effect of changed thrust line.

Alan Baker wrote:
In a glide in a low wing aircraft:

Total aerodynamic force (lift and drag!)
^
|
|
M (Centre of Mass)
|
C (Centre of Aerodynamic Pressure)
|
|
Weight (no down arrow head... ...sorry)

Now remember, the aircraft must be descending to make this work.

The above diagram is simplified too soon in the analysis. You may as well
have dispensed with the weight and aerodynamic forces too, as they
contribute nothing to your subsequent argument since you never vary them.

Now if you add thrust at the "drag line" (the line through the CoP
parallel to the aircraft's motion):

Total aerodynamic force
^
|
|
M (Centre of Mass)
|
(Thrust)--C (Centre of Aerodynamic pressure)
|
|
Weight

You can align the engine any way you want and it will still create a
pitch up, right?

Sure - and the object will rotate about M until it reaches a rotation
speed in equilibrium with air drag (by definition, the only point where
we are allowed to add that drag component is at point C):

Total aerodynamic force
^
|
|
M (Centre of Mass)
|
(Thrust)--C--(air drag) (Centre of Aerodynamic pressure)
|
|
Weight


But:

Total aerodynamic force
^
|
|
(Thrust)--M (Centre of Mass)
|
C (Centre of Aerodynamic Pressure)
|
|
Weight

Add the thrust at the centre of mass, and you get no pitching moment.

The diagram above is of a system that isn't in equilibrium. Furthermore,
there is no vector we can anchor at C that brings it into equilibrium -
if we add a vector so that we get a pure couple, like so:

Total aerodynamic force
^
|
|
(Thrust)--M (Centre of Mass)
|
C--(air drag) (Centre of Aerodynamic Pressure)
|
|
Weight

....then the _couple_ rotates the aircraft around M in a counterclockwise
direction (i.e. pitch down!) Your force diagram is flawed because it
makes incorrect assumptions about the location of C at equilibrium and
the direction of the total aerodynamic forces.

Running the thrust line through M does _not_ guarantee you wont get any
couple.

In fact none of the diagrams you or I drew are complete and do not
accurately capture the reality. Center of mass changes with each flight
and even during flight, and center of pressure changes with aircraft
orientation.

In article ,
Jim Logajan wrote:

Alan Baker wrote:
In a glide in a low wing aircraft:

Total aerodynamic force (lift and drag!)
^
|
|
M (Centre of Mass)
|
C (Centre of Aerodynamic Pressure)
|
|
Weight (no down arrow head... ...sorry)

Now remember, the aircraft must be descending to make this work.

The above diagram is simplified too soon in the analysis. You may as well
have dispensed with the weight and aerodynamic forces too, as they
contribute nothing to your subsequent argument since you never vary them.

No, it's not.

It represents all the forces on an aircraft in a trimmed glide: total
aerodynamic force perfectly balancing weight.


Now if you add thrust at the "drag line" (the line through the CoP
parallel to the aircraft's motion):

Total aerodynamic force
^
|
|
M (Centre of Mass)
|
(Thrust)--C (Centre of Aerodynamic pressure)
|
|
Weight

You can align the engine any way you want and it will still create a
pitch up, right?

Sure - and the object will rotate about M until it reaches a rotation
speed in equilibrium with air drag (by definition, the only point where
we are allowed to add that drag component is at point C):

It will never reach such an equilibrium. That's the problem. With the
increased thrust, the aircraft will both: pitch up and gain airspeed.
Remember: drag is notional. It is just the component of the total
aerodynamic force anti-parallel to the motion of the aircraft. In this
situation of a low wing aircraft, if you add thrust at the CoA, the
aircraft will pitch up, and that will rotate the craft and you'll have
to trim the aircraft. No waiting for drag to grow will do it.


Total aerodynamic force
^
|
|
M (Centre of Mass)
|
(Thrust)--C--(air drag) (Centre of Aerodynamic pressure)
|
|
Weight


But:

Total aerodynamic force
^
|
|
(Thrust)--M (Centre of Mass)
|
C (Centre of Aerodynamic Pressure)
|
|
Weight

Add the thrust at the centre of mass, and you get no pitching moment.

