effect of changed thrust line.

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

wrote:
On Sun, 16 Nov 2008 00:58:06 -0800, John Kimmel
wrote:


wrote:


Spreadsheet is not quite right.Prop center is BELOW the CG by about 13
inches.
One inch change in prop height according to your spreadsheat makes a
change of 2.14 degrees.
I cannot buy that. Particularly since it would go from 1.5 down to .64
up.

Here is REV A, with a couple math errors fixed:
http://www.spiretech.com/~guynoir/sl...thrustreva.xls



That looks a lot closer.

What is it based on?

"Vector Mechanics For Engineers: Statics and Dynamics", by Beer and Johnston, 5th edition, Chapter 4: "Equilibrium of Rigid Bodies",
page 126, equation 4.1. Equation 4.1 can be easily found on the internet, just type "statics" into Google and go to the first hit,
which should be Wikipedia.

All my spreadsheet does is keep the distance from the thrust line to the CG the same with different engine locations, so that the sum
of moments about the CG remains the same (as in: Zero. See equation 4.1).


--
John Kimmel


I think it will be quiet around here now. So long.

John Kimmel wrote:

wrote:

On Sun, 16 Nov 2008 00:58:06 -0800, John Kimmel
wrote:


wrote:


Spreadsheet is not quite right.Prop center is BELOW the CG by about 13
inches.
One inch change in prop height according to your spreadsheat makes a
change of 2.14 degrees.
I cannot buy that. Particularly since it would go from 1.5 down to .64
up.


Here is REV A, with a couple math errors fixed:
http://www.spiretech.com/~guynoir/sl...thrustreva.xls




That looks a lot closer.

What is it based on?


"Vector Mechanics For Engineers: Statics and Dynamics", by Beer and
Johnston, 5th edition, Chapter 4: "Equilibrium of Rigid Bodies", page
126, equation 4.1. Equation 4.1 can be easily found on the internet,
just type "statics" into Google and go to the first hit, which should be
Wikipedia.

All my spreadsheet does is keep the distance from the thrust line to the
CG the same with different engine locations, so that the sum
of moments about the CG remains the same (as in: Zero. See equation 4.1).



I tried the first sheet.
Set the offset to 4 inches and got 8 something degrees angle!

That may have been a right answer for the problem you've set up.

But no way it would work on the airplane.



--

Richard

(remove the X to email)

cavelamb himself wrote:


I tried the first sheet.
Set the offset to 4 inches and got 8 something degrees angle!

That may have been a right answer for the problem you've set up.

But no way it would work on the airplane.



I checked the math on my rev a spread sheet more carefully than on the original, and included a check feature within the spreadsheet
itself. I also checked the spreadsheet graphically using a cad program. The spread sheet as currently posted will work for all
aircraft, but if you don't have a good understanding of the vector mechanics underlying it in the references I included in my previous
post, you should not use it.

If you think there is an error in my spread sheet, you are welcome to identify it, correct it, and re-post it, or you can show me the
problem graphically with a free body diagram and I'll make the correction, but your post above doesn't give me enough information.

--
John Kimmel


I think it will be quiet around here now. So long.

On Mon, 17 Nov 2008 20:41:30 -0800, John Kimmel
wrote:

wrote:
On Sun, 16 Nov 2008 00:58:06 -0800, John Kimmel
wrote:


wrote:


Spreadsheet is not quite right.Prop center is BELOW the CG by about 13
inches.
One inch change in prop height according to your spreadsheat makes a
change of 2.14 degrees.
I cannot buy that. Particularly since it would go from 1.5 down to .64
up.

Here is REV A, with a couple math errors fixed:
http://www.spiretech.com/~guynoir/sl...thrustreva.xls



That looks a lot closer.

What is it based on?

"Vector Mechanics For Engineers: Statics and Dynamics", by Beer and Johnston, 5th edition, Chapter 4: "Equilibrium of Rigid Bodies",
page 126, equation 4.1. Equation 4.1 can be easily found on the internet, just type "statics" into Google and go to the first hit,
which should be Wikipedia.

All my spreadsheet does is keep the distance from the thrust line to the CG the same with different engine locations, so that the sum
of moments about the CG remains the same (as in: Zero. See equation 4.1).


Only problem I see is you still have it set up with the thrust above
the CG. The center line of the prop is significantly lower than the
CG. Do I just use negative numbers? Or should the spreadsheet still be
corrected?

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.

What is that the center of mass of, exactly? The empty airframe only (i.e.
no engine)? Or is it the center of mass of a complete airplane (sans fuel,
pilot, passenger, and baggage) with the engine mounted at the original
point?

I'm also curious as to how you determined the location in two dimensions. I
can think of a few ways of determining it, but they all involve a bunch of
effort.

On Tue, 18 Nov 2008 12:05:40 -0600, Jim Logajan
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.

What is that the center of mass of, exactly? The empty airframe only (i.e.
no engine)? Or is it the center of mass of a complete airplane (sans fuel,
pilot, passenger, and baggage) with the engine mounted at the original
point?

I'm also curious as to how you determined the location in two dimensions. I
can think of a few ways of determining it, but they all involve a bunch of
effort.


Center of mass for the completed plane when built to specs. Provided
by the designer.

wrote:


Only problem I see is you still have it set up with the thrust above
the CG. The center line of the prop is significantly lower than the
CG. Do I just use negative numbers? Or should the spreadsheet still be
corrected?

If you downloaded the corrected version, the numbers you see when you open the spreadsheet a
60 -13 -1.5 -14 -2.408899631

CG is 60" aft of propeller center. Propeller center is 13" below CG. Thrust angle is 1.5 degrees down.
If you move propeller center down 1" to 14" below CG, the new thrust angle will be 2.4 down.
--
John Kimmel


I think it will be quiet around here now. So long.

On Tue, 18 Nov 2008 21:20:43 -0800, John Kimmel
wrote:

wrote:


Only problem I see is you still have it set up with the thrust above
the CG. The center line of the prop is significantly lower than the
CG. Do I just use negative numbers? Or should the spreadsheet still be
corrected?

If you downloaded the corrected version, the numbers you see when you open the spreadsheet a
60 -13 -1.5 -14 -2.408899631

CG is 60" aft of propeller center. Propeller center is 13" below CG. Thrust angle is 1.5 degrees down.
If you move propeller center down 1" to 14" below CG, the new thrust angle will be 2.4 down.


That's what I suspected - and it seams to work. The resulting
reduction in the angle is counterintuitive however.