I still think the pitch line vel. is not so simple.
When you have a moving planet gear and planet gear
carrier moving around the sun gear at the the same
circumferential vel and direction as the teeth on the
sun gear the planet gear would not rotate at all.
The pitch line velocity would be zero.
Paul Lamar
Hello Paul,
No agony here, just pure joy.
And no hard feelings, it?s only meant as a contribution.
I think pitch line velocity is zero when nothing moves.
With rotation pitch line velocity is pitch diameter * pi * rpm / 12 in
fpm. (rpm = relative rpm).
For the ring gear rpm is output speed, for the sun gear rpm is the
differential rpm of input and output (input rpm ? output rpm), and for
the planet gear the rpm is calculated: rpm out * pitch diameter of ring
gear / pitch diameter of planet gear. All three velocities turn out to
be the same.
As to tangential contact force between teeth along the pitch line the
simplest is to calculate from the sun gear. F = T / r.
Tooth contact force is Input torque in ft lbs * 12 / radius of sun gear.
Again, the contact force is the same for all three gears.
The bearing force for the planet gear is twice the contact force plus
centrifugal force added quadratically. Bearing loads on the input shaft
and output shaft are zero due to torque when at least two planet gears
evenly spaced are being used.
Now, you don?t have to post that. Likely no one cares anyway.
Since I have learned so much in other areas, I ought to give
something back.
This one is mechanical where I feel confident.
Example:
Ring gear 7.2 in diameter, sun gear 4 in diameter, planet gears 1.6 in
diameter, 6 planet gears.
Ratio is then 1 + (7.2 / 4) = 2.8 : 1.
Input is 7,000 rpm, out put 7,000 / 2.8 = 2,500 rpm, planet gear = 2,500
* 7.2 / 1.6 = 11,250 rpm.
Ring gear velocity = 7.2 * pi * 2,500 rpm / 12 = 4712 fpm,
Sun gear velocity = 4 * pi * (7000 ? 2500) /12 = 4712 fpm,
Planet gear velocity = 1.6 * pi * 11,250 /12 = 4712 fpm.
Say input is 300 hp, input torque is 300 * 5252 / 7,000 = 225 ft lbs.
Tooth contact force = T / r = 225 ft lbs * 12 / (4? /2) = 1350 lbs.
Having 6 planets, 1350 / 6 = 225 lbs tooth contact per each planet gear.
The tangential bearing load on the planet carrier is twice that = 2 *
225 lbs = 450 lbs.
This is simply because the planet gear engages two other gears at
opposite sides and both gear loads of 225 lbs apply, as both of these
reaction forces are acting in the same direction on the bearing.
Proof calculation: Diameter of planet carrier is diameter of ring gear ?
diameter of planet gear.
Radius of planet carrier = (7.2 ? 1.6) / 2 = 2.8 inches.
Force = T / r. Torque is output torque = 300 hp * 5252 / 2,500 rpm = 630
ft lbs.
Also: 225 ft lbs * 2.8 ratio = 630 ft lbs.
Force = T / r = 630 ft lbs * 12 / 2.8 inch sun gear radius = 2,700 lbs.
Divide by 6 = 450 lbs for each carrier.
Normal to the tangential force is the centrifugal force in radial
direction that should be added to the 450 lbs like this: (tangential
force^2 (here 450) + radial force^2) ^.5. The tooth separating forces
from contact with the sun gear and contact with the ring gear are close
to being equal and essentially cancel each other out.
If one wants to be very precise, due to the pressure angle the tooth
contact forces generate a further component acting on the bearing race.
However, things like that should be covered by a blanket safety margin
to be used.
Another interesting point is that the torque on the ring gear is in
opposite direction to the input torque from the sun gear. Consequently,
it is the difference between the output torque and input torque.
Torque on ring gear is 630 ft lbs output ? 225 ft lbs input = 405 ft
lbs.
Proof calculation: T = F * r,
Tooth contact force = 225 * 6 planets = 1350 lbs. (see above)
T = F * r = 1350 lbs * (7.2 / 2) = 405 ft lbs.
The planetary gearing is quite amazing.
Regards
Rolf
Yes it is but I think you are missing my point.
Assume for the moment there is no ring gear.
Also assume the planet gear is locked to the planet
carrier and the planet carrier is free to rotate.
What happens?
The planet gear and the planet carrier rotate
in concert with the sun gear tooth. The pitch line velocity
of the sun gear relative to a fixed point is indeed
finite (is something). However relative to the planet
gear it is zero. The planet gear is just going along
for the ride. Do you buy that?
