Overdrive and its operation

Why it matters

If a car's top gear is chosen for maximum speed, the engine turns faster than necessary at ordinary cruising speeds, wasting fuel and making noise. An overdrive – a ratio in which the gearbox output turns faster than the engine – lets the engine cruise slowly and at higher load, where it is more efficient. Knowing how overdrive units work and what they cost in pulling power is part of every transmission course.

Key ideas

• Definition. Overdrive is any gear ratio below 1:1 (output faster than input) between engine and propeller shaft. Today it is usually built into the gearbox as a fifth or sixth gear with a ratio of roughly 0.7–0.85 (and automatics' top two or three gears). Historically it was a separate overdrive unit bolted behind a four-speed gearbox.
• Why it saves fuel. At a given road speed and power demand, a lower engine speed means a wider throttle opening (smaller pumping losses in a petrol engine), lower friction losses and operation nearer the engine's best specific fuel consumption. Noise, vibration and wear also fall.
• Price paid. Tractive effort at the wheels falls in proportion to the ratio, so surplus power for overtaking, gradients or towing is smaller; the driver (or controller) must change down more often. Overdrive is therefore a cruising gear, not a performance gear. Top speed is normally reached in the direct (or next-lower) gear, where the engine can reach its maximum-power speed.
• Epicyclic overdrive unit (Laycock de Normanville type). A simple epicyclic train:
  • The planet carrier is driven by the gearbox main shaft (input); the ring (annulus) is the output; the sun gear can be locked to the ring or held to the casing by a double-sided cone clutch. A one-way (roller) clutch between the input shaft and the annulus is also fitted.
  • Direct drive (overdrive off): springs hold the cone clutch against the annulus, locking sun to annulus, so the whole train turns as one (ratio 1). In normal drive the one-way clutch also carries the drive; when coasting, the locked cone clutch transmits engine braking.
  • Overdrive on: an electrically controlled solenoid valve admits oil pressure (from a pump driven off the input shaft) to pistons that move the cone clutch onto a brake ring in the casing, holding the sun stationary. With carrier input, ring output and sun held, the ring overspeeds the carrier; the one-way clutch simply overruns.
  • Ratio = Z_R / (Z_S + Z_R), typically about 0.75–0.8.
  • Controls: a driver's switch, with inhibitor switches so it can work only in top (and sometimes third) gear and not in reverse, plus a kickdown/throttle switch in some versions. Changes are made without the clutch pedal because the unit has its own clutch.
• Gearbox-integrated overdrive. In a five- or six-speed manual, the overdrive gear is simply a gear pair in which the layshaft gear is larger than the main-shaft gear (overall gearbox ratio below 1). In automatics, it is obtained by driving a planetary carrier while holding the sun, as above, and the TCU selects it automatically (some older automatics had an "O/D off" button to lock it out when towing or on hills).
• Matching. The final-drive ratio is often chosen so the direct gear gives maximum vehicle speed at maximum-power engine speed, and the overdrive then gives low cruising engine speed. Some designs instead use overdrive top gear to reach maximum speed with a numerically higher final drive.

Formulas

i_OD = N_in / N_out < 1

• i_OD: overdrive ratio (input speed ÷ output speed, –); N_in, N_out: input and output speeds (rev/min).

i_OD = Z_R / (Z_S + Z_R)

• Epicyclic overdrive with carrier input, ring output, sun held. Z_S, Z_R: teeth on sun and ring.

N_e = N_w · i_g · i_OD · i_f, with N_w = 60 · v / (2π · r)

• N_e: engine speed (rev/min); N_w: wheel speed (rev/min); v: vehicle speed (m/s); r: rolling radius (m); i_g: gearbox ratio; i_f: final-drive ratio.

F_t = T_e · i_g · i_OD · i_f · η / r

• Tractive effort (N) in overdrive; T_e engine torque (N·m); η transmission efficiency (–).

