Gear Ratio & Transmission
Determine the ratio i of a gear stage from tooth counts, pitch diameters, the worm gear (number of starts and worm-wheel teeth) or a speed pair, and reflect the load-side quantities to the drive side with efficiency η: speed, torque (motoring and generating), power and the reduced moment of inertia - live with every input.
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Formulas and fundamentals
Definition of the ratio i
The ratio of a stage is defined as a speed ratio, where the drive side (in) is the fast motor side and the output side (out) is the slow load side:
For the four common ways to determine it: from tooth counts, for a worm gear (a single-start worm acts like a pinion with one tooth), from the pitch diameters of belt or friction drives, and directly from a measured speed pair:
A value of i > 1 is a reduction to slower speed, the normal case in a drivetrain.
Speed and torque reflection
Angular quantities are reflected to the drive side in proportion to i:
Torque is transformed inversely and additionally corrected by the stage efficiency η - differently depending on the power-flow direction. In the driving (motoring) case, in which the motor moves the load:
Efficiency increases the required drive torque. In the generating case, in which the load drives the input (braking, lowering a hoist load):
Here efficiency reduces the torque arriving at the drive shaft.
Power and moment of inertia
Power is not transformed by the ratio, only corrected for losses, motoring and generating:
The load's moment of inertia appears on the fast drive shaft reduced by the square of the ratio:
This reduction is stated loss-free (efficiency enters the dynamic torque separately via the angular acceleration). A higher ratio strongly lowers the reflected load inertia and is therefore the central lever for matching motor and load inertia.
Worked example
A spur-gear stage has z1 = 20 teeth on the driving gear and z2 = 40 teeth on the driven gear, with an efficiency of η = 0.9. This gives the ratio i = z2/z1 = 2. The load rotates at n2 = 500 rpm and demands a torque M2 = 200 Nm and a power P2 = 630 W at the output, and it has a moment of inertia J2 = 10 kg·m².
The drive speed is n1 = n2·i = 500·2 = 1000 rpm. The required drive torque in the driving case is M1 = M2/(i·η) = 200/(2·0.9) = 111.11 Nm; if instead the load drives (generating), only M1 = M2·η/i = 200·0.9/2 = 90 Nm arrives at the drive shaft. The drive power is P1 = 630/0.9 = 700 W motoring and P1 = 630·0.9 = 567 W generating.
The moment of inertia reduced to the drive shaft is J1 = J2/i² = 10/4 = 2.5 kg·m². The comparison shows the typical picture: the reduction to slower speed halves the speed and (approximately, without losses) doubles the torque, but quarters the reflected load inertia - which is why the ratio is the most effective lever for matching the load inertia to the motor inertia.
Frequently asked questions
Is the ratio i a step-up or a step-down?
i = n_in/n_out is the ratio of drive to output speed. A value of i > 1 means the drive turns faster than the output - the stage reduces to slower speed and increases torque. That is the normal case in a drivetrain. A value of i < 1 would be a step-up to higher speed. The inconsistent wording often causes confusion; what matters is solely the ratio n_in/n_out.
Why do the motoring and generating torques differ?
The efficiency η describes the losses of the stage and always acts against the driving side. When the motor drives the load (motoring), it must additionally cover the losses: M_in = M_out/(i·η) becomes larger. When the load drives the input (generating, e.g. when braking or lowering), the losses eat into the arriving torque: M_in = M_out·η/i becomes smaller. At η = 1 both values coincide.
Why is the moment of inertia divided by i²?
The reflected inertia follows from energy conservation: the kinetic energy ½·J·ω² must be equal on both shaft sides. Since ω_in = i·ω_out, we need J_in = J_out/i² for the energy to match. Intuitively: a slowly rotating, heavy load appears much less inertial on the fast drive shaft - by the square of the ratio.
How do I determine i for a worm gear?
In a worm gear the worm acts like a pinion whose number of teeth equals the number of starts. A single-start worm has one start, a double-start worm has two. The ratio is therefore i = z2/number of starts, where z2 is the tooth count of the worm wheel. Worm gears thus reach very high ratios in a single stage (e.g. z2 = 40 with 2 starts gives i = 20).
Does the calculation also apply to multiple stages?
The calculator handles a single stage. A multi-stage gearbox multiplies the individual ratios (i_total = i1·i2·…) and the efficiencies (η_total = η1·η2·…). You can compute the stages in sequence by using the drive-side output quantities of one stage as the load-side input quantities of the next. For the overall ratio it is enough to enter i_total and η_total in one stage.
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