I see no one reacted to this post. So, either we are all geniuses or all of us were too embarrassed to admit that the maths used by

@Dairyfarmer is shall we say quite overwhelming. Therefore in the interest of us slobs who were rather bamboozled at the time and maybe still are, here is an explanation of what he was up to.

The calculation he did is based on sound principles.

So maybe there a bit of room to explain it @Dairfarmer if you have no objections I hope

?

1. Assuming there are no losses and conversion losses, efficiencies and power losses involved then,

a 7Ah, 12V battery will have an energy capacity of 7 x 12 = 84 Wh [Ah x V= A x V x h] = Wh.

2. The load is 20 W

3. So then the run time would be Wh (battery capacity)/W (load) = Rt (hours); 84/20 = 4.2 Hours,

IF the battery is completely discharged to 100% DoD.

4. Now IF you want to only discharge to the battery to 50%, then the available Capacity would be

84/2 = 42 Wh.

Then the Rt would be 42/20 = 2.1 hours.

{50% of battery capacity[Wh]/load [W] = Rt [hrs]}

5. If you now assume that all the losses etc mean that 15% of the capacity would be lost,

then only 85% of the capacity will be available to drive the load.

ie 42 [Wh] x 85% / 20 = Rt = 1.785 hours = 1h 47min.

6.

**For a 20w load you need approximately 1ah of every 15 minutes of backup time (12 / 20 / 2 x 0.85**
The line at the top, therefore, is a shortcut calculation designed to confuse the masses that allows for both a 50%DoD and a loss allowance or efficiency of 85% and comes up with hours required per battery Ah.

12 [V] / 20 [W] x 50% DoD x 85% [efficiency] = 15.3 minutes/Ah

7Ah = 7 x 15.3 = 1h 47.1 min ( rounded to 1 h 45min, to confuse the masses some more)