decided to try paralleling the load-recovery "source" voltage with the batteries. There's a bit to consider. 

For one thing, we've got a flyback inductor releasing power back to the input. At a 0.6A@5V dummy-load, that's 3W... But, of course, the, say 90%-efficient, converter needing loading is driven from a 40V battery-source, and thus the current going into that converter is much less than going out. 3.3W, 40V in, 3W, 5V out, so, roughly 0.6/8, less than 100mA. 

So, now, we charge up an inductor from its 5V output at, say, 1A [because we need a 0.6A average load, but some of the time will be spent with the inductor discharging]. Now, when discharging the inductor we need to dump 1A as quickly as possible. But it's discharging back into the converter's input, which is far less than 1A, more like 100mA. 

So, the thought, then, is to dump the majority of that current into charging a capacitor, then since the converter's input-range is 36-75V, that capacitor can charge to 75V. But, of course, this isn't free-energy, so once the capacitor's charged-up, it will begin to discharge, during that whole process dumping roughly 2.4W [assuming my circuit is 80% efficient] into the converter. But the converter needs 3.3W, so the inductor/capacitor obviously can't power the converter continuously. Once the cap discharges to 40V, the batteries take over.

Meanwhile, the inductor's being charged back up to 1A, and the cycle will repeat.

Now, all those numbers are rough guesses, and I've never designed something like this before, so it's coming on time to simulate. [And I don't have a clue how to calculate the necessary inductor/capacitor requirements] Like... how do I assure the cap doesn't exceed 75V, but that all our input power also goes out? Or, worse, how can this whole thing be properly-timed such that our inductor-charging and discharging times work, as well as the capacitor discharge time?

so, I thought an easy test would begin with falstad's simulator, best thing I could find for simulating the 3.3W converter input is an ideal voltage-controlled current source. P=VI, right? Nah, I=P/V, and it *really* wants to try V=0. So, now I've a bit to learn about the tools.

Ultimately, the point of the project, which led to this tangent, is to have a reliable and long-term power-source for my computer, among other things, so continuing down this tangent is a bit chicken/egg. That simulation's going to take some time to setup, meaning some time on the computer, which needs power. The 0.6A minimum load, realistically, is probably not a necessary concern to get this thing running.