Rotation is measured three ways

Angular displacement is how far something has turned. Its symbol is θ (theta), and there are three units in everyday use for it. They measure the same rotation; they differ only in how much turning counts as one unit.

Revolution (r)

One complete turn. This is what a machine nameplate counts.

Degree (°)

One revolution cut into 360 parts. This is what a drawing and a protractor use.

Radian (rad)

The angle whose arc is exactly as long as the radius. This is the SI unit, and it is the one formulas such as v = ωr expect.

The two factors worth keeping to hand: 1° = 0.01745329 rad and 1 rad = 180°/π = 57.29578°.

The degree sign sits tight against the number — 90°, not 90 °. Every other unit symbol takes a space: 1.57 rad, 45 r/min. The degree, minute, and second of plane angle are the only exceptions to that rule.

Supporting detail: Addendum 1 — Why a radian is barely a unit at all.

Angular acceleration is how fast the turning changes

Angular acceleration uses the symbol α (alpha), and it is change in angular velocity divided by the time it took:

αavg = Δω / Δt

Its SI unit is rad/s2. Everything true of straight-line acceleration on Acceleration and Squared Time is true here as well: it is a change, so a shaft slowing down has angular acceleration opposite its angular velocity (negative if it is turning in the positive direction), and the squared second means a squared factor whenever the time unit changes.

Worked example: a motor speeding up

A motor shaft is turning at 17 rpm. Power to the motor is increased, and 1.1 s later it is turning 8 rpm faster. What is the angular acceleration in rad/s2?

1. Only the change matters, so convert the 8 rpm increase: 8 r/min × 0.1047198 rad/s per r/min = 0.837758 rad/s.
2. Divide by the time: 0.837758 rad/s ÷ 1.1 s = 0.7616 rad/s2.

Worked example: a spindle running up

A machining-centre spindle goes from rest to 8,000 rpm in 2.5 s.

1. 8,000 r/min × 0.1047198 rad/s per r/min = 837.758 rad/s.
2. (837.758 rad/s − 0 rad/s) ÷ 2.5 s = 335.10 rad/s2.

That is the average over the run-up. A drive that ramps gently at each end has a changing instantaneous angular acceleration inside the same interval, exactly as a linear drive does.

Worked example: a conveyor drive roller

A conveyor belt runs over a drive roller 200 mm in diameter, turning at 45 rpm. What is the belt speed, and how long does a package take to travel 15 m?

1. Halve the diameter: r = 100 mm = 0.100 m.
2. Convert the speed: 45 r/min × 0.1047198 rad/s per r/min = 4.712389 rad/s.
3. Multiply: v = 4.712389 rad/s × 0.100 m = 0.4712 m/s.
4. Divide the distance by the speed: 15 m ÷ 0.4712389 m/s = 31.83 s.

Worked example: a roll-forming feed rate

Forming rolls 150 mm in diameter rotate at 25 rpm. What is the feed rate of the sheet in m/min, and how long does 20 m of sheet take?

1. r = 75 mm = 0.075 m.
2. 25 r/min × 0.1047198 rad/s per r/min = 2.617994 rad/s.
3. v = 2.617994 rad/s × 0.075 m = 0.19635 m/s.
4. The question asks for metres per minute, so convert the time unit: 0.19635 m/s × 60 s/min = 11.78 m/min.
5. 20 m ÷ 11.78 m/min = 1.70 min, which is about 1 min 42 s.

Note what step 4 did and did not do. Changing seconds to minutes in a speed takes one factor of 60. If the quantity had been an acceleration, the same change would take 60² — see Acceleration and Squared Time.

Worked example: running the conversion backwards

Sometimes the linear quantity is what you know. The formulas simply rearrange: ω = v/r and α = a/r.

A wheel, from angular to linear

A tyre 28 in in diameter turns at 3.5 rad/s. How fast is the vehicle moving, in miles per hour?

1. r = 14 in × 0.0254 m/in = 0.3556 m.
2. v = 3.5 rad/s × 0.3556 m = 1.2446 m/s.
3. 1.2446 m/s ÷ 0.44704 m/s per mph = 2.78 mph.

A wheel, from linear to angular

A vehicle accelerates at 5 m/s² on tyres 78 cm in diameter. What is the angular acceleration of a wheel?

1. r = 39 cm = 0.39 m.
2. α = 5 m/s2 ÷ 0.39 m = 12.8 rad/s2.

Check the unit on the answer, not just the number. Dividing m/s2 by m leaves 1/s2, and because the radian is a ratio of two lengths, rad/s2 is the same thing named for what it describes.

The unit, and the word order that identifies it

Case matters, and this one bites. The SI torque unit is N·m with a lower-case m. Written N·M it is not a valid SI torque unit: M is the prefix for 10⁶, but a prefix cannot stand alone as a unit. The same trap catches Kg for kg and S for s. See Measurement Language.

Worked example: can the motor hold the arm up?

A robot-arm joint motor produces 15 N·m. The arm segment is 0.35 m long and carries a 25 N load at its end. Will the motor hold the arm horizontal?

1. The load acts straight down, at right angles to a horizontal arm, at the full 0.35 m: τ = 0.35 m × 25 N = 8.75 N·m.
2. Compare with what the motor produces: 8.75 N·m < 15 N·m.

