Surveying II – Week 2

Intersections
Intersections are the group of planar surveying calculations where we use two control points (three in the case of
resection) with known coordinates and certain angle/distance measurements to compute the coordinates of an
unknown point.

1. Intersections

1.1. Intersection using inner angles

Given information: coordinates of points 𝐴𝐴 and 𝐵𝐵.
Measured quantities: 𝛼𝛼 and 𝛽𝛽 inner angles.

Figure 1. Intersection using inner angles.

Computing the coordinates of the unknown point 𝑃𝑃:

1. Using the II. fundamental task of surveying (abbreviated further as II. FTS), we compute the distance
𝑑𝑑𝐴𝐴𝐴𝐴 and the whole circle bearings (WCB) WCB𝐴𝐴𝐴𝐴 and WCB𝐵𝐵𝐵𝐵 .
2. Using the measured inner angles, we compute the WCBs between the control points and the unknown
point WCB𝐴𝐴𝐴𝐴 and WCB𝐵𝐵𝐵𝐵 . According to the figure:
WCB𝐴𝐴𝐴𝐴 = WCB𝐴𝐴𝐴𝐴 − 𝛼𝛼
WCB𝐵𝐵𝐵𝐵 = WCB𝐵𝐵𝐵𝐵 + 𝛽𝛽
3. Using the sine theorem, we can compute the distances between the control points and point 𝑃𝑃:
sin(𝛼𝛼) 𝑑𝑑𝐵𝐵𝐵𝐵 sin(𝛼𝛼)
= ⇒ 𝑑𝑑𝐵𝐵𝐵𝐵 = ∙ 𝑑𝑑
sin(𝛼𝛼 + 𝛽𝛽) 𝑑𝑑𝐴𝐴𝐴𝐴 sin(𝛼𝛼 + 𝛽𝛽) 𝐴𝐴𝐴𝐴
sin(𝛽𝛽) 𝑑𝑑𝐵𝐵𝐵𝐵 sin(𝛽𝛽)
= ⇒ 𝑑𝑑𝐵𝐵𝐵𝐵 = ∙ 𝑑𝑑
sin(𝛼𝛼 + 𝛽𝛽) 𝑑𝑑𝐴𝐴𝐴𝐴 sin(𝛼𝛼 + 𝛽𝛽) 𝐴𝐴𝐴𝐴

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 Figure 2. Computation of the WCB’s.

4. We use the I. fundamental task of surveying (abbreviated further as I. FTS) to find the coordinates of the
unknown point 𝑃𝑃. We can do this computation from using both 𝐴𝐴 and 𝐵𝐵, so we can check our results:
𝐸𝐸𝑃𝑃 = 𝐸𝐸𝐴𝐴 + 𝑑𝑑𝐴𝐴𝐴𝐴 ∙ sin(WCB𝐴𝐴𝐴𝐴 )
𝑁𝑁𝑃𝑃 = 𝑁𝑁𝐴𝐴 + 𝑑𝑑𝐴𝐴𝐴𝐴 ∙ cos(WCB𝐴𝐴𝐴𝐴 )
or
𝐸𝐸𝑃𝑃 = 𝐸𝐸𝐵𝐵 + 𝑑𝑑𝐵𝐵𝐵𝐵 ∙ sin(WCB𝐵𝐵𝐵𝐵 )
𝑁𝑁𝑃𝑃 = 𝑁𝑁𝐵𝐵 + 𝑑𝑑𝐵𝐵𝐵𝐵 ∙ cos(WCB𝐵𝐵𝐵𝐵 )
Example 1:
Coordinates of the control points:

Point E [m] N [m]

A 658 077.70 247 431.38
B 657 310.23 247 123.54
Measured angles:
𝛼𝛼 = 81-34-45
𝛽𝛽 = 66-45-57
The clockwise order of the points is 𝐵𝐵, 𝑃𝑃, 𝐴𝐴.
According to the coordinates and the clockwise order, we can create the following figure:

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 Figure 3. The layout of the points in Example 1.

