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Banked track: from the survey stakes to the cornering force

SPH4U Grade 12 Physics · Dynamics

A closed test oval is being commissioned, and the survey crew's last job of the morning is the banked north turn. The surveyor sets a level across a single 3.50 m lane of fresh pavement and reads the staff at each edge: the outer edge of the pavement sits 1,400 mm above the inner edge. The turn itself is struck from a centre stake in the infield on a 175 m radius, and the stakes still trace the arc through the grass. In the afternoon the commissioning engineer will send an instrumented 1,520 kg test car around the finished turn, and wants the numbers worked out ahead of time. Find the bank angle, the speed a car could hold through the turn with no help at all from sideways friction, the centripetal acceleration at that speed, and the inward force the pavement must supply to the car.

21.8°3.50 m lane, levelledrise 1,400 mmr = 175 m turndesign speed 26.2 m/s1,520 kg

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • w = 3.5 m — Lane width the surveyor levels across
  • rise = 1,400 mm — Outer edge above the inner edge
  • r = 175 m — Radius of the turn
  • m = 1,520 kg — Test car
Determine
  1. (a)the bank angle of the turn
  2. (b)the design speed, with no help from sideways friction
  3. (c)the centripetal acceleration at that speed
  4. (d)the inward force the pavement must supply to the car
Step 1 of 4(a) · solve for Acute angle

The bank angle is pure geometry — rise over horizontal run — and it has to come first, because every dynamics question about this turn is really a question about this one angle. The rise comes off the level in millimetres and the lane width in metres, so watch the conversion: 1,400 mm across 3.50 m is a ratio of 0.4, not 400.

θoa
Rearranged for θ
θ=arctan⁡(oa)\theta = \arctan\left(\frac{o}{a}\right)
Your values, in your units
θ=arctan⁡((1,400 mm)(3.5 m))\theta = \arctan\left(\frac{\left( 1{,}400\ \text{mm} \right)}{\left( 3.5\ \text{m} \right)}\right)
Converted to base units
θ=arctan⁡((1.4 m)(3.5 m))\theta = \arctan\left(\frac{\left( 1.4\ \text{m} \right)}{\left( 3.5\ \text{m} \right)}\right)
Answer
θ=21.801 ∘\theta = 21.801\ ^{\circ}

Carried onward at full precision, not this rounded figure.

Open the Right-Triangle Tangent Ratio (TOA) solver →

Step 2 of 4(b) · solve for Design speed

Now the physics. At the design speed the tilted normal force alone supplies mv²/r, so tan θ = v²/rg and the angle from step 1 hands over a speed. Note the mass never enters — the same tilt works for a motorbike.

θrv
Rearranged for v
v=rgtan⁡θv = \sqrt{r g \tan\theta}
21.801 °carried from step 1
Your values, in your units
v=(175 m) (9.80665 m/s2) tan⁡(0.380506 rad)v = \sqrt{\left( 175\ \text{m} \right) \, \left( 9.80665\ \text{m/s}^{2} \right) \, \tan \left( 0.380506\ \text{rad} \right)}
Converted to base units
v=(175 m) (9.80665 m/s2) tan⁡(21.8014 ∘)v = \sqrt{\left( 175\ \text{m} \right) \, \left( 9.80665\ \text{m/s}^{2} \right) \, \tan \left( 21.8014\ ^{\circ} \right)}
Answer
v=26.2 m/sv = 26.2\ \text{m/s}

Carried onward at full precision, not this rounded figure.

Open the Banked Curve Angle solver →

Step 3 of 4(c) · solve for Centripetal acceleration

Feed the design speed back into a = v²/r. The result is g tan θ, which is a useful thing to recognise on sight.

rvac
Rearranged for ac
ac=v2ra_c = \frac{v^2}{r}
26.2 m/scarried from step 2
Your values, in your units
ac=(26.2005 m/s)2(175 m)a_c = \frac{\left( 26.2005\ \text{m/s} \right)^2}{\left( 175\ \text{m} \right)}
Answer
ac=3.9227 m/s2a_c = 3.9227\ \text{m/s}^{2}

Carried onward at full precision, not this rounded figure.

Open the Centripetal Acceleration (a = v²/r) solver →

Step 4 of 4(d) · solve for Force

Newton's second law converts that acceleration into the force the road has to deliver. This is the number a structural engineer needs; the driver only feels it.

mFa
Rearranged for F
F=maF = m a
3.9227 m/s²carried from step 3
Your values, in your units
F=(1,520 kg) (3.92266 m/s2)F = \left( 1{,}520\ \text{kg} \right) \, \left( 3.92266\ \text{m/s}^{2} \right)
Answer
F=5.9624 kNF = 5.9624\ \text{kN}

Carried onward at full precision, not this rounded figure.

Open the Newton's Second Law solver →

Answer

Therefore the north turn banks at 21.8°, its friction-free design speed is 26.2 m/s — about 94 km/h — the car at that speed accelerates inward at 3.92 m/s² (exactly 0.4 g), and the pavement must supply 5.96 kN to hold the 1,520 kg car on its arc.

Why this order

The order here is the whole lesson. A bank angle is a geometric fact about a piece of concrete, measurable with a level and a tape before a single car drives on it, so it goes first. Only then does dynamics get a say: tan θ = v²/rg turns that fixed geometry into one particular speed, the design speed, at which the normal force does all the work and friction does none. Below it a car tends to slide down the bank; above it, friction has to make up the difference, and when the tyres run out the car goes over the top. A single bank angle is not a safety margin — it is a single speed, with tolerance on either side supplied by grip.

Two things trip students between step 1 and step 2. The first is units, and it bites twice. The level reading arrives in millimetres and the lane width in metres, so a ratio taken off the raw numbers is 400 instead of 0.4 and the bank comes out at 89.86°; then the answer to step 1 is an angle, which this site carries in radians even though it shows you degrees, and a calculator left in the wrong mode produces a design speed that is wrong by nothing recognisable. The second is the disappearing mass. Students expect a heavy car to need a steeper bank, but writing N sin θ = mv²/r beside N cos θ = mg and dividing kills the m outright — which is why step 2 needs no mass at all and step 4 suddenly does. Mass decides how much force, never how much tilt. Railway engineers hit the same result in the 1830s and still call it cant; Daytona's 31° and Talladega's 33° are this equation built at the limit of what a paving machine can hold.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.