wrote:

So if I normally commence a loop at 100 knotts but get the
entry speed wrong and start at 105 knots ....

Anyone care to formulate what happens when speed ( or "G")
are not constant?

Your speed and G are NEVER constant during a loop. A
vertical maneuver is always low and fast, then high and slow,
then low and fast again, etc.

You continually convert your kinetic energy at the bottom,
to potential energy at the top, then back to kinetic energy
on the downline. A hhead (aka stall turn) is a perfect example
of this. You go straight up until you stop, then pivot, and
fly down and gain airspeed again.

Given a constant density altitude, additional entry speed
implies additional G to make the same radius, assuming you
fly at (or near) the stalling AOA which generates Clmax.

Think of it this way: given that you fly at Clmax:

1) the radius of the vertical maneuver is a function of the
aircraft stall speed (Vs), and

2) The G you must pull or push is a function of the entry speed.

Does that make sense? It's not completely true - it will not
withstand a rigorous proof, but practically speaking, it's
what you really need to know to yank and bank down low.

The t-bird F-16 (famous canopy pic) that dug a hole this year
doing a vanilla reverse inside 1/2 cuban-eight is a perfect
example of this: he blew his gate - he was 1000 feet low. It
didn't matter how much G he pulled, or what speed he flew, the
F-16 was simply not capable of that tight a radius.

However, the Pitts with it's lower stall speed would have been
easily capable of it - 1500 feet is plenty for me, because my
stall speed is roughly half of his.

This is worth repeating: for a gate at the top of a vertical
maneuver (with downwards energy vector) such as a split-s
or reverse cuban-eight:

1) the altitude determines whether or not you hit the ground
(your radius is a function of your true stall speed, which
in turn is a function of density altitude) and,

2) the entry speed determines how much G you will have to pull
(or push, if it's outside) to attain Clmax which results in
the minimum radius. Hopefully this G is less than ultimate load!

I am assuming that everybody reading this is familiar with
a fundamental equation of aerodynamics, which is crucial for
understanding aerobatics:

Vs(G) = sqrt(G) x Vs(1G)

There is a fundamental relationship between speed and how
much G you can fly. A wing stalls at the pretty much the
same AOA, which can be attained at a variety of airspeeds.

N.B. The above is easily derived from the lift equation.
I will do so for the lowly price of a beer

P.S. One doesn't need a Phd (piled higher and deeper) to
understand this stuff. It's mostly high school physics with
a smattering of first year college/university mechanics.
Remember all that crap about weightless ropes and frictionless
pulleys? :-) I had forgotten about it too, until years later
I ran across a brainless pulley, but that's another story :-)

P.P.S. Good luck trying to find an aerobatic instructor
who has both a solid practical background and an understanding
of the theory involved, and who can explain both. They are few
and far between!

--
ATP www.pittspecials.com