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Terramechanics and off-road mobility

Idealised track steering resistance

Resisting yaw moment when a tracked vehicle pivots on the spot, assuming uniform pressure and fully developed sideways (Coulomb) sliding of both tracks.

Ref
TER-19
Rev
v1
Field
Terramechanics and off-road mobility
Result
Mres [kN·m]

μy is the lateral friction coefficient of the track on the ground, W the total weight and ℓc the track contact length. For symmetric pivoting with uniform pressure over two equal rectangular tracks and fully developed lateral Coulomb sliding, the resisting yaw moment has the magnitude μy W ℓc/4. The moment balance for the vehicle is Iz ṙ = (B/2)(HR − HL) − Mres, so the steering thrust difference must exceed Mres to start the turn. Soft soil and turns of finite radius need a distributed shear model.

Resisting yaw moment when a tracked vehicle pivots on the spot, assuming uniform pressure and fully developed sideways (Coulomb) sliding of both tracks. It is the moment the track thrust difference must overcome.

Calculator

Solve for
kN·m
0.11
kN
30600
m
27
  • Resisting moment in newton-metres98,000 N·m
  • Moment relative to weight times length, Mres / (W ℓc)0.125
  • Thrust difference needed for a 2 m track spacing, 2Mres / B98 kN

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.

Steering resistance moment (kN·m) current value 98 kN·m
WTotal vehicle weightKey driver
15300
μyLateral friction coefficient
19.6196
ℓcTrack contact length
49172
  • A 20 t vehicle (196 kN) on tracks 4 m long, with μy = 0.5, needs a yaw moment of 98 kN·m to pivot. With a 2 m track spacing the two tracks must differ by 98 kN of thrust, about 50 % of the weight.
  • Steering resistance rises with track length, so a vehicle with tracks of 5 m instead of 4 m needs 122.5 kN·m. This is why the length-to-width ratio of a tracked vehicle is kept below about 1.8 to 2: the longer it is, the harder it is to steer.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (kN·m)Local stepChange in result
W Total vehicle weight30 to 600 kN15300285+10%+10 %
μy Lateral friction coefficient0.1 to 1 19.6196176.4+10%+10 %
ℓc Track contact length2 to 7 m49171.5122.5+10%+10 %

Sweep of Track contact length

Track contact length (m)Steering resistance moment (kN·m)
249
2.62564.31
3.2579.63
3.87594.94
4.5110.3
5.125125.6
5.75140.9
6.375156.2
7171.5

Sweep one variable

Steering resistance moment (kN·m)Hover, touch, or use the arrow keys on the chart
406080100120140160180 234567 Track contact length (m) now 98

Variables and how each one contributes

Mres

Steering resistance moment

Result · kN·m

What the equation gives you. Everything else feeds it.

μy

Lateral friction coefficient

· typical 0.1 to 1

The lateral friction coefficient of the track on the ground: about 0.4 to 0.9 on firm ground, and more for tracks with grousers. Directly proportional: a high-grip surface resists pivoting more. Raising it by 10 % (to 0.55) moves the result by +10 %, from 98 to 107.8 kN·m.

W

Total vehicle weight

kN · typical 30 to 600

The total weight of the vehicle, in kN. Directly proportional. Raising it by 10 % (to 215.6) moves the result by +10 %, from 98 to 107.8 kN·m.

ℓc

Track contact length

m · typical 2 to 7

The length of each track in contact with the ground, in m. Directly proportional: long tracks must be dragged sideways through more ground, which is why long tracked vehicles are hard to pivot. Raising it by 10 % (to 4.4) moves the result by +10 %, from 98 to 107.8 kN·m.

Worked example with the values above

  1. Weight times length: W ℓc = 196 × 4 = 784 kN·m.
  2. Multiply by the friction coefficient: μy W ℓc = 0.5 × 784 = 392 kN·m.
  3. Mres = 392 / 4 = 98 kN·m.

Assumptions and limits

  • Symmetric pivot turn with uniform pressure over two equal rectangular tracks, and fully developed lateral Coulomb sliding.
  • Firm ground without bulldozing or sinkage effects, with the weight evenly carried by the tracks.

Common mistakes

  • Using the formula for a turn of large radius. It applies to the pivot turn, and a finite radius needs a smaller moment.
  • Forgetting that the steering thrust difference acts at half the track spacing: the force needed is 2Mres / B.

Statistics and uncertainty

The lateral friction coefficient dominates the uncertainty: it varies by ±20 to 40 % with the surface, vegetation and the soil tendency to build up in front of the track side, and it changes during a turn. The pressure distribution is also not uniform. The calculated moment is therefore best treated as an estimate within a factor of 1.5, adequate to size the steering drive but not to predict a particular turn.

Where it is used

  • Sizing steering drive power and judging whether a heavy tracked vehicle can pivot on a given surface.
  • Comparing track length–width proportions for manoeuvrability.

Maths you need

  • Coulomb friction: sliding friction force μ N
  • Moment of a distributed force: integrating force × distance

Related formulas

Practise this formula