Optimized gearbox design - Wind Power Generation and Wind Turbine Design - page 519

effective centroid of its tangential loads would still be at the midpoint of the spindle.
However, the associated radial components of tooth load due to pressure angle create
a tilt in the radial plane, which reduces the tipping couple and the mal-distribution of
load by which it is generated, so that the planet adopts a skewed equilibrium attitude
with a low mal-distribution commensurate with its very low angular rigidity. The
crossed helix effect created by having non-parallel axes leads to a notional point
contact rather than line contact on the tooth faces. Since the crossed helix angle is
very small it relieves the tendency for edge contacts in a manner analogous to crown-
ing. The fl exible spindle has proved conclusively that it can compensate for helix
errors of different magnitude and hand in sun, planet and annulus by tilting in a
complex way to a position of minimum strain energy to enable the planet wheel to
avoid the load mal-distributions that are imposed by a more rigid support. In simple
terms, the planet dictates where it wants the spindle to be rather than vice versa.
Unlike a fl exible annulus, the planet spindles are all independent of one another so
they are all free to do their own thing and because they are cantilevers, the only limit
on the number of planets is the clearance of their adjacent tip diameters and the
annulus to sun ratio. As this ratio varies from 2.15 to 5.2 the number of planets
reduces from 8 to 4. For bigger ratios than 5.2 only 3 can be accommodated.
A larger number of equally loaded planets directly reduce the overall volume of
an epicyclic geartrain. This is shown by deriving a similar volumetric expression
as that shown above for a simple parallel shaft pinion and wheel viz.
It can be seen in Fig. 5 that the relationship of three pitch diameters can be
expressed as

dd
d=

−

(7 )

a s

p

2

Epicyclic analogy gives

d

R − 1

p

n= =

(8 )

d 2

s

Noting that

1 2 R+ 1

(9 )

1 +=+ 1 =

n R − 1 R − 1

Thus,

R+ C
N= RK

1
1

⎛ ⎞

(10 )

⎜ ⎟

s

⎝ ⎠

−

R+ C
N=NR=R RK

1
1

⎛ ⎞

(11 )

⎜ ⎟

a s

⎝ ⎠

−

T R+ 1

1

T

⎛ ⎞

2 s c

fd = = (12 )

⎜ ⎟

s

⎝ ⎠

QK R − QK R ( − 1)