The position of the centre of gravity (C/G, balance point) on
control line speed models has always been a mystery to me.
As a F/F guy in the main, to find the C/G close up by the leading
edge is a real puzzle. The same applies in spades to monoline;
also to F2C team race models.
In analysing aircraft, it is common to treat the machine as being
supported by lift at the aerodynamic centre, which is usually at
the 25% chord position, where also operates a pitching moment
caused primarily by the wing camber. This latter pitching moment
is usually balanced by an opposing moment generated by the
tailplane. The drag associated with the participation of the
tailplane is called "trim" drag, which is undesirable. Placing the
balance point at the 25% position can minimise the trim drag.
But with controlline speed models, we have a strange business
indeed. The airfoil section is commonly close to symmetrical, which
means low camber and low pitching moment. So you would think
that the C/G would logically be placed near the 25% chord position,
with the tailplane loafing along behind.
But it is not. Indeed, the C/G can be within 5% of the leading edge.
This apparent noseheavy condition should require lots of upelevator in flight, with lots of trim drag as a result. But that does
not happen. My observation is that at takeoff and landing, lots
of upelevator is needed, but not in level flight. This is all very
weird, a puzzle for those driving 5000 km from Perth to the 63rd MAAA
AlburyWodonga Nats in 2009.
Indeed, on the dead straight 90 mile section, one does not even
need to turn the steering wheel. So with the A/C directed at my face
and with Vera Lynn on the C/D player, I ignored the 40C plus heat
outside on the Nullabor and gave my brain free rein on this problem.
This is my analysis. The rotating masses of the propeller and
crankshaft have angular momentum, thereby producing the
behaviour of a gyroscope. Gyroscopes are wonderful things, with
weird behaviour. So you see, we are on the right track thinking
this way. The weirdness is this. If you push on the gyroscope's
axis of rotation, the wretched thing does not move the way you
want. Indeed, the axis wants, perversely, to move at right angles
to the push. This is strange indeed.
Here is the rule of gyroscopes. The reaction to a push on the axis
of rotation takes place 90 degrees later in the direction of rotation.
The resultant motion is called precession. Now back to our speed model.
The axis of rotation of our propeller is changing direction all
the time. I mean, it is going around in a circle, and pretty quickly
at that! The result is a nose up pitching moment. Thats right. Just
because the model is flying in a circle and has a gyroscope up
front, the nose wants to lift. You would need to give down elevator
to stop this, but that then introduces trim drag, which we don't want.
So by making the model noseheavy, with the C/G well forward, we can
counter the pitchup without using the elevator. Now that is really
neat.
But is it true? As it happens, we can put some numbers into this
hypothesis to see if we are on the right track. There is a simple
formula for the moment M of a gyroscope. With I the moment of
inertia, w1 the angular velocity of the gyro and w2 the angular
velocity of the gyro axis, we have:
M = I . w1 . w2
Hmm, what is all that about. Well I is to rotation as mass is to
linear motion. That is, moment of inertia is like mass, but the object
is spinning. As an approximation, for a propeller of mass m and
length L we can pretend the propeller is a slender rod. Then the
propeller has moment of inertia:
I = m . L . L / 12
For a rough calculation for an F2A setup, we can take the mass
of the rod (propeller) to be .002 Kg ( 2 grams), and the length
0.15 m (150 mm). Then, ignoring the weight of the crankshaft:
I = 0.002 x 0.15 x 0.15 / 12
= 0.00000375 Kg.m.m
Now F2A engines turn about 40000 RPM, so:
w1 = 2 * pi * RPM / 60
= 4200 radians per second
Like wise, the model sweeps thru 2 * pi radians every 1.4 seconds
(roughly 290 kph), so that:
w2 = 2 * pi / 1.4
= 4.49 radians per second
(2 * pi radians is 360 degrees)
Finally, the gyroscopic turning moment is:
M = 0.00000375 * 4200 * 4.49
= 0.071 Kg/m
So far so good. We have an estimate of the noseup tendency
resulting from the gyroscopic effect. I have no sense of the size
of this number, except that we need a similar sort of number to
counteract and hence balance this moment.
The counterbalancing moment comes from having the C/G well
forward, say a distance "x" from the aerodynamic centre. Let us
call this moment B for want of something better. Then the
condition for balance is:
M = B
Whoa, aint maths marvellous!
Well B is a moment, like a lever, so we then have:
B = mg . x
Here mg is the weight force, being the mass of the airplane times
the acceleration due to gravity. I don't know what an F2A airplane
weighs, but lets guess its mass is 0.42 Kg, and recall that g is
9.8 m/s/s, so that:
mg = 0.42 * 9.8
= 4.12 Newtons (force units)
Going back to M = B and substituting,
4.12 * x = .071
x = 0.017 m
= 17 mm
So there needs to be a noseheavy condition, placed 17mm forward to,
balance the gyro noseup tendency. Well, I reckon that places the
C/G close to the leading edge. Indeed, with a heavy prop and spinner,
the C/G could even be in front of the leading edge!
So there we have it. The C/G miles forward, the nose held up
by gyroscopic force and tailplane loafing along behind, doing
nothing! Fun things, these speed models.
Now I ask you. Is that not cool?
