Mounting a turn coordinator on the tail?

[Tim Auckland]

(Things are so slow in this group at the moment, I thought I'd post
this to get your input on something I've been mulling over...)

If you put a plane into a skidding left turn (wings level) with left
rudder, the ball on the panel goes to the right.

However, I've been trying to work out what would happen to the ball if
it was mounted on the tail.
The rudder is pushing the tail of the plane to the right, so I think
the ball would go to the left.
Can anyone confirm this?

Tim.

[Stubby]

I don't think so. My understanding is the ball rests in a slightly
curved tube, arranged so that gravity tends to center it. Centrifugal
force tries to move it to the outside of the skid, up the curved tube.
Neither of these forces depends on where you mount the turn coordinator.


Tim Auckland wrote:
> (Things are so slow in this group at the moment, I thought I'd post
> this to get your input on something I've been mulling over...)
>
> If you put a plane into a skidding left turn (wings level) with left
> rudder, the ball on the panel goes to the right.
>
> However, I've been trying to work out what would happen to the ball if
> it was mounted on the tail.
> The rudder is pushing the tail of the plane to the right, so I think
> the ball would go to the left.
> Can anyone confirm this?
>
> Tim.

[Tim Auckland]

Stubby,

Thanks, I think I've got it now.

As the turn is initiated, the tail has to go right (this is what
causes the plane to turn) and my guess is that the ball initially goes
very slightly left.

However, once a constant-radius turn has been achieved, the ball will
be on the right (outside) of the turn.

Tim.

On Sat, 29 Jul 2006 18:28:28 -0400, Stubby
> wrote:

>I don't think so. My understanding is the ball rests in a slightly
>curved tube, arranged so that gravity tends to center it. Centrifugal
>force tries to move it to the outside of the skid, up the curved tube.
>Neither of these forces depends on where you mount the turn coordinator.
>
>
>Tim Auckland wrote:
>> (Things are so slow in this group at the moment, I thought I'd post
>> this to get your input on something I've been mulling over...)
>>
>> If you put a plane into a skidding left turn (wings level) with left
>> rudder, the ball on the panel goes to the right.
>>
>> However, I've been trying to work out what would happen to the ball if
>> it was mounted on the tail.
>> The rudder is pushing the tail of the plane to the right, so I think
>> the ball would go to the left.
>> Can anyone confirm this?
>>
>> Tim.

[Stubby]

What causes a plane to turn is the horizontal component of the lift
vector. It certainly does not depend on the turn coordinator.
What counts is the center of gravity of the plane, not the tail.



Tim Auckland wrote:
> Stubby,
>
> Thanks, I think I've got it now.
>
> As the turn is initiated, the tail has to go right (this is what
> causes the plane to turn) and my guess is that the ball initially goes
> very slightly left.
>
> However, once a constant-radius turn has been achieved, the ball will
> be on the right (outside) of the turn.
>
> Tim.
>
> On Sat, 29 Jul 2006 18:28:28 -0400, Stubby
> > wrote:
>
>> I don't think so. My understanding is the ball rests in a slightly
>> curved tube, arranged so that gravity tends to center it. Centrifugal
>> force tries to move it to the outside of the skid, up the curved tube.
>> Neither of these forces depends on where you mount the turn coordinator.
>>
>>
>> Tim Auckland wrote:
>>> (Things are so slow in this group at the moment, I thought I'd post
>>> this to get your input on something I've been mulling over...)
>>>
>>> If you put a plane into a skidding left turn (wings level) with left
>>> rudder, the ball on the panel goes to the right.
>>>
>>> However, I've been trying to work out what would happen to the ball if
>>> it was mounted on the tail.
>>> The rudder is pushing the tail of the plane to the right, so I think
>>> the ball would go to the left.
>>> Can anyone confirm this?
>>>
>>> Tim.
>

[Tim Auckland]

I've never quite bought in to the "horizontal component of lift"
explanation for what causes a plane to turn.

I started this discussion talking about a "wings level skidding turn".
There's no horizontal component of lift generated by the wings if the
wings are level. There is however a couple comprising of the
rightwards force from the rudder, and the induced leftwards force of
wind resistance acting further forward along the fuselage. It's the
couple which causes the plane to turn.

