I have one technical remark regarding Dave's presentation.

That's a good point.(..) In my post #174 i used a rough curve fit and then solved for the several cutoff voltages and then calculated the ratios

Another thing worth noting is that the integral of u(t) on a power=const curve does not represent any meaningful quantity (that I know of). The area is proportional to the integral of reciprocal of i(t).

The energy left that you/we were interested in is simply proportional to the distance left on the time axis on that chart - no need to integrate anything (that is the core idea of that chart anyway). I have just noticed that you follow Dave's mistake (13:00). Oops.

If you would like the area to be proportional to something meaningful, I'd suggest integrating either voltage of constant current discharge curve (then the area is proportional to the energy) or simply integrating current on a current=f(something) chart (then the area is proportional to the charge).

Still, my previous offset remark applies.

Hi,

I am not sure why but you must have misinterpreted my post, or else i did not explain what i did clearly enough.

First, i never integrated anything. All i did was do a quick curve fit of the constant power curve, then solved for specific voltage set points that are easy to relate to such as 1.4, 1.3, 1.2, and 1.1 volts. So there was no area calculation of any kind, just one calculation for the exact voltage set point, then another calculation for the ratio of two time values in hours.

The first calculation is just like using the inverse function of the constant power discharge curve, which given a voltage, yields a time in hours. The ratio is then the time in hours to discharge to 0.8v divided by the time to discharge to the given voltage set point (such as 1.3v). This ratio is then a number like 1.6, and because that is a ratio that expresses the entire gain the actual gain over a normal run without any circuit would be 1, so the gain with a circuit (that works that is) would be 1.6-1 and of course this equals 0.6 and that is the percent gain in time expressed as a fraction so this would mean 60 percent more run time.

Also just to be clear, this would be with a boost circuit that is 100 percent efficient with any input voltage. We'd have to cut this back of course for a boost circuit that ranges from 50 to 90 percent efficient, and if they got 95 percent efficiency from their circuit i'd be very surprised seeing as they dont know how to calculate anything else <little chuckle> :-)

Short form of the math i used:

For v(t) = approximate constant power discharge curve,

t = T(v), where T is the inverse function of v(t), and t in hours, v in volts,

t1=T(0.8 ), the time to discharge to 0.8v, in hours,

then the total gain ratio is:

r(t)=t1/t for any given time t, or:

r(v)=t1/T(v) for any given cutoff voltage.

I'll also point out that the curve from 0.8v to 0.6v is almost a straight line down so there is very little gain in time to add there if we went all the way down to 0.6v.

So we see ratios of about 1.0 to 12 or so, but many of these cutoff points are not realistic that's what makes this whole product seem like a bunch of bool crap :-)