Distance and WCB between the control points from II: FTS:

𝑑𝑑𝐴𝐴𝐴𝐴 = 826.907 m
WCB𝐴𝐴𝐴𝐴 = 248-08-38
WCB𝐵𝐵𝐵𝐵 = 68-08-38
The WCBs from the control points to point 𝑃𝑃:

WCB𝐴𝐴𝐴𝐴 = WCB𝐴𝐴𝐴𝐴 + 𝛼𝛼 = 329-43-23

WCB𝐵𝐵𝐵𝐵 = WCB𝐵𝐵𝐵𝐵 − 𝛽𝛽 = 1-22-41
The distances between the control points and point 𝑃𝑃:
sin(𝛽𝛽)
𝑑𝑑𝐴𝐴𝐴𝐴 = ∙ 𝑑𝑑 = 1447.866 m
sin(𝛼𝛼 + 𝛽𝛽) 𝐴𝐴𝐴𝐴
sin(𝛼𝛼)
𝑑𝑑𝐵𝐵𝐵𝐵 = ∙ 𝑑𝑑 = 1558.655 m
sin(𝛼𝛼 + 𝛽𝛽) 𝐴𝐴𝐴𝐴
The coordinates of 𝑃𝑃 computed from point 𝐴𝐴:

𝐸𝐸𝑃𝑃 = 𝐸𝐸𝐴𝐴 + 𝑑𝑑𝐴𝐴𝐴𝐴 ∙ sin(WCBAP ) = 657 347.714 m ≈ 657 347.71 m

𝑁𝑁𝑝𝑝 = 𝑁𝑁𝐴𝐴 + 𝑑𝑑𝐴𝐴𝐴𝐴 ∙ cos(WCBAP ) = 248 681.754 m ≈ 248 681.75 m

The coordinates of 𝑃𝑃 computed from point 𝐵𝐵:

𝐸𝐸𝑃𝑃 = 𝐸𝐸𝐵𝐵 + 𝑑𝑑𝐵𝐵𝐵𝐵 ∙ sin(WCBBP ) = 657 347.714 m ≈ 657 347.71 m

𝑁𝑁𝑝𝑝 = 𝑁𝑁𝐵𝐵 + 𝑑𝑑𝐵𝐵𝐵𝐵 ∙ cos(WCBBP ) = 248 681.754 m ≈ 248 681.75 m

The two sets of coordinates match at least up to centimeter precision.

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1.2. Intersection using whole circle bearings
If there is no line of sight between the control points, we cannot directly measure the inner angles mentioned in
section 1.1. In such a case, we instead measure the angles 𝜑𝜑𝐴𝐴 and 𝜑𝜑𝐵𝐵 in the figure between the unknown point 𝑃𝑃
and two control points 𝐶𝐶 and 𝐷𝐷.

Figure 4. Intersection using whole circle bearings.

Given information: coordinates of points 𝐴𝐴, 𝐵𝐵, 𝐶𝐶, 𝐷𝐷

Measured quantites: angles 𝜑𝜑𝐴𝐴 and 𝜑𝜑𝐵𝐵
Computing the coordinates of point 𝑃𝑃:
1. Using the II. FTS, we can compute the 𝑑𝑑𝐴𝐴𝐴𝐴 and WCB𝐴𝐴𝐴𝐴 , WCB𝐵𝐵𝐵𝐵 , WCB𝐴𝐴𝐴𝐴 and WCB𝐵𝐵𝐵𝐵 .
2. Using the WCBs and the measured angles, we can compute the WCBs from 𝐴𝐴 and 𝐵𝐵 to the unknown
point:
WCB𝐴𝐴𝐴𝐴 = WCB𝐴𝐴𝐴𝐴 − 𝜑𝜑𝐴𝐴
WCB𝐵𝐵𝐵𝐵 = WCB𝐵𝐵𝐵𝐵 + 𝜑𝜑𝐵𝐵
3. We can compute the inner angles at 𝐴𝐴 (𝛼𝛼) and 𝐵𝐵 (𝛽𝛽) from the calculated WCBs:
𝛼𝛼 = WCB𝐴𝐴𝐴𝐴 − WCB𝐴𝐴𝐴𝐴
𝛽𝛽 = WCB𝐵𝐵𝐵𝐵 − WCB𝐵𝐵𝐵𝐵
4. The computation is now a simple intersection with inner angles (see section 1.1.)

2. Side section
In case of the side section, one of the control points is inaccessible, meaning that we cannot set up our station on
that point. To compensate for this, we set up the instrument on the unknown point 𝑃𝑃 and measure the inner angle
here.

