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Gears

Reference data and engineering information about gears for mechanics applications.

gears

Overview

Engineering reference data for Gears in mechanics.

Key Formulas

Newton's Second Law

F=maF = ma

Force = mass × acceleration.

Work

W=FdcosθW = Fd\cos\theta

Work = force × displacement × cos(angle).

Kinetic Energy

Ek=12mv2E_k = \frac{1}{2}mv^2

Energy of motion.

Potential Energy

Ep=mghE_p = mgh

Gravitational potential energy.

Variables

SymbolDescriptionUnit
FFForceN
mmMasskg
aaAccelerationm/s²
vvVelocitym/s

Example Calculation: Three-Wheel Gear Transmission

The effort force in a gear transmission can be calculated using the general formula:

F=Wr1r2rnR1R2RnF = W \frac{r_1 \cdot r_2 \cdots r_n}{R_1 \cdot R_2 \cdots R_n}

Consider a system with three gear wheels and an applied load force (W) of 1000 N.

3 rows
Gear wheel dimensions for the transmission example.
Wheel
Inside Diameter (r)(m)
Outside Diameter (R)(m)
10.050.12
20.060.13
30.070.14

Source: engineeringtoolbox.com

Plugging the values into the formula:

F=(1000N)(0.05m)(0.06m)(0.07m)(0.12m)(0.13m)(0.14m)F = (1000 \, \text{N}) \frac{(0.05 \, \text{m})(0.06 \, \text{m})(0.07 \, \text{m})}{(0.12 \, \text{m})(0.13 \, \text{m})(0.14 \, \text{m})} F=(1000N)0.00021m30.002184m396N=0.1kNF = (1000 \, \text{N}) \frac{0.00021 \, \text{m}^3}{0.002184 \, \text{m}^3} \approx 96 \, \text{N} = 0.1 \, \text{kN}

The resulting effort force is 96 N (0.1 kN).

Interactive Charts

Gears transmission - force and line of action

References