A transportation problem uses Vertical curve length (sight-distance controlled, crest). Given algebraic grade difference (A) = 1.5000 %; sight distance (S) = 400.0 ft, determine the curve length (L) in ft.

Start with the thinking

• The governing relation printed in this handbook section is Vertical curve length (sight-distance controlled, crest).
• Everything except L is given, so isolate L 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:

   $$L = A S^2 / 2158$$

2. Step 2 — Rearrange the relation so that L stands alone on the left-hand side.

3. Step 3 — List the givens: algebraic grade difference (A) = 1.5000 %, sight distance (S) = 400.0 ft.

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

5. Step 5 — Evaluate:

   $$L = 111.2\ \text{ft}$$

6. Step 6 — Check: returning L = 111.2 ft to

   $$L = A S^2 / 2158$$

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

Why the other options are there

• 222.4 — kept a factor of two that cancels in the correct rearrangement.
• 55.6070 — dropped that same factor in the other direction.
• 122.3 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Transportation → Vertical Curves

a sag vertical curve at a highway underpass Given elevation at PVC (y_0) = 201.5 m; initial grade (g_1) = -0.0100; distance from PVC (x) = 40.0000 m; final grade (g_2) = 0.0150; curve length (L) = 255.0 m, determine the elevation at station x (y) in m.

Start with the thinking

• The governing relation printed in this handbook section is Vertical Curves (parabolic elevation).
• Everything except y is given, so isolate y 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.
• Vertical curves connect two roadway grades with a parabolic profile to provide smooth elevation transitions.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$y = y_0 + g_1 x + \dfrac{(g_2-g_1)}{2L} x^2$$

2. Step 2 — Rearrange the relation so that y stands alone on the left-hand side.

3. Step 3 — List the givens: elevation at PVC (y_0) = 201.5 m, initial grade (g_1) = -0.0100, distance from PVC (x) = 40.0000 m, final grade (g_2) = 0.0150, curve length (L) = 255.0 m.

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

5. Step 5 — Evaluate:

   $$y = 201.2\ \text{m}$$

6. Step 6 — Check: returning y = 201.2 m to

   $$y = y_0 + g_1 x + \dfrac{(g_2-g_1)}{2L} x^2$$

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

Why the other options are there

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

Reference: FE Handbook — Vertical Curves

A transportation problem uses Vertical curve length (sight-distance controlled, crest). Given sight distance (S) = 690.0 ft; curve length (L) = 120.0 ft, determine the algebraic grade difference (A) in %.

Start with the thinking

• The governing relation printed in this handbook section is Vertical curve length (sight-distance controlled, crest).
• Everything except A is given, so isolate A 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:

   $$L = A S^2 / 2158$$

2. Step 2 — Rearrange the relation so that A stands alone on the left-hand side.

3. Step 3

   $$List the givens: sight distance (S) = 690.0 ft, curve length (L) = 120.0 ft$$

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

5. Step 5 — Evaluate:

   A = 0.5439\ \text{%}

6. Step 6 — Check: returning A = 0.5439 % to

   $$L = A S^2 / 2158$$

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

Why the other options are there

• 1.0878 — kept a factor of two that cancels in the correct rearrangement.
• 0.2720 — dropped that same factor in the other direction.
• 0.5983 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Transportation → Vertical Curves

vertical curves designed for adequate stopping sight distance Given initial grade (g_1) = -0.0400; distance from PVC (x) = 52.0000 m; final grade (g_2) = -0.0200; curve length (L) = 105.0 m; elevation at station x (y) = 114.2 m, determine the elevation at PVC (y_0) in m.

Start with the thinking

• The governing relation printed in this handbook section is Vertical Curves (parabolic elevation).
• Everything except y_0 is given, so isolate y_0 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.
• Vertical curves connect two roadway grades with a parabolic profile to provide smooth elevation transitions.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$y = y_0 + g_1 x + \dfrac{(g_2-g_1)}{2L} x^2$$

2. Step 2 — Rearrange the relation so that y_0 stands alone on the left-hand side.

3. Step 3 — List the givens: initial grade (g_1) = -0.0400, distance from PVC (x) = 52.0000 m, final grade (g_2) = -0.0200, curve length (L) = 105.0 m, elevation at station x (y) = 114.2 m.

