motorgliders as towplanes

Weight is a force????? Did I miss something in HS Physics? Weight = Mass.

Mike Schumann

"KevinFinke" wrote in message
...
Ian, "The Real Doctor" Out of curiosity, what exactly do you have a
doctorate in?

Aside from that.... In order to seek clarity in all of these
discussions I suspect that we have a mis-understanding because we are
trying to discuss these using two different reference frames. If
that's the case, then that would explain a lot.

I hope that we are all in agreement about the three forces acting on a
glider. For simplicity they are lift(L), drag(D) and weight(W=mg). As
has been corrected by Darryl, I agree that it is correct that lift, by
definition, is perpendicular to the airflow. However, for a glider in
steady state gliding flight, airflow and direction of motion are
parallel. Any body have any problems so far? I'm hoping this will get
me out of the hen house...

If we align the axis system such that weight is vertical and the
descent angle is theta. The equilibrium equations a

Vert. Axis 0 = L*cos(theta) + D*sin(theta) - W
Horz. Axis 0 = L*sin(theta) - D*cos(theta)

I'm guessing this is the source of the Lift providing the horizontal
motion argument. Clearly there is no gravity term in that component.
But the motion isn't strictly horizontal or vertical with these
equations. It is both, and therefore I would advocate a simplified set
where the direction of motion is the basis for the axis system.
Therefore....

If the axis system is aligned along the lift vector the equations
simplify to: (For the sliding block this tends to be the convention
that most books I own present) Replace L with N for Normal.

Lift Axis 0 = L - W*cos(theta)
Drag Axis 0 = D - W*sin(theta)

Any objections so far? I sure hope not. I can't imagine how....

The nice thing about convention 2 is that the lift and drag vectors
are isolated variables in the equation, and the weight is already
known so it's easy to solve the other values.

L = W*cos(theta) and D=W*sin(theta)

I can even rearrange the equations in set 1 and get the same
relationships. So, what in the world am I missing when I say Lift =
Weight * cos (glide angle)? Ian, you are the real doctor. I'll confess
my ignorance. I don't want to guess, cause I just don't know what
answer you're looking for, but what did I forget?

The other advantage of using convention 2 is in describing the motion
of the system. The object is constrained to the plane, and therefore
you can get rid of the "vertical" axis in this example and look at the
equation with one dimension. Because the lift force or normal force
constrains the object to the plane, you'll have no accelerations or
displacements in this direction, for a steady state example. In this
case that would be it's glide path. The only equation left is D = W*sin
(theta) So again I argue, Lift, because it is perpendicular to the
direction of motion, can not provide the motive force! The motive
force is governed by a balance between gravity, drag, and the glide
angle. Don't get me wrong, I'm not saying lift isn't important. It is
very important to making the glider stay on a glide path. Maybe this
is just a chicken before the egg argument. I can see the circularity
of the discussion. Why do those chickens keep coming up....

This would be a whole lot easier to explain with pictures. So I'll
cite a reference...If anybody has a copy of the BGA Manual: "Gliding:
Theory of Flight", please reference the discussion of forces on flight
in Chapter 4. The book goes through a very good explanation of how
gravity provides the motive force for gliding. It's an excellent book
and I highly recommend it. If only it had a discussion of forces on
tow....

-Kevin

PS I think we need some good flying weather so that we all get out of
the house and away from the computer....