[amsat-bb] Re: probably simple
ko6th_greg at hotmail.com
Thu Jan 7 21:41:45 PST 2010
Oh, my. You are correct; I was picturing a satellite running between two stations, versus running overhead from one and on to the other. In that case, in the limit, one station would be at AOS, and the other at LOS, and neither at zero doppler.
Sorry, and thanks for your patience!
> Date: Fri, 8 Jan 2010 00:30:44 -0500
> From: glasbrenner at mindspring.com
> To: ko6th_greg at hotmail.com
> CC: w7lrd at comcast.net; amsat-bb at amsat.org
> Subject: Re: [amsat-bb] Re: probably simple
> > But any pass where you are really stretching the footprint is going to
> > be a low elevation pass. The more you stretch, the lower the pass.
> > In the limit, I think Bob's ultimate pass has a peak at .001-degrees
> > for both stations. If you're doing that, then you're at TCA, and zero
> > doppler.
> > That's all I meant to convey,
> > Greg KO6TH
> I understand, but with all due respect your assumption is incorrect that
> long distance QSOs are always at TCA and zero Doppler, even
> theoretically . This is only the case when the two stations are at near
> right angles to the track of the satellite.
> Use the following example. If I want to work LU5BOJ/O in FG75 from EL88
> on HO-68, my only windows are at either LOS or AOS, depending on whether
> it is an ascending or descending pass. Neither pass will be a low pass
> for either station, and neither QSO will occur at TCA or zero Doppler.
> In Bob's case, let's look at his next possible window with Paul, 2E1EUB
> in IO92. Bob is in CN76. At the beginning of the 1 minute window
> tomorrow at 1251Z, Bob's Doppler shift on 432 is -3.68 khz.
> On the next mutual window at 1452Z, the beginning Doppler is -7.93 khz.
> On the next, at 2020Z, it's -8.3 khz. None of these windows are over 2
> degrees elevation, and none are at TCA for either station.
> When you have 60s to make the QSO, being right dead on frequency is
> Drew KO4MA
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