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Centrifugal Force & Lateral G-Force Calculator

Calculate cornering centrifugal force in pounds and Newtons, lateral acceleration in G-forces, roadway banking angles, and tire friction grip requirements.

Calculated Results

Lateral Acceleration (G-Force)
0.60 Gs (Aggressive Cornering 🏎️)
Centripetal Force: 2,283 lbs-force (10,155 Newtons)
Centripetal Force (lbf)
2,283 lbf (10.15 kN)
Tire Grip Demand (Friction μ)
μ = 0.60 (Dry Asphalt Standard)
Neutral Steer Speed (for Bank)
0.0 MPH (Flat Turn)
Velocity in ft/s and m/s
88.0 ft/s (26.8 m/s)
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Frequently Asked Questions

What is the difference between centrifugal and centripetal force?

Centripetal force is the real inward force (provided by tire friction or roadway banking) required to pull an object toward the center of a circular turn ($F = mv^2 / r$). Centrifugal force is the fictitious outward inertia experienced by occupants inside the rotating reference frame.

What is lateral G-force and what are standard vehicle limits?

1.0 G equals Earth's gravitational acceleration (9.81 m/s²). Standard passenger cars lose tire grip around 0.70 to 0.85 Gs; high-performance sports cars corner at 1.0 to 1.15 Gs on sticky summer tires; Formula 1 cars exceed 4.0 to 5.0 lateral Gs due to aerodynamic downforce.

How does highway curve banking eliminate lateral sliding?

Super-elevation (banking angle θ) tilts the normal force inward. At the 'ideal neutral speed' $v = \sqrt{r \cdot g \cdot an heta}$, the inward component of vehicle weight perfectly balances centripetal acceleration, meaning tires require zero lateral friction to stay on course.

Why does doubling cornering speed quadruple the force?

Because velocity is squared in the centripetal equation ($F = m v^2 / r$). Entering a curve at 60 MPH instead of 30 MPH quadruples the cornering force required from your tires to avoid sliding off the roadway.