Sliding mesh, constant mesh and synchromesh gearboxes

Why it matters

An engine produces useful torque only over a limited speed band, while a vehicle must move from walking pace to highway speed and climb hills. The manual gearbox provides a set of torque-multiplying ratios plus neutral and reverse. How those ratios are engaged – by sliding gears, by dog clutches or through synchronisers – decides how easy, quick and quiet gear changes are, and is the evolutionary story every automotive engineer is expected to explain.

Key ideas

• Layout of a layshaft (countershaft) gearbox. Three parallel elements: the primary (clutch) shaft, which carries the main drive pinion; the layshaft, driven permanently from that pinion through the constant-mesh pair; and the main (output) shaft, in line with the primary shaft. Each indirect gear passes torque through two meshes (primary → layshaft → main shaft). In top gear the primary and main shafts are locked together for a direct drive (ratio 1:1, highest efficiency). Reverse adds an idler gear between a layshaft gear and a main-shaft gear; the idler reverses direction but does not change the ratio.
• Sliding-mesh gearbox. The main-shaft gears are spur gears splined to the shaft and are slid by selector forks into mesh with the layshaft gears. Simple and cheap, but the teeth themselves must be brought into mesh while turning at different peripheral speeds, so changes are noisy, slow and damage tooth ends unless speeds are matched by double declutching (upshift: declutch to neutral, re-engage clutch and let the layshaft slow, declutch again and select; downshift: same, but blip the throttle in neutral to speed the layshaft up). Spur teeth are needed because helical teeth cannot easily be slid into mesh. Now found only in some tractors and in reverse gears.
• Constant-mesh gearbox. All forward gear pairs are always in mesh. The main-shaft gears run free on bearings, and a gear is selected by sliding a dog clutch (splined to the shaft) into engagement with dogs on the side of the gear. Because the gears never slide, they can be helical – quieter and stronger – and damage is limited to the dogs. Speeds of the dog clutch and gear still have to be matched, so double declutching is still needed for clean changes. Used in heavy trucks and motorcycles (where sequential dog boxes are quick).
• Synchromesh gearbox. A constant-mesh box in which each dog clutch has a synchroniser: a friction cone (or multi-cone) that first brings the gear and shaft to the same speed, and only then lets the toothed sleeve slide over the dog teeth. In the common baulk-ring (blocker-ring) synchroniser, the friction torque on the cone holds the ring indexed so its chamfered teeth block the sleeve until the speeds are equal; the speed difference then falls to zero and the sleeve passes through. Synchromesh on all forward gears is now universal in cars; reverse is often unsynchronised (select only at rest).
• Selector mechanism. Selector rails and forks move the sliding gears or sleeves; detent balls and springs hold each position, and an interlock prevents two gears being selected at once (covered further in the gear-shifting topic).
• Synchroniser sizing. Cone torque depends on the axial shift force, the friction coefficient, the mean cone radius and the cone angle (a small cone angle multiplies the normal force). Synchronising time equals the inertia to be speeded up or slowed (clutch disc, primary shaft, layshaft and free gears, referred to the synchroniser) times the speed difference, divided by the cone torque.
• Efficiency. Each mesh costs roughly 1–2% of power; indirect gears (two meshes) are slightly less efficient than direct top gear.

Formulas

i = (z_L1 / z_P) · (z_M / z_L)

• i: gear ratio for an indirect gear (input speed ÷ output speed, –); z_P: teeth on the primary (main drive) pinion; z_L1: teeth on the layshaft constant-mesh gear; z_L: teeth on the layshaft gear of the selected pair; z_M: teeth on the main-shaft gear of that pair. An idler in the reverse train does not appear in the ratio.

N_out = N_in / i, T_out = T_in · i · η

• N: speed (rev/min); T: torque (N·m); η: gearbox efficiency in that gear (–).

T_c = μ · F_a · r_m / sin α

• T_c: synchroniser cone friction torque (N·m); μ: cone friction coefficient (–, about 0.08–0.12 for brass or carbon in oil); F_a: axial force on the sleeve (N); r_m: mean cone radius (m); α: semi-cone angle (typically about 6–7°).

t_s = I_e · Δω / T_c

• t_s: synchronising time (s); I_e: equivalent moment of inertia referred to the synchroniser (kg·m²); Δω: initial speed difference (rad/s), with Δω = 2π·ΔN/60. Drag torques are neglected.

