October 2012 - AMSAT-BB - mailman.amsat.org

Japan AO-7A
by Bill W1PA 04 Oct '12

04 Oct '12

In the right place, but no rigs or reciprocal licensing. :-) Bill W1PA/JA (shin-yokohama)

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Euroluna Google Lunar X-Prize Quarterly Update
by Trevor . 03 Oct '12

03 Oct '12

The President of Euroluna, Palle Haastrup OZ1HIA, provides a video update on the progress of the Euroluna Google Lunar X-Prize team who are aiming to land a rover on the Moon. Their first amateur radio spacecraft Romit-1 is a 2U CubeSat that will transmit on 437.505 MHz. It will be fitted with an Ion Motor capable of raising the orbit from 310 km to 700 km. Watch the videos at http://www.uk.amsat.org/?p=10731 73 Trevor M5AKA AMSAT-UK http://www.amsat-uk.org/ ----

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temperamental AO-7
by Bob- W7LRD 03 Oct '12

03 Oct '12

Just an observation...this last pass on AO-7 1650Z. I noticed that on a downlink of 145.957 is significantly stronger than on 145.950. I don't know why, just my observation. Maybe .957 could be a "watering hole " for DX. 73 Bob W7LRD Seattle

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SATPC32 using Icom 9100
by n4csitwo＠bellsouth.net 03 Oct '12

03 Oct '12

Can anyone confirm whether or not the current version of SATPC32 plays well, or at all, with an Icom 9100? If not, is there a fix for this? Dave, AA4KN

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Japan AO-7A
by Bob- W7LRD 02 Oct '12

02 Oct '12

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

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Time Is Running Out
by Martha 02 Oct '12

02 Oct '12

*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

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SatPC32 & Flex5000A
by Richard Lawn 02 Oct '12

02 Oct '12

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

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maximum AO-7 distance
by Bob- W7LRD 02 Oct '12

02 Oct '12

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

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Re: maximum AO-7 distance
by Stephen Melachrinos 02 Oct '12

02 Oct '12

On 10/01/12, Bob- W7LRD<w7lrd(a)comcast.net> 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-t…

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Re: ISS CubeSat deployment Oct 4
by Trevor . 01 Oct '12

01 Oct '12

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

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