The diagram above is of a system that isn't in equilibrium. Furthermore,
there is no vector we can anchor at C that brings it into equilibrium -
if we add a vector so that we get a pure couple, like so:

So, what do you expect an aircraft in a stable glide to do when you add
thrust: accelerate. The natural consequence of a system that isn't in
equilibrium.


Total aerodynamic force
^
|
|
(Thrust)--M (Centre of Mass)
|
C--(air drag) (Centre of Aerodynamic Pressure)
|
|
Weight

...then the _couple_ rotates the aircraft around M in a counterclockwise
direction (i.e. pitch down!) Your force diagram is flawed because it
makes incorrect assumptions about the location of C at equilibrium and
the direction of the total aerodynamic forces.

Sorry, but no.

By definition, an aircraft in a stable glide has a *total* aerodynamic
force acting on it that must be precisely equal to the aircraft's weight
and *must* be acting through the centre of mass. You're suddenly adding
a new force as if it isn't accounted for in the previous diagram.


Running the thrust line through M does _not_ guarantee you wont get any
couple.

It guarantees you won't get a couple from the thrust. You say you have a
B.SC: from where?


In fact none of the diagrams you or I drew are complete and do not
accurately capture the reality. Center of mass changes with each flight
and even during flight, and center of pressure changes with aircraft
orientation.

So? The point I've been trying to make is that if you're trying to keep
the aircraft's flight characteristics, what you need to consider is
orientation of the thrust line with respect to the CoM. For the
purposes of argument, I've been using a thrust line through the centre
of mass to illustrate my point, but at no time have I argued that it is
the only place you can have the thrust line and have a stable aircraft.

But by using the zero point, I can illustrate it well. If you have an
airframe with an engine installation where the thrust line goes through
the centre of mass, then you're noting going to have a pitching moment
generated by thrust, period. So if you install a new engine and have to
adjust it's mounting point such that it maintains the CoM in the same
location, but moves the thrust line up or down, all of sudden you *will*
have a pitching moment generated by changes in thrust.

That is a change in the aircraft's flying characteristics, period.

To remove that change, simply reangle the engine to once again have the
thrust line pass through the CoM. Then once again, you will have no
thrust induced pitch changes.

Do the same reasoning for an aircraft with a thrust line above the CoM,
where a new engine lowers it to coincide with the CoM. You'll once again
change the flying characteristics from one where increased thrust causes
a pitch up, to one where thrust does not. Reangle the engine and you'll
restore the original flying characteristics.

Period.

--
Alan Baker
Vancouver, British Columbia
http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg

Alan Baker wrote:
Jim Logajan wrote:
Running the thrust line through M does _not_ guarantee you wont get
any couple.

It guarantees you won't get a couple from the thrust.

I think I see one of your problems. How many forces are needed for a
couple? Can one of those forces pass through the center of mass?

In article ,
Jim Logajan wrote:

Alan Baker wrote:
Jim Logajan wrote:
Running the thrust line through M does _not_ guarantee you wont get
any couple.

It guarantees you won't get a couple from the thrust.

I think I see one of your problems. How many forces are needed for a
couple? Can one of those forces pass through the center of mass?

One. There only has to be a one force on a body in order for it to
experience to angular acceleration. That force just has to act on a line
that is not through the centre of mass.

Questions for you:

Do you believe that in the absence of any other forces, if the
aerodynamic forces act on a line that is not through the centre of mass,
will the aircraft experience angular acceleration?

If you add thrust acting through the centre of mass to the situation,
will it change the angular acceleration?

--
Alan Baker
Vancouver, British Columbia
http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg

In article
,
Alan Baker wrote:

In article ,
Jim Logajan wrote:

Alan Baker wrote:
Jim Logajan wrote:
Running the thrust line through M does _not_ guarantee you wont get
any couple.

It guarantees you won't get a couple from the thrust.

I think I see one of your problems. How many forces are needed for a
couple? Can one of those forces pass through the center of mass?

One. There only has to be a one force on a body in order for it to
experience to angular acceleration. That force just has to act on a line
that is not through the centre of mass.

I apologize, I was using the term "couple" incorrectly.