Worked examples

Example 1 (standard). A four-speed layshaft gearbox has a 20-tooth main drive pinion meshing with a 35-tooth layshaft gear. The other pairs (layshaft/main shaft) are: first 14/28, second 20/25, third 26/21; fourth is direct. Reverse uses a 12-tooth layshaft gear, an idler and a 30-tooth main-shaft gear. Find all ratios and the output speed in each gear at an engine speed of 3000 rev/min.

1. Constant-mesh ratio: 35/20 = 1.75.
2. First: i₁ = 1.75 × 28/14 = 3.50; N = 3000/3.50 = 857 rev/min.
3. Second: i₂ = 1.75 × 25/20 = 2.19; N = 3000/2.1875 = 1371 rev/min.
4. Third: i₃ = 1.75 × 21/26 = 1.41; N = 3000/1.4135 = 2122 rev/min.
5. Fourth: i₄ = 1.00; N = 3000 rev/min.
6. Reverse: i_R = 1.75 × 30/12 = 4.375 (idler ignored); N = 3000/4.375 = 686 rev/min, opposite direction.

Answer: ratios 3.50, 2.19, 1.41, 1.00 and 4.375 (reverse); output speeds 857, 1371, 2122, 3000 and 686 rev/min.

Example 2 (GATE level). During a 2–3 upshift, the speed difference across the third-gear synchroniser is 600 rev/min. The equivalent inertia to be slowed is 0.012 kg·m². The cone has mean radius 30 mm, semi-cone angle 7° and μ = 0.10; the driver applies an axial force of 300 N at the sleeve. Find the cone torque and synchronising time.

1. T_c = μ · F_a · r_m / sin α = 0.10 × 300 × 0.030 / sin 7° = 0.90 / 0.12187 = 7.385 N·m.
2. Δω = 2π × 600 / 60 = 62.83 rad/s.
3. t_s = I_e · Δω / T_c = 0.012 × 62.83 / 7.385 = 0.102 s.

Answer: T_c ≈ 7.4 N·m; t_s ≈ 0.10 s. Halving the shift force doubles the synchronising time, which is why heavy-truck gearboxes use double or triple cones.

Common mistakes

• Writing the ratio as driver teeth over driven teeth; ratio is driven/driver for each mesh, multiplied through the train.
• Forgetting the constant-mesh pair in a layshaft gearbox, or counting the reverse idler in the ratio.
• Thinking a constant-mesh gearbox needs no speed matching: the dogs still clash unless speeds are matched by double declutching.
• Saying a synchromesh gearbox is a different gear layout; it is a constant-mesh box with synchronisers added.
• Using sin α with α in degrees in a calculator set to radians, or using the full cone angle instead of the semi-angle.
• Assuming top gear is always the direct drive; many front-wheel-drive two-shaft gearboxes have no direct drive at all.

For GATE ME

Expect gear-train numericals: overall ratio of a layshaft gearbox from tooth numbers, output speed and torque in a given gear, the effect of the idler, and occasionally cone-clutch (synchroniser) torque and synchronising time. Conceptual questions compare sliding-mesh, constant-mesh and synchromesh boxes, ask why helical gears suit constant-mesh designs, and test double declutching. Practise multiplying ratios through two meshes without inverting them.

Quick check

1. A pinion of 18 teeth drives a layshaft gear of 36; the selected pair is 15 (layshaft) to 30 (main shaft). What is the gear ratio?
2. Why can sliding-mesh gearboxes not use helical gears easily?
3. What does the baulk ring do in a synchroniser?
4. Does a reverse idler change the magnitude of the ratio?
5. Why is direct top gear the most efficient?

Answers: 1. (36/18) × (30/15) = 4.0. 2. Helical teeth would have to slide and rotate together to enter mesh. 3. It blocks the sleeve until friction on its cone has equalised the speeds of gear and shaft. 4. No, only the direction. 5. Torque passes straight through without any gear mesh.