12.3 Areas, DMD, and Earthwork Volumes
Key Takeaways
- Double meridian distance (DMD) area of the four-sided closed traverse (lats +30, +50, −20, −60 ft and deps +80, +20, −70, −30 ft) is 3,500 ft²; the double-area sum is 7,000 ft².
- Average-end-area volume is V = L(A1+A2)/2. Between 50-ft sections of 108 ft² and 432 ft² the cut is 13,500 ft³ = 500 cy.
- Prismoidal volume is V = L(A1+4Am+A2)/6. With mid-depth area 252 ft² on the same 50 ft, V = 12,900 ft³.
- When a profile goes from cut to fill, interpolate the grade (zero-cut) station and split the interval; do not average a cut end area with a fill end area as one net prism.
- Convert cubic feet to cubic yards by dividing by 27, not by 25. Side-slope areas use A = d(W + s d) for level original ground and s:1 (H:V) slopes.
12.3 Areas, DMD, and Earthwork Volumes
Domain III letter J is calculating areas, with double meridian distance (DMD) named in the 2022 CES test plan. BPELSG also publishes a retired sample item that asks for an earthwork quantity from a profile and typical section. That sample is not reproduced here. This section teaches the same office method with different widths, depths, and stations, and it does not use the retired cubic-yard choices.
Earthwork is the natural next calculation after end areas: once a cross-section has an area of cut or fill, two stations and a length produce a volume. CES candidates should be fluent in both the area tool (DMD) and the volume tools (average-end-area and prismoidal).
DMD area for a closed traverse
Latitude is the northing component of a course (positive north). Departure is the easting component (positive east). For a closed figure, Σ latitude = 0 and Σ departure = 0 before you trust an area. If the traverse does not close, adjust it first (Chapter 8); DMD on raw unbalanced courses is not a closure substitute.
DMD rules:
- DMD of the first course = departure of the first course.
- DMD of each following course = DMD of the previous course + departure of the previous course + departure of the current course.
- Double area of a course = DMD × latitude (keep the sign of the latitude).
- Area = |Σ double area| / 2.
Starting on a different first course changes the DMD column but not the final area.
Worked four-sided example
| Course | Latitude (ft) | Departure (ft) |
|---|---|---|
| 1-2 | +30 | +80 |
| 2-3 | +50 | +20 |
| 3-4 | −20 | −70 |
| 4-1 | −60 | −30 |
| Sum | 0 | 0 |
The traverse closes. Vertices in order: (N, E) = (0, 0), (30, 80), (80, 100), (60, 30), back to (0, 0).
| Course | Lat | Dep | DMD | Double area = DMD × Lat |
|---|---|---|---|---|
| 1-2 | +30 | +80 | 80 | 80 × 30 = 2,400 |
| 2-3 | +50 | +20 | 80+80+20 = 180 | 180 × 50 = 9,000 |
| 3-4 | −20 | −70 | 180+20+(−70) = 130 | 130 × (−20) = −2,600 |
| 4-1 | −60 | −30 | 130+(−70)+(−30) = 30 | 30 × (−60) = −1,800 |
| Sum | 7,000 |
Area = 7,000 / 2 = 3,500 ft².
Shoelace check using easting then northing: Σ x_i y_{i+1} = 12,400, Σ y_i x_{i+1} = 5,400, |12,400 − 5,400| / 2 = 3,500 ft². Same result.
Traps: reporting 7,000 ft² (forgot to divide by 2); omitting the two negative-latitude courses so the sum is only 2,400 + 9,000 = 11,400 and the area comes out 5,700 ft² (dropping the minus signs instead gives 15,800 and 7,900 ft² — also wrong); using DMD × departure instead of DMD × latitude.
DMD is an area method for a closed figure of latitudes and departures. It is not a volume method. Earthwork uses the section areas from Chapter 12.1 and the formulas below.
End areas from a profile and typical section
A retired BPELSG sample asks for earthwork from a profile plus a typical section. The method, with different numbers, is:
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At each station, read existing ground and proposed grade from the profile. Depth d = |ground − grade|. Label it cut if ground is above grade, fill if ground is below grade.
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From the typical section, compute the end area of that cut or fill. For level original ground, roadbed width W, and side slopes s:1 (H:V):
A = d (W + s d)
That is a trapezoid: roadbed W, catch width s d on each side, so the wide side is W + 2 s d, and the average width times depth is d × (W + s d).
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Between two stations a distance L apart, apply average-end-area or prismoidal as the item requires.
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Convert cubic feet to cubic yards by dividing by 27.
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If the profile crosses grade (cut becomes fill), split at the interpolated zero-cut station. Never average a cut area with a fill area as if they were the two ends of one material.
Worked typical section. Level original ground. Finished roadbed W = 30 ft. Cut slopes 2:1 (H:V), so s = 2. Stations 50 ft apart.
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Sta 14+00: CL cut d1 = 3.0 ft A1 = 3 × (30 + 2 × 3) = 3 × 36 = 108 ft² Catch each side = 6 ft; top width = 42 ft; trapezoid (30+42)/2 × 3 = 108 ft².