This is the shape of most torque questions: work out what the load demands, then check it against what the drive can supply, and be explicit about what you have left out.

Worked example: torque from a tangential force

A wheel 250 mm in diameter needs 3.5 kN of tangential force to overcome resistance. What torque must the motor supply to the axle?

1. r = 125 mm = 0.125 m.
2. 3.5 kN = 3,500 N.
3. τ = 0.125 m × 3,500 N = 437.5 N·m.

The same formula, rearranged

When the torque is what you know and the force is what you want, F = τ/r. That single rearrangement explains a family of everyday machines: a longer wrench, a bigger steering wheel, and a larger pulley all trade radius against force. Double the radius and the same torque delivers half the force — and, at the same rotational speed, the rim travels twice as far, which is why Power and Efficiency comes out unchanged.

Count the teeth to find the speed

The driver supplies motion; the driven gear receives it. For one external gear mesh, each tooth passing on the driver advances the driven gear by one tooth space. A gear with twice as many teeth needs twice as many passing teeth to complete a turn.

This lesson uses reduction ratio i = input speed ÷ output speed, with speed magnitudes. A 2:1 reduction means two input turns for one output turn. Ratios above 1 reduce speed; ratios below 1 increase it. At zero speed, use tooth counts to find the ratio; do not divide 0 by 0.

Notation map: this page's original equations use N for teeth and n for speed. The course uses n for teeth, ω for angular velocity and τ for torque. Thus page N = course n; page speed n is written ω in the additions below. GR and i name the same reduction ratio.

In course notation, using magnitudes: GR = nout/nin = dout/din = ωin/ωout = τout/τin for an ideal single pair. Here d is pitch diameter; the torque equality assumes no losses. For a compound train, multiply the stage ratios; the first and last tooth counts alone do not give the overall ratio.

Supporting detail: Addendum 7 — Why tooth counts work, and what the dashed circles mean.

An idler and a compound shaft do different jobs

Same shaft means the same angular speed. A large B and a small C fixed to one shaft make the same number of turns in the same direction. Their rim speeds differ because their radii differ. They do not mesh with each other.

In the compound example, 20 → 40 teeth halves the speed. A second 20 → 40 stage halves it again: 120 → 60 → 30 rpm. The two external meshes make the final output turn in the same direction as the input.

An idler is not another reduction stage. Drop its tooth count when finding the end-to-end ratio, but count every external mesh for direction: odd means opposite; even means the same, viewed from the same end of the parallel shafts. Do not drop a gear that shares a shaft with a different gear in a compound train.

More torque costs speed; it does not create power

Torque is turning effort. In this ideal, lossless train at steady load, output torque magnitude = input torque magnitude × reduction ratio. A 2:1 reduction changes 120 rpm and 2 N·m into 60 rpm and 4 N·m. The 4:1 compound example gives 30 rpm and 8 N·m.

Rotational power is torque × angular speed: P = T × 2πn/60 when T is in N·m and n is in rpm. Both sides of either example carry 25.13 W ideally. Real friction and other losses leave less output power. Tooth strength, bearings, shafts, lubrication and the motor all limit usable torque.

At zero speed, mechanical power is zero. A loaded, stationary gear train can still carry torque. The lab shows ideal static torque balance in that case; it does not model motor stall behavior or promise a safe load.

Supporting detail: Addendum 8 — Primary references and model boundaries.

Connect the gears to Angular Motion, Rotation and Linear Speed and Torque.

Stage efficiencies multiply; losses accumulate

Each stage receives only the power that survived the stage before it. Use decimal efficiencies and multiply: η_total = η₁ × η₂ × η₃. Never add or average them. See Power and Efficiency for the general meaning of efficiency.

At 98% per stage, three stages give 0.98³ = 0.941192: about 5.88% total loss, not 6%. Fewer, higher-quality stages often beat many cheap stages for efficiency, provided they can meet the ratio and load requirements.

Supporting detail: Addendum 9 — Why adding losses is only an approximation.

Positive drives keep the speed ratio; losses reduce torque

A positive drive keeps input and output linked in a fixed ratio without slipping. Gears, chains and timing belts are positive drives. A slipping flat belt or V-belt is not. Follow Rotation and Linear Speed for the connection from rotation to belt travel.

The teeth still set the speed exactly in this model. Efficiency reduces output torque from its ideal value. It does not reduce output rpm as well. Use matching torque units on both sides: lbf·ft or N·m. The course conversion is 1 rpm ≈ 0.1047 rad/s when angular speed in rad/s is needed.

Supporting detail: Addendum 10 — Follow both stages of one loaded train.

Count stages for ratio, meshes for direction and stated losses

For a simple train A → I → B, the idler cancels from (NI/NA) × (NB/NI) = NB/NA. The two external meshes still reverse direction twice.

Follow the loss assumption in the problem. Real meshes lose some power, but a problem may explicitly ignore it. If every actual mesh is 96% efficient, A → I → B has ηtotal = 0.96². If a whole stage including its idler is assigned 96%, use that stage value once. If η = 1.00 is stated, ignore losses. Do not infer the loss count from the reduction-stage count when an idler is present.