Banked Curves

Set the radius of the bend, how steeply it is banked, the grip between tyre and road, the speed and the mass. The page resolves the forces, says which way friction has to push and whether the tyres can give it, and lists every speed the bend will hold.

Worked example — Banked Curves

Bank as a road engineer writes it	8.00%
Angle of friction	38.66°
Design speed, where no friction is needed	45.09 km/h
Slowest speed that holds	0.00 km/h
Fastest speed that holds	154.59 km/h
Acceleration towards the centre	3.125 m/s²
Road's push at right angles to the surface	15,037 N
Friction the tyres must supply	3,500 N down the slope

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Banked Curves

A banked curve is a bend whose surface is tilted towards the centre by an angle θ, so that the road's push on the vehicle, which is always at right angles to the surface, leans inwards and supplies some or all of the force that turns it. At one speed, the design speed v₀ = √(r g tan θ), it supplies all of it and no friction is needed. Friction up to μ times the road's push widens that single speed into a window, r g tan(θ − φ) ≤ v² ≤ r g tan(θ + φ), where φ = arctan μ is the angle of friction. The mass appears in none of these.

What each symbol means

r: radius of the bend, measured horizontally from its centre to the vehicle
θ: bank angle from the horizontal; road engineers give tan θ as a percentage, the superelevation
μ: the most friction the tyres can supply, as a fraction of the road's push at right angles
φ: angle of friction, arctan μ: the steepest slope a vehicle can stand on
v_0: design speed, at which the bank does all the turning and friction is zero

Assumes: A vehicle treated as one point, on a bend of constant radius and constant bank, at constant speed on a horizontal circle. Standard gravity, 9.80665 m/s². One friction coefficient for every direction; the published numbers are braking measurements at 17% slip, and a rolling tyre's sideways grip is taken to be of the same size. No air: no downforce, no drag, no crosswind.
Breaks when: A vehicle with a high centre of mass can tip over before it slides, and on a banked curve that limit depends on the width of the wheel track as well as on the bank; a lorry reaches it first. Real roads design with side friction far below the measured grip, to keep drivers comfortable and to leave a margin for worn tyres. A racing car's wings press it into the road with a force that grows as the square of the speed, which moves the ceiling up. And a real bend is entered along a transition curve whose radius shrinks gradually, not at a single radius.

Below 45.09 km/h, friction on this bend pushes outwards 🖖

Press Dry bend, 30 km/h and then Dry bend, 90 km/h. At the design speed, 45.09 km/h for 200 m at 8%, the bank does all the work: the road pushes at right angles to its surface, that push leans towards the centre, and its horizontal part is exactly the m v²/r the turn needs. Friction is zero. Go slower and the leaning push is more than the turn needs; left to itself the car would slide down into the inside of the bend, so friction holds it back, pointing up the slope, 654 N at 30 km/h. Go faster and the push is not enough, and friction points down the slope, 3,500 N at 90 km/h. The picture of friction pulling a car into a bend is true on a flat road, and on this one it is wrong at every speed below 45.09 km/h. Now press 40 t lorry, 90 km/h: every force is 26.67 times the car's and the grip it needs, 0.2327, is the same, because the mass multiplies both sides of every equation here. A lorry needs no steeper bank than a car. What it has that a car does not is a high centre of mass, and on a dry road a laden lorry can tip over before it slides, which this page does not model.

On wet black ice, a car that stops on a 10% bank slides down it 🖖

A bank has a slowest speed as well as a fastest. A car standing still needs friction of mg sin θ up the slope against a push of mg cos θ, so it stays put only while tan θ is no more than μ. Carl-Gustaf Wallman and Henrik Åström's 2001 review for the Swedish road research institute VTI puts wet black ice at 0.05 to 0.10. Press Stopped on icy 10%: standing still the car needs 0.1000 of grip, the ice offers 0.05, and it slides into the inside of the bend. Anything slower than 35.56 km/h slides with it, and the second panel shows the floor of the window lifting off zero. Washington State's highway design manual allows banks of up to 10%, and where snow and ice accumulate it limits them to 6%, because “vehicles moving at slow speeds or stopped on curves with supers greater than 6% tend to slide inward”. Set the bank to 3.434°, which is 6%: the car now stands on any grip better than 0.06, and on the very worst ice, 0.05, it still slides. On that ice a stopped car holds only where the bank is 5% or less.

