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Terramechanics and off-road mobility

Track thrust with finite slip

Thrust of one flat track on soil with uniform pressure, when the soil is sheared gradually along the contact.

Ref
TER-18
Rev
v1
Field
Terramechanics and off-road mobility
Result
H [kN]

For one flat track with uniform normal pressure and soil parameters, driving straight, the shear displacement along the contact is j = s x, where x is the distance from the front of the contact. Integrating the Janosi–Hanamoto law τ = τmax(1 − e−j/Ks) over the length gives H = (c bc ℓc + Wc tan φs)[1 − (Ks/(s ℓc))(1 − e−sℓc/Ks)]. As s → 0 the thrust tends to zero, and the slip must be greater than zero for the formula to be used. Ks is entered in mm. Cohesion in kPa, widths and lengths in m and the load in kN give the thrust in kN.

Thrust of one flat track on soil with uniform pressure, when the soil is sheared gradually along the contact. It rises towards the maximum soil thrust as slip grows and is smaller at low slip.

Calculator

Solve for
kN
kPa
020
m
0.251.2
m
27
kN
30600
°
545
mm
5100
0.010.6
  • Maximum (fully mobilised) thrust, c bc ℓc + Wc tan φs117.2 kN
  • Fraction of the maximum developed93.75 %
  • Shear displacement at the rear of the contact, s ℓc400 mm

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.

Track thrust (kN) current value 109.8 kN
WcLoad carried by this trackKey driver
20329
φsSoil friction angle
19.8188
sTrack slip
58.7116
cSoil cohesion
106144
KsShear deformation parameter
88.4116
ℓcTrack contact length
101116
bcTrack width
108115
  • At 10 % slip the track develops 93.8 % of its 117.2 kN maximum, 109.8 kN. At 2 % slip it develops only 70 % (82.04 kN) and at 40 % slip 115.3 kN.
  • The thrust reaches its plateau when s ℓc is a few times Ks. A long track reaches the plateau at a lower slip: halving the length to 2 m lowers the thrust at 10 % slip to 100.8 kN, because the soil at the front of the contact has sheared less.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (kN)Local stepChange in result
Wc Load carried by this track30 to 600 kN19.99328.5308.5+10%+9.66 %
φs Soil friction angle5 to 45 °19.83187.5167.7+10%+12.1 %
s Track slip0.01 to 0.6 58.72115.957.22+10%+0.606 %
c Soil cohesion0 to 20 kPa106.1143.637.5+10%+0.341 %
Ks Shear deformation parameter5 to 100 mm88.41115.727.29+10%−0.667 %
ℓc Track contact length2 to 7 m100.8115.915.1+10%+0.95 %
bc Track width0.25 to 1.2 m108115.17.125+10%+0.341 %

Sweep of Track slip

Track slipTrack thrust (kN)
0.0158.72
0.08375108.4
0.1575112.5
0.2313114
0.305114.8
0.3787115.2
0.4525115.5
0.5262115.8
0.6115.9

Sweep one variable

Track thrust (kN)Hover, touch, or use the arrow keys on the chart
60708090100110120 0.10.20.30.40.50.6 Track slip now 109.8

Variables and how each one contributes

H

Track thrust

Result · kN

What the equation gives you. Everything else feeds it.

c

Soil cohesion

kPa · typical 0 to 20

The soil cohesion, in kPa. Raises the maximum thrust through the area term bc ℓc c, a small part of the total at this load. Raising it by 10 % (to 2.2) moves the result by +0.341 %, from 109.8 to 110.2 kN.

bc

Track width

m · typical 0.25 to 1.2

The width of the track, in m. Raises only the cohesive part (c bc ℓc), so it has little effect on frictional soil. Raising it by 10 % (to 0.55) moves the result by +0.341 %, from 109.8 to 110.2 kN.

ℓc

Track contact length

m · typical 2 to 7

The length of the track in contact with the ground, in m. Raises the cohesive part and also the shear displacement s ℓc at the rear of the contact, so the soil is better mobilised. Raising it by 10 % (to 4.4) moves the result by +0.95 %, from 109.8 to 110.9 kN.

Wc

Load carried by this track

kN · typical 30 to 600

The load carried by this track, in kN. Raises the frictional part in direct proportion. It is the biggest term for frictional soil. Raising it by 10 % (to 215.6) moves the result by +9.66 %, from 109.8 to 120.4 kN.

φs

Soil friction angle

° · typical 5 to 45

The soil friction angle, in degrees. Raises the frictional part through its tangent. Raising it by 10 % (to 33) moves the result by +12.1 %, from 109.8 to 123.1 kN.

Ks

Shear deformation parameter

mm · typical 5 to 100

The shear deformation parameter of the soil, in mm. Lowers the thrust at a given slip, because the soil needs more shear displacement to develop its strength. Raising it by 10 % (to 27.5) moves the result by −0.667 %, from 109.8 to 109.1 kN.

s

Track slip

· typical 0.01 to 0.6

The track slip, from 0 to 1. Raises the thrust, quickly at first, then gradually as it approaches the maximum. Here the default slip is already near the plateau. Raising it by 10 % (to 0.11) moves the result by +0.606 %, from 109.8 to 110.5 kN.

Worked example with the values above

  1. Maximum thrust: c bc ℓc + Wc tan φs = 4 + 113.2 = 117.2 kN.
  2. Displacement ratio: s ℓc / Ks = 0.1 × 4 / 0.025 = 16. Bracket: 1 − (1 − e−16) / 16 = 0.9375.
  3. H = 117.2 × 0.9375 = 109.8 kN.

Assumptions and limits

  • One flat track with uniform normal pressure, uniform soil parameters, straight driving and j = s x along the contact.
  • A monotonic Janosi–Hanamoto soil law with no softening, and s greater than zero.

Common mistakes

  • Using zero slip. The formula has no value at s = 0, where the thrust is zero.
  • Using Ks in metres instead of millimetres.

Statistics and uncertainty

The thrust inherits the uncertainty of the maximum soil strength (±15 to 30 %) and of Ks (±30 to 50 %). At the plateau only the first one matters, but at low slip the term in Ks dominates and the thrust varies nearly in direct proportion to s ℓc / Ks. The assumption of uniform pressure is also optimistic, since real tracks have pressure peaks under the road wheels.

Where it is used

  • Estimating track thrust at a realistic slip rather than assuming the full soil strength.
  • Choosing the track length and the slip at which a tracked vehicle can deliver a given drawbar pull.

Maths you need

  • Integration of an exponential: ∫ (1 − e^(−x/K)) dx
  • Mohr–Coulomb strength: maximum shear stress

Related formulas

Practise this formula