Cornering Physics
How radius, grip and speed decide whether a steady turn is possible.
Physical principle
To follow a curve, the car must accelerate toward the centre. Only tyre friction can provide that force, and a simple Coulomb model caps it at μ times the car's weight. Speed at which demand equals supply is the steady-state cornering limit.
Equations
a_lat = v² / Ra_lat,max = μ gv_max = √(μ g R)- v
- speed (m/s; shown in mph)
- R
- corner radius (m)
- μ
- effective tyre–road friction
- g
- 9.81 m/s²
Your results
At R = 60 m and μ = 1.00, the tyres can supply 1.00 g, so v_max = 54.3 mph. At 50 mph the corner needs 0.85 g — 85% of the grip.
Doubling the radius to 120 m would raise v_max by 41% to 76.8 mph — not 100%, because v_max grows with √R. Meanwhile doubling speed would quadruple the lateral demand.
Try this
- Increase the corner radius while keeping μ constant. Watch the marker climb the √R curve — each extra metre is worth less speed.
- Find the speed where utilization is exactly 100%, then add 10% to the speed. Utilization jumps by ≈ 21%, because a_lat ∝ v².
- Drop μ from 1.0 to 0.5 (a wet road). How much radius do you need to keep the same v_max?
Model limits: flat, unbanked road; constant μ; point-mass car; no aerodynamic downforce; steady state only. A real car at the limit transfers load, its tyres slip progressively and its handling depends on suspension and driver input — none of which this experiment simulates.