CivilSolve

Transportation & Surveying

Vertical Curve Elevations (Symmetric Parabola)

Symmetric parabolic vertical curve between grades g1 and g2: BVC/EVC stations and elevations, K value, elevation at any station, and the high/low (turning) point — reported only when it actually lies on the curve. Accepts curve length L or rate of vertical curvature K. Use for: 'find the elevation at station...', high/low point of a sag or crest curve, BVC/EVC layout, K = L/A conversions. Not for: choosing the REQUIRED curve length from sight distance (design charts), unsymmetrical curves.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Load a sample problem:
pvi station
stations of interestAdditional stations at which to compute elevations
stations of interest 1

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

US crest: g1 = +3%, g2 = −2%, L = 600 ft, PVI 25+00 at 100.00 ft

Given

g1percent
3
g2percent
-2
length
600 ft
pvi station
25+00
pvi elevation
100 ft

Assumptions

  • Symmetric (equal-tangent) parabolic vertical curve: BVC and EVC each lie L/2 horizontally from the PVI; x is horizontal distance from the BVC.
  • A = g2 − g1 < 0: this is a CREST curve (any interior turning point is a high point).
  • Stationing is US survey stationing in feet ("25+00" = 2500 ft), increasing in the direction of travel.

Solution steps

  1. Grade change and rate of vertical curvature

    A is the algebraic grade change; K is the horizontal length needed per 1% of grade change (larger K = flatter curve).

    A = g2 − g1; K = L/|A|

    A = -2 − (3) = -5 percent; K = 600 ft / 5 = 120 ft per percent

    = 120 ft

  2. Locate BVC and EVC

    The curve begins and ends half its length either side of the PVI; end elevations follow the tangent grades from the PVI.

    yBVC = yPVI − (g1/100)·(L/2); yEVC = yPVI + (g2/100)·(L/2)

    BVC at 2200 ft, elevation 91 ft; EVC at 2800 ft, elevation 94 ft

    = 91 ft

  3. Elevation equation

    The symmetric parabola runs tangent to g1 at the BVC and to g2 at the EVC; x is measured horizontally from the BVC.

    y(x) = yBVC + (g1/100)·x + ((g2 − g1)/100)·x²/(2L)

    with yBVC = 91 ft, g1 = 3 percent, A = -5 percent, L = 600 ft

  4. High point of the crest

    The turning point sits where the curve's grade passes through zero; it lies ON the curve only when x* lands between the BVC and EVC.

    x* = −g1·L/(g2 − g1), valid only for 0 ≤ x* ≤ L

    x* = 360 ft from the BVC → station 2560 ft, elevation 96.4 ft

    = 96.4 ft

Vertical curve profileVertical curve profile96.40022002800Station (ft)Elevation (ft)BVC (2200, 91)EVC (2800, 94)high point (2560, 96.4)

Results

Grade change A = g2 − g1 (percent)

-5

Rate of vertical curvature K (length per 1% of A)

120ft

Curve length L

600ft

BVC station 21+100.00

2200ft

BVC elevation

91ft

EVC station 28+00.00

2800ft

EVC elevation

94ft

High point station 25+60.00

2560ft

High point elevation

96.4ft

Where this answer was checked
source
Symmetric parabolic curve closed forms (surveying-handbook route) — BVC/EVC elevations, K, and the interior high point
verified by
hand-recomputed
derivation
A = −2 − 3 = −5 (crest); K = 600/5 = 120 ft/%. BVC = 2500 − 300 = 2200 ft (22+00), yBVC = 100 − 0.03·300 = 91.00 ft. EVC = 2800 ft, yEVC = 100 − 0.02·300 = 94.00 ft. x* = −3·600/(−5) = 360 ft ∈ [0, 600] → station 2560 (25+60); y* = 91 + 0.03·360 + (−5/100)·360²/(2·600) = 91 + 10.8 − 0.05·129600/1200 = 91 + 10.8 − 5.4 = 96.40 ft.

Example 2

SI sag: g1 = −2.5%, g2 = +1.5%, L = 200 m, PVI at 1200 m, elev 150 m

Given

g1percent
-2.5
g2percent
1.5
length
200 m
pvi station
1200 m
pvi elevation
150 m

Assumptions

  • Symmetric (equal-tangent) parabolic vertical curve: BVC and EVC each lie L/2 horizontally from the PVI; x is horizontal distance from the BVC.
  • A = g2 − g1 > 0: this is a SAG curve (any interior turning point is a low point).

Solution steps

  1. Grade change and rate of vertical curvature

    A is the algebraic grade change; K is the horizontal length needed per 1% of grade change (larger K = flatter curve).