But that doesn't matter to my argument. So ignore the first bit and
answer my questions below:


Questions for you:

Do you believe that in the absence of any other forces, if the
aerodynamic forces act on a line that is not through the centre of mass,
will the aircraft experience angular acceleration?

If you add thrust acting through the centre of mass to the situation,
will it change the angular acceleration?

--
Alan Baker
Vancouver, British Columbia
http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg

Alan Baker wrote:
Alan Baker wrote:
Jim Logajan wrote:
Alan Baker wrote:
Jim Logajan wrote:
Running the thrust line through M does _not_ guarantee you wont
get any couple.

It guarantees you won't get a couple from the thrust.

I think I see one of your problems. How many forces are needed for
a couple? Can one of those forces pass through the center of mass?

One. There only has to be a one force on a body in order for it to
experience to angular acceleration. That force just has to act on a
line that is not through the centre of mass.

I apologize, I was using the term "couple" incorrectly.

Not surprising, given that most freshman level college physics textbooks
don't even mention the concept (a pity). Most introductory mechanics
(statics and dynamics) textbooks, on the other hand, introduce the couple
concept in the first or second chapters.

But that doesn't matter to my argument.

It does matter in the sense that it is an important (perhaps essential)
concept needed to properly set up and solve these sorts of problems.
Another concept needed is the fictitious inertia force (m*A) that is
anchored at the center of mass and always points opposite the net
external force. That means there is always at least a second force around
to create a couple.

So ignore the first bit and answer my questions below:


Questions for you:

Do you believe that in the absence of any other forces, if the
aerodynamic forces act on a line that is not through the centre of
mass, will the aircraft experience angular acceleration?

It will undergo both angular and translational acceleration. The motion
of the system over time is generally not obvious and holds a few
surprises. For example:

There is a related problem that often goes by the label "center of
percussion" (try Googling it) that shows that at the instantaneous point
in time that the system is placed at rest with respect a reference frame,
the instantaneous center of rotation in that reference frame is not at
the center of mass.

If you add thrust acting through the centre of mass to the situation,
will it change the angular acceleration?

When moving in a vacuum? No. When moving in a fluid? Yes it could. It is
not hard to imagine scenarios where that can happen.

In article ,
Jim Logajan wrote:

Alan Baker wrote:
Alan Baker wrote:
Jim Logajan wrote:
Alan Baker wrote:
Jim Logajan wrote:
Running the thrust line through M does _not_ guarantee you wont
get any couple.

It guarantees you won't get a couple from the thrust.

I think I see one of your problems. How many forces are needed for
a couple? Can one of those forces pass through the center of mass?

One. There only has to be a one force on a body in order for it to
experience to angular acceleration. That force just has to act on a
line that is not through the centre of mass.

I apologize, I was using the term "couple" incorrectly.

Not surprising, given that most freshman level college physics textbooks
don't even mention the concept (a pity). Most introductory mechanics
(statics and dynamics) textbooks, on the other hand, introduce the couple
concept in the first or second chapters.

But that doesn't matter to my argument.

It does matter in the sense that it is an important (perhaps essential)
concept needed to properly set up and solve these sorts of problems.
Another concept needed is the fictitious inertia force (m*A) that is
anchored at the center of mass and always points opposite the net
external force. That means there is always at least a second force around
to create a couple.

No. It is a *useful* concept, but it is in no way essential. Any problem
you can solve by taking couples and pure moments, you can solve without
reference to couples.

And the fictitious force you suggest would, if used, make the concept
conflict with its own definition. The very point of a couple is that it
can be represented by a pure vector moment, which by definition, cannot
accelerate a body linearly.

So if you say there is a couple when you talk about a system with a
single external force that isn't aligned with the centre of mass of the
body it is acting on, then your solution would suggest that the mass
doesn't undergo linear acceleration.


So ignore the first bit and answer my questions below:


Questions for you:

Do you believe that in the absence of any other forces, if the
aerodynamic forces act on a line that is not through the centre of
mass, will the aircraft experience angular acceleration?

It will undergo both angular and translational acceleration. The motion
of the system over time is generally not obvious and holds a few
surprises. For example:

A yes or no would have sufficed.


