Home / Calculators / Vertical Curve Elevations (Symmetric Parabola) Transportation & Surveying
Calculate elevations and stations on a symmetric parabolic vertical curve. Find BVC/EVC, K value, and any high or low point within the curve.
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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 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
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
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
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 profile Vertical curve profile 96.4 0 0 2200 2800 Station (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
BVC station 21+100.00
2200ft
EVC station 28+00.00
2800ft
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 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
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
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
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 profile Vertical curve profile 152.5 0 0 1100 1300 Station (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
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. 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 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
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
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
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
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 profile Vertical curve profile 509 0 0 3700 4300 Station (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
BVC station 37+00.00
3700ft
EVC station 43+00.00
4300ft
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.