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

5. Step 5 — Evaluate:

   $$y_{0} = 116.0\ \text{m}$$

6. Step 6 — Check: returning y_0 = 116.0 m to

   $$y = y_0 + g_1 x + \dfrac{(g_2-g_1)}{2L} x^2$$

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

Why the other options are there

• 232.0 — kept a factor of two that cancels in the correct rearrangement.
• 58.0112 — dropped that same factor in the other direction.
• 127.6 — rounded an intermediate value before the final step.

Reference: FE Handbook — Vertical Curves

A transportation problem uses Vertical curve length (sight-distance controlled, crest). Given algebraic grade difference (A) = 8.0000 %; curve length (L) = 769.0 ft, determine the sight distance (S) in ft.

Start with the thinking

• The governing relation printed in this handbook section is Vertical curve length (sight-distance controlled, crest).
• Everything except S is given, so isolate S 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:

   $$L = A S^2 / 2158$$

2. Step 2 — Rearrange the relation so that S stands alone on the left-hand side.

3. Step 3 — List the givens: algebraic grade difference (A) = 8.0000 %, curve length (L) = 769.0 ft.

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

5. Step 5 — Evaluate:

   $$S = 455.5\ \text{ft}$$

6. Step 6 — Check: returning S = 455.5 ft to

   $$L = A S^2 / 2158$$

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

Why the other options are there

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

Reference: FE Reference Handbook — Transportation → Vertical Curves

a crest vertical curve on a highway summit Given elevation at PVC (y_0) = 272.0 m; initial grade (g_1) = 0.0500; distance from PVC (x) = 25.0000 m; final grade (g_2) = -0.0600; curve length (L) = 160.0 m, determine the elevation at station x (y) in m.

Start with the thinking

• The governing relation printed in this handbook section is Vertical Curves (parabolic elevation).
• Everything except y is given, so isolate y 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.
• Vertical curves connect two roadway grades with a parabolic profile to provide smooth elevation transitions.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$y = y_0 + g_1 x + \dfrac{(g_2-g_1)}{2L} x^2$$

2. Step 2 — Rearrange the relation so that y stands alone on the left-hand side.

3. Step 3 — List the givens: elevation at PVC (y_0) = 272.0 m, initial grade (g_1) = 0.0500, distance from PVC (x) = 25.0000 m, final grade (g_2) = -0.0600, curve length (L) = 160.0 m.

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

5. Step 5 — Evaluate:

   $$y = 273.0\ \text{m}$$

6. Step 6 — Check: returning y = 273.0 m to

   $$y = y_0 + g_1 x + \dfrac{(g_2-g_1)}{2L} x^2$$

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

Why the other options are there

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

Reference: FE Handbook — Vertical Curves

A transportation problem uses Vertical curve length (sight-distance controlled, crest). Given algebraic grade difference (A) = 9.0000 %; sight distance (S) = 360.0 ft, determine the curve length (L) in ft.

Start with the thinking

• The governing relation printed in this handbook section is Vertical curve length (sight-distance controlled, crest).
• Everything except L is given, so isolate L 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:

   $$L = A S^2 / 2158$$

2. Step 2 — Rearrange the relation so that L stands alone on the left-hand side.

3. Step 3 — List the givens: algebraic grade difference (A) = 9.0000 %, sight distance (S) = 360.0 ft.

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

5. Step 5 — Evaluate:

   $$L = 540.5\ \text{ft}$$

6. Step 6 — Check: returning L = 540.5 ft to

   $$L = A S^2 / 2158$$

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

Why the other options are there

• 1,081 — kept a factor of two that cancels in the correct rearrangement.
• 270.3 — dropped that same factor in the other direction.
• 594.6 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Transportation → Vertical Curves

a sag vertical curve at a highway underpass Given initial grade (g_1) = -0.0450; distance from PVC (x) = 75.0000 m; final grade (g_2) = -0.0400; curve length (L) = 195.0 m; elevation at station x (y) = 151.9 m, determine the elevation at PVC (y_0) in m.