I can also generate a substantial amount of horizontal force from the
wings without the plane turning (think forward slip); it just moves
sideways through the air.

I believe what causes a plane to turn is the couple comprised of
opposing forces which aren't aligned.

For example, in a coordinated left turn, the wings generate a
horizontal force to the left. The tail generates a horizontal force
to the right. The forces aren't aligned, so a couple is generated,
and the airplane turns left.

Without a force to the right, the "horizontal component of lift" to
the left generated by the banked wings would merely cause the plane to
move left -- it wouldn't turn.

Just my $0.10 worth, and I'll happily listen to contradictory
arguments.

Tim.

On Sat, 29 Jul 2006 20:52:00 -0400, Stubby
> wrote:

>What causes a plane to turn is the horizontal component of the lift
>vector. It certainly does not depend on the turn coordinator.
>What counts is the center of gravity of the plane, not the tail.
>
>
>
>Tim Auckland wrote:
>> Stubby,
>>
>> Thanks, I think I've got it now.
>>
>> As the turn is initiated, the tail has to go right (this is what
>> causes the plane to turn) and my guess is that the ball initially goes
>> very slightly left.
>>
>> However, once a constant-radius turn has been achieved, the ball will
>> be on the right (outside) of the turn.
>>
>> Tim.
>>
>> On Sat, 29 Jul 2006 18:28:28 -0400, Stubby
>> > wrote:
>>
>>> I don't think so. My understanding is the ball rests in a slightly
>>> curved tube, arranged so that gravity tends to center it. Centrifugal
>>> force tries to move it to the outside of the skid, up the curved tube.
>>> Neither of these forces depends on where you mount the turn coordinator.
>>>
>>>
>>> Tim Auckland wrote:
>>>> (Things are so slow in this group at the moment, I thought I'd post
>>>> this to get your input on something I've been mulling over...)
>>>>
>>>> If you put a plane into a skidding left turn (wings level) with left
>>>> rudder, the ball on the panel goes to the right.
>>>>
>>>> However, I've been trying to work out what would happen to the ball if
>>>> it was mounted on the tail.
>>>> The rudder is pushing the tail of the plane to the right, so I think
>>>> the ball would go to the left.
>>>> Can anyone confirm this?
>>>>
>>>> Tim.
>>

[Jose[_1_]]

> I started this discussion talking about a "wings level skidding turn".
> There's no horizontal component of lift generated by the wings if the
> wings are level. There is however a couple comprising of the
> rightwards force from the rudder, and the induced leftwards force of
> wind resistance acting further forward along the fuselage. It's the
> couple which causes the plane to turn.

There is also the fact that the thrust vector is more aligned with the
direction of desired flight.

> I believe what causes a plane to turn is the couple comprised of
> opposing forces which aren't aligned.

This is always true, for any acceleration not in the direct line of
flight. There is no "one thing" which causes anything in aviation
(except at the most fundamental level, where all flight is controlled by
money). In a coordinated turn, there are several forces, as you pointed
out. However, not all turns are coordinated. What makes a car turn?
Are there analogs in aviation of these forces?

Jose
--
The monkey turns the crank and thinks he's making the music.
for Email, make the obvious change in the address.

[Dave S]

Tim Auckland wrote:
> (Things are so slow in this group at the moment, I thought I'd post
> this to get your input on something I've been mulling over...)
>
> If you put a plane into a skidding left turn (wings level) with left
> rudder, the ball on the panel goes to the right.
>
> However, I've been trying to work out what would happen to the ball if
> it was mounted on the tail.
> The rudder is pushing the tail of the plane to the right, so I think
> the ball would go to the left.
> Can anyone confirm this?
>
> Tim.

If you are coordinated, the ball will be centered
If you are slipping, or skidding, the ball will be just the same as if
it was in the panel.