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2.1. Side section using inner angles
Given information: coordinates of points 𝐴𝐴, 𝐵𝐵
Measured quantities: angles: 𝛼𝛼 and 𝛾𝛾 inner angles

Figure 5. Side section using inner angles.

Computing the coordinates of point 𝑃𝑃:

1. Using the II. FTS, we compute the distance 𝑑𝑑𝐴𝐴𝐴𝐴 , WCB𝐴𝐴𝐴𝐴 and WCB𝐵𝐵𝐵𝐵 .
2. Using the formula for the sum of the inner angles of a triangle, we can find the value of 𝛽𝛽 at point 𝐵𝐵:
𝛽𝛽 = 180° − 𝛼𝛼 − 𝛾𝛾
3. Compute the intersection with inner angles (see section 1.1.).

2.2. Side section using a distance measurement

Given information: coordinates of points 𝐴𝐴, 𝐵𝐵
Measured quantities: angles: 𝛾𝛾 inner angle and distance 𝑑𝑑𝐴𝐴𝐴𝐴

Figure 6. Side section using a distance measurement.

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Computing the coordinates of point 𝑃𝑃:
1. Using the II. FTS, we compute the distance 𝑑𝑑𝐴𝐴𝐴𝐴 , WCB𝐴𝐴𝐴𝐴 and WCB𝐵𝐵𝐵𝐵 .
2. Using the sine theorem, we can compute the inner angle at point 𝐵𝐵:
sin(𝛽𝛽) 𝑑𝑑𝐴𝐴𝐴𝐴 𝑑𝑑𝐴𝐴𝐴𝐴
= ⇒ 𝛽𝛽 = arcsin � ∙ 𝑠𝑠𝑠𝑠𝑠𝑠(𝛾𝛾)�
sin(𝛾𝛾) 𝑑𝑑𝐴𝐴𝐴𝐴 𝑑𝑑𝐴𝐴𝐴𝐴
3. We can compute the third inner angle, at point 𝐴𝐴, using the other two inner angles:
𝛼𝛼 = 180° − 𝛽𝛽 − 𝛾𝛾
4. Compute the intersection with inner angles (see section 1.1.).

3. Arc section
In case of the arc section, instead of angles, we measure distances between the control points and the unknown
point. We also have to specify the clockwise order of the points, as only using the two distances results in two
possible locations of point 𝑃𝑃.
Given information: coordinates of points 𝐴𝐴, 𝐵𝐵, the clockwise order of the points
Measured quantites: distances 𝑑𝑑𝐴𝐴𝐴𝐴 and 𝑑𝑑𝐵𝐵𝐵𝐵

Figure 7. Arc section.

Computing the coordinates of point 𝑃𝑃:

1. Using the II. FTS, we compute the distance 𝑑𝑑𝐴𝐴𝐴𝐴 , WCB𝐴𝐴𝐴𝐴 and WCB𝐵𝐵𝐵𝐵 .
2. Using the cosine theorem, we can find the two inner angles 𝛼𝛼 and 𝛽𝛽:
2 2 2
2 2 2
𝑑𝑑𝐵𝐵𝐵𝐵 − 𝑑𝑑𝐴𝐴𝐴𝐴 − 𝑑𝑑𝐴𝐴𝐴𝐴
𝑑𝑑𝐵𝐵𝐵𝐵 = 𝑑𝑑𝐴𝐴𝐴𝐴 + 𝑑𝑑𝐴𝐴𝐴𝐴 − 2 ∙ 𝑑𝑑𝐴𝐴𝐴𝐴 ∙ 𝑑𝑑𝐴𝐴𝐴𝐴 ∙ cos(𝛼𝛼) ⇒ 𝛼𝛼 = arccos � �
−2 ∙ 𝑑𝑑𝐴𝐴𝐴𝐴 ∙ 𝑑𝑑𝐴𝐴𝐴𝐴
2 2 2
𝑑𝑑𝐴𝐴𝐴𝐴 − 𝑑𝑑𝐴𝐴𝐴𝐴 − 𝑑𝑑𝐵𝐵𝐵𝐵
𝛽𝛽 = arccos � �
−2 ∙ 𝑑𝑑𝐴𝐴𝐴𝐴 ∙ 𝑑𝑑𝐵𝐵𝐵𝐵
3. Compute the intersection with inner angles (see section 1.1.).