Start with the thinking

• The governing relation printed in this handbook section is Vertical Curves (parabolic elevation).
• Everything except y_0 is given, so isolate y_0 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.
• Vertical curves connect two roadway grades with a parabolic profile to provide smooth elevation transitions.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$y = y_0 + g_1 x + \dfrac{(g_2-g_1)}{2L} x^2$$

2. Step 2 — Rearrange the relation so that y_0 stands alone on the left-hand side.

3. Step 3 — List the givens: initial grade (g_1) = -0.0450, distance from PVC (x) = 75.0000 m, final grade (g_2) = -0.0400, curve length (L) = 195.0 m, elevation at station x (y) = 151.9 m.

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

5. Step 5 — Evaluate:

   $$y_{0} = 155.2\ \text{m}$$

6. Step 6 — Check: returning y_0 = 155.2 m to

   $$y = y_0 + g_1 x + \dfrac{(g_2-g_1)}{2L} x^2$$

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

Why the other options are there

• 310.4 — kept a factor of two that cancels in the correct rearrangement.
• 77.6014 — dropped that same factor in the other direction.
• 170.7 — rounded an intermediate value before the final step.

Reference: FE Handbook — Vertical Curves

A transportation problem uses Vertical curve length (sight-distance controlled, crest). Given sight distance (S) = 750.0 ft; curve length (L) = 362.0 ft, determine the algebraic grade difference (A) in %.

Start with the thinking

• The governing relation printed in this handbook section is Vertical curve length (sight-distance controlled, crest).
• Everything except A is given, so isolate A 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:

   $$L = A S^2 / 2158$$

2. Step 2 — Rearrange the relation so that A stands alone on the left-hand side.

3. Step 3

   $$List the givens: sight distance (S) = 750.0 ft, curve length (L) = 362.0 ft$$

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

5. Step 5 — Evaluate:

   A = 1.3888\ \text{%}

6. Step 6 — Check: returning A = 1.3888 % to

   $$L = A S^2 / 2158$$

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

Why the other options are there

• 2.7776 — kept a factor of two that cancels in the correct rearrangement.
• 0.6944 — dropped that same factor in the other direction.
• 1.5277 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Transportation → Vertical Curves

vertical curves designed for adequate stopping sight distance Given elevation at PVC (y_0) = 121.0 m; initial grade (g_1) = 0.0300; distance from PVC (x) = 56.0000 m; final grade (g_2) = -0.0050; curve length (L) = 250.0 m, determine the elevation at station x (y) in m.

Start with the thinking

• The governing relation printed in this handbook section is Vertical Curves (parabolic elevation).
• Everything except y is given, so isolate y 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.
• Vertical curves connect two roadway grades with a parabolic profile to provide smooth elevation transitions.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$y = y_0 + g_1 x + \dfrac{(g_2-g_1)}{2L} x^2$$

2. Step 2 — Rearrange the relation so that y stands alone on the left-hand side.

3. Step 3 — List the givens: elevation at PVC (y_0) = 121.0 m, initial grade (g_1) = 0.0300, distance from PVC (x) = 56.0000 m, final grade (g_2) = -0.0050, curve length (L) = 250.0 m.

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

5. Step 5 — Evaluate:

   $$y = 122.5\ \text{m}$$

6. Step 6 — Check: returning y = 122.5 m to

   $$y = y_0 + g_1 x + \dfrac{(g_2-g_1)}{2L} x^2$$

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

Why the other options are there

• 244.9 — kept a factor of two that cancels in the correct rearrangement.
• 61.2302 — dropped that same factor in the other direction.
• 134.7 — rounded an intermediate value before the final step.

Reference: FE Handbook — Vertical Curves