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Levels between surveyed points

What is worked out: a level where nobody held a staff. CSE does it in three places:

  • Along a section, between two of its points: a straight line.
  • Between surveyed chainages in a field book (Interpolate): triangles across chainage and offset, and a straight line along the side of a triangle.
  • On a From coordinates route (Chainages at an interval…): a TIN of the surveyed points in plan, with breaklines along the codes, and the plane of the triangle a point falls in.

Each is linear: no curve is fitted, and no level is guessed beyond the survey except where this page says so.

1. Along a section

A cross section's ground is a straight line from each point to the next. The level at an offset x between two points (x₁, L₁) and (x₂, L₂) is:

L = L₁ + (L₂ − L₁) × (x − x₁) ÷ (x₂ − x₁)

This straight line is the ground wherever CSE needs it between points:

Beyond the outermost point on either side there is no line to follow. When a template is laid on the ground, the ground is held level at the outermost point's level from there on. A side slope that runs out past the survey therefore meets a level ground: survey wide enough that the toes and the tops of the cuttings fall inside your points.

2. Between surveyed chainages in a field book: Interpolate

Interpolate (the Field Book panel, under Readings) puts in sections at chainages you did not survey, from the ones you did. Choose From, To and Interval (m). See Enter a field book.

The chainages

  • The new chainages are From + Interval, From + 2 × Interval, and so on, up to but not including To. Each is worked out from From (not added one to the next), so a step of 0.1 m does not drift, and each is rounded to 0.001 m.
  • A chainage already surveyed (within half a millimetre) is not added again.
  • A chainage is left out when either surveyed chainage beside it has no points with both a Distance and an R.L.
  • Interpolated rows already between From and To are taken out first and worked out again.
  • Only surveyed rows are used: an interpolated row never feeds another interpolation.
  • More than 5,000 chainages is refused: That makes more than 5,000 chainages: choose a longer interval.

The dialog says what it will add before you click: Adds 4 chainages between 80 and 160 (3 are already surveyed).

The levels

  1. The surveyed points of the chainages either side are joined into triangles (a Delaunay triangulation), drawn as if the field book were a plan: chainage along, offset across.
  2. Two extra points go in at every chainage, 5 m beyond the widest and the narrowest offset of all those points, at those two points' levels. They give the triangles near the edges something to lean on; no new row is made beyond the surveyed offsets.
  3. At each new chainage, a line is drawn across. Every place where it crosses the side of a triangle gives one new point: its offset is where the side is crossed, and its level is read along that side, in a straight line between the side's two ends.
  4. Only points between the smallest and largest offset of the two neighbouring chainages are kept, one per offset, from left to right. Offsets and levels are rounded to 0.001.

Where both chainages were surveyed at the same offset, a triangle side joins those two points, so that offset comes out in the new section with its level straight between the two. A side that runs diagonally from one offset to another gives a point between them. Which way a diagonal runs is the triangulation's choice; it is the same each time for the same points.

The rows

  • Each chainage's rows go after the last row of the surveyed chainage before it, in the sheet's order, with the chainage on the first row only.
  • In a level book (B.S., I.S., F.S.), each row is an I.S. on the set-up in use where it goes: I.S. = H.I. − R.L., rounded to 0.001, and the R.L. is written as H.I. − I.S., so Calculate R.L.s gives it back unchanged and the level check still closes.
  • In a field book of levels only (no staff readings), each row gets the R.L. only.
  • The Remark is Interpolated (Ch 100–120), naming the two surveyed chainages, and the row is shaded.

Interpolate refuses a level book whose R.L.s do not go with the readings (Calculate R.L.s first), one where a row has no staff reading above other readings, and one sorted past a change point (The rows aren't in the order they were read): in each case the H.I. at the place a row would go cannot be known.

3. On a From coordinates route: the TIN

Chainages at an interval… (the Interval sections fold of the From Coordinates panel) places points at chosen offsets on sections every so many metres, and gives each a level from the surveyed points. See Sections every N metres with offsets.

The points of the TIN

The TIN (triangulated irregular network) is made of the route's own points in plan, by Easting and Northing, each with its R.L.:

  • every point in a surveyed section, and every point kept on the plan only;
  • not a point of a code under Leave out the codes, not a point merged into another (under 5 cm from it), and not an Interpolated row;
  • of two points at the same E and N, the first.

Points on the plan only are ground to the TIN. A point that is not ground (a tree, a pole, the base station) should be left out by its code.

Breaklines along the codes

Lines are drawn along the codes from each surveyed section to the next, and no triangle may cross them:

  • In each surveyed section, the points of each code are numbered outwards from the alignment on each side: the first RE on the left, the second RE on the left, and so on. The centre-line code (the one under Join the points coded) is numbered the same way, as a side of its own.
  • Point k of a code on a side is joined to point k of the same code, on the same side, in the next surveyed section. The line stops at a section that has no such point; one starting further on is a new line.
  • Codes are matched ignoring case and the spaces around them. A point without a code joins nothing. Interval sections are never used for breaklines.
  • Where two breaklines cross in plan, both are split there, and the crossing takes the mean of their two levels at that place.
  • Where the triangulation has to add a point on a breakline to keep it, that point takes the breakline's level at that place, along the line.

