Importance of Balancing
So you probably read some stuff about balancing those blades, while the manufacturer sells
the blades 'fully balanced', so why bother?
To understand the importance of balancing them, we look here at the magnitude of
the forces, and the consequences of a few grams or millimeters error in the balance.
To find this, we have to do some mathematics.
Please just skip to the next section if you can't follow it.
The centrifugal force F=m ( 2 p f ) 2 r,
where m is the mass of the blade (in kg),
p=3.1415, f the rotor speed in Hz (rotations per second) and
r the distance between the center of mass
of the blade and the center of rotation (the axis) in meters. The result F is the centrifugal force
of the blades in Newton (1 N is roughly equal to the gravity force of the earth pulling at 100g).
Assuming 1800 rpm = 30 Hz (which is a little on the high side but not unrealistic during
stunt flying), we find that ( 2 p f ) 2 = 35000,
and thus
F=35000 m r
As you can see, the mass and radius have a BIG impact on the forces.
If we set the mass of the left blade equal to that of the right blade plus a little bit
Dm, and similar we set the
distance to the center of gravity of the left blade equal to that of the right blade plus
a little bit Dr,
and we subtract the force of the right blade 35000 m r from that on the left blade
35000 (m+ Dm) (r+Dr), we get
for the difference in force DF:
DF =
35000 ( r Dm +
( m + Dm ) Dr)
~ 35000
( r Dm +
m Dr) (result in Newtons)
If the following still is too technical for you, please skip to the next section.
Let's assume a typical blade of 100g, and a length of 55cm. The center of gravity will lay at about
25 cm, plus maybe 5 cm extra from the rotor head, gives say 30 cm from the rotation center.
The formula then becomes
DF=
Dm + 0.36 * Dr (result is in kg force).
Extra mass (grams) Dm and
extra distance to the COG Dr (millimeters)
should both be measured on the same blade.
If the blade you picked has its COG further out, but its mass is lower than the
other blade, its Dr should be positive but its
Dm should be negative.
So, assume that you have a blade that is 1g heavier and has its COG 1 mm more outside.
Then, we get F=1 + 0.36 = 1.36 kg force. So your heli will feel a force of 1.36 kg pulling the
head to the right 30 times per second (and 30 times to the left as well)!
Another example, assume you a have a blade where the COG is 3mm more to the outside,
but it weighs 2 grams lower than the other blade, we get F = -2 + 0.36 * 3 = -0.92 kg.
The force is negative, ie the other blade is pulling harder.
Note that a COG and mass inbalance can compensate each other.
Roughly, every gram of weight difference and every 3 mm of error on the center of
gravity will add another kilogram of force on your helicopters mechanism.
Concluding, an extra check to make sure COG and mass are equal for the blades is worth the efford.
© W.Pasman, 9/4/1