"Greg Esres" wrote in message
...
"As seen in this example, for steady climbing flight, L (hence Cl) is
smaller, and thus induced drag is smaller. Consequently, total drag
for climbing flight is smaller than for level flight at the same
velocity."

I'm not questioning whether thrust contributes to lift, and thus reduces the
total lift requirement. It is patently obvious to me that a force directed
at least partially downward contributes to lift. That quote says nothing
more than that. What I am questioning is whether for a given performance
scenario there are multiple drag scenarios.

That is, he's proposing that at the same speed, there are multiple
steady states that produce different amounts of drag.

There are precedents. A banked aircraft at a given airspeed will have
a larger AOA than a non-banked one, and thus incur larger amounts of
induced drag.

It's clear that I continue to fail to state my objection properly. Let me
try again...

David's post implies that for a given performance scenario (straight and
level flight, for example) you can nudge the airplane into a "new steady
state" where drag is lower. Your examples of climbing and turning don't
address that issue; they are entirely different performance scenarios (that
is, the airplane is doing something different) than the scenario to which
drag is being compared.

According to David's original post (if I read it correctly), there are
multiple drag scenarios for a given path of flight. Each time you come up
with an example, it starts out by assuming a new path of flight compared to
the "base case".

I envision that a climbing airplane is essentially a lighter one,
since thrust will support a small amount of weight.

Seems reasonable to me. But what if you don't want to climb? And in
particular, if we're talking about comparing one airplane in straight and
level flight to another in straight and level flight, introducing a climb to
the discussion doesn't help much.

Pete