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Example 2.
Coordinates of the control points:

Point E [m] N [m]

A 654 653.23 232 456.39
B 654 234.92 232 167.47
Measured distances:

𝑑𝑑𝐴𝐴𝐴𝐴 = 967.34 m
𝑑𝑑𝐵𝐵𝐵𝐵 = 846.45 m
The clockwise order of the points is 𝐵𝐵, 𝑃𝑃, 𝐴𝐴.
According to the coordinates and the clockwise order, we can create the following figure:

Figure 8. The layout of the points in Example 2.

Distance and WCB between the control points from II: FTS:
𝑑𝑑𝐴𝐴𝐴𝐴 = 508.388 m
WCB𝐴𝐴𝐴𝐴 = 235-22-04
WCB𝐵𝐵𝐵𝐵 = 55-22-04
The inner angles from the cosine theorem:
2 2 2
𝑑𝑑𝐵𝐵𝐵𝐵 − 𝑑𝑑𝐴𝐴𝐴𝐴 − 𝑑𝑑𝐴𝐴𝐴𝐴
𝛼𝛼 = arccos � � = 60-56-28
−2 ∙ 𝑑𝑑𝐴𝐴𝐴𝐴 ∙ 𝑑𝑑𝐴𝐴𝐴𝐴
2 2 2
𝑑𝑑𝐴𝐴𝐴𝐴 − 𝑑𝑑𝐴𝐴𝐴𝐴 − 𝑑𝑑𝐵𝐵𝐵𝐵
𝛽𝛽 = arccos � � = 87-23-35
−2 ∙ 𝑑𝑑𝐴𝐴𝐴𝐴 ∙ 𝑑𝑑𝐵𝐵𝐵𝐵
The WCBs from the control points to point 𝑃𝑃:

WCB𝐴𝐴𝐴𝐴 = WCB𝐴𝐴𝐴𝐴 + 𝛼𝛼 = 296-18-32

WCB𝐵𝐵𝐵𝐵 = WCB𝐵𝐵𝐵𝐵 − 𝛽𝛽 = 327-58-39
The coordinates of 𝑃𝑃 computed from point 𝐴𝐴:

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 𝐸𝐸𝑃𝑃 = 𝐸𝐸𝐴𝐴 + 𝑑𝑑𝐴𝐴𝐴𝐴 ∙ sin(WCBAP ) = 653 786.089 m ≈ 653 786.09 m
𝑁𝑁𝑝𝑝 = 𝑁𝑁𝐴𝐴 + 𝑑𝑑𝐴𝐴𝐴𝐴 ∙ cos(WCBAP ) = 232 885.125 m ≈ 232 885.12 m

The coordinates of 𝑃𝑃 computed from point 𝐵𝐵:

𝐸𝐸𝑃𝑃 = 𝐸𝐸𝐵𝐵 + 𝑑𝑑𝐵𝐵𝐵𝐵 ∙ sin(WCBBP ) = 653 786.087 m ≈ 653 786.09 m
𝑁𝑁𝑝𝑝 = 𝑁𝑁𝐵𝐵 + 𝑑𝑑𝐵𝐵𝐵𝐵 ∙ cos(WCBBP ) = 232 885.124 m ≈ 232 885.12 m

The two sets of coordinates match to at least centimeter precision.

4. Resection
In case of the resection, we only set up the instrument on the unknown point 𝑃𝑃 and measure the angles subtended
by the directions from the unknown point to exactly 3 control points.

Figure 9. Two possible layouts of the points in the resection problem.

In the figure above, the angles 𝜉𝜉 and 𝜂𝜂 are measured. There are multiple solutions to the resection problem,
graphical ones include the solutions by Collins and Sossna, examples for analytical solutions would be Tienstra’s
method or Runge’s method and there are numerical solutions as well. In the following, Collins’ method is briefly
introduced.
Suppose the layout given in Figure 10 below. First, we draw a circle around the points 𝐴𝐴, 𝐶𝐶 and 𝑃𝑃. The line
connecting 𝑃𝑃 and 𝐵𝐵 intersect the circle at point 𝑆𝑆.
According to the inscribed angle theorem, the angle at 𝐴𝐴 in the triangle 𝐴𝐴𝐴𝐴𝐴𝐴 is