Library / Driveline and transmission / DRV-07

Driveline and transmission

Gearbox ratio span

Ratio between the first and top gear ratios needed to climb the steepest slope and still reach the top speed.

Ref: DRV-07
Rev: v1
Field: Driveline and transmission
Result: S

S = \frac{i_{1}}{i_{top}} = \frac{F_{climb} v_{\max}}{T_{\max} ω_{top} η}

The span S is the ratio of the largest gear ratio (first) to the smallest (top). Using i₁ = F_climb r/(T_max η) and i_top = ω_top r/v_max, the rolling radius r cancels, leaving S = F_climb v_max/(T_max ω_top η). The product T_max ω_top is only a sizing product; it need not be an actual engine operating point. The assumptions are the same radius and the same efficiency in both calculations.

Ratio between the first and top gear ratios needed to climb the steepest slope and still reach the top speed. The wheel radius cancels, so the span depends on force, speed, torque and engine speed only.

Calculator

Solve for

SRatio spanComputed

F_climbClimbing force

5100

v_maxTarget top speed

m/s

580

T_maxMaximum engine torque

N m

2004,000

ω_topEngine speed at top speed

rad/s

100600

ηDriveline efficiency

0.51

Minimum first-gear ratio at r = 0.55 m14.26
Top-gear ratio at r = 0.55 m5.06

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Ratio span () current value 2.818

T_maxMaximum engine torqueKey driver

1.0621.1

v_maxTarget top speed

0.5649.02

F_climbClimbing force

0.4038.05

ω_topEngine speed at top speed

1.086.48

ηDriveline efficiency

2.545.07

A climb force of 35 kN, a top speed of 25 m/s and an engine of 1,500 N m at 230 rad/s give a span of 2.818. With a rolling radius of 0.55 m that is a first gear near 14.3 and a top near 5.07: the radius does not change the span, only the individual ratios.
A wider span needs more gears or a larger step. A narrower span, from more engine torque or a lower top speed, could use a simpler gearbox.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing ()	Local step	Change in result
T_max Maximum engine torque	200 to 4,000 N m	1.057	21.14	20.08	+10%	−9.09 %
v_max Target top speed	5 to 80 m/s	0.5636	9.018	8.454	+10%	+10 %
F_climb Climbing force	5 to 100 kN	0.4026	8.052	7.649	+10%	+10 %
ω_top Engine speed at top speed	100 to 600 rad/s	1.08	6.481	5.401	+10%	−9.09 %
η Driveline efficiency	0.5 to 1	2.536	5.072	2.536	+10%	−9.09 %

Sweep of Climbing force

Climbing force (kN)	Ratio span
5	0.4026
16.88	1.359
28.75	2.315
40.63	3.271
52.5	4.227
64.38	5.183
76.25	6.139
88.13	7.095
100	8.052

Sweep one variable

Vary

Ratio span ()Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

Ratio span

Result ·

What the equation gives you. Everything else feeds it.

F_climb

Climbing force

kN · typical 5 to 100

The climbing force, in kN. Raises the span in proportion: a steeper climb needs a lower first gear. Raising it by 10 % (to 38.5) moves the result by +10 %, from 2.818 to 3.1.

v_max

Target top speed

m/s · typical 5 to 80

The target top speed, in m/s. Raises the span in proportion: a higher top speed needs a lower top-gear ratio. Raising it by 10 % (to 27.5) moves the result by +10 %, from 2.818 to 3.1.

T_max

Maximum engine torque

N m · typical 200 to 4,000

The engine’s maximum torque, in N m. Lowers the span in inverse proportion: a stronger engine needs a narrower range. Raising it by 10 % (to 1,650) moves the result by −9.09 %, from 2.818 to 2.562.

ω_top

Engine speed at top speed

rad/s · typical 100 to 600

The engine angular speed at top speed, in rad/s. 230 rad/s is about 2200 rpm. Lowers the span in inverse proportion: a higher engine speed at top speed shrinks the span. Raising it by 10 % (to 253) moves the result by −9.09 %, from 2.818 to 2.562.

Driveline efficiency

· typical 0.5 to 1

The driveline efficiency, between 0 and 1. Lowers the span slightly: a more efficient driveline needs less range. Raising it by 0.05 (to 0.95) moves the result by −5.26 %, from 2.818 to 2.67.

Worked example with the values above

Numerator: F_climb v_max = 35,000 N × 25 m/s = 875,000 W.
Denominator: T_max ω_top η = 1,500 × 230 × 0.9 = 310,500 W.
S = 2.818.

Assumptions and limits

The same rolling radius and the same efficiency for first and top gear.
Steady low-speed climbing for the first gear and a steady top-speed condition for the top gear.

Common mistakes

Treating T_max ω_top as the engine’s actual power at top speed.
Using different radii in the two calculations.

Statistics and uncertainty

The span is a ratio of products of uncertain quantities, so relative errors combine in quadrature: 10 % errors in force, torque and efficiency, and 5 % in speed and engine speed, give about ±17 % in S. Choose a few per cent extra span.

Where it is used

First step in deciding how many gears a gearbox needs.
Comparing the gearing needs of two vehicles.

Maths you need

Overall ratio: engine speed over wheel speed
Torque and power: power equals torque times angular speed

Related formulas

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