I have often found, when trying to stay above buildups and flying right at the max ceiling of the aircraft for the conditions, that when it is no longer possible to hold altitude because one is really close to the stall, it takes only a very slight descent, say -100fpm, to gain quite a bit of useful speed.
And -100fpm means that you will lose just 1000ft over 10 mins.
This happens most often when crossing the Alps, and is usually due to a localised downdraught. In the summer, the air can be really warm e.g. -5C at FL160, and in those conditions you will get rapid icing if you dip down into the cloud.
Is there some formula relating VS to IAS? The complication is that one is likely on the back of the curve at this point.
L/D in highest cruise speeds is higher than 5 for most SEP in clean config, engine power just shift the VSI back to zero, if you fly near some (150kts) VNE power off you would see -3000fpm (30kts) if the VSI can show it, you can get same number or less toward -2000fpm with gear down at max gear speed
I gather -100fpm should get you +5kts on ASI near cruise and +10kts near best glide
Just thinking load here.
Let’s say your aircraft weighs 1000 kg. You descend at constant speed in vertical direction. The vertical speed is 100 fpm, which is 0.5 m/s. Then you constantly trade potential energy for alt (+ horizontal speed eventually) at a rate of dE/dt = mg*dh/dt = 1000*9.82*0.5 = 4910 J/s = 4.9 kW
You can think of this as the engine suddenly gain additional 4.9 kW of power, or 6.6 hp. Now, that power is used to gain speed, obviously as there is nothing else the power can be converted to, considering constant 100 fpm descend. But how much? That is entirely dependent on where on the back of the curve you started. Maybe the POH has some graphs that can be used? The principle will be that you gain much more speed than those 6.6 hp would suggests, since the aircraft also continuously looses drag when the speed increases. But those 6.6 hp is there continuously, like a jet engine, supplying clean thrust. Anyway, without the curves, it is impossible to say, at least this “early” in the morning 
I think that’s a great way to look at the issue, and 6.6HP is a LOT of power in percentage terms, because at say FL160 the engine is making only something like 50% power, 125HP, so it is 5% more HP.
I gather -100fpm should get you +5kts on ASI near cruise and +10kts near best glide
How would that be calculated? It is certainly in the right ballpark.
It’s actually a bit more. That power doesn’t have to go through the prop. The efficiency of the prop is 0.85 perhaps. Converted to shaft hp it will be 7.8 hp, which is 6.2 % increase in engine shaft hp.
100 fpm will give you 4.9 kW of power. Power is also force times speed. 125 hp is just enough to keep you flying at 100 knots? IAS, or 50 m/s. 125 hp at the shaft is about 107 hp from the prop, or 80 kW. P = F*v then F = P/v = 80000/50 = 1600 N. The drag is approximately 1600 N. Considering the drag is constant (which it isn’t, but let’s assume it is for now), then since v = P/F = (80 kW + 4.9 kW)/1600 = 53 m/s, which is around 106 knots. About 6 knots increase.
Taking into account the drag goes down a bit, starting from the left side of the curve, then +10 knots doesn’t sound very far off IMO. Using the correct mass and correct initial speed, it should be able to calculate this rather accurately, + that curve of course.
You don’t need to go via power: just wing L/D will link delta(VSI) to delta(ASI), you need will need power to link VSI to ASI though 
With low speed you fly a high angle of attack, resulting in high lift and drag.
Lowering the nose will improve that and lowering drag. It may even be the case that you end up flying level after some time, when you have built up enough speed.
I guess. But then I’m a “power engineer” working with power generation
Still, 100 fpm constant descent will give you a constant added power, regardless of L/D, only varying with the mass of the plane, at least in still air. What is not clear to me is what constitutes max alt. Which speed? I would think it is a bit lower than max L/D, more in the range of min sink or max endurance ? Intuitively I would think accelerating to max L/D would be a better approach than constant 100 fpm descend if the aim is to get farthest possible by losing the least amount of alt. You would lose alt faster, but get farther across the ground. hmm
Surely in conventional sailplane technique it goes something like this: (?)
Minimum sink speed is slower that best L/D (glide angle) speed. If you’re in sinking air, you need to get out of it, the faster the better, it doesn’t matter how efficiently. Then select speed based on whether you need to get somewhere or just stay in the air.
Wouldn’t that translate directly to the power plane situation?
Maybe this graph will shed some light.
I think Peter is flying way up there, and is therefore flying right at the bottom of the power required curve. The tangent to the curve at the bottom is horizontal, then starts to deflect as speed increases or decreases. The point is, at the bottom of the curve, only a slight increase in propulsive power will make a large increase in speed. That curve above is probably fictional, but nevertheless representative. Minimum power required is at 125 knots. Adding about 6% more power (100 fpm descent), and one can suddenly fly at about 150 knots. This is way below max L/D, but still a substantial increase in speed, 20 %, traded for a loss of only 1000 feet over 10 minutes.
This corresponds to glider polars of speed and sink, for instance below:
IMO, the extra power you get from 100 fpm descend is easy to calculate, but what that represents in horizontal speed is impossible without that curve (at least I have no idea how that is done without the curve). It’s probably better just try it a few times for different descent rates (in still air preferably). It’s also one of those things one can investigate with X-plane, considering an OK model of the aircraft exists.
