How does battery voltage translate to watts stored? Does it?

Adding to what Marc says... For a first estimate, Lead Acid Batteries are fairly close to 100% "amp*hour" efficient... If you have a 100 AH battery and pull 50 Amp*hours out of it (assuming good battery, normal room temperature, etc.), you will have used ~50% of the Amp*Hour capacity.

However, Power (Watts) and Energy (Watt*Hours) are both on Voltage, current, and time...

• Amp*Hours = Amps * Hours (time)
• Watts = Amps * Volts [power]
• Watt*Hours = Amps * Volts * Hours (time) = Watts * Hours

So, if your battery starts out at 12.8 volts full, and goes to 10.5 volts Dead (not good thing to do), The voltage * current (measured every few minutes, gives you Watts*Hours.

Just for an easy example. Say your load is 1/20th (5%) of the battery's Amp*Hour capacity (a standard test condition--And works well for Off Grid calculations). Say starting voltage is 12.5 Volts under load, and ending is 11.5 Volts at 50% of battery capacity. And you are using (actual numbers will be different, but close enough for our estimates here):

• 100 AH @ 12 volt battery
• 100 AH * 5% rate of discharge (1/20 hour rate) = 5 amp draw.
• Say battery discharges evenly from 12.5 to 11.5 volts... Average would be 12 volts. And this goes for 10 Hours (1/2 of 20 hour discharge capacity)
• 12 volts * 50 AH = 600 Watt*Hours of energy used

Now if you were to draw at 1/10 capacity, the "apparent capacity of the battery may be ~80 AH... So 50% discharge would be something like:

• 80 AH capacity * 10% discharge rate = 8 amps
• 5 hour * 8 amp discharge  * 12.25 battery average discharge voltage = 490 Watt*Hours to 50% capacity (just a rough estimate for example).

There are battery monitor systems that are designed to give you a better estimate of state of charge/available AH/WH from the battery bank (not cheap, and are not perfect):

https://www.solar-electric.com/bogart-engineering-tm-2030-a-battery-monitor.html (less expensive, maybe less user friendly setup)
https://www.solar-electric.com/victron-energy-bmv-712-smart-battery-monitor.html (very nice monitor, with Bluetooth/alarms/etc.)

And there are some very interesting and cheap monitors these days (not sure I would call them battery monitors, but they do measure you loads). True battery monitor systems take time, temperature, battery capacity, battery efficiency, etc. into account... These AH/WH meters do not (in general?):

https://www.amazon.com/s?k=dc+ah+wh+meters&ref=nb_sb_noss

Seeing your second post... "K" is just a 1,000x Factor... 1 kAmp = 1,000 Amps, 1kAH = 1,000 AH, 1 kW = 1,000 kW etc. 

Our on grid homes use so much energy, the utility charges in kWH... An example for a medium size off grid power system:

• 3,300 WH per day = 3.3 kWH per day
• 3,300 WH per day * 20 Days = 99,000 WH per month
• 3.3 kWH per day * 30 Days = 99 kWH per month 

Amps and Watts are both "rates"... Amps is like Gallons per Seconds (or hour) (an amount). Watts is like Gallons per second (or hour) pumped at 100 psi (power).

So, A 12 Watt lamp running for 2 hours: 

• P=V*I (power or "rate")
• I=P/V= 12 volts / 12 volts = 1 Amp
• Amp*Hour = 1 Amp * 2 hours = 2 Amp*Hours (assuming 12 volts)--A "rate", but does not include voltage, so is not "power".
• Power = Volts * Amps = 12 volts * 1 Amp = 12 Watts (power, or "rate" of energy usage)
• Energy = Volts * Amps * Time = Watts * Time = 12 volts * 1 Amp * 2 hours = 24 Watt*hour of consumption.
• 1 amp * 2 hours = 2 hours from 12 volt battery
• 2 Amp*Hours / 100 AH capacity (at 20 hour rate) = 0.02 = 2% of battery AH capacity used (or 98% left)

For solar power systems, because we are usually using a mix of voltages (like 12 volt battery bus, and 120 VAC power from AC inverter)--We tend toi use Watts/Watt*Hours for our power/energy measurements to avoid confusion... Watts/WH are a "complete unit" Amps/AH are a partial unit (does not have Volts).

For things like a 12 volt Boat were all the loads (and generation) are done at 12 volts... Amps and Amp*Hours works great because everything is assumed to be at 12 VDC... 100 AH battery, draw 2 amps * 4 hours = 8 AH gives:

• 100 AH - 8 AH load = 92 AH left in battery (suggest to not draw down below ~50AH or 1/2 of capacity for longer battery life, and recharge at soon a pracible.

For us, say you have a 12 Watt lamp... Is that 12 Watts at 12 volts or is that 12 Watts at 120 Volts (AC):

• P=V*I
• I=P/V= 12 Watts / 12 volts (DC battery bus) = 1 amp @ 12 VDC
• I=P/V= 12 Watts / 120 volts (AC inverter output) = 0.1 Amps @ 120 VAC

You can see, that depending on how you power that 12 Watt lamp (12 VDC or 120 VAC), the amperage draw is a factor of 10x different.

But when we simply use 12 Watts, that is an easy number and we don't have to keep track of 12 VDC bus or 120 VAC house wiring... The "power/energy usage" is the same either way.

Not that you can use 1,000 WH in 1 hour, 10 hours or 10 minutes, the Average Load (in Watts) will be different:

• 1,000 WH / 1 hour = 1,000 Watt average load (for 1 hour)
• 1,000 WH / 10 hour = 100 Watt average load (for 10 hours)
• 1,000 WH / 0.1 hours (6 minutes out of 60 minute hour) = 10,000 Watt average load (for 6 minutes)

Note that draws of "non-standard voltages" such as 12 Watts at 7.5 Volts.. It does depend on how you make that 7.5 volts... Use a resistor:

• 12 watts / 7.5 volts = 1.6 Amp draw (at 12 volt)
• 1.6 amps * 12 volts = 19.2 Watt draw (extra power is lost as heat in resistor)
• 12 Watts / 19.2 Watts = 0.625 = 62.5% efficient conversion

But if you use a switch mode ballast (very common these days), then the usage would be:

• 12 volts * 1/0.95 switch mode ballast = 12.6 Watts used (ballast would lose about 5% of energy to heat).

You are asking lots of good questions, but it can be confusing to try and answer "everything" in one post, and not make things more confusing.

I suggest that you give us a "problem" that you want to solve, and we go through the design step by step (and probably several posts).

Also, many times we give answers that are "close enough" for discussion... Solar power, an answer that is within 10% of the "right answer" is usually close enough for our needs.

-Bill