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submitted 8 hours ago by Iconoclast@feddit.uk to c/pics@lemmy.world

I'm only half kidding. I'm a bit of a prepper and I have lots of powerbanks and devices that charge from USB but besides idling my truck I really had no other way to charge any of them in case of a long-term power outage which seemed a bit of an oversight on my part.

Not like this solves the issue. 30 watts (under ideal conditions) isn't much but it's a start.

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[-] danekrae@lemmy.world 28 points 7 hours ago

Have you calculated how long it takes/took to pay for itself?

[-] IcedRaktajino@startrek.website 38 points 6 hours ago* (last edited 3 hours ago)

All of the labor has been DIY, and I've got about $9,000 invested in the system components.

Back of the napkin math, at current electric rates, puts my ROI at just under 9 years. If electric rates keep going up, it'll pay back even faster.

Edit: Depending on how you look at it, ROI could be in as few as 1.5 to 2 years. This system is in lieu of a whole house generator that would have cost us about $7,000. Had we gone with that instead, it would never have paid for itself (other than peace of mind) and we would have been perfectly content. So if I'm only considering the price difference between this and the generator, it'd only have to pay back $2,000.

[-] mrsilkworm@piefed.social 12 points 5 hours ago

A hobby pays itself just by enjoying it. I've never calculated how long it will take to pay off my gaming rig

[-] Dookieman12@piefed.social 20 points 7 hours ago

If you enjoy it, it already has.

[-] huppakee@lemmy.world 5 points 6 hours ago

I get you but it was worth the money you spent ≠ it earned back the money you spent.

[-] ripcord@lemmy.world 3 points 6 hours ago
[-] Eyekaytee@aussie.zone 4 points 6 hours ago

have battery system and solar since jan 2023, still love it

[-] Dookieman12@piefed.social 1 points 5 hours ago* (last edited 5 hours ago)

Time, in hours (H), equals average solar kilowatt hours per month (K), multiplied by the price of one kilowatt hour (P), divided by the total cost (C) of all the purchased components.

H = (K * P) / C

I'm sure if you were patient and dedicated enough, you could approximate each of those numbers using the info OP has already posted and get a general idea (weeks, months, or years).

[-] wunami@lemmy.world 0 points 4 hours ago* (last edited 4 hours ago)

Time, in hours (H), equals average solar kilowatt hours per month (K), multiplied by the price of one kilowatt hour (P), divided by the total cost (C) of all the purchased components.

H = (K * P) / C

So you're dividing the average saving per month by the total cost and expecting to get hours?

If we generously estimate a very high 3000kWh/month generated and high $0.40/kWh price and it cost OP $9000, then you formula is (3000×0.40)/9000 = 0.133 hours.

Breaking even after less than 1 hour?!??! Extraordinary!

[-] Dookieman12@piefed.social 1 points 1 hour ago

It's pretty easy to cry about bad math, but it's a lot harder to figure out the right math.

Don't worry, I'll try to do it for you again a second time.

Power consumption varies. Use the average monthly power draw from the solar array, let's assume for demonstration purposes 1,000 kWh/month.

Multiply that by the cost of 1 kWh from the power company, let's say 20 cents.

In one month, that means you saved $200.

Let's assume the solar equipment costs $1,000.

The answer is 5 months, or 5,000 kWh.

Sorry, I'll make sure the free work I do for you is better quality next time.

[-] ripcord@lemmy.world 0 points 5 hours ago

Or I could read the comment where he said

[-] Dookieman12@piefed.social 0 points 4 hours ago

Then do that instead of asking random people who aren't in any position to know.

this post was submitted on 18 Jul 2026
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