Dave

[Jay Somerset]

On Sat, 29 Jul 2006 15:52:39 -0600, Tim Auckland > wrote:

> (Things are so slow in this group at the moment, I thought I'd post
> this to get your input on something I've been mulling over...)
>
> If you put a plane into a skidding left turn (wings level) with left
> rudder, the ball on the panel goes to the right.
>
> However, I've been trying to work out what would happen to the ball if
> it was mounted on the tail.
> The rudder is pushing the tail of the plane to the right, so I think
> the ball would go to the left.
> Can anyone confirm this?
>
> Tim.

In the situation you describe, the ball should first go to the left, as the
tail is pushed out to the right by the action of the rudder. Once the skid
is stable, and the ball will go over to the right, to reflect the fact that
the aircraft (including its tail) is turning left.

[Judah]

Tim Auckland > wrote in news:eglnc2tk7l96o5qq3mqajbujfip02iaaol@
4ax.com:

> (Things are so slow in this group at the moment, I thought I'd post
> this to get your input on something I've been mulling over...)
>
> If you put a plane into a skidding left turn (wings level) with left
> rudder, the ball on the panel goes to the right.
>
> However, I've been trying to work out what would happen to the ball if
> it was mounted on the tail.
> The rudder is pushing the tail of the plane to the right, so I think
> the ball would go to the left.
> Can anyone confirm this?
>
> Tim.

More importantly, if you put the TC in the tail, how much rudder pressure
would the pilot have to apply in order to turn the tail quickly enough around
to see it?

I believe cats and dogs have the ability to achieve the necessary speeds,
almost even to the point that some may catch their tails in their mouths as
they look to focus on their balls. But I don't think airplanes are quite
limber enough for this capability.

[Ron Natalie]

Stubby wrote:
> What causes a plane to turn is the horizontal component of the lift
> vector. It certainly does not depend on the turn coordinator.
> What counts is the center of gravity of the plane, not the tail.

BZZT. Incorrect over simplification and psuedoscientific drivel
given by the FAA. If you believed the FAA description, then the
airplane would just translate sidewise rather than turning in a
circle. While the horizontal component of lift is what pulls
you to the interior of the turn, the tail is VERY important to
actually "turn" the aircraft direction so that the horizontal
component continually gets pointed to the center of the turn.

[Tim Auckland]

On Sun, 30 Jul 2006 23:36:47 GMT, Judah > wrote:


>More importantly, if you put the TC in the tail, how much rudder pressure
>would the pilot have to apply in order to turn the tail quickly enough around
>to see it?

There's a Tony Hancock radio sketch where he gets ejected from a
plane, lands on the tail, lassoos the controls, and steers it home.

>
>I believe cats and dogs have the ability to achieve the necessary speeds,
>almost even to the point that some may catch their tails in their mouths as
>they look to focus on their balls. But I don't think airplanes are quite
>limber enough for this capability.

[Tim Auckland]

On Sun, 30 Jul 2006 02:12:48 GMT, Jose >
wrote:

>> I started this discussion talking about a "wings level skidding turn".
>> There's no horizontal component of lift generated by the wings if the
>> wings are level. There is however a couple comprising of the
>> rightwards force from the rudder, and the induced leftwards force of
>> wind resistance acting further forward along the fuselage. It's the
>> couple which causes the plane to turn.
>
>There is also the fact that the thrust vector is more aligned with the
>direction of desired flight.
>
>> I believe what causes a plane to turn is the couple comprised of
>> opposing forces which aren't aligned.
>
>This is always true, for any acceleration not in the direct line of
>flight. There is no "one thing" which causes anything in aviation
>(except at the most fundamental level, where all flight is controlled by
>money). In a coordinated turn, there are several forces, as you pointed
>out. However, not all turns are coordinated. What makes a car turn?

Same thing that makes a plane turn: unbalanced forces not acting
through the centre of gravity.

(In the case of a car, the initial sideways forces are generated by
the front tires.)

By the way, I'm beginning to realize that it only takes one force to
turn an object, as long as that force is not acting through the centre
of gravity.
Interestingly enough, if the "vertical component of lift" were the
only sideways force acting on the plane, it would cause adverse yaw.
The wing's centre of lift is behind the plane's centre of gravity, so
a pull to the left would cause the plane to turn right.

