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Driveline and transmission
Minimum friction coefficient to climb a slope
Least tyre or track friction needed to climb a steady slope when every wheel is driven and carries its share of the weight.
- Ref
- DRV-06
- Rev
- v1
- Field
- Driveline and transmission
- Result
- μreq
If every wheel is driven and carries its share of the weight, the traction force the ground can supply is μ m g cos θ. The climbing force is m g (f cos θ + sin θ). Dividing, the least friction needed is μreq = f + tan θ, independent of the vehicle mass. If only some wheels drive, the load on the driven wheels is only part of the weight and the requirement is higher. The friction coefficient model is a hard-ground approximation; on deformable terrain use a soil model.
Least tyre or track friction needed to climb a steady slope when every wheel is driven and carries its share of the weight. The slope contributes tan θ and rolling resistance adds f.
Calculator
- Grade, 100 tan θ30.57 %
- Largest slope for μ = 0.6 (dry road or firm soil)29.68 °
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.
- A 17° slope (30.6 % grade) with f = 0.03 needs a friction coefficient of at least 0.3357. That is more than wet grass or loose gravel typically offers (about 0.3 to 0.4), so the vehicle slips even if the gearing could deliver the force.
- On a dry road (μ about 0.8) the slope limit is 37.6°; on ice (μ about 0.1) it is only 4°.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing () | Local step | Change in result |
|---|---|---|---|---|---|---|
| θ Slope angle | 0 to 40 ° | 0.03 | 0.8691 | 0.8391 | +10% | +9.76 % |
| f Rolling resistance coefficient | 0.01 to 0.2 | 0.3157 | 0.5057 | 0.19 | +10% | +0.894 % |
Sweep of Slope angle
| Slope angle (°) | Friction coefficient required |
|---|---|
| 0 | 0.03 |
| 5 | 0.1175 |
| 10 | 0.2063 |
| 15 | 0.2979 |
| 20 | 0.394 |
| 25 | 0.4963 |
| 30 | 0.6074 |
| 35 | 0.7302 |
| 40 | 0.8691 |
Sweep one variable
Variables and how each one contributes
Friction coefficient required
What the equation gives you. Everything else feeds it.
Rolling resistance coefficient
The rolling resistance coefficient, a plain number, about 0.01 to 0.2. Raises the friction needed by the same amount, one for one. Raising it by 0.01 (to 0.04) moves the result by +2.98 %, from 0.3357 to 0.3457.
Slope angle
The slope angle in degrees, up to 40°. A 30 % grade is 16.7°. Raises the friction needed, more steeply as the slope increases, because tan θ grows faster than θ. Raising it by 5° (to 22°) moves the result by +29.3 %, from 0.3357 to 0.434.
Worked example with the values above
- Slope term: tan θ = tan 17° = 0.3057.
- Add the rolling resistance: 0.03 + 0.3057.
- μreq = 0.3357.
Assumptions and limits
- Every wheel driven, each carrying its share of the normal load.
- A steady slope, with negligible air drag and acceleration.
Common mistakes
- Applying the all-wheel-drive result to a vehicle with only some wheels driven.
- Using the slope angle in the formula as if it were a percentage grade.
Statistics and uncertainty
Friction coefficients vary widely with the surface, moisture and tyre, often by ±30 %. Because the requirement is almost linear in tan θ, a margin of 20 to 30 % between the available and required friction is a sensible minimum.
Where it is used
- Checking that the ground can supply the force the gearing could deliver.
- Judging which surfaces a vehicle can climb.
Maths you need
- Friction force: μ times the normal force
- Tangent of an angle: rise over run