How a transmission mechanism is calculated
In kinematics tasks of this type the bodies are linked by gear teeth or by a belt. The key rule: the contact points of the two bodies move with the same speed. If wheel 1 of radius r₁ rotates with angular velocity ω₁, the contact point has the speed v = ω₁r₁. For wheel 2 of radius r₂ the same speed is ω₂r₂, so ω₂ = ω₁·r₁/r₂.
Direction of rotation
External gear mesh reverses the rotation, an internal mesh and an open belt keep it, a crossed belt reverses it. In the calculation angular velocities carry a sign: plus means the same sense as body 1, minus means the opposite one.
Accelerations
The gear ratio is constant, so angular accelerations obey the same factors as angular velocities: ε₂ = ε₁·r₁/r₂. The tangential acceleration of the contact point is the same for both wheels. For a point at distance ρ from the axis: v = ωρ, aτ = ερ, aₙ = ω²ρ, a = √(aτ² + aₙ²).
Stepped wheels and a load on a thread
Two gears on one shaft rotate with the same ω. In the contacts table such a wheel takes two rows: the radius at which it meshes with the previous body and the radius for the next one. A load on an inextensible thread moves like the rim point on which the thread is wound: v = ωr, a = εr, and its path is s = φr.
Using the calculator
The example: wheel 1 of radius 10 cm rotates with ω₁ = 12 rad/s and ε₁ = 6 rad/s²; it meshes with a wheel of 30 cm, then a belt runs between pulleys of 20 and 40 cm, and one more mesh joins wheels of 15 and 30 cm. Press Example and check the result with the formulas above.