Levelling loops
Profile levelling reduced to a profile and a grade line
Reduce a profile level run with intermediate foresights to ground elevations at six stations, then compute a grade line at -1.20 percent and the cut or fill at each station.
Given
- A profile level run along a proposed roadway centerline from station 10+00 to station 15+00, run from BM-7 and closed on BM-8. Distances in feet; stations are 100 ft apart.
- Elevation of BM-7 = 528.64 ft. Elevation of BM-8 is unknown and is to be carried through.
- Setup 1: backsight 6.42 on BM-7. Intermediate foresights (ground rod readings): 4.8 at 10+00, 5.6 at 11+00, 7.1 at 12+00. Foresight 8.94 to TP-1.
- Setup 2: backsight 3.17 on TP-1. Intermediate foresights: 2.4 at 13+00, 4.0 at 14+00, 5.5 at 15+00. Foresight 6.28 on BM-8.
- Rounding convention: benchmarks, turning points and heights of instrument to 0.01 ft; ground rod readings and ground elevations to 0.1 ft, which is all a rod held on soil can support.
- The proposed grade line passes through elevation 530.30 ft at station 10+00 and falls at 1.20 percent.
Required
- The reduced profile notes, including the elevation of TP-1 and BM-8.
- Ground elevation at each of the six stations.
- Grade elevation and the cut or fill at each station.
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
Intermediate foresights do not move the instrument
Profile levelling differs from differential levelling in one respect only: between the backsight and the closing foresight of a setup, the rod is read on a series of ground points. Those readings are intermediate foresights. They are subtracted from the same height of instrument, they do not create a new height of instrument, and they never carry the line forward.
That distinction is what keeps a rough ground shot from contaminating the level line. A ground rod reading taken on soft turf may be a tenth of a foot in error and it does not matter, because the next setup is established from the turning point, which was chosen to be a firm, well-defined object and read to 0.01 ft.
HI(1) = 528.64 + 6.42 = 535.06
ground elevation = HI - intermediate foresight
TP-1 = 535.06 - 8.94 = 526.12
HI(2) = 526.12 + 3.17 = 529.29
BM-8 = 529.29 - 6.28 = 523.01The reduced note page
Written out, the page keeps intermediate foresights in their own column so that the page check can still be worked on the turning-point foresights alone. Mixing them into the foresight column is the single most common way to wreck the check on a profile page.
| Station | BS (+) | HI | IFS | FS (-) | Elevation |
|---|---|---|---|---|---|
| BM-7 | 6.42 | 535.06 | - | - | 528.64 |
| 10+00 | - | - | 4.8 | - | 530.3 |
| 11+00 | - | - | 5.6 | - | 529.5 |
| 12+00 | - | - | 7.1 | - | 528.0 |
| TP-1 | 3.17 | 529.29 | - | 8.94 | 526.12 |
| 13+00 | - | - | 2.4 | - | 526.9 |
| 14+00 | - | - | 4.0 | - | 525.3 |
| 15+00 | - | - | 5.5 | - | 523.8 |
| BM-8 | - | - | - | 6.28 | 523.01 |
| Sums | 9.59 | - | - | 15.22 | - |
Page check
Only the backsights and the turning-point and benchmark foresights enter the check, because only they change the height of instrument or carry the line. The intermediate foresights are read off an existing height of instrument and cancel out of the identity entirely.
sum BS = 6.42 + 3.17 = 9.59
sum FS = 8.94 + 6.28 = 15.22
sum BS - sum FS = -5.63
BM-8 - BM-7 = 523.01 - 528.64 = -5.63 (agrees)The grade line
A grade line is a straight line in profile, so its elevation changes by the grade times the horizontal distance. A grade of 1.20 percent falling means 1.20 ft of drop per 100 ft of station, so each full station drops 1.20 ft exactly.
Grade elevations are carried to 0.01 ft because they are computed, not observed, and there is no reason to throw away precision in a design figure. The cut or fill is then reported to 0.1 ft, matching the ground elevations, since a difference can be no better than its worse term.
grade elevation = 530.30 - 0.0120 x (station - 1000)
11+00: 530.30 - 0.0120 x 100 = 529.10
12+00: 530.30 - 0.0120 x 200 = 527.90
13+00: 530.30 - 0.0120 x 300 = 526.70
14+00: 530.30 - 0.0120 x 400 = 525.50
15+00: 530.30 - 0.0120 x 500 = 524.30Cut and fill at each station
Cut is ground above grade and fill is grade above ground, so the signed difference ground minus grade is positive in cut and negative in fill. Reading down the column shows the roadway starting exactly at grade, running in shallow cut for three stations and passing into fill somewhere between 13+00 and 14+00.
| Station | Ground elev (ft) | Grade elev (ft) | Cut (+) / Fill (-) (ft) |
|---|---|---|---|
| 10+00 | 530.3 | 530.30 | 0.0 |
| 11+00 | 529.5 | 529.10 | +0.4 |
| 12+00 | 528.0 | 527.90 | +0.1 |
| 13+00 | 526.9 | 526.70 | +0.2 |
| 14+00 | 525.3 | 525.50 | -0.2 |
| 15+00 | 523.8 | 524.30 | -0.5 |
Where the grade line crosses the ground
The point of no cut and no fill lies between 13+00 and 14+00, where the signed difference changes from +0.2 to -0.2. Interpolating linearly between the two stations puts it at the midpoint, station 13+50, which is close enough for a preliminary profile given that the ground elevations are only good to 0.1 ft.
In practice a grade-crossing station computed from tenth-of-a-foot ground shots should be treated as approximate and staked from a cross-section run at the location rather than from the profile alone.
grade point = 1300 + 100 x (0.2 / (0.2 + 0.2)) = 1300 + 50 = station 13+50Answer
- TP-1 = 526.12 ft; BM-8 = 523.01 ft
- Ground elevations: 10+00 = 530.3, 11+00 = 529.5, 12+00 = 528.0, 13+00 = 526.9, 14+00 = 525.3, 15+00 = 523.8 ft
- Grade elevations: 530.30, 529.10, 527.90, 526.70, 525.50, 524.30 ft
- Cut/fill: 0.0, +0.4 cut, +0.1 cut, +0.2 cut, -0.2 fill, -0.5 fill
- Grade crosses existing ground at approximately station 13+50
Check
Page check: sum BS 9.59 - sum FS 15.22 = -5.63 ft, and BM-8 minus BM-7 = 523.01 - 528.64 = -5.63 ft.
Grade-line check: the total drop from 10+00 to 15+00 is 530.30 - 524.30 = 6.00 ft over 500 ft, which is 1.200 percent, the design grade.
Ground-slope check: the ground falls 530.3 - 523.8 = 6.5 ft over the same 500 ft, an average of 1.30 percent, slightly steeper than grade. That is consistent with the profile passing from cut into fill, as the table shows.
More levelling loops
All in this category- Differential level circuit reduced from field notesReduce a four-setup differential level run from BM-1 to BM-2 through three turning points, then prove the page with the sum of backsights against the sum of foresights.
- Closed level loop tested against a tolerance and adjustedA four-leg level loop returns to its starting benchmark 0.060 ft out. Test the misclosure against a C times root-M tolerance, then distribute it in proportion to leg length and compute adjusted elevations.
- Level net of three loops adjusted by inspectionThree benchmarks and an interior junction point are connected by six level lines forming three independent loops. Adjust the net by inspection, distributing each loop misclosure by line length and iterating until every loop closes.