>Are there analogs in aviation of these forces?
Only when taxiing a tricycle(:-)
>
>Jose

[Tim Auckland]

On Mon, 31 Jul 2006 08:06:07 -0400, Ron Natalie >
wrote:

>While the horizontal component of lift is what pulls
>you to the interior of the turn, the tail is VERY important to
>actually "turn" the aircraft direction
----------------------------------------
>so that the horizontal
>component continually gets pointed to the center of the turn.
----------------------------------------
Now that's a VERY interesting way of thinking about it.

What got me thinking about all this was the observation that I found
it difficult to keep the ball exactly centered in a 50-degree bank, so
I started thinking about whether it's even possible to do this in a
60-degree bank. In theory, in a 60-degree bank, the angle of attack,
controlled by the elevator, has to be such that 2g of lift is
generated. However, in a steep bank, the rate of turn is mostly
controlled by the elevator. The rudder's forces are mostly acting
vertically, so it has a large effect on whether the nose is pointing
up or down.
If you keep the nose roughly horizontal with the rudder, and 2g of
lift with the elevator, you've no controls left to affect the rate of
turn.
Do the forces in this case work out such that the ball is centered?

Any areobatic piliots out there? Is the ball typically centered in a
60-degree bank?

Tim.

[Jose[_1_]]

> By the way, I'm beginning to realize that it only takes one force to
> turn an object, as long as that force is not acting through the centre
> of gravity.

That's not quite true (unless by "turn" you mean "yaw"). A single force
not through the CG will cause the object to rotate about the CG. To get
an airplane to actually =turn= however requires also a change in
direction, else the airplane would simply skid sideways the rest of the
flight.

Let's say you are flying straight and level, due North, and simply stomp
on the (left) rudder pedal. The first thing is that the tail will swing
to the right, because the force from the rudder is not through the CG.
But then other things happen. As the tail swings, the right wing ends
up going faster and the left wing slower through the air. So the right
wing provides more lift, and the aircraft banks to the left. Since the
airplane is still going due North, but pointing slightly west, the
airflow on the side of the plane will push the airplane somewhat to the
left, and the propeller (now pointing slightly west) will also help pull
the airplane west. The bank also introduces some net leftward force.

So, there's a lot going on. (and no, stomping on the rudder pedals isn't
usually the best way to turn).

Jose
--
The monkey turns the crank and thinks he's making the music.
for Email, make the obvious change in the address.

[Stubby]

Ron Natalie wrote:
> Stubby wrote:
>> What causes a plane to turn is the horizontal component of the lift
>> vector. It certainly does not depend on the turn coordinator.
>> What counts is the center of gravity of the plane, not the tail.
>
><...opinion deleted...>

> While the horizontal component of lift is what pulls
> you to the interior of the turn, the tail is VERY important to
> actually "turn" the aircraft direction so that the horizontal
> component continually gets pointed to the center of the turn.
>
The elevator/rudder mechanism is for applying the torques to the plane
so it rolls and yaws. Also, as I remember from the first day of
physics class, a physical body behaves as a point mass at the center of
gravity with 3 translational forces and 3 rotational torques that can be
applied to it.

The horizontal component of lift behaves like a string tied to a rock
being swung around. The string does indeed apply force to the center of
gravity of the rock and "points" to the center of the turn. If you put
a paint spot on the rock and want to make the spot always face you, the
rock will have to yaw at the same rate as you are rotating it around
you; consequently something like a rudder will be needed.

[Dave Butler[_1_]]

Tim Auckland wrote:

> The wing's centre of lift is behind the plane's centre of gravity, so

Eh? If this were true, there would be a torque acting about the pitch axis and
forcing the nose downward.

[Jose[_1_]]

>> The wing's centre of lift is behind the plane's centre of gravity, so
> Eh? If this were true, there would be a torque acting about the pitch axis and forcing the nose downward.

There is. That's what the tail is for - it pushes down (behind the CG)
providing the balancing torque.

Canard aircraft are different, the main wing is behind the CG, and the
canard is in front; both provide lift in the same (upward) direction.

Jose
--
The monkey turns the crank and thinks he's making the music.
for Email, make the obvious change in the address.

