Banking of Road Formula: tan θ = v²/rg, Angle of Banking

The banking of road formula is $\tan\theta = \frac{v^2}{rg}$ , where θ is the angle the road surface makes with the horizontal, v is the speed, r is the radius of the curve and g is 9.8 m/s². A road tilted at this angle lets a vehicle take the bend at speed v without relying on friction at all.

On a flat curve, only friction between tyres and road can supply the inward centripetal force. Friction falls on wet or icy roads, so engineers raise the outer edge. The tilted road pushes on the car at an angle, and the horizontal part of that push does the job friction would otherwise have to do. The same idea tilts railway tracks, cycle velodromes and the path of an aircraft in a turn.

Quick facts

Formula
$\tan\theta = \frac{v^2}{rg}$
Design speed
$v = \sqrt{rg\tan\theta}$
Max speed with friction
$v_{max} = \sqrt{\frac{rg(\mu + \tan\theta)}{1 - \mu\tan\theta}}$
Mass
Cancels, so the angle suits every vehicle at that speed
Angle unit
degree or radian; tan θ has no unit
Syllabus
Class 11 laws of motion, JEE, NEET

Deriving the banking of road formula

Take a car of mass m on a frictionless road banked at θ. Two forces act: the weight mg straight down and the normal reaction N at right angles to the road. Resolve N into a vertical part $N\cos\theta$ and a horizontal part $N\sin\theta$ pointing toward the centre of the curve.

Vertical balance: $N\cos\theta = mg$ .
Horizontal, toward the centre: $N\sin\theta = \frac{mv^2}{r}$ .
Divide the second by the first: $\tan\theta = \frac{v^2}{rg}$ .

The mass has gone, which is why one bank angle serves a bicycle and a lorry alike, provided they travel at the design speed.

State the formula for angle of banking: with and without friction

Banked curve results
Case	Formula	Meaning
No friction	$\tan\theta = \frac{v^2}{rg}$	Ideal angle for one speed
With friction μ, fastest safe speed	$v_{max}^2 = \frac{rg(\mu + \tan\theta)}{1 - \mu\tan\theta}$	Friction acts down the slope
With friction μ, slowest safe speed	$v_{min}^2 = \frac{rg(\tan\theta - \mu)}{1 + \mu\tan\theta}$	Friction acts up the slope
Flat road (θ = 0)	$v_{max} = \sqrt{\mu rg}$	Only friction holds the car

If a board question asks you to state the formula of angle of banking, write $\theta = \tan^{-1}\left(\frac{v^2}{rg}\right)$ and name every symbol with its unit. Mention that it assumes no friction.

Worked examples

Example 1: finding the angle

A curve of radius 100 m is designed for 20 m/s. Find the angle of banking. Take g = 9.8 m/s².

$\tan\theta = \frac{20^2}{100 \times 9.8} = \frac{400}{980} \approx 0.408$

$\theta = \tan^{-1}(0.408) \approx 22.2^\circ$ . The calculator above returns the same angle.

Example 2: finding the radius, with km/h

A track is banked at 10° for vehicles moving at 54 km/h. What radius of curve is needed?

Speed in SI: 54 km/h becomes 15 m/s. Also, tan 10° ≈ 0.1763.

$r = \frac{v^2}{g\tan\theta} = \frac{225}{9.8 \times 0.1763} \approx 130\ \text{m}$

Example 3: top speed with friction, exam style

A curve of radius 50 m is banked at 15°. The coefficient of friction is 0.3. Find the maximum safe speed.

tan 15° ≈ 0.268. Numerator: $50 \times 9.8 \times (0.3 + 0.268) = 490 \times 0.568 \approx 278.3$ .

Denominator: $1 - 0.3 \times 0.268 \approx 0.920$ . So $v_{max}^2 \approx 302.6\ \text{m}^2/\text{s}^2$ .

$v_{max} \approx 17.4\ \text{m/s} \approx 62.6\ \text{km/h}$

What happens away from the design speed

Go faster than $\sqrt{rg\tan\theta}$ and the horizontal part of N is too small; the car tends to slide up and out, and friction must act down the slope. Go slower and the car tends to slide down toward the inside, so friction acts up the slope. The two friction formulas in the table above give the band of speeds that are safe for a given μ. Friction values are covered on the coefficient of static friction page.

A cyclist leaning into a turn obeys the same rule: the lean angle from the vertical satisfies tan θ = v²/(rg).

Slips that lose marks

Working in km/h. Convert to m/s before squaring.
Using the diameter of the curve as r.
Calculator in radian mode when the answer should be in degrees.
Treating the bank angle as the slope along the road. θ is measured across the road, from the inner edge up to the outer edge.
Swapping sin θ and cos θ when resolving N. The vertical part balances weight, so it is N cos θ.
Forgetting that the formula assumes zero friction unless the question says otherwise.

A rider who feels pushed outward on the curve is describing the centrifugal force of the turning frame. For the inward force itself see the centripetal force formula, and for the bigger picture the circular motion hub.

Frequently asked questions

State the formula for angle of banking.

tan θ = v² / (rg), so θ = tan⁻¹(v² / rg). Here v is the speed in m/s, r the radius of the curve in m, and g is 9.8 m/s². The result holds for a frictionless road.

Why are roads banked on curves?

Banking lets the normal force from the road supply the inward force a turning vehicle needs, so the car does not depend only on tyre friction, which can be low on wet roads.

Does the angle of banking depend on the mass of the vehicle?

No. Mass cancels in the derivation, so the same angle works for any vehicle travelling at the design speed.

What is the maximum speed on a banked road with friction?

v max = √[rg(μ + tan θ) / (1 − μ tan θ)], where μ is the coefficient of static friction between tyres and road.

What is the safe speed on a flat curve?

With no banking, friction alone supplies the force, so v max = √(μrg).