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Horizontal Curves

Transportation · FE Reference Handbook section

Transportation
33 formulas
10 exam-style examples
~60 min
All Transportation lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • Sight Distance (to see around obstruction)

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
Minimum radius for superelevation — solve for minimum radius — Horizontal Curves

A transportation problem uses Minimum radius for superelevation. Given design speed (V) = 60.0000 mph; superelevation rate (e) = 0.0500; side friction factor (f) = 0.1550, determine the minimum radius (R) in ft.

Given

  • designspeed(V)=60.0000mphdesign speed (V) = 60.0000 mph
  • superelevationrate(e)=0.0500superelevation rate (e) = 0.0500
  • sidefrictionfactor(f)=0.1550side friction factor (f) = 0.1550

Find

minimum radius (R), in ft

Start with the thinking

  • The governing relation printed in this handbook section is Minimum radius for superelevation.
  • Everything except R is given, so isolate R symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Transportation items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=V2/(15(e+f))R = V^2 / (15 (e + f))
  2. Step 2 — Rearrange the relation so that R stands alone on the left-hand side.

  3. Step 3 — List the givens: design speed (V) = 60.0000 mph, superelevation rate (e) = 0.0500, side friction factor (f) = 0.1550.

  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    R=1171 ftR = 1171\ \text{ft}
  6. Step 6 — Check: returning R = 1,171 ft to

    R=V2/(15(e+f))R = V^2 / (15 (e + f))

    reproduces the given quantities, and both sides carry the same units.

Answer:
R=1171 ftR = 1171\ \text{ft}

Why the other options are there

  • 2,341 — kept a factor of two that cancels in the correct rearrangement.
  • 585.4 — dropped that same factor in the other direction.
  • 1,288 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Transportation → Horizontal Curves

Example 2
Minimum radius for superelevation — solve for design speed — Horizontal Curves (2)

A transportation problem uses Minimum radius for superelevation. Given superelevation rate (e) = 0.0750; side friction factor (f) = 0.1400; minimum radius (R) = 1,288 ft, determine the design speed (V) in mph.

Given

  • superelevationrate(e)=0.0750superelevation rate (e) = 0.0750
  • sidefrictionfactor(f)=0.1400side friction factor (f) = 0.1400
  • minimumradius(R)=1,288ftminimum radius (R) = 1,288 ft

Find

design speed (V), in mph

Start with the thinking

  • The governing relation printed in this handbook section is Minimum radius for superelevation.
  • Everything except V is given, so isolate V symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Transportation items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=V2/(15(e+f))R = V^2 / (15 (e + f))
  2. Step 2 — Rearrange the relation so that V stands alone on the left-hand side.

  3. Step 3 — List the givens: superelevation rate (e) = 0.0750, side friction factor (f) = 0.1400, minimum radius (R) = 1,288 ft.

  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    V=64.4500 mphV = 64.4500\ \text{mph}
  6. Step 6 — Check: returning V = 64.4500 mph to

    R=V2/(15(e+f))R = V^2 / (15 (e + f))

    reproduces the given quantities, and both sides carry the same units.

Answer:
V=64.4500 mphV = 64.4500\ \text{mph}

Why the other options are there

  • 128.9 — kept a factor of two that cancels in the correct rearrangement.
  • 32.2250 — dropped that same factor in the other direction.
  • 70.8950 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Transportation → Horizontal Curves

Example 3
Minimum radius for superelevation — solve for superelevation rate — Horizontal Curves (3)

A transportation problem uses Minimum radius for superelevation. Given design speed (V) = 50.0000 mph; side friction factor (f) = 0.1450; minimum radius (R) = 2,516 ft, determine the superelevation rate (e).

Given

  • designspeed(V)=50.0000mphdesign speed (V) = 50.0000 mph
  • sidefrictionfactor(f)=0.1450side friction factor (f) = 0.1450
  • minimumradius(R)=2,516ftminimum radius (R) = 2,516 ft

Find

superelevation rate (e)

Start with the thinking

  • The governing relation printed in this handbook section is Minimum radius for superelevation.
  • Everything except e is given, so isolate e symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Transportation items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=V2/(15(e+f))R = V^2 / (15 (e + f))
  2. Step 2 — Rearrange the relation so that e stands alone on the left-hand side.

