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Gear Power

04/29/2009 9:24 PM
GearSunPlanetring
Teeth61230
train locked111
carrier lockedminus (Ns/Np)Ns/Np-1
speed120ccw300ccw600ccw
Gear Ratio 10:7:3

How do i calculate power required to drive gear. Force applied on rod that the gear is used for is 6000 lbs

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#1

Re: Gears

04/30/2009 4:24 AM

You'll need to describe that out drive/rod assembly a bit more yes expand upon it with impunity.

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#2

Re: Gear Power

04/30/2009 7:16 PM

Well, power is calculated by means of torque and speed. Torque in a reduction series is multiplied by the gears reduction, ok, minus the losses, but... So, the power required to drive the train will be output torque x output speed x transmission efficiency or, in the same maner, input torque x input speed x transmission efficiency Do the calculations using consistent units. I prefer SI system: power in W, when torque is in Nm and speed is in rad/s

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#3

Re: Gear Power

05/01/2009 5:07 AM

You need to be more specific, what gear is driving and what gear is driven and what component/gear is fixed. Your table seems to be inconsistent as it on the one hand indicates you are driving the sun with output on the carrier (ring gear fixed) but at the same time you describe a rotation of the ring gear. If the carrier was fixed then the direction of rotation of ring and sun would be opposite and the ratio's would be different.

To work the ratio calculation you need to describe all members in the table, first lock all components and rotate as a whole ("Train lock", 1 rotation for unity). Then hold the member that is the output (i.e. carrier, hence "Carrier lock") and return the member that should have been stationary to the original position (ring gear -1). Determine the ratio's of the members that are dependent on these two "rotations" (returning the ring one rotation while keeping the carrier stationary). Then you add up the two for each member to get each members absolute rotation and you can calculate relative rotation of member to member:

The ratio between input and output then becomes:

1 [Carrier-Output] / (1+Zring/Zsun) [Sun-Input]

Now you know the ratio and because you know the torque that is applied on one member you can calculate the power on that member e.g.: Speed [Rad/s] * Torque [Nm] = Power [Watt]

If you assume 100% efficiency then the power on the output would of course be the same, otherwise you have to multiply by the efficiency of each mesh (sun-planet & planet-ring) to get the output power. When you know that you can also calculate the torque for the output member.

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