FISEKON GmbH – Fischer Elektrokonstruktion

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27 — Calculators

Calculate the required motor rating

The driven machine needs a torque at a speed — from that follows the power. In between sit the gearbox and a margin, and at the end stands not an arbitrary number but one of the standard sizes. This calculation walks the whole way in one go.

Input

Around 97 percent per stage for helical gears, markedly less for worm gears.

Covers starting torque, wear and variation in the load.

Result

Next standard motor rating
—kW
Power required at the driven machine
—kW
Required motor rating with gearbox and margin
—kW
Utilisation of the standard motorBelow roughly 50 % the motor works in the poor efficiency region.
—%

A guide, not a design to standard. We give no warranty for the correctness of the results, the applicable standards and case-by-case verification govern.

Worked example

The default case the calculator starts with, worked through once. Change the values above and it recalculates immediately.

Inputs

Torque required at the driven machine
480 Nm
Speed of the driven machine
190 1/min
Efficiency of the gearbox
94 %
Margin allowance
15 %

Result

Next standard motor rating
15 kW
Power required at the driven machine
9.55 kW
Required motor rating with gearbox and margin
11.7 kW
Utilisation of the standard motor
77.9 %

Formula

  • P = M · 2π · n / 60
  • P_motor = P / η_gearbox · (1 + margin)
  • rounded up to the next standard rating

Assumptions and standards

  • The steady-state power demand is calculated. Starting torque and inertia are only covered in bulk by the margin and deserve separate checking for heavy starts.
  • The gearbox efficiency is entered here. It depends on type and ratio; worm gears are markedly below helical gears.
  • The ratings are the usual frame sizes for three-phase motors.

Frequently asked

How large should the margin be?

With steady load and easy starting, ten percent is enough. With varying load, contamination or expected wear, twenty makes sense. More is rarely better — a markedly oversized motor runs permanently in the poor efficiency region.

Why show the utilisation?

Because the standard size almost always exceeds the demand, and the question is by how much. Landing at forty percent utilisation makes it worth looking at the smaller size or a different gearbox — efficiency is better there, and the power factor with it.

Does this apply to hoists?

Only to the steady-state part. Hoists add acceleration, braking and the lowering load case; there the design is a calculation of its own, not a rule of thumb.

Calculating is the easy part.

A formula gives you a number. Designing a plant also demands installation method, grouping, discrimination, the standards in force and a look at the installed base. That is what we take on.

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