There is a related problem that often goes by the label "center of
percussion" (try Googling it) that shows that at the instantaneous point
in time that the system is placed at rest with respect a reference frame,
the instantaneous center of rotation in that reference frame is not at
the center of mass.

If you add thrust acting through the centre of mass to the situation,
will it change the angular acceleration?

When moving in a vacuum? No. When moving in a fluid? Yes it could. It is
not hard to imagine scenarios where that can happen.

No, it couldn't. The other effects that result in the thrust *might*
cause a change in the angular acceleration (increased/changed airflow
over the aerodynamic surfaces), that that isn't the thrust doing it.

--
Alan Baker
Vancouver, British Columbia
http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg

OK - got some more info.

The center of mass is something like 34 inches behind the firewall
and roughly 7 inches above the top engine mount point on the firewall.
so roughly speeking 13 inches above the prop centerline.
The prop flange with the O200 is 29.75 inches from the firewall.
This means it is 63.75 inches from the prop flange to the CM.(center
of mass)

This means there is NO WAY the thrust line is aligned anywhere close
to the center of mass.
This would require a downward displacement of almost 15 degrees.

THAT is not going to fly - PERIOD. We are hitting about 5.5 inches
BELOW the center of mass

If we aim for the middle of the rear stabilizer, about 183 inches from
the prop flange, 1.5 degrees down is 5.5 inches above the prop center,
which is about the middle of the rear of the fuselage and roughly 10
inches below the center of the rear horizontal stabilizer .

If I want to hit the same spot with the engine down 1.5 inches, i need
to change the angle to 1.875 degrees.
2 inches goes to 2 degrees.
2.5 inches would be 2.15 degrees, +/-
3 inches would be 2.31 degrees
4 inches would be 2.58 degrees.

Does this make any sense??
It sounds right to me.

wrote:
OK - got some more info.

The center of mass is something like 34 inches behind the firewall
and roughly 7 inches above the top engine mount point on the firewall.
so roughly speeking 13 inches above the prop centerline.
The prop flange with the O200 is 29.75 inches from the firewall.
This means it is 63.75 inches from the prop flange to the CM.(center
of mass)

This means there is NO WAY the thrust line is aligned anywhere close
to the center of mass.
This would require a downward displacement of almost 15 degrees.

THAT is not going to fly - PERIOD. We are hitting about 5.5 inches
BELOW the center of mass

If we aim for the middle of the rear stabilizer, about 183 inches from
the prop flange, 1.5 degrees down is 5.5 inches above the prop center,
which is about the middle of the rear of the fuselage and roughly 10
inches below the center of the rear horizontal stabilizer .

If I want to hit the same spot with the engine down 1.5 inches, i need
to change the angle to 1.875 degrees.
2 inches goes to 2 degrees.
2.5 inches would be 2.15 degrees, +/-
3 inches would be 2.31 degrees
4 inches would be 2.58 degrees.

Does this make any sense??
It sounds right to me.



I'd fly it.

--

Richard

(remove the X to email)

On Sat, 15 Nov 2008 23:11:33 -0600, cavelamb himself wrote:

wrote:
OK - got some more info.

The center of mass is something like 34 inches behind the firewall
and roughly 7 inches above the top engine mount point on the firewall.
so roughly speeking 13 inches above the prop centerline.
The prop flange with the O200 is 29.75 inches from the firewall.
This means it is 63.75 inches from the prop flange to the CM.(center
of mass)

This means there is NO WAY the thrust line is aligned anywhere close
to the center of mass.
This would require a downward displacement of almost 15 degrees.

THAT is not going to fly - PERIOD. We are hitting about 5.5 inches
BELOW the center of mass

If we aim for the middle of the rear stabilizer, about 183 inches from
the prop flange, 1.5 degrees down is 5.5 inches above the prop center,
which is about the middle of the rear of the fuselage and roughly 10
inches below the center of the rear horizontal stabilizer .

If I want to hit the same spot with the engine down 1.5 inches, i need
to change the angle to 1.875 degrees.
2 inches goes to 2 degrees.
2.5 inches would be 2.15 degrees, +/-
3 inches would be 2.31 degrees
4 inches would be 2.58 degrees.

Does this make any sense??
It sounds right to me.


I'd fly it.

You would. But you have NEVER been known for your intelligence. ))))