All EM diagrams must specify a weight, altitude, power setting, and
configuration...

I was trying to relate the diagrams to real situation. As you put it, too
much factors come in to play as far as performance is concerned. The bank
angle and whether your opponent is sustaining level turn is one thing you
can see in ACM.

The absolute max turn rate for a given weight and altitude would be with
the aircraft inverted and pulling max available g...

That is the tactical egg right? If that is the case, Split S maneuver to
reverse flight direction beats max turn rate to turn 180 degrees?

Emilio.

"Andy Bush" wrote in message
...
Possibly.

Energy maneuvering (EM) diagrams that show turn rates usually assume a
level
flight condition (no altitude gain or loss). The sustained turn rate is
defined as a line that starts at the left hand boundary ("stall" line) and
goes across the chart to the right until it meets the max airspeed line.
This line, known as the "zero Ps" line, goes up and down as a function of
speed and "g".

The peak of the zero Ps line is the max sustained turn rate and is defined
as the intersection of a turn rate and airspeed value. In most cases, this
point is below the max attainable turn rate point. All EM diagrams must
specify a weight, altitude, power setting, and configuration...therefore
there is no such thing as one max sustainable speed.

Bank angle enters the picture as a function of available g and the need to
maintain level flight. If an aircraft can pull additional g while
maintaining level flight, then it is not at its max zero Ps point.

The absolute max turn rate for a given weight and altitude would be with
the
aircraft inverted and pulling max available g...the additional one g of
gravity would increase the turn rate as long as the lift vector was
oriented
below the horizon.
"Emilio" wrote in message
...
Is it correct to say that maximum sustained turn rate of an aircraft is
equivalent to steepest bank angle aircraft can sustain in a turn?

Emilio