Levelling loops
Cross sections reduced to end areas and an earthwork volume
Two roadway cross sections in cut are reduced from slope-stake notes to end areas by the coordinate method, and the volume between them is computed by average end area.
Given
- Two cross sections taken 100.00 ft apart on a roadway in cut, at stations 12+00 and 13+00. All dimensions in feet.
- Subgrade template: level, 24.00 ft wide, extending 12.00 ft either side of centerline. Backslopes are 1.5 horizontal to 1 vertical.
- Cross-section notes at 12+00, given as cut above subgrade at offset from centerline (negative left, positive right): left slope stake 6.2 at -21.3; left subgrade edge 3.4 at -12.0; centerline 2.6 at 0.0; right subgrade edge 4.1 at +12.0; right slope stake 7.0 at +22.5.
- Cross-section notes at 13+00: left slope stake 4.4 at -18.6; left subgrade edge 2.0 at -12.0; centerline 1.5 at 0.0; right subgrade edge 3.2 at +12.0; right slope stake 5.6 at +20.4.
- Areas are carried to 0.1 square foot and volumes to the nearest cubic yard.
Required
- Verification that the slope stakes are consistent with a 1.5 to 1 backslope.
- The end area of each cross section.
- The volume of excavation between the two sections by average end area, in cubic yards.
Work it through yourself before reading on — the solution below shows every step, so there is no way to skim it without giving the answer away.
Worked solution
Check the slope stakes before computing anything
A slope stake is set where the design backslope meets existing ground, so the horizontal run from the subgrade edge to the stake divided by the cut at the stake must equal the design slope ratio. Checking this first costs a minute and catches the commonest field blunder, which is a slope stake set from the wrong subgrade edge or with the wrong slope ratio.
All four stakes check exactly at 1.500, so the notes are internally consistent and the sections can be treated as closed figures.
12+00 left: (21.3 - 12.0) / 6.2 = 9.3 / 6.2 = 1.500
12+00 right: (22.5 - 12.0) / 7.0 = 10.5 / 7.0 = 1.500
13+00 left: (18.6 - 12.0) / 4.4 = 6.6 / 4.4 = 1.500
13+00 right: (20.4 - 12.0) / 5.6 = 8.4 / 5.6 = 1.500Turn each section into a closed polygon
The excavation at a station is the area between the original ground above and the finished template below. Treating offset as one coordinate and height above subgrade as the other turns that area into a closed polygon, which can then be computed by the coordinate (shoelace) method with no trapezoid bookkeeping at all.
Walk the boundary in order: start at the left subgrade edge, climb the left backslope to the left slope stake, follow the original ground across to the right slope stake, come down the right backslope to the right subgrade edge, and close along the flat subgrade. The subgrade edges are at height zero by definition, because heights are being measured from the subgrade.
| Vertex | Offset x (ft) | Height y (ft) | Description |
|---|---|---|---|
| 1 | -12.0 | 0.0 | left subgrade edge |
| 2 | -21.3 | 6.2 | left slope stake |
| 3 | -12.0 | 3.4 | ground at left subgrade edge |
| 4 | 0.0 | 2.6 | ground at centerline |
| 5 | +12.0 | 4.1 | ground at right subgrade edge |
| 6 | +22.5 | 7.0 | right slope stake |
| 7 | +12.0 | 0.0 | right subgrade edge |
End area at 12+00 by the coordinate method
The coordinate method accumulates the cross products x(i) times y(i+1) minus x(i+1) times y(i) around the closed figure; half the absolute value of that sum is the area. The sign of the sum only records the direction the boundary was walked and carries no information about the section.
Working the seven products in order gives a doubled area of -227.07, so the end area is 113.535 square feet, reported as 113.5 square feet.
1-2: (-12.0)(6.2) - (-21.3)(0.0) = -74.40
2-3: (-21.3)(3.4) - (-12.0)(6.2) = +1.98
3-4: (-12.0)(2.6) - (0.0)(3.4) = -31.20
4-5: (0.0)(4.1) - (12.0)(2.6) = -31.20
5-6: (12.0)(7.0) - (22.5)(4.1) = -8.25
6-7: (22.5)(0.0) - (12.0)(7.0) = -84.00
7-1: (12.0)(0.0) - (-12.0)(0.0) = 0.00
sum = -227.07 ; A(12+00) = 227.07 / 2 = 113.535 -> 113.5 sq ftEnd area at 13+00
The same seven vertices, with the 13+00 notes, give a smaller section: the cut has shallowed by roughly a foot on centerline and the slope stakes have moved in.
vertices: (-12.0, 0.0) (-18.6, 4.4) (-12.0, 2.0) (0.0, 1.5) (+12.0, 3.2) (+20.4, 5.6) (+12.0, 0.0)
1-2: (-12.0)(4.4) - (-18.6)(0.0) = -52.80
2-3: (-18.6)(2.0) - (-12.0)(4.4) = +15.60
3-4: (-12.0)(1.5) - (0.0)(2.0) = -18.00
4-5: (0.0)(3.2) - (12.0)(1.5) = -18.00
5-6: (12.0)(5.6) - (20.4)(3.2) = +1.92
6-7: (20.4)(0.0) - (12.0)(5.6) = -67.20
7-1: 0.00
sum = -138.48 ; A(13+00) = 138.48 / 2 = 69.240 -> 69.2 sq ftVolume by average end area
The average end area rule multiplies the mean of the two end areas by the distance between them. It is exact only when the section varies linearly along the alignment, and it overstates the volume whenever the section is changing shape as well as size, which is the usual case in a transition from deeper to shallower cut.
Working with the rounded areas keeps the reported volume consistent with the precision of the field notes.
V = L x (A1 + A2) / 2
V = 100.00 x (113.5 + 69.2) / 2
V = 100.00 x 91.35 = 9,135 cubic feet
V = 9,135 / 27 = 338.3 cubic yardsHow much the rule is likely to be overstating
The prismoidal correction gives a feel for the size of the bias. For sections of this shape the prismoidal formula uses a middle area computed from the mean of corresponding dimensions rather than the mean of the end areas, and it always returns the smaller figure when one section is markedly larger than the other.
Here the two end areas differ by a factor of 1.6, so the average end area result is high by something on the order of one to two percent, or three to six cubic yards. For a pay quantity on a long alignment that bias accumulates and is worth carrying; for a single 100 ft interval it is well inside the uncertainty of the ground shots themselves.
Answer
- End area at 12+00 = 113.5 sq ft
- End area at 13+00 = 69.2 sq ft
- Volume between the sections by average end area = 9,135 cubic feet = 338 cubic yards of excavation
Check
Independent check on 12+00 by trapezoids: the area under the ground line from -21.3 to +22.5 is 44.64 + 36.00 + 40.20 + 58.275 = 179.115 sq ft, and the area under the template over the same width is 28.83 + 0 + 36.75 = 65.58 sq ft. The difference is 113.535 sq ft, matching the coordinate method exactly.
Independent check on 13+00 by trapezoids: 21.12 + 21.00 + 28.20 + 36.96 = 107.28 sq ft under the ground line, less 14.52 + 23.52 = 38.04 sq ft under the template, gives 69.24 sq ft.
Order-of-magnitude check on the volume: mean cut on centerline is about 2.05 ft over a roadbed 24 ft wide plus backslopes, over 100 ft. That is roughly 300 to 400 cubic yards, consistent with 338.
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