Past 51.34° on a dry road, no speed is too fast 🖖

Faster needs more friction, but it also presses the car harder into the road, and on a steep bank the second wins. As the speed grows without limit, the grip needed tends to 1/tan θ, so once 1/tan θ is less than μ no speed ever needs more than the road can give. For dry road, μ = 0.8, that happens at 51.34°: in the table below, the 200 m bend's ceiling is 478.30 km/h at 45° and gone at 60°. At 90° the road has become a wall, and the Wall of death is the extreme case. Friction now has to hold up the whole weight, and what it can give is μ times the wall's push, m v²/r, so there is a floor, √(rg/μ), and no ceiling at all. A wall 5 m in radius needs 28.18 km/h. At 40 km/h the rider uses 0.3972 of their grip, and the wall presses on them with 2.518 times their weight.

Problem solved in full

What does a banked turn do to an aircraft? 6 steps
A light aircraft flies a level turn at 100 knots, 51.44 m/s, banked at 60°. Find the radius of the turn, how long a full circle takes, how hard the seat pushes on the pilot, the steepest bank its 3.8 g structure allows, and what happens to its 50-knot stall speed. Press Dry bend, 90 km/h to check the first step.

Open this problem in the tool
1. An aircraft in a level turn is the road with the friction taken away and the wing's lift in place of the road's push. It therefore always flies at its design speed, tan θ = v²/(rg). The page's 200 m road at 8% satisfies the same equation at 45.09 km/h:
  
  $tan θ = \frac{v _{0}^{2}}{r g} : \frac{( 12 . 53 m/s ) ^{2}}{200 \times 9 . 80665} = 0.0800 = 8 . 00% \Rightarrow v_{0} = 45.09 km/h$
2. Solved for the radius at 60°:
  
  $r = \frac{v ^{2}}{g tan 6 0 ^{\circ}} = \frac{51 . 4 4 ^{2}}{9 . 80665 \times 1 . 732} = 155.8 m$
3. Going round that circle at 51.44 m/s takes
  
  $t = \frac{2 π r}{v} = \frac{2 π \times 155 . 8}{51 . 44} = 19.03 s ω = 18.9 2^{\circ} / s$
4. The vertical part of the lift has to carry the weight on its own, so the lift is mg/cos θ. That is the road's push at the design speed, and there is no speed in it:
  
  $L cos θ = m g \Rightarrow n = \frac{L}{m g} = \frac{1}{cos 6 0 ^{\circ}} = 2.000$
5. The structure's 3.8 g limit is reached at the bank angle whose cosine is 1/3.8. The table's 75° row prints 3.864, already past it:
  
  $\frac{1}{cos θ} = 3.8 \Rightarrow θ = arccos \frac{1}{3 . 8} = 74.7 4^{\circ} \frac{1}{cos 7 5 ^{\circ}} = 3.864$
6. A wing stalls at a fixed angle of attack, so the speed at which it can still make the lift goes up as the square root of the load:
  
  $v_{s}^{'} = v_{s} n = 50 kn \times 2 = 70.71 kn$
Answer

A circle of 155.8 m in 19.03 s, with the seat pushing on the pilot at twice their weight whatever the airspeed. The FAA's Airplane Flying Handbook gives the same 2.0 at 60° and 1.41 at 45°, and notes that a level turn at 75° exceeds the 3.8 a standard-category light aircraft is designed for; the exact limit is 74.74°. The stall speed rises from 50 to 70.71 knots, which the handbook rounds to 70.

$θ = 6 0^{\circ}, 100 kn : r = 155.8 m, t = 19.03 s, n = 2.000, θ_{m a x} = 74.7 4^{\circ}, v_{s}^{'} = 70.71 kn$

References (3)

Measured tyre-road friction by surface, from Swedish skiddometer surveys: dry bare road 0.8 to 1.0, wet 0.7 to 0.8, packed snow 0.20 to 0.30, wet black ice 0.05 to 0.10, the values behind the page's grip hint and its presets: C.-G. Wallman and H. Åström, "Friction measurement methods and the correlation between road friction and traffic safety. A literature review." VTI meddelande 911A, Swedish National Road and Transport Research Institute, Linköping, 2001, p. 36.
Why roads that see ice are banked gently: the steepest bank allowed is 10%, and where snow and ice accumulate it is limited to 6%, because slow or stopped vehicles on steeper banks slide inward: Washington State Department of Transportation, Design Manual M 22-01.23, Chapter 1250 "Cross Slope and Superelevation", section 1250.03, September 2024.
A banked aircraft's load factor depends on the bank alone, 1.41 at 45° and 2.0 at 60° for any airspeed or aeroplane, a standard-category light aircraft is built for 3.8, and the stall speed rises as the square root of the load: Federal Aviation Administration, Airplane Flying Handbook (FAA-H-8083-3C), Chapter 10 "Performance Maneuvers", section on steep turns.

By the 42Math team · Reviewed against published sources. Corrections welcome.