Rick Cavallaro wrote:

On Oct 9, 3:15 am, Ronald Raygun wrote:

I've been looking forward to your assessment of my analyses.

Alrighty then... it looks like I'm back. Yes, I agreed with each part
of your analysis.

I'd welcome any additional comments you'd care to make.

Unfortunately, I don't recall what comments I did attempt to make over
the past few days. Both JB and I were shut out from posting for some
reason.

I think one of the things which confused me initially was the notion
of propeller efficiency. I thought what was meant was something
similar to the efficiency of any other device or machine intended to
convert one form of energy into another, like the generator, the motor,
and any associated bits and pieces of couplings.

But that is exactly what is meant by propeller efficiency.

But not in the sense of what I said below, that energy is lost to
heat *during* the actual conversion.

Any inefficiency would mean a "loss" of energy, a conversion into
undesirable form, usually heat. I assumed that in the same sense
a less than 100% efficient propeller would also be losing energy
somehow.

That's right. Consider a Cessna sitting still in the run-up area with
brakes locked and prop spinning at full revs. It's blowing a lot of
air, but the plane is not moving forward. All of that air motion
ultimately becomes waste heat.

Yes, ultimately, but we're not really interested in what happens ultimately,
we're interested in what happens initially. Perhaps I should say "I" rather
than "we", because I think I was interpreting efficiency differently from
the way you were, and are. Remember, what I'm trying to do now is explain
how I think my understanding of prop efficiency differed from yours, so if
I say "I thought it was X" it confuses me more if you then say "but that's
what it does mean".

Let's change the skater analogy to match your Cessna. Suppose our hero
has a rope tied around his waist which is tied to a stake in the ground
behind him. This time the plebs are coming towards him at 1 m/s from the
front, and he pushes back on one of them for a second at the same 60 N as
before, accelerating him to 2 m/s. During the second of contact the pleb
travels 1.5 m, so our hero does 90 J of work. The 60 kg pleb's kinetic
energy increases from 30 J to 120 J.

The hero's arm may conceivably have been 100% efficient at converting 90 J
of chemical energy into a 90 J kinetic energy increase for the pleb. In
the same way, the Cessna's prop could conceivably be 100% efficient at
converting engine power into kinetic air power. We don't care what happens
to the accelerated pleb after he's left the grasp of the hero's arm, in
terms of how that pleb might vaporize when it collides with another pleb.
Likewise we shouldn't need to care whether the accelerated air gets hot
as it slows down a few tens of yards downstream.

I was thinking about the prop's efficiency as how its kinetic output power
relates to the mechanical input power. I think you're considering prop
efficiency in terms of what the plane gets out of the deal. In your Cessna
with the brakes on, I presume you are considering the prop as running at 0%
efficiency because its thrust is doing no work *on the plane*. I guess you
don't care whether the loss ends up as heat or not, you simply consider any
work done on the air as wasted. But that won't help you calculate how much
fuel the engine's going to need.

If I cut the skater's rope so that he can accelerate too (corresponding to
your Cessna releasing the brakes), he will travel forward 0.5 m during the
second, and so he will have done 120 J of work. 90 J of that is still going
into the pleb, and the remaining 30 J will have accelerated himself. In my
sense his arm is still 100% efficient, in another sense he could be thought
of as 25% efficient because, out of 120 J expended, 30 J have gone into
useful KE, and 90 J into useless KE. But in your sense I think he's being
50% efficient because 30 J/s (30 W) is 50% of the best he could get by
exerting 60 N at 1 m/s (60 W). Am I right?