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, its 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 dont 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.
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