In machine design one often needs to incorporate a Power Transmission between an energy source and the desired output motion. Examples of Power Transmission elements include: gears, friction drives, timing belts, flat belts, levers, and screw drives
The Power Transmission often includes a Gear Ratio or Mechanical Advantage. A Gear Ratio can increase the output torque or output speed of a mechanism, but not both. A classical example is the gears on a bicycle. One can use a low gear that allows one to pedal easily up hill, but with a lower bicycle speed. Conversely a high gear provides a higher bicycle speed, but more torque is required to turn the crank arm of the pedal. This tradeoff is fundamentally due to the law of energy conservation and is the key concept of Mechanical Advantage. With a given power source you can either achieve high velocity output or high force/torque output but not both.
Mechanical Advantage refers to an increase in torque or force that a mechanism achieves through a power transmission element. For rotary devices the term Gear Ratio is used to define the Mechanical Advantage. The term Mechanical Advantage is used to describe components that include translation. The analysis below shows how one calculates the Gear Ratio and Mechanical Advantage of a Power Transmission component.
Energy and Power Equations
The law of energy conservation dictates that one can never get more energy in the output motion than provided by the energy source. Indeed, one always has some energy loss in a Power Transmission. Energy loss rates can vary from 5% for a flat belt drive to up to 80% for a multi-stage gear transmission (higher and lower rates can occur too).
Before the analysis we first define some notation:
For basic analysis of Gear Ratios we initially neglect frictional losses, and then incorporate their effect separately. With this assumption we can set the power in to equal to the power out.
Pin = Pout
Power is defined as the change in energy divided by the change in time.
P = δE/δt
In a mechanism energy is transferred by mechanical work. For translational motion work is given by:
The corresponding definition of work for rotational motion is given by:
Work = Torque X Rotary Motion
In a Power Transmission the Work is the source of the change in energy, and thus:
P = δE/δt = W/δt
Substituting the rotary definition of work into the above equation, and noting that rotation velocity is given by ω=δθ/δt, the power transfer in a rotary device is given by:
P = W/δt = τ δθ/δt
In a similar fashion, for translational motion the power transfer is given by:
A gear train is shown below with an input gear on the left and an output gear on the right. For the purposes of this analysis we assume the input gear may is attached to a motor, and the output gear is attached to a shaft on a machine that performs a desired function.
As shown the input gear is rotating counterclockwise with an angular velocity of ωin and the output gear is rotating clockwise with an angular velocity of ωout. An input torque, τin, is applied by the motor onto the input gear, and an opposing output torque, τout, is applied by the machine onto the output gear. The radius of the gears is shown at the Pitch Circle of the gear, which is between the top and bottom of the gear tooth, and represents the radius at which contact occurs between the two gears.
The shape of the gear teeth are the some on both the input and output gears, and thus the larger gear has more teeth on it. The Pitch distance, Pd, is the distance between gears. Thus the number of teeth on the gear, n, times the Pitch is equal to the circumference of the gear. Accordingly,
The gear pair is analyzed with the following assumptions:
Quasi-static analysis (it is assumed that the gears are rotating at a constant speed, and thus acceleration torques can be neglected)
Frictional losses are neglected (friction can be significant, and should be considered separately!)
The gear teeth mesh with each other (no jumping of gears!)
Since there are no frictional losses, the input and output power can be set equal to each other as:
We now need to consider the relative velocity of the two gears, which is determined by the meshing of the teeth. Since the teeth mesh, we know that the same number of teeth must mesh from both gears. For each revolution of the input gear the following number of teeth pass through the mesh area, where nrevin is the number of revolutions of the input gear:
number of teeth that mesh = nrevin 2 π rin / Pd
Applying the same equation to the output gear, and setting the number of meshed teeth equal to each other provides:
nrevout 2 π rout / Pd = nrevin 2 π rin / Pd
The above equation simplifies to:
nrevout / nrevin = rin / rout
If we multiple the number of revolutions by 2π, we get the angle of rotation of both gears in radians, which gives:
rin δθin = rout δθout
If we divide the angle of rotation by time, δt, then we get the ratios of angular velocities in radians per second
ωout / ωin = rin / rout
An alternative interpretation is that the angular velocity at the mesh point is the same for both gears. Since velocity of a point on a rotating object is given by rω. The velocity equality at the mesh point is given by:
rin ωin = rout ωout
And we see that the two previous equations are identical.
Since the radius of a gear is proportional to the number of teeth, the velocity relationship can be given in terms on numbers of teeth on the input and output gears. Simply substitute into the above equation that nPd=2πr for both gears, to give:
ωout / ωin = nin / nout
We can now combine the power equation with the velocity equation to get the ratio of input and output torques:
Thus when the input gear is smaller than the output gear:
The output torque is higher than the input torque
The output velocity is lover than the input velocity (i.e. the smaller gear needs to make more revolutions than the larger gear)
The fundamental equations for a gear pair are:
The Gear Ratio is defined as the input speed relative to the output speed. It is typically written as:
Gear Ratio = ωin : ωout