East River turning radius

I got into this debate here several years ago. I was of the opinion (not
having a G meter to test it) that your G load would be less if you were
descending, but most seemed to think that 60 degrees was a 2 G turn
regardless of whether you were holding altitude or not.

Which is it? I still think G forces would be reduced by descending, but
can't come up with a good explanation why.

mike

"Grumman-581" wrote in message
.. .
"d&tm" wrote in message
...
I have heard guys on this group regulary mention 60 degree
or 2 g turns, but in my training steep turns were 45 degrees
maximum.

A 60 degree turn is only 2Gs if you you maintain altitude during the
turn...

"Grumman-581" writes:

A 60 degree turn is only 2Gs if you you maintain altitude during the turn...

More generally, it has to be a coordinated turn.

--
Transpose mxsmanic and gmail to reach me by e-mail.

In article ,
"mike regish" wrote:

I got into this debate here several years ago. I was of the opinion (not
having a G meter to test it) that your G load would be less if you were
descending, but most seemed to think that 60 degrees was a 2 G turn
regardless of whether you were holding altitude or not.

Which is it? I still think G forces would be reduced by descending, but
can't come up with a good explanation why.

Consider the vectors. Consider the portion of the lift vector that is vertical.
If you bank, the vertical portion of the lift vector is decreased and you will
descend (assuming level flight initially).

--
Bob Noel
Looking for a sig the
lawyers will hate

Meaning that a descending turn will produce a smaller G force, correct?

mike

"Bob Noel" wrote in message
...
In article ,

Consider the vectors. Consider the portion of the lift vector that is
vertical.
If you bank, the vertical portion of the lift vector is decreased and you
will
descend (assuming level flight initially).

--
Bob Noel
Looking for a sig the
lawyers will hate

"mike regish" wrote in message
...
"Grumman-581" wrote in message
.. .
"d&tm" wrote in message
...
I have heard guys on this group regulary mention 60 degree
or 2 g turns, but in my training steep turns were 45 degrees
maximum.

A 60 degree turn is only 2Gs if you you maintain altitude during the
turn...
I got into this debate here several years ago. I was of the opinion (not
having a G meter to test it) that your G load would be less if you were
descending, but most seemed to think that 60 degrees was a 2 G turn
regardless of whether you were holding altitude or not.

Which is it? I still think G forces would be reduced by descending, but
can't come up with a good explanation why.

You need to look at the acceleration, not the velocity. If you're
descending at a constant vertical velocity (zero acceleration), then the
forces are the same as with constant altitude, so you still get a 2G turn
with a 60-degree bank. (Emergency-descent spirals are often flown that way.)
If you accelerate downward though, you get less than 2G during the
acceleration.

This is a special case of the Galilean/Newtonian relativity principle that
there is no absolute frame of reference for velocity--no detectable
difference (except relative to other objects) between being at rest and
being at (constant) motion. Here, we're just applying that principle to the
vertical component of motion.

--Gary

mike regish writes:

Meaning that a descending turn will produce a smaller G force, correct?

You would have to be accelerating downward, and not just descending at
a constant speed. The greater the rate of acceleration, the smaller
the acceleration due to gravity, and the smaller the overall G force.

The smaller the acceleration due to gravity, the greater the bank
angle for a given turn. If there is no gravity at all (i.e., the
aircraft is vertically in free fall), all turns will have a bank angle
of 90°.

--
Transpose mxsmanic and gmail to reach me by e-mail.

OK. That makes sense.

mike

"Gary Drescher" wrote in message


You need to look at the acceleration, not the velocity. If you're
descending at a constant vertical velocity (zero acceleration), then the
forces are the same as with constant altitude, so you still get a 2G turn
with a 60-degree bank. (Emergency-descent spirals are often flown that
way.) If you accelerate downward though, you get less than 2G during the
acceleration.

This is a special case of the Galilean/Newtonian relativity principle that
there is no absolute frame of reference for velocity--no detectable
difference (except relative to other objects) between being at rest and
being at (constant) motion. Here, we're just applying that principle to
the vertical component of motion.

--Gary

"Grumman-581" wrote in message
.. .
"d&tm" wrote in message
...
I have heard guys on this group regulary mention 60 degree
or 2 g turns, but in my training steep turns were 45 degrees
maximum.

A 60 degree turn is only 2Gs if you you maintain altitude during the
turn...


Close, but not exactly. The geometry might cause a to be trivially more in
a descending turn and trivially more in a climbing turn. However, during a
constant rate climb or descent, a 60 degree bank will be very close to
2Gs--for a typical standard or utility category airplane which can maintain
only a modest angle of climb or descent.

Maneuvering in any manner which trades altitude against airspeed causes
radical changes in the relationship of G loading and bank angle, and most
are easier to demonstrate than describe.

Even in the simplest case (which really exceeds my ability to fully
describe), consider an aircraft flying a circular path in a plane angled 30
degrees from the horizontal (or plane of gravity). At the bottom of the
maneuver, a 30 degree bank relative to the horizontal would be 60 degrees
relative to the plane of flight and would significantly exceed 2Gs, but may
still not be sufficient. At that point, a 60 degree bank would be knife
edge relative to the plane of flight, so it would be quite easy to reach the
structural limit of standard category within the maneuvering limits of
standard category. Conversely, at the top of the maneuver, a 60 degree bank
is only 30 degrees relative to the plane of flight; but a very short turning
radius is easily achieved and the loading can easily be 1G, or even slightly
less. Meanwhile, halfway up or down during the maneuver, the relationship
of bank angle to loading should be identical to what would be expected in
level flight.

Crop dusters and banner towers routinely exploit portions of the above, plus
some usefull additions, but a Google search did not yeild any usefull links
to include for a graphical description.

I hope this helps.

Peter

On Oct 15, 11:34 am, "Peter Dohm" wrote:
Close, but not exactly. The geometry might cause a to be trivially more in
a descending turn and trivially more in a climbing turn. However, during a
constant rate climb or descent, a 60 degree bank will be very close to
2Gs--for a typical standard or utility category airplane which can maintain
only a modest angle of climb or descent.

I don't have a G-meter on my plane, so all I can tell is from the
feeling in the seat of my pants, but if I'm doing a 90 degree turn
banked at 60 degrees and I'm willing to lose 500 ft in the process, it
feels like quite a bit lesss than what I experience when I don't want
to lose 500 ft in the process... I do this quite often when coming back
to my home airport from the south... Fly over midfield at 1500 ft until
I nearly intercept the downwind leg... If no traffic conflicts, drop
the left wing to 60 degrees or more while making a 90 degree turn to
downwind, losing 500 ft in the process...

"Grumman-581" wrote in message
oups.com...
On Oct 15, 11:34 am, "Peter Dohm" wrote:
However, during a
constant rate climb or descent, a 60 degree bank will be very close to
2Gs--

I don't have a G-meter on my plane, so all I can tell is from the
feeling in the seat of my pants, but if I'm doing a 90 degree turn
banked at 60 degrees and I'm willing to lose 500 ft in the process, it
feels like quite a bit lesss than what I experience when I don't want
to lose 500 ft in the process...

Right, because you're accelerating downward. But if you perform a
constant-rate descent instead, you'll feel the usual 2 Gs.

--Gary