barrel roll in 172

On Sun, 23 Jul 2006 14:00:37 GMT, Matt Whiting
wrote in ::

Larry Dighera wrote:

On Sun, 23 Jul 2006 01:54:35 GMT, Matt Whiting
wrote in ::


Flight load factor

Flaps up = +4.4 G's and -1.76 G's


So, in inverted flight a C-172 has only a 76% of a G margin to carry
additional G force. That isn't much.

Thanks.

No, 176% of a G.


No. A _margin_ of only a 76% of a G to carry G forces in addition to
the one G natural force.

When you are pulling negative G, there is no one natural G force.

While you are in the vicinity of the Earth, your aircraft is being
acted upon by the Earth's one G gravitational force.

If you are flying straight and level while inverted, the airframe is
experiencing -1G, not 0 G.

That provides the remaining 0.76 (76%) of a G of the C-172's negative
load factor specification of -1.76 to carry the load of any
acceleration that may subsequently occur.

It takes -1 G of acceleration to counter gravity and get you to 0 G.

Agreed.

You can then add -1.76 G of additional acceleration and still be within load
limits.

That analysis presumes the aircraft is not inverted.

The negative G load factor is referenced to 0 G, not 1 G
straight and level.

Are not both the positive and negative load factors referenced to 0 G?
You don't set your G-meter to 0 when you are on the taxiway; you set
it to read 1 G, right?

We were discussing the negative load that might be encountered during
the inverted recovery from a barrel roll, so the earth's gravity would
add to any accelerative force while the aircraft is inverted, right?

"john smith" wrote in message
...
In article ,
Matt Whiting wrote:

When you are pulling negative G, there is no one natural G force. It
takes -1 G of acceleration to counter gravity and get you to 0 G. You
can then add -1.76 G of additional acceleration and still be within load
limits. The negative G load factor is referenced to 0 G, not 1 G
straight and level.

Is negative G an up force or a down force?

I'll disagree, here.

Straight level flight, right side up is 1 G

Free-fall, so you are not touching the seat or seatbelt is O G

Straight level flight upside down, is -1G

If the plane is upside down, and pushes up elevator, until a100KG weight
weighs 176KG, that is a -1.76 G factor. Not much extra for upside down
flight, is it.
--
Jim in NC

Larry Dighera wrote:


We were discussing the negative load that might be encountered during
the inverted recovery from a barrel roll

If the barrel roll is performed correctly it is a +G maneuver
throughout, even while inverted.

On Sun, 23 Jul 2006 14:18:23 -0600, RomeoMike
wrote in
::

Larry Dighera wrote:


We were discussing the negative load that might be encountered during
the inverted recovery from a barrel roll

If the barrel roll is performed correctly it is a +G maneuver
throughout, even while inverted.

While that may be true, it isn't pertinent to this branch of this
message thread:

From: "Jim Macklin"
Newsgroups: rec.aviation.piloting
Subject: barrel roll in 172
Message-ID: 2SQvg.78944$ZW3.35114@dukeread04
Date: Thu, 20 Jul 2006 14:31:05 -0500

The issue is recovery from a botched barrel roll, such as in
inverted stall followed by a high speed dive and too strong
a pull on recovery.

"Jim Macklin" wrote in message
news:_aMvg.78918$ZW3.55764@dukeread04...
Didn't Lorena Bobbitt say that?

Not sure if we've ever seen the entire quote as her post was "snipped."

Jay B

john smith wrote:

In article ,
Matt Whiting wrote:


When you are pulling negative G, there is no one natural G force. It
takes -1 G of acceleration to counter gravity and get you to 0 G. You
can then add -1.76 G of additional acceleration and still be within load
limits. The negative G load factor is referenced to 0 G, not 1 G
straight and level.


Is negative G an up force or a down force?

It is an acceleration opposite the normal acceleration due to gravity,
so it would be downward with respect to the airframe.

Matt

Larry Dighera wrote:

On Sun, 23 Jul 2006 14:00:37 GMT, Matt Whiting
wrote in ::


Larry Dighera wrote:


On Sun, 23 Jul 2006 01:54:35 GMT, Matt Whiting
wrote in ::



Flight load factor

Flaps up = +4.4 G's and -1.76 G's


So, in inverted flight a C-172 has only a 76% of a G margin to carry
additional G force. That isn't much.

Thanks.

No, 176% of a G.


No. A _margin_ of only a 76% of a G to carry G forces in addition to
the one G natural force.

When you are pulling negative G, there is no one natural G force.


While you are in the vicinity of the Earth, your aircraft is being
acted upon by the Earth's one G gravitational force.

If you are flying straight and level while inverted, the airframe is
experiencing -1G, not 0 G.

That provides the remaining 0.76 (76%) of a G of the C-172's negative
load factor specification of -1.76 to carry the load of any
acceleration that may subsequently occur.

If you had said 76% of a negative 1 G, I'd have agreed with you. :-)
Just saying of a G, implies positive 1 G by most conventions.

Matt

Matt Whiting wrote:

It is an acceleration opposite the normal acceleration due to gravity,
so it would be downward with respect to the airframe.

What about the case of a fighter jet climbing on afterburners at a steep
nose-high attitude quickly rolling forward through the horizon to a steep
nose-low attitude? Wouldn't the negative G force be considered an upward
force with respect to the airframe?


--
Peter

On Tue, 25 Jul 2006 18:21:54 -0400, "Peter R."
wrote in ::

Matt Whiting wrote:

It is an acceleration opposite the normal acceleration due to gravity,
so it would be downward with respect to the airframe.

What about the case of a fighter jet climbing on afterburners at a steep
nose-high attitude quickly rolling forward through the horizon to a steep
nose-low attitude? Wouldn't the negative G force be considered an upward
force with respect to the airframe?


Of course.

In article ,
Matt Whiting wrote:

Just saying of a G, implies positive 1 G by most conventions.

More accurately, doesn't it depend on your frame of reference?