[amsat-bb] Re: probably simple

[Edward Cole]

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.
>
>
>73,
>
>Mark N8MH
>
>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
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