Also assume for the moment the planet carrier is
fixed and the planet gear is free to rotate
influenced by the sun gear. The pitch line velocity
is now the same for both gears. Do you buy that?
I think you do because that is what you have stated.
Well if you buy both of those statements
then there MUST be a happy medium where the planet
carrier is rotating but not at the same speed of
the sun gear tooth and the pitch line velocity between
the sun gear and the planet gear cannot be both
zero and finite at the same time.
I wish I had and animation program.
Paul Lamar ...No rotor no motor.
Paul,
I fully agree with both your points above, except for the conclusion.
If there is no ring gear, and the planet carrier rotates at the same
speed as the sun gear, there is no rotation between the sun gear and the
planet gear and pitch line velocity between the two is zero, agreed.
That is because there is no ?relative? rotational speed between the two.
At the outset above I stated in brackets that rpm is to be ?relative?
rpm between two rotating bodies.
If the planet carrier is stationary and both the sun gear and planet
gear rotate, then the pitch line velocity of the sun gear and the planet
gear is the diameter of the sun gear * pi * rpm of sun gear / 12, just
as usual.
Note that the ?rotation? of the planet carrier is zero in this case, and
the differential rpm between the two is the rpm of sun gear minus the
rpm of planet carrier which is zero, so it?s the rpm of the sun gear
that counts.
Now, if the planet carrier assumes some rotation, either backward or
forward, meaning in same or in opposite direction of the sun gear, then
the relative rotation between both, the sun gear and planet carrier is
the difference between the two rotations.
Let?s say the sun gear rotates 3 rpm clockwise and the planet carrier 1
rpm clockwise then the difference between both is 2 rpm, and the pitch
line velocity is calculated on 2 rpm of the sun gear.
You could also look at it that way, that if the total assembly is
rotated counter-clockwise by 1 rpm, then the rpm of the planet carrier
becomes zero and the rotation of the sun gear becomes 2 rpm clockwise,
which is the 2 rpm relative rotation we are looking for.
If the sun gear rotates 3 rpm clockwise and the planet carrier 1 rpm
counter-clockwise, then the differential rpm is 4 rpm, and the pitch
line velocity would be double to the case before with clockwise rotation
of the planet carrier.
It is this differential rpm that is used to calculate the pitch line
velocity between the two gears, where the pitch line velocity of both
gears is the same. It is the velocity of how fast the pitch line moves
in respect to the imaginary line between the two gear centres. The line
between the two gear centres can be stationary or can be rotating. It
does not alter the pitch line velocity between the gears. It is not the
movement is space, but the rotation between the two relative to each
other that determines pitch line velocity.
Just be definition, the pitch line velocity between two gears is the
velocity obtained by their relative rotation to each other.
In case of the planetary gearing where the planet gear rotates in space
around the sun gear, this space rotation has to be deducted to bring it
to the ?space rotation? of the sun gear which is calculated from zero
space rotation.
I hope it is understandable what I have said? Perhaps my last sentence
is a bit weird. Just trying to explain that rotation in space needs to
be accounted for. Only the relative rpm between the gears determines
their relative rpm and their relative pitch line speed.
Greetings
Rolf
Paul,
I think I know what is driving your intuition about what is happening
when the planet gears are moving. Let's define a term that I'll call
"frequency of tooth contact" or maybe "duty cycle". Although the pitch
line velocity is determined by the rotation and diameter of the sun gear
and is invariant, the "duty cycle" of a given tooth on the sun gear is
changed by movement of the planet. When the planet gear moves in the
direction of the sun gear, a given tooth on the sun gear contacts the
planet less frequently than it would if the planet were stationary.
Let's use the Bell 47 unit as an example with its 46 tooth sun, its 23
tooth planets and its 92 tooth ring gear. Think of one of the planets
in the Bell 47 unit starting at the 12 o'clock position. Mark the 12
o'clock tooth on the sun gear as tooth #1. Now rotate the sun gear
through one revolution, or 46 teeth. If the planet gear were
stationary, the #1 tooth would be back at 12 o'clock and make contact
again with the planet gear. But, the planet gear has rolled through 46
teeth on the ring gear and is now at the 6 o'clock position. So, since
the #1 tooth on the sun is "chasing" the planet gear around the ring
gear, the "duty cycle" for tooth #1 is less than it would be if the
planet gear were stationary. And, your intuition is vindicated! If
Doug paints some dots and takes photos, I think it will be very helpful.
Aubrey
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