[amsat-bb] Re: probably simple
kl7uw at acsalaska.net
Fri Jan 8 10:36:22 PST 2010
The rate of change of Doppler is a sinusoidal function, so...
Think of a sine wave: the change in amplitude is highest close to the
zero crossing and smallest around the plus and minus maximums.
73, Ed - KL7UW
At 09:28 AM 1/8/2010, Mark L. Hammond wrote:
>I think part of the subject that confuses most of us (me included!) is
>that while at TCA the offset is 0, the RATE of change on either side
>of TCA is higher than at AOS or LOS. So, the actual frequency
>offset is BIGGEST at LOS and AOS, but the rate of change is slow (it
>doesnt' change much). It's easier to tune. At TCA you should hear
>the satellite at the "actual" frequency, but not for long because the
>change in frequency is fastest right before and right after...you tune
>like crazy to keep up.
>Maybe that will make sense to somebody...but I won't promise :)
>Also---don't forget---a 10kHz offset at 2M is 30kHz at 70cm (and one
>is increasing in freq while the other is decreasing...)
>Use SatPC32 and you just about forget about having to do anything, and
>watch it happen--good chance to see what is really going on.
>On Fri, Jan 8, 2010 at 1:11 PM, Pete Rowe <ptrowe at yahoo.com> wrote:
> > Hi Greg
> > I'm confused. It seems to me that the place when Doppler is zero
> is when the satellite is about half way through the pass. This is
> when the Doppler goes from positive (coming at me) to negative
> (going away from me). It seems like this place would be the time of
> closest approach (TCA) and not way out on the horizon when I see
> maximum Doppler.
> > What am I missing?
> > 73,
> > Pete
> > WA6WOA
> > --- On Thu, 1/7/10, Greg D. <ko6th_greg at hotmail.com> wrote:
> > From: Greg D. <ko6th_greg at hotmail.com>
> > Subject: [amsat-bb] Re: probably simple
> > To: glasbrenner at mindspring.com
> > Cc: amsat-bb at amsat.org
> > Date: Thursday, January 7, 2010, 7:34 PM
> > Hi Drew,
> > SatPC32 is probably an excellent program (I'm on Linux here, so
> can't use it), and if you've got the automation available, that's
> certainly the best way to go. And, by definition, the lower the
> elevation, the farther away the satellite is, so your DX contacts
> are going to be at the edges of the pass.
> > 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
> >> Date: Thu, 7 Jan 2010 05:53:13 -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
> >> Greg D. wrote:
> >> > Hi Bob,
> >> >
> >> > Whatever the satellite, if you're trying to stretch the
> footprint, your sked is always going to be at the peak of
> elevation, for that fleeting moment when the satellite is a few
> degrees above the horizon. That means that you're always going to
> be at zero doppler shift, and the math will always be the
> same. Find yourself once at TCA on any pass, and lock them
> in. The numbers will be the same for your sked.
> >> >
> >> > Good luck,
> >> >
> >> > Greg KO6TH
> >> >
> >> >
> >> I wouldn't agree with that statement at all. Most of my long haul
> >> contacts on AO-7, FO-20 and 29, and now HO-68, are right after AOS or
> >> just before LOS, certainly not at TCA. Use my recent QSOs on HO-68 with
> >> Argentina as an example. Even when I work Europe on AO-7 it is at the
> >> beginning or end of a pass...not the middle.
> >> Bob, SatPC32 will show you the frequency with Doppler shift, and the
> >> Doppler shift. A little subtraction or addition and you have
> what you want.
> >> 73, Drew KO4MA
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73, Ed - KL7UW, WD2XSH/45
BP40IQ 500 KHz - 10-GHz www.kl7uw.com
500-KHz/CW, 144-MHz EME, 1296-MHz EME
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