[Mark Hansen]

On 07/31/06 14:13, Dave Butler wrote:
> Tim Auckland wrote:
>
>> The wing's centre of lift is behind the plane's centre of gravity, so
>
> Eh? If this were true, there would be a torque acting about the pitch axis and
> forcing the nose downward.

Isn't there? Isn't this what the downward force produced by the horizontal
stabilizer is trying to equalize?



--
Mark Hansen, PP-ASEL, Instrument Airplane
Cal Aggie Flying Farmers
Sacramento, CA

[Tim Auckland]

On Mon, 31 Jul 2006 18:57:30 GMT, Jose >
wrote:

>> By the way, I'm beginning to realize that it only takes one force to
>> turn an object, as long as that force is not acting through the centre
>> of gravity.
>
>That's not quite true (unless by "turn" you mean "yaw").
Yes, when I made the "one force" statement, I was thinking of yaw.


>A single force
>not through the CG will cause the object to rotate about the CG. To get
>an airplane to actually =turn= however requires also a change in
>direction, else the airplane would simply skid sideways the rest of the
>flight.
>
>Let's say you are flying straight and level, due North, and simply stomp
>on the (left) rudder pedal. The first thing is that the tail will swing
>to the right, because the force from the rudder is not through the CG.
>But then other things happen. As the tail swings, the right wing ends
>up going faster and the left wing slower through the air. So the right
>wing provides more lift, and the aircraft banks to the left.
In my original scenario, I stated a "skidding left turn (wings
level)". I was picturing doing whatever is required with the ailerons
to keep the wings level. Come to think of it, that'll mean
down-aileron on the left side, which'll introduce more drag on the
left, which may or may not be balanced by the increased drag from the
faster moving right wing and up-aileron on the right side.....

>Since the
>airplane is still going due North, but pointing slightly west, the
>airflow on the side of the plane will push the airplane somewhat to the
>left, and the propeller (now pointing slightly west) will also help pull
>the airplane west. The bank also introduces some net leftward force.
>
>So, there's a lot going on.
I agree absolutely. This isn't a simple issue.

>(and no, stomping on the rudder pedals isn't
>usually the best way to turn).
but it can produce some interesting results (:-)

Tim.

[Ron Natalie]

Stubby wrote:

>
> The horizontal component of lift behaves like a string tied to a rock
> being swung around.

No, it doesn't. The lift vector points in a direction (roughly)
perpendicular to the wing. Nothing causes it to point to the
a "center" other than the other aerodynamic surfaces .

[Stubby]

Ron Natalie wrote:
> Stubby wrote:
>
>>
>> The horizontal component of lift behaves like a string tied to a rock
>> being swung around.
>
> No, it doesn't. The lift vector points in a direction (roughly)
> perpendicular to the wing. Nothing causes it to point to the
> a "center" other than the other aerodynamic surfaces .

The lift vector(s) point perpendicular to the wing (s). With a dihedral
angle in the two wings there will be horizontal components of the lift
vectors and in level flight they will be equal and opposite, canceling.
In a turn with the plane banked, it's easy to see how one lift vector
will point entirely straight up while the other doubles its horizontal
component. This pulls the plane toward the center of the turn. (It
sure is easier to explain with a blackboard!)

[[email protected]]

Didn't anybody read "Stick and Rudder"?

To make an object travel on a curved path, you need a "centripetal"
force. This isn't a "centrifugal force", which is a made-up construct
that helps to understand what it feels like to be in a vehicle that's
moving you in a circle, a centripetal force is a real force that's
acting upon an object that pulls it towards the center of a circle.
Now, if you're traveling in a circle, the force that's pulling you
toward the center of the circle acts in a direction perpendicular to
your direction of motion, and since your direction of motion is always
changing (you're going around a circle, after all), the direction of
that centripetal force is necessarily continually changing as well.

If you bank the wings to the left, the lift vector will start dragging
you to the left. You'll begin "slipping", drifting sideways through the
air. If this was the end of it, you'd never end up traveling in a
circle, you'd just keep drifting slightly sideways, you nose would keep
pointing where it was pointing before, and your track over the ground
would still be a straight line, but just slightly "diagonal", at an
angle to what it was before.