  3. Step 3 — List the givens: design speed (V) = 50.0000 mph, side friction factor (f) = 0.1450, minimum radius (R) = 2,516 ft.

  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    e=−0.0788e = -0.0788
  6. Step 6 — Check: returning e = -0.0788 to

    R=V2/(15(e+f))R = V^2 / (15 (e + f))

    reproduces the given quantities, and both sides carry the same units.

Answer:
e=−0.0788e = -0.0788

Why the other options are there

  • -0.1575 — kept a factor of two that cancels in the correct rearrangement.
  • -0.0394 — dropped that same factor in the other direction.
  • -0.0866 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Transportation → Horizontal Curves

Example 4
Minimum radius for superelevation — solve for minimum radius (case 2) — Horizontal Curves (4)

A transportation problem uses Minimum radius for superelevation. Given design speed (V) = 50.0000 mph; superelevation rate (e) = 0.0950; side friction factor (f) = 0.1350, determine the minimum radius (R) in ft.

Given

  • designspeed(V)=50.0000mphdesign speed (V) = 50.0000 mph
  • superelevationrate(e)=0.0950superelevation rate (e) = 0.0950
  • sidefrictionfactor(f)=0.1350side friction factor (f) = 0.1350

Find

minimum radius (R), in ft

Start with the thinking

  • The governing relation printed in this handbook section is Minimum radius for superelevation.
  • Everything except R is given, so isolate R symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Transportation items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=V2/(15(e+f))R = V^2 / (15 (e + f))
  2. Step 2 — Rearrange the relation so that R stands alone on the left-hand side.

  3. Step 3 — List the givens: design speed (V) = 50.0000 mph, superelevation rate (e) = 0.0950, side friction factor (f) = 0.1350.

  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    R=724.6 ftR = 724.6\ \text{ft}
  6. Step 6 — Check: returning R = 724.6 ft to

    R=V2/(15(e+f))R = V^2 / (15 (e + f))

    reproduces the given quantities, and both sides carry the same units.

Answer:
R=724.6 ftR = 724.6\ \text{ft}

Why the other options are there

  • 1,449 — kept a factor of two that cancels in the correct rearrangement.
  • 362.3 — dropped that same factor in the other direction.
  • 797.1 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Transportation → Horizontal Curves

Example 5
Minimum radius for superelevation — solve for design speed (case 2) — Horizontal Curves (5)

A transportation problem uses Minimum radius for superelevation. Given superelevation rate (e) = 0.0850; side friction factor (f) = 0.1300; minimum radius (R) = 2,955 ft, determine the design speed (V) in mph.

Given

  • superelevationrate(e)=0.0850superelevation rate (e) = 0.0850
  • sidefrictionfactor(f)=0.1300side friction factor (f) = 0.1300
  • minimumradius(R)=2,955ftminimum radius (R) = 2,955 ft

Find

design speed (V), in mph

Start with the thinking

  • The governing relation printed in this handbook section is Minimum radius for superelevation.
  • Everything except V is given, so isolate V symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Transportation items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=V2/(15(e+f))R = V^2 / (15 (e + f))
  2. Step 2 — Rearrange the relation so that V stands alone on the left-hand side.

  3. Step 3 — List the givens: superelevation rate (e) = 0.0850, side friction factor (f) = 0.1300, minimum radius (R) = 2,955 ft.

  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    V=97.6211 mphV = 97.6211\ \text{mph}
  6. Step 6 — Check: returning V = 97.6211 mph to

    R=V2/(15(e+f))R = V^2 / (15 (e + f))

    reproduces the given quantities, and both sides carry the same units.

