11.2 High and Low Points and Intermediate Elevations

11.2 High and Low Points and Intermediate Elevations

Quick Answer: From the PVC, x_ext = −g1 L / A. If 0 < x_ext < L, that station is the high point (crest) or low point (sag) on the curve. On the §11.1 crest (g1 = +2.0%, g2 = −1.5%, L = 400 ft, PVC 20+00 at 520.00 ft): x_ext = 228.57 ft, station 22+28.57, elevation 522.29 ft. The PVI at 22+00 is not that summit.

CES Domain III item F lists high/low point and intermediate point by name. This section is those two calculations on the equal-tangent parabola from §11.1. Line and grade from a plan-profile sheet is §11.3.

Where the Turning Point Is

Along the parabola the instantaneous grade is the derivative of elevation with respect to x:

g(x) = g1 + (A / L) x

(The rate of change of grade is r = A/L per foot if A and L are decimal and feet.) Set g(x) = 0 to find a horizontal tangent:

0 = g1 + (A / L) x_ext → x_ext = −g1 L / A

g1 and A must be in the same units (both decimal or both percent). x_ext then has the same length unit as L.

Percent form (identical x): x_ext = −g1% × L / A%.

On-curve test: the turning point is a usable high/low on this curve only if 0 < x_ext < L. That happens when g1 and g2 have opposite signs — the profile actually goes through a horizontal tangent between PVC and PVT.

• A < 0 (crest) and x_ext on the curve → high point (maximum elevation).
• A > 0 (sag) and x_ext on the curve → low point (minimum elevation).

If x_ext < 0 or x_ext > L, there is no interior stationary point. The highest or lowest elevation on the curve segment you were given is then at an end: PVC or PVT. Compute both end elevations and pick the extreme. Do not report a high point 50 ft before the PVC just because the formula went negative.

Elevation at the turning point (either plug x_ext into the PVC formula, or use the compact form):

elev_ext = elev_PVC − (g1² L) / (2A)

Same unit discipline: g1 decimal with A decimal, L in feet → elevation in feet.

Worked High Point (Same Crest as §11.1)

g1 = +0.020, A = −0.035, L = 400 ft, PVC 20+00, elev_PVC 520.00 ft.

x_ext = −(0.020)(400) / (−0.035) = −8.00 / −0.035 = 228.571 ft → 228.57 ft
Percent check: x = −(2.0)(400) / (−3.5) = 800 / 3.5 = 228.57 ft.

0 < 228.57 < 400, so the turning point is on the curve. A < 0 → it is a high point.

Station = 20+00 + 2+28.57 = 22+28.57.

Compare with the PVI at 22+00 (x = 200). The high point is 28.57 ft past the PVI, toward the PVT, because |g2| = 1.5% is flatter than g1 = 2.0%. The parabola has to travel farther to bleed off the steeper incoming climb.

Elevation:
g1 x = 0.020 × 228.571 = 4.571 ft
A/(2L) = −0.035 / 800 = −0.00004375
(A/(2L)) x² = −0.00004375 × 52,244.9 = −2.286 ft
elev = 520.00 + 4.571 − 2.286 = 522.286 ft → 522.29 ft

Compact check: elev_ext = 520.00 − (0.020² × 400) / (2 × −0.035)
= 520.00 − (0.0004 × 400) / (−0.070)
= 520.00 − (0.160 / −0.070)
= 520.00 − (−2.286) = 522.29 ft.

Curve at the PVI station was 522.25 ft. The high point is 0.04 ft higher and 28.57 ft farther along. On a grading plan that 0.04 ft is a hundredths decision; on the exam it is the difference between 522.25, 522.29, and the PVI’s 524.00.

The PVI Is Not Automatically the High Point

On an equal-tangent curve the PVI station is x = L/2. Set x_ext = L/2:

−g1 L / A = L/2 → −2 g1 = A = g2 − g1 → g2 = −g1.

So the high or low point falls at the PVI station only when the incoming and outgoing grades are equal in magnitude and opposite in sign (a symmetric crest or sag). The worked curve has +2.0% and −1.5%, not a match, so the summit is not at 22+00.

Exam trap: “It is a crest, so the high point is the PVI.” False unless g1 = −g2. The PVI is a tangent intersection, not a turning point of the parabola.

If g1 = +2.0% and g2 = −2.0% with the same L = 400 ft, then x_ext = 200 ft, the high point would sit under the PVI in station, and E = AL/8 would also be the offset from that summit to the PVI. That is the special case, not the rule.

Intermediate Elevations at 50-ft Stations

CES “intermediate point” means: pick a station between PVC and PVT, set x = station − PVC station, and run

elev = 520.00 + 0.020 x − 0.00004375 x².

Do not interpolate linearly between PVC and PVT — that is the chord, not the parabola. Do not use the tangent elevations except as a check that y is the offset from the g1 tangent.

Read the table as a crest: elevations climb from 520.00 through 522.25 at 22+00, peak at 522.29 at 22+28.57, then ease to 521.00 at the PVT. Station 22+50 is already past the high point (522.27 < 522.29).

Worked intermediate at 21+00 (x = 100):
y = (−3.5)(100)² / (200 × 400) = −35,000 / 80,000 = −0.4375 ft
elev = 520.00 + 2.00 − 0.44 = 521.56 ft.

Worked intermediate at 23+00 (x = 300):
y = (−3.5)(300)² / (200 × 400) = −315,000 / 80,000 = −3.9375 ft
elev = 520.00 + 6.00 − 3.94 = 522.06 ft.

When Grades Do Not Reverse

Example (not the main worked curve): g1 = +1.0%, g2 = +3.0%, L = 400 ft. A = +2.0% > 0 (sag shape, curve above the tangents), but both grades climb.

x_ext = −(1.0)(400) / (+2.0) = −200 ft (before the PVC). There is no low point on the curve. The lowest on-curve elevation is the PVC; the PVT is higher. Reporting x = −200 as “the low point” is a formula without the on-curve test.

Exam Traps

1. High point = PVI always — only when g1 = −g2 on an equal-tangent curve.
2. Using curve elevation 522.25 at 22+00 as the summit — close, still wrong; summit is 522.29 at 22+28.57.
3. Linear interpolation PVC to PVT — ignores the parabola; 21+00 would come out 520.25 instead of 521.56.
4. x_ext off the curve reported anyway — if grades do not reverse, the extreme is an end point.
5. HDM K used to find the high point — K = L/A does not locate x_ext. Use −g1 L / A.