But when you slip, there's now a net component of drag acting sideways
against the side of the plane. Since most of our surface area is behind
the CG, when you blow really hard against the side of a plane, it tends
to yaw into the wind. As the plane yaws to the left, the wings also
turn to the left, and the direction that the lift vector is pointed in
turns as well. Bingo, now you've got a real "centripetal force", one
that continually changes direction to point at a 90 degree angle to
your curving path. Now your ground track can form a circular path.

You need that yaw to turn the wings to continually update the direction
of lift, to create circular motion. Now you're a little closer to that
rock on a string model.

Alternately, if you're intentionally doing a slip, like if you're
landing in a cross-wind, you could always apply opposite rudder. This
acts counter to the "weather-vaning" moment, and prevents the plane
from yawing, which prevents the wings from "rotating", preventing the
lift vector from changing direction. By stopping the nose from changing
direction, you stop the plane from moving in a circular path, and you
just "slip" sideways up against the wind (possibly just to resist a
cross-wind and maintain a straight path along the ground).
-harry

[Tim Auckland]

On 1 Aug 2006 13:50:21 -0700, wrote:

>Didn't anybody read "Stick and Rudder"?
>
>To make an object travel on a curved path, you need a "centripetal"
>force. This isn't a "centrifugal force", which is a made-up construct
>that helps to understand what it feels like to be in a vehicle that's
>moving you in a circle, a centripetal force is a real force that's
>acting upon an object that pulls it towards the center of a circle.
>Now, if you're traveling in a circle, the force that's pulling you
>toward the center of the circle acts in a direction perpendicular to
>your direction of motion, and since your direction of motion is always
>changing (you're going around a circle, after all), the direction of
>that centripetal force is necessarily continually changing as well.
>
>If you bank the wings to the left, the lift vector will start dragging
>you to the left. You'll begin "slipping", drifting sideways through the
>air. If this was the end of it, you'd never end up traveling in a
>circle, you'd just keep drifting slightly sideways, you nose would keep
>pointing where it was pointing before, and your track over the ground
>would still be a straight line, but just slightly "diagonal", at an
>angle to what it was before.
Well, no, actually, if this was the end of it and the "horizontal
component of lift" was the ONLY horizontal force acting on the plane
(no drag of any kind, no thrust from the engine, etc) the plane as a
whole would end up travelling in the same direction it was originally
going, but spinning around a vertical axis passing though the plane's
centre of gravity. What's more, because the horizontal force is
acting through a point behind the CG, the plane would spin in the
opposite direction to the force.
The plane would spin, rather than drift, because the horizontal force
changes direction as the plane rotates.
It's not easy to imagine this, but think of the space shuttle
stationary in space, then having a sideways horizontal force applied
behind the CG,

>
>But when you slip, there's now a net component of drag acting sideways
>against the side of the plane. Since most of our surface area is behind
>the CG, when you blow really hard against the side of a plane, it tends
>to yaw into the wind. As the plane yaws to the left, the wings also
>turn to the left, and the direction that the lift vector is pointed in
>turns as well. Bingo, now you've got a real "centripetal force", one
>that continually changes direction to point at a 90 degree angle to
>your curving path. Now your ground track can form a circular path.
>
Well, no, actually, if you're introducing wind resistance, you need to
also consider the increased drag from the outer wing being raised as
the plane is banked. The turning force from this is often
substantially greater than the weather-vaning effect, and is why many
planes demonstrate adverse yaw when they're initially banked.

>You need that yaw to turn the wings to continually update the direction
>of lift, to create circular motion. Now you're a little closer to that
>rock on a string model.
>
>Alternately, if you're intentionally doing a slip, like if you're
>landing in a cross-wind, you could always apply opposite rudder. This
>acts counter to the "weather-vaning" moment, and prevents the plane
>from yawing, which prevents the wings from "rotating", preventing the
>lift vector from changing direction. By stopping the nose from changing
>direction, you stop the plane from moving in a circular path, and you
>just "slip" sideways up against the wind (possibly just to resist a
>cross-wind and maintain a straight path along the ground).
>-harry

[[email protected]]

> ... What's more, because the horizontal force is
> acting through a point behind the CG, the plane would spin in the
> opposite direction to the force.