Answer:
V=97.6211 mphV = 97.6211\ \text{mph}

Why the other options are there

  • 195.2 — kept a factor of two that cancels in the correct rearrangement.
  • 48.8105 — dropped that same factor in the other direction.
  • 107.4 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Transportation → Horizontal Curves

Example 6
Minimum radius for superelevation — solve for superelevation rate (case 2) — Horizontal Curves (6)

A transportation problem uses Minimum radius for superelevation. Given design speed (V) = 20.0000 mph; side friction factor (f) = 0.1050; minimum radius (R) = 133.0 ft, determine the superelevation rate (e).

Given

  • designspeed(V)=20.0000mphdesign speed (V) = 20.0000 mph
  • sidefrictionfactor(f)=0.1050side friction factor (f) = 0.1050
  • minimumradius(R)=133.0ftminimum radius (R) = 133.0 ft

Find

superelevation rate (e)

Start with the thinking

  • The governing relation printed in this handbook section is Minimum radius for superelevation.
  • Everything except e is given, so isolate e symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Transportation items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=V2/(15(e+f))R = V^2 / (15 (e + f))
  2. Step 2 — Rearrange the relation so that e stands alone on the left-hand side.

  3. Step 3 — List the givens: design speed (V) = 20.0000 mph, side friction factor (f) = 0.1050, minimum radius (R) = 133.0 ft.

  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    e=0.0955e = 0.0955
  6. Step 6 — Check: returning e = 0.0955 to

    R=V2/(15(e+f))R = V^2 / (15 (e + f))

    reproduces the given quantities, and both sides carry the same units.

Answer:
e=0.0955e = 0.0955

Why the other options are there

  • 0.1910 — kept a factor of two that cancels in the correct rearrangement.
  • 0.0478 — dropped that same factor in the other direction.
  • 0.1051 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Transportation → Horizontal Curves

Example 7
Minimum radius for superelevation — solve for minimum radius (case 3) — Horizontal Curves (7)

A transportation problem uses Minimum radius for superelevation. Given design speed (V) = 50.0000 mph; superelevation rate (e) = 0.0700; side friction factor (f) = 0.0950, determine the minimum radius (R) in ft.

Given

  • designspeed(V)=50.0000mphdesign speed (V) = 50.0000 mph
  • superelevationrate(e)=0.0700superelevation rate (e) = 0.0700
  • sidefrictionfactor(f)=0.0950side friction factor (f) = 0.0950

Find

minimum radius (R), in ft

Start with the thinking

  • The governing relation printed in this handbook section is Minimum radius for superelevation.
  • Everything except R is given, so isolate R symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Transportation items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=V2/(15(e+f))R = V^2 / (15 (e + f))
  2. Step 2 — Rearrange the relation so that R stands alone on the left-hand side.

  3. Step 3 — List the givens: design speed (V) = 50.0000 mph, superelevation rate (e) = 0.0700, side friction factor (f) = 0.0950.

  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    R=1010 ftR = 1010\ \text{ft}
  6. Step 6 — Check: returning R = 1,010 ft to

    R=V2/(15(e+f))R = V^2 / (15 (e + f))

    reproduces the given quantities, and both sides carry the same units.

Answer:
R=1010 ftR = 1010\ \text{ft}

Why the other options are there

  • 2,020 — kept a factor of two that cancels in the correct rearrangement.
  • 505.1 — dropped that same factor in the other direction.
  • 1,111 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Transportation → Horizontal Curves

Example 8
Minimum radius for superelevation — solve for design speed (case 3) — Horizontal Curves (8)

A transportation problem uses Minimum radius for superelevation. Given superelevation rate (e) = 0.0900; side friction factor (f) = 0.1050; minimum radius (R) = 1,615 ft, determine the design speed (V) in mph.

Given

  • superelevationrate(e)=0.0900superelevation rate (e) = 0.0900
  • sidefrictionfactor(f)=0.1050side friction factor (f) = 0.1050
  • minimumradius(R)=1,615ftminimum radius (R) = 1,615 ft

Find

design speed (V), in mph

Start with the thinking

  • The governing relation printed in this handbook section is Minimum radius for superelevation.
  • Everything except V is given, so isolate V symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Transportation items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=V2/(15(e+f))R = V^2 / (15 (e + f))
  2. Step 2 — Rearrange the relation so that V stands alone on the left-hand side.

