OK, thanks - that's interesting. Couple of questions if I may:
How do you arrive at the 7Ah figure? So the 0.165A draw is from what you're calling the serial/Ethernet hub (is that a device server or something else?) So that's 4Ah. So does the cell modem live on just 3Ah? Thats an average draw of 0.125A or 1.5W @ 12v, which seems quite good. Is the cell modem powered up continuously in the sense of not sleeping between transmits? But presumably the current draw does fluctuate to some extent when an active transmission is under way? So how do you measure the average draw in practice?
Rather than trust the spec sheets, I set everything up on the bench and measured the current individually for the hub, and for the router both in transmit and in receive, and measured the transmit time for one transaction. Then I used that to calculate the total current draw per day.
I was very careful about this, as the site already had its solar power system in place, 3 or 4 panels on the roof and 400 AH of storage. I wanted to make sure I wasn't tapping too much power out of the battery bank.
And, second, have you ever calculated how many Wh of sunshine you need to generate your 7Ah (Did you say that your total panel power was 150W?) I guess this is only going to be very rough comparison because it will depend on the panel angle. I'm thinking that I should have my panels more vertical than horizontal because you need to maximise the winter power generation (at 52N) - I don't really care about summer power generation because I've got to size the panel array to get through the winter - summer generation is almost irrelevant as long as it exceeds winter. (This sort of argument tends to throw the folk installing panels for power generation because they're keen to maximise summer or year-round generation whereas all I want to do is to get through the winter reliably.) Actually what I would like to know is how to calculate the optimum panel angle to maximise midwinter power generation
Yes, I used to do this a lot at my previous job. Let me see if I can remember everything.
Add up all the amperages of the stuff you are trying to run and multiply x 24 to find out how many amp-hours (AH) you need per day.
(1) Let's say our system draws an average of 0.3 amps x 24 hours = 7.2 AH.
For 35 degree N latitude and this central California coast climate, in the winter, we have at most 5 hours of good sun on panels that are not tracked.
So, I need enough current from the panels to supply 7.2 AH plus 5 hours x 0.3 A = 1.5 AH (to run the equipment during sunny hours/charging), or 8.7 AH.
8.7 AH / 5 hours = a panel that can deliver 1.74 peak amps, or about a 20 watt 12 volt panel (12 x 1.74 = 20.88).
Note: This is all "perfect world" design. Round everything UP and add fudge factors.
For your 52N lat you may need to double that. There are charts that show things like this on the net, but like your experience with solar installers, everything I can find is related to that sort of work, not this.
Then, the battery bank. Lead acids of any sort last longest if never discharged below half their capacity.
Therefore, in a perfect world, your battery should be at least 2x the size of the AH you calculated in (1), or 17 AH.
To allow for decrease in cold weather capacity, and for operation in extended overcast, I'd make the battery 4-5 x bigger.
The tilt angle rule for the panel is usually to set it up from flat to an angle equal to your lat, pointed due south in the northern hemisphere, so I'd try setting your panel up at 52 degrees to start.
So, my conservative design for this system would be a 2x bigger panel (40 watts), and a battery 70 AH or bigger.