T_out = T_in · i_OD (ideal) and T_h = T_in − T_out

• Output torque falls in the same ratio as speed rises; T_h is the reaction torque on the held sun gear (N·m).

Worked examples

Example 1 (standard). An epicyclic overdrive has a sun with 18 teeth and an annulus with 66 teeth. It is fitted behind a gearbox whose top gear is direct (1:1); the final drive is 4.1 and the rolling radius 0.30 m. Find the overdrive ratio and the engine speed at 100 km/h with overdrive off and on.

1. i_OD = 66 / (18 + 66) = 0.786.
2. Wheel speed: N_w = 60 × (100/3.6) / (2π × 0.30) = 884.2 rev/min.
3. Overdrive off: N_e = 884.2 × 1 × 4.1 = 3625 rev/min.
4. Overdrive on: N_e = 3625 × 0.786 = 2848 rev/min.

Answer: i_OD ≈ 0.786; engine speed falls from about 3625 to 2848 rev/min (a 21% reduction).

Example 2 (GATE level). The car in Example 1 has mass 1200 kg, f = 0.015, C_d·A = 0.66 m², ρ = 1.2 kg/m³ and driveline efficiency 0.90. Assume the engine delivers 140 N·m at both engine speeds found above. Can it hold a steady 100 km/h on (a) a 6% grade and (b) an 8% grade in overdrive and in direct top gear?

1. Tractive effort, overdrive: F_t = 140 × 0.786 × 4.1 × 0.90 / 0.30 = 1353 N; direct: F_t = 140 × 4.1 × 0.90 / 0.30 = 1722 N.
2. Air resistance at 27.78 m/s: R_a = 0.5 × 1.2 × 0.66 × 27.78² = 305.6 N.
3. 6% grade (θ = 3.43°): R_r = 176.3 N, R_g = 705.1 N, total = 1186.9 N.
4. 8% grade (θ = 4.57°): R_r = 176.0 N, R_g = 938.8 N, total = 1420.3 N.
5. Compare: on 6%, both 1353 N and 1722 N exceed 1187 N. On 8%, 1353 N < 1420 N but 1722 N > 1420 N.

Answer: On the 6% grade the car holds 100 km/h in either gear. On the 8% grade it slows in overdrive (about 67 N short) and must change down to direct top. This is the practical meaning of "overdrive reduces tractive effort".

Common mistakes

• Defining ratio as output ÷ input in one step and input ÷ output in the next. With the usual convention, overdrive ratios are below 1 and output speed = input speed ÷ ratio.
• Calling any "high gear" an overdrive; direct top (1:1) is not overdrive.
• Using the ring-held formula 1 + Z_R/Z_S for an overdrive unit; with sun held, carrier in and ring out, the ratio is Z_R/(Z_S + Z_R).
• Thinking overdrive increases power. It only lowers engine speed; available tractive effort falls.
• Leaving overdrive engaged while towing or climbing long grades, which causes lugging or frequent hunting between gears.

For GATE ME

Expect numericals on overdrive ratio from epicyclic tooth numbers, engine speed at a given road speed with and without overdrive, and tractive effort or gradeability in overdrive versus direct drive. Short theory questions ask why overdrive saves fuel and how the cone clutch and one-way clutch work in an overdrive unit. Practise combining this with the tractive-effort and epicyclic-train topics.

Quick check

1. An overdrive has an input speed of 2800 rev/min and ratio 0.7. What is its output speed?
2. Which member is held in an epicyclic overdrive unit?
3. Sun 20 teeth, ring 60 teeth: what is the overdrive ratio?
4. Why is a separate overdrive unit usually inhibited in the lower gears?
5. Does overdrive raise or lower the tractive effort at a given engine torque?

Answers: 1. 2800 / 0.7 = 4000 rev/min. 2. The sun gear (via the cone clutch on the brake ring). 3. 60/80 = 0.75. 4. In the lower gears the torque passing through the unit would be far higher than its cone clutch and gears are designed for, so it is wired to work only in the upper gears. 5. Lowers it, in proportion to the ratio.