xx
Subject: [amsat-bb] Japan AO-7A
Message-ID:
<2103637495.84209.1349201896008.JavaMail.root@xxxxxxx.xxxxxxxxxx.xx.xxxx.xxxxx
xx.xxx>
Content-Type: text/plain; charset=utf-8
This last pass of AO-7, I could hear my downlink down to about 3 degrees.
From CN87wk (near Seattle) that covered most of Japan. So any AO-7 mode A Ja
station who would like to stretch the footprint on AO-7 set up a sked with me.
73 Bob W7LRD
CN87wk
------------------------------
_______________________________________________
Sent via amsat-bb@xxxxx.xxx. Opinions expressed are those of the author.
Not an AMSAT member? Join now to support the amateur satellite program!
http://amsat.org/mailman/listinfo/amsat-bb
End of AMSAT-BB Digest, Vol 7, Issue 327
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