If the triangulation cannot keep the breaklines, it is built from the points alone. If it cannot be built at all, the dialog says The levels cannot be worked out: the surveyed points could not be made into a TIN.

The level at a point

A new point is placed from the alignment, at its chainage and offset. Its level is the plane of the triangle it falls in, through the triangle's three corners. With corners A, B, C and the point P, in plan:

L(P) = w(A) × L(A) + w(B) × L(B) + w(C) × L(C)

where w(A) is the area of triangle PBC divided by the area of ABC, w(B) the area of PCA divided by it, and w(C) the area of PAB divided by it. The three weights add up to 1. A point on a triangle's side gets the same level from either triangle. The level is rounded to 0.001 in the row, with N and E.

What gets no level, and why

A point with no level is left out: it makes no row and gets no name, and a chainage whose points all have no level makes no section. The dialog counts them by offset code before you add them.

  • Outside the survey. The TIN reaches only as far as the straight-edged outline round all its points; a point beyond it has no triangle. The dialog says … points fall outside the survey and get no level: ….
  • Wider than the survey on its side. A point further from the centre line than the widest point on that side of both surveyed sections either side of it (the one at or before its chainage, and the one at or after it). Before the first or after the last surveyed section, the nearest section stands for both. A point exactly at the widest point's offset is within. The TIN may have a level there, from long triangles reaching across from further along the route, but it would be a guess. The centre line is never left out this way. The dialog says … points are wider than the surveyed sections on both sides and get no level: ….
  • No TIN. Fewer than three points, or all of them in one line, make no triangles, and no point gets a level.

On a route whose levels came from a contour job, the levels come from the contour surface instead (see Contour surface), and there is no Interval sections fold.

Conventions and rounding

What How
Along a section Straight line between neighbouring points; level beyond the outermost point when a template is laid
Interpolate: chainages From + k × Interval, each rounded to 0.001 m; one within 0.0005 m of a surveyed chainage is not added
Interpolate: offsets and levels Where the cross line meets a triangle's side; rounded to 0.001
Interpolate in a level book I.S. = H.I. − R.L., rounded to 0.001; R.L. = H.I. − I.S.
Interval sections: level The plane of the TIN triangle the point falls in
Interval sections: rows Chainage, Distance, N, E and R.L. rounded to 0.001
Breaklines that cross Split at the crossing, at the mean of their two levels there

Worked example 1: a straight line between surveyed points

A field book of levels only, two chainages 20 m apart:

Chainage Distance R.L.
100 −5 100.500
0 101.000
5 100.400
120 −5 100.900
0 101.600
5 101.000

Along the section. At Ch 100, the ground 2.5 m right of the centre line lies between the points at 0 and 5:

L = 101.000 + (100.400 − 101.000) × (2.5 − 0) ÷ (5 − 0) = 101.000 − 0.600 × 0.5 = 100.700

Interpolate, From 100, To 120, Interval 5 m, adds Ch 105, 110 and 115. At Ch 110, offset 0 lies on the triangle side from (Ch 100, 0, 101.000) to (Ch 120, 0, 101.600), halfway along it:

L = 101.000 + (101.600 − 101.000) × (110 − 100) ÷ (120 − 100) = 101.000 + 0.600 × 0.5 = 101.300

In the same way offset −5 gets 100.700 and offset 5 gets 100.700. A diagonal side from (Ch 100, 0, 101.000) to (Ch 120, −5, 100.900) is crossed halfway too, at offset 0 + (−5 − 0) × 0.5 = −2.5, level 101.000 + (100.900 − 101.000) × 0.5 = 100.950. Ch 110 gets five rows:

Distance R.L.
−5 100.700
−2.5 100.950
0 101.300
2.5 101.000
5 100.700

Each has the Remark Interpolated (Ch 100–120).

Worked example 2: a level from a triangle

Four surveyed points at the corners of a 10 m square, E and N in metres from a local origin:

Point E N R.L.
A 0 0 100.000
B 10 0 102.000
C 10 10 110.000
D 0 10 104.000

Suppose A–C and B–D are breaklines (each joins two points of one code). They cross at X (5, 5). Along A–C the level there is halfway from 100.000 to 110.000, 105.000; along B–D it is halfway from 102.000 to 104.000, 103.000. X takes the mean: (105.000 + 103.000) ÷ 2 = 104.000. The TIN is four triangles round X.

The level at P (E 5, N 2) comes from triangle A (0, 0), B (10, 0), X (5, 5), the one P falls in.

  • Area of ABX = 10 × 5 ÷ 2 = 25
  • Area of PBX = 7.5, so w(A) = 7.5 ÷ 25 = 0.3
  • Area of PXA = 7.5, so w(B) = 7.5 ÷ 25 = 0.3
  • Area of PAB = 10 × 2 ÷ 2 = 10, so w(X) = 10 ÷ 25 = 0.4

L(P) = 0.3 × 100.000 + 0.3 × 102.000 + 0.4 × 104.000 = 30.000 + 30.600 + 41.600 = 102.200

The same plane written out: L = 100.000 + 0.2 × E + 0.6 × N, which gives 100.000 at A, 102.000 at B, 104.000 at X and 100.000 + 1.000 + 1.200 = 102.200 at P.

On the edge of the TIN, at (10, 5) halfway between B and C, the level is 106.000. Just outside it, at (10.001, 5), there is no level: such a point falls outside the survey.