  3. Step 3 — List the givens: superelevation rate (e) = 0.0900, side friction factor (f) = 0.1050, minimum radius (R) = 1,615 ft.

  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    V=68.7305 mphV = 68.7305\ \text{mph}
  6. Step 6 — Check: returning V = 68.7305 mph to

    R=V2/(15(e+f))R = V^2 / (15 (e + f))

    reproduces the given quantities, and both sides carry the same units.

Answer:
V=68.7305 mphV = 68.7305\ \text{mph}

Why the other options are there

  • 137.5 — kept a factor of two that cancels in the correct rearrangement.
  • 34.3652 — dropped that same factor in the other direction.
  • 75.6035 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Transportation → Horizontal Curves

Example 9
Minimum radius for superelevation — solve for superelevation rate (case 3) — Horizontal Curves (9)

A transportation problem uses Minimum radius for superelevation. Given design speed (V) = 35.0000 mph; side friction factor (f) = 0.1650; minimum radius (R) = 2,720 ft, determine the superelevation rate (e).

Given

  • designspeed(V)=35.0000mphdesign speed (V) = 35.0000 mph
  • sidefrictionfactor(f)=0.1650side friction factor (f) = 0.1650
  • minimumradius(R)=2,720ftminimum radius (R) = 2,720 ft

Find

superelevation rate (e)

Start with the thinking

  • The governing relation printed in this handbook section is Minimum radius for superelevation.
  • Everything except e is given, so isolate e symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Transportation items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=V2/(15(e+f))R = V^2 / (15 (e + f))
  2. Step 2 — Rearrange the relation so that e stands alone on the left-hand side.

  3. Step 3 — List the givens: design speed (V) = 35.0000 mph, side friction factor (f) = 0.1650, minimum radius (R) = 2,720 ft.

  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    e=−0.1350e = -0.1350
  6. Step 6 — Check: returning e = -0.1350 to

    R=V2/(15(e+f))R = V^2 / (15 (e + f))

    reproduces the given quantities, and both sides carry the same units.

Answer:
e=−0.1350e = -0.1350

Why the other options are there

  • -0.2700 — kept a factor of two that cancels in the correct rearrangement.
  • -0.0675 — dropped that same factor in the other direction.
  • -0.1485 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Transportation → Horizontal Curves

Example 10
Minimum radius for superelevation — solve for minimum radius (case 4) — Horizontal Curves (10)

A transportation problem uses Minimum radius for superelevation. Given design speed (V) = 60.0000 mph; superelevation rate (e) = 0.1000; side friction factor (f) = 0.1450, determine the minimum radius (R) in ft.

Given

  • designspeed(V)=60.0000mphdesign speed (V) = 60.0000 mph
  • superelevationrate(e)=0.1000superelevation rate (e) = 0.1000
  • sidefrictionfactor(f)=0.1450side friction factor (f) = 0.1450

Find

minimum radius (R), in ft

Start with the thinking

  • The governing relation printed in this handbook section is Minimum radius for superelevation.
  • Everything except R is given, so isolate R symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Transportation items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=V2/(15(e+f))R = V^2 / (15 (e + f))
  2. Step 2 — Rearrange the relation so that R stands alone on the left-hand side.

  3. Step 3 — List the givens: design speed (V) = 60.0000 mph, superelevation rate (e) = 0.1000, side friction factor (f) = 0.1450.

  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    R=979.6 ftR = 979.6\ \text{ft}
  6. Step 6 — Check: returning R = 979.6 ft to

    R=V2/(15(e+f))R = V^2 / (15 (e + f))

    reproduces the given quantities, and both sides carry the same units.

Answer:
R=979.6 ftR = 979.6\ \text{ft}

Why the other options are there

  • 1,959 — kept a factor of two that cancels in the correct rearrangement.
  • 489.8 — dropped that same factor in the other direction.
  • 1,078 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Transportation → Horizontal Curves

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