And when the plane is flying straight and level, since the wing is
behind the CG, shouldn't the wing, similarly, be continually pulling
the tail up over the nose, leading the plane to tumble through space?
Why doesn't it? Because while the wing is aft of the CG, the tail
provides a downward force to balance it. Since the tail is so much
further away from the CG than the wing is, it can leverage a relatively
smaller force to balance out the torque, while allowing the much larger
force at the wing to provide a net upwards force to support the plane's
weight.

When the plane is in a bank, yes, the wing is providing a horizontal
force from behind the CG again. But nothing has changed from the
straight and level case, there's still an upside-down wing back there
at the tail, and it's still balancing out the torque produced by the
wing being behind the CG. When the wing is banked, the tail is banked
too.

> It's not easy to imagine this, but think of the space shuttle
> stationary in space, then having a sideways horizontal force applied
> behind the CG

But imagine another sideways horizontal force applied in the opposite
direction at the tail, exactly strong enough to balance the "torque".
If your left force is applied aft of the CG, but fairly close to it
(like where the wings are), and your counter-balancing right force is
applied at the tail, much further from the CG, then a relatively small
force at the tail will be enough to counter the much larger force at
the wing, due to leverage.

So our torques are equal, preventing any rotation, but there's still a
net force to the left (the force up near the wing is much larger), so
there will be a net translational force to the left. Your space ship
will accellerate to the left without rotating.
-harry

[Tim Auckland]

This makes sense. Thanks.

On 1 Aug 2006 19:16:28 -0700, wrote:

>> ... What's more, because the horizontal force is
>> acting through a point behind the CG, the plane would spin in the
>> opposite direction to the force.
>
>And when the plane is flying straight and level, since the wing is
>behind the CG, shouldn't the wing, similarly, be continually pulling
>the tail up over the nose, leading the plane to tumble through space?
>Why doesn't it? Because while the wing is aft of the CG, the tail
>provides a downward force to balance it. Since the tail is so much
>further away from the CG than the wing is, it can leverage a relatively
>smaller force to balance out the torque, while allowing the much larger
>force at the wing to provide a net upwards force to support the plane's
>weight.
>
>When the plane is in a bank, yes, the wing is providing a horizontal
>force from behind the CG again. But nothing has changed from the
>straight and level case, there's still an upside-down wing back there
>at the tail, and it's still balancing out the torque produced by the
>wing being behind the CG. When the wing is banked, the tail is banked
>too.
>
>> It's not easy to imagine this, but think of the space shuttle
>> stationary in space, then having a sideways horizontal force applied
>> behind the CG
>
>But imagine another sideways horizontal force applied in the opposite
>direction at the tail, exactly strong enough to balance the "torque".
>If your left force is applied aft of the CG, but fairly close to it
>(like where the wings are), and your counter-balancing right force is
>applied at the tail, much further from the CG, then a relatively small
>force at the tail will be enough to counter the much larger force at
>the wing, due to leverage.
>
>So our torques are equal, preventing any rotation, but there's still a
>net force to the left (the force up near the wing is much larger), so
>there will be a net translational force to the left. Your space ship
>will accellerate to the left without rotating.
>-harry

[Greg Esres]

<<Now that's a VERY interesting way of thinking about it.>>

It's the proper and textbook way to think about it.

<<difficult to keep the ball exactly centered in a 50-degree bank, so
I started thinking about whether it's even possible to do this in a
60-degree bank. >>

Of course it is.

<<However, in a steep bank, the rate of turn is mostly controlled by
the elevator. >>

No. The rate of turn is controlled by load factor at a given
airspeed. You can increase the rate of turn by banking more, or you
can pull back on the yoke to increase the load factor (temporarily).

In a 60 degree banked turn, you will get a load factor of 2 regardless
of what you do with the elevator. If you don't increase the AOA
during turn entry however, the aircraft will accelerate and the 60
degree banked turn will occur at a higher airspeed, in a rapid
descent.