    A = g2 − g1; K = L/|A|

    A = 1.5 − (-2.5) = 4 percent; K = 200 m / 4 = 50 m per percent

    = 50 m

  2. Locate BVC and EVC

    The curve begins and ends half its length either side of the PVI; end elevations follow the tangent grades from the PVI.

    yBVC = yPVI − (g1/100)·(L/2); yEVC = yPVI + (g2/100)·(L/2)

    BVC at 1100 m, elevation 152.5 m; EVC at 1300 m, elevation 151.5 m

    = 152.5 m

  3. Elevation equation

    The symmetric parabola runs tangent to g1 at the BVC and to g2 at the EVC; x is measured horizontally from the BVC.

    y(x) = yBVC + (g1/100)·x + ((g2 − g1)/100)·x²/(2L)

    with yBVC = 152.5 m, g1 = -2.5 percent, A = 4 percent, L = 200 m

  4. Low point of the sag

    The turning point sits where the curve's grade passes through zero; it lies ON the curve only when x* lands between the BVC and EVC.

    x* = −g1·L/(g2 − g1), valid only for 0 ≤ x* ≤ L

    x* = 125 m from the BVC → station 1225 m, elevation 150.938 m

    = 150.938 m

Vertical curve profileVertical curve profile152.50011001300Station (m)Elevation (m)BVC (1100, 152.5)EVC (1300, 151.5)low point (1225, 150.9)

Results

Grade change A = g2 − g1 (percent)

4

Rate of vertical curvature K (length per 1% of A)

50m

Curve length L

200m

BVC station

1100m

BVC elevation

152.5m

EVC station

1300m

EVC elevation

151.5m

Low point station

1225m

Low point elevation

150.938m

Where this answer was checked
source
Symmetric parabolic curve closed forms (metric) — BVC/EVC and the interior low point of a sag
verified by
hand-recomputed
derivation
A = 1.5 − (−2.5) = +4 (sag); K = 200/4 = 50 m/%. BVC = 1100 m, yBVC = 150 + 0.025·100 = 152.50 m; EVC = 1300 m, yEVC = 150 + 0.015·100 = 151.50 m. x* = 2.5·200/4 = 125 m ∈ [0, 200] → station 1225 m; y* = 152.5 − 0.025·125 + (4/100)·125²/(2·200) = 152.5 − 3.125 + 0.04·15625/400 = 152.5 − 3.125 + 1.5625 = 150.9375 m.

Example 3

K given: K = 150 ft, g1 = −1%, g2 = +3%, PVI 40+00 at 500.00 ft, elevation at 39+50

Given

g1percent
-1
g2percent
3
k value
150 ft
pvi station
40+00
pvi elevation
500 ft
stations of interest
  1. 1.39+50

Assumptions

  • Symmetric (equal-tangent) parabolic vertical curve: BVC and EVC each lie L/2 horizontally from the PVI; x is horizontal distance from the BVC.
  • A = g2 − g1 > 0: this is a SAG curve (any interior turning point is a low point).
  • Stationing is US survey stationing in feet ("25+00" = 2500 ft), increasing in the direction of travel.

Solution steps

  1. Grade change and rate of vertical curvature

    A is the algebraic grade change; K is the horizontal length needed per 1% of grade change (larger K = flatter curve).

    A = g2 − g1; K = L/|A|

    A = 3 − (-1) = 4 percent; K = 600 ft / 4 = 150 ft per percent

    = 150 ft

  2. Locate BVC and EVC

    The curve begins and ends half its length either side of the PVI; end elevations follow the tangent grades from the PVI.

    yBVC = yPVI − (g1/100)·(L/2); yEVC = yPVI + (g2/100)·(L/2)

    BVC at 3700 ft, elevation 503 ft; EVC at 4300 ft, elevation 509 ft

    = 503 ft

  3. Elevation equation

    The symmetric parabola runs tangent to g1 at the BVC and to g2 at the EVC; x is measured horizontally from the BVC.

    y(x) = yBVC + (g1/100)·x + ((g2 − g1)/100)·x²/(2L)

    with yBVC = 503 ft, g1 = -1 percent, A = 4 percent, L = 600 ft

  4. Low point of the sag

    The turning point sits where the curve's grade passes through zero; it lies ON the curve only when x* lands between the BVC and EVC.

    x* = −g1·L/(g2 − g1), valid only for 0 ≤ x* ≤ L

    x* = 150 ft from the BVC → station 3850 ft, elevation 502.25 ft

    = 502.25 ft

  5. Elevations at the requested stations

    Each station's offset x from the BVC goes into the elevation equation; stations off the curve use the tangent grades instead.

    station 3950 ft (x = 250 ft): 502.583 ft

    = 502.583 ft

Vertical curve profileVertical curve profile5090037004300Station (ft)Elevation (ft)BVC (3700, 503)EVC (4300, 509)low point (3850, 502.3)

Results

Grade change A = g2 − g1 (percent)

4

Rate of vertical curvature K (length per 1% of A)

150ft

Curve length L

600ft

BVC station 37+00.00

3700ft

BVC elevation

503ft

EVC station 43+00.00

4300ft

EVC elevation

509ft

Low point station 38+50.00

3850ft

Low point elevation

502.25ft

Elevation at station 39+50.00

502.583ft

Where this answer was checked
source
Symmetric parabolic curve closed forms via K = L/|A| — L from K, low point, and elevation at an intermediate station
verified by
hand-recomputed
derivation
A = 3 − (−1) = 4; L = K·|A| = 150·4 = 600 ft. BVC = 4000 − 300 = 3700 ft (37+00), yBVC = 500 + 0.01·300 = 503.00 ft; EVC = 4300 ft, yEVC = 500 + 0.03·300 = 509.00 ft. Low point: x* = 1·600/4 = 150 ft → station 38+50; y* = 503 − 0.01·150 + 0.04·150²/(2·600) = 503 − 1.5 + 0.75 = 502.25 ft. At 39+50: x = 3950 − 3700 = 250 ft; y = 503 − 0.01·250 + 0.04·250²/1200 = 503 − 2.5 + 2.0833 = 502.5833 ft.