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Sta 14+50: CL cut d2 = 9.0 ft A2 = 9 × (30 + 2 × 9) = 9 × 48 = 432 ft² Catch each side = 18 ft; top width = 66 ft; trapezoid (30+66)/2 × 9 = 432 ft².
If original ground is not level, you cannot use A = d(W + s d) from CL depth alone. Build the existing-ground polyline from contours or from rod readings, superimpose the typical section, and take the polygonal cut/fill area. The retired-sample skill still applies; only the area-getting step changes.
Average-end-area volume
The trapezoidal or average-end-area (AEA) formula is:
V = L (A1 + A2) / 2
With L = 50 ft, A1 = 108 ft², A2 = 432 ft²:
V = 50 × (108 + 432) / 2 = 50 × 540 / 2 = 50 × 270 = 13,500 ft³
In cubic yards: 13,500 / 27 = 500 cy of cut.
AEA is exact when area varies linearly between the two ends (a true prism or wedge with straight rulings). It is approximate when area includes a d² term from side slopes and depth changes along the alignment.
Prismoidal volume
The prismoidal formula is:
V = L (A1 + 4 Am + A2) / 6
Am is the area at mid-station, not automatically (A1+A2)/2. For linearly varying depth, compute mid-depth and run the typical-section formula again.
Mid-station 14+25: d_m = (3.0 + 9.0) / 2 = 6.0 ft Am = 6 × (30 + 2 × 6) = 6 × 42 = 252 ft²
Note that (A1+A2)/2 = (108+432)/2 = 270 ft², which is not Am. Area A = 30d + 2d² is quadratic in d, so the mid-depth area is smaller than the average of the end areas.
V = 50 × (108 + 4 × 252 + 432) / 6 = 50 × (108 + 1,008 + 432) / 6 = 50 × 1,548 / 6 = 50 × 258 = 12,900 ft³
12,900 / 27 = 477.78 cy, about 478 cy to the nearest cubic yard. AEA at 500 cy is 22 cy high on this interval because it treats area as if it varied linearly.
If an item does not give a mid-area or a mid-depth, it is usually asking for AEA. If it gives a mid-section, use prismoidal. Do not mix the two formulas in one interval.
| Quantity | Value |
|---|---|
| A1 at 3 ft cut | 108 ft² |
| Am at 6 ft cut | 252 ft² |
| A2 at 9 ft cut | 432 ft² |
| L | 50 ft |
| AEA volume | 13,500 ft³ = 500 cy |
| Prismoidal volume | 12,900 ft³ ≈ 478 cy |
Cut/fill interpolation (grade point)
When one end is cut and the other is fill, there is a grade point (zero cut/fill) between them. Interpolate station by depth, not by area.
Worked grade point. Sta 22+00 is 3.0 ft cut. Sta 23+00 is 5.0 ft fill. Interval = 100 ft. Total depth change from cut to fill = 3.0 + 5.0 = 6.0 ft.
Distance from 22+00 to grade = 100 × 3.0 / 6.0 = 37.5 ft → station 22+37.5. Fill occupies the remaining 62.5 ft.
Suppose the cut end area at 22+00 is 90 ft² and the fill end area at 23+00 is 160 ft². Split AEA:
- Cut: L = 37.5 ft, areas 90 and 0 → V_cut = 37.5 × (90+0)/2 = 1,687.5 ft³
- Fill: L = 62.5 ft, areas 0 and 160 → V_fill = 62.5 × (0+160)/2 = 5,000 ft³
Wrong: 100 × (90 − 160)/2, or 100 × (90+160)/2 treated as a single material. Those mix cut with fill and skip the grade point.
Unit and method traps
- Dividing by 25 instead of 27: 13,500 / 25 = 540 cy, a clean wrong number.
- Forgetting the 1/2 in AEA: 50 × (108+432) = 27,000 ft³ = 1,000 cy.
- Using A = W d and dropping side slopes: A1 = 90, A2 = 270, V = 50 × 360 / 2 = 9,000 ft³ = 333 cy.
- Reporting prismoidal 12,900 ft³ as 500 cy, or AEA 13,500 ft³ as 478 cy.
- Copying cubic-yard choices from a retired Board sample. Compute from the figure in front of you.
Independent OpenExamPrep practice for CES Domain III.J is DMD for closed figures and these volume formulas for earthwork. Shrinkage, swell, and pay-quantity rules appear only if the item states them; do not apply a 90% shrinkage habit that was not given.
A closed four-sided traverse has latitudes +30, +50, −20, −60 ft and departures +80, +20, −70, −30 ft. What is the DMD area?
A 30-ft roadbed with 2:1 cut slopes has end areas 108 ft² and 432 ft² at sections 50 ft apart (level original ground, depths 3 ft and 9 ft). What is the average-end-area cut volume?
For the same 50-ft interval, mid-depth is 6 ft and the mid-area from A = d(30 + 2d) is 252 ft². What is the prismoidal volume in cubic feet?