Hi Joe, Check out this : https://qso365.co.uk/2018/04/updated-guide-to-setting-up-an-aprs-rx-only-igate-using-a-raspberry-pi-rtl-sdr-dongle-and-a-pre-built-image/

Right now I am only concerned with receiving and decoding to make sure payload(s) are running correctly before release. I have zero need to be transmitting at this time.

Also look at byonics -----------------------------------------From: "Christopher Rose" To: GPSL@groups.io Cc: Sent: Tuesday August 10 2021 11:58:57AM Subject: Re: [GPSL] APRS Raspberry Pi, Direwolf

Raspberry Pi, Direwolf software sound card, add gps for position and altitude and whatever radio u want. Argent data makes trackers with gps with combined transmitters as well.

It has been a LONG LONG time since I have done anything with APRS. With the state of modern computers now days, I would expect that the need of the TNC's and

Oops. I think I replied to a post that has been out for quite a while. Not only that, but I see I misunderstood the scenario. Glad that so many jumped into the thread! Cheers! DE N8PK --

Hello Joe, A “first order “ approximation might be calculated fairly easily. Since the shadow/terminator runs around the Earth once per day, just make a good guess* as to the circumference (your

I have studied the code a bit but not yet used it. I am planning on building the instrument it is intended for but I am going to convert the spreadsheet to some python code so that it can be more

Jim, Have you used many of the features of that spreadsheet? It's full of bells and whistles. Hank

I think you're really solving too far when you want to go past a table of apparent sun angular elevations and get into speeds.. Because the sun's vertical speed varies with sun elevation (and also

It might make more sense if it's minus .27 degrees/minute where the definition is downward movement of the setting sun is considered negative. A positive virtual sun movement component is caused by

This spreadsheet was intended for a different purpose but it will calculate the height of the earth's shadow directly above the specified location vs time along with a lot of other

At the moment of sunset and near where you are (Milwaukee, WI), the sun is moving at .27 degrees per minute at this time of year. This means the sun's shadow is moving at 1.6 kilometers/minute (60

I'm gonna attack this in a reverse way like this, Ok, Say the Sun has set, and it is now 1 degree below the horizon. How high would you have to get, to be able to see

I found what you found, and the example is for the equator only. This is way off for your latitude.

I found this, The variation with altitude is approximately linear, and so we conclude that sunset is later by 1 minute for every 1.5 kilometres in altitude, and that

Who will be the first to explain why the logic (quoted below) and the resulting speed are wrong? The division is actually 329.875 which is 330 MPH rounded off. The diameter at the equator can be

Joe, This wording of the question explains the reason you're asking the question, and also makes a quicker (easier to obtain) and more practical (less abstract) answer possible. Watching the Olympics

You're thinking of another "angle" on the shadow and another speed. The solution calls for a speed in the vertical direction if you just think about it a little. Not horizontal speed along the

The question now is defined and is solvable. It can be solved at the 1. moment of sunset, and it can be solved as a 2. function of time elapsed after sunset. For solution 1, the