Help with calculating and understanding

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Your calculations are a bit off, because batteries are rated at the 20 hour rate, but you are assuming about a 120 hour discharge time, so the actual capacity will be somewhat more - around 20%.
But that won't really affect the basic calculations - all it does really is give some safety factor.
The panel size is a bit more complicated, but not rocket science. First you need to determine how important the load is - will it be a major event if whatever you are running goes offline. If so, then you need to factor in a pretty large safety factor - 35% or more - for those rare occasions when you get much longer than usual cloudy periods. Basically, 100% uptime will cost you a lot more than 99% uptime.
But ignore that for now - the basic calcuation is that you want your solar panel to supply what you use each day + around 20%. That is for the WORST case conditions - such as you might see seasonally with lots of clouds.
Assuming you look at the insolation charts, and you get worst case in December of 4 hours full sun per day. You are drawing .5 amps over 24 hours, or 12 amp-hours. You need to replace that 12 AH in during the 4 hours of sun you get, so you need a panel that will supply 12/4 = 3 amps. Add 20-25% or so safety factor, and you come up with around 3.5 amps. Most panels are rated at 17 volts or so, so 3.5 amps x 17 volts = about a 60 watt panel.
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If cloudy days were coinflips, storing enough energy for 1 day would make 50% uptime possible, 2 would make 75%, with 3 for 88%, 4 for 94%, 5 for 97%, 6 for 98%, and 7 for 99%... 100% would be impossible. Nick
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wrote:

But they are not, so the rest of your supposition is false. Anyone that watches the 6:00 o'clock local news and weather can tell that when you have one cloudy day, the chances of the next day also being cloudy are higher. Similarly, when one day is sunny, the probability of the next day also being sunny are higher than 50/50.
(and yes, I have evidence for that 'article of faith'. Look on NOAA's web site for any given weather reporting station and do the stats on cloudy days for a year or two)

daestrom
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You are using a common math fallacy that proves nothing.
We have installed many many 100% uptime systems over the past 30 years or so. It is quite simple, it is just more expensive.
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wrote:

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He did say "IF"
wrote:

than 99% uptime.

energy for 1 day would

for 88%, 4 for 94%,

be impossible.

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Hogwash. *Nothing* works *every* time.
--
Regards,
Doug Miller (alphageek at milmac dot com)
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Try stopping breathing for 15 minutes. It works ***EVERYTIME***
Wash your pigs someplace else.

again.
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The sad truth is that solar weather is probabilistic, like 100-year floods or coin flips. No matter how much energy storage we provide, if we wait long enough, there will always be a long enough cloudy day sequence to exhaust our energy store and leave the house cold or the lights out, unless we have some other backup system, eg a woodstove or a generator.
Nick
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snipped-for-privacy@ece.villanova.edu wrote:

Just an FYI, in the Detroit area, we just went from Dec 20 through Jan 5 (inclusive) with 14 minutes total sunshine.
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It has been one dark dreary depressing year end season eh?
We saw some sun for an hour this am then about 30 minutes too late at 4pm today.
Yuk! My PV panels are running dry and shrivelling up from lack of use.
Steve(Crocket)Spence must be on the parafin candles by now...LOL
wrote:

more than 99% uptime.

energy for 1 day would

for 88%, 4 for 94%,

be impossible.

Dec 20 through Jan

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Tony Wesley wrote:

I think what that means is that the clouds parted and exposed the solar disk for 14 minutes total over that time period (and having spent my holidays there, I cannot doubt its accuracy). It's not representative of the total amount of solar energy that penetrated the clouds and reached the surface of the earth. A little bit of math provides a quick sanity check:
14 minutes of full sun = 233 Wh/m^2 233 Wh/m^2 = 13.7 Wh/m^2/day over 17 days 13.7 Wh/m^2/day = approximately 2 W/m^2 during daylight hours
That's a tad darker than Barrow, Alaska (in the Arctic Circle) in late January, and no way was Detroit that dark. Indeed, if you check NREL's data on the subject, Detroit averages 1300 Wh/m^2/day in December despite the fact that long cloudy spells like the one you described are the norm for the area at that time of year. You can also check NREL's 30 years of hourly data to see that peak horizontal insolation in Detroit is usually greater than 200 W/m^2 and nearly always exceeds 100 W/m^2, even on Detroit's cloudiest days.
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This is a multi-part message in MIME format. --------------010907060709000304090302 Content-Type: text/plain; charset=ISO-8859-15; format=flowed Content-Transfer-Encoding: 7bit
Jim Baber
Tony Wesley wrote:

Another FYI, 'Total sunshine' is not required to generate appreciable power.
Today it has been cloudy for the 9 hours (so far) of daylight. My 10 kW system has been far from it's most productive for this time of year today. It has only produced 21 kWh today Feb. 4, 2006. On on the other side of the coin, on Feb. 1, 2006 it made 44 kWh on a beautiful Fresno winter day. So far this year the least I have produced was 0.568 kWh (truly terrible) on one cold rainy dark day in January. (Which is why I'm on the grid.) :-)
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begin:vcard fn:Jim Baber n:Baber;Jim adr:;;1350 W. Mesa Ave.;Fresno;CA;93711-2008;USA email;internet: snipped-for-privacy@baber.org tel;home:(559)435-9068 tel;cell:(559)905-2204 x-mozilla-html:TRUE url:http://www.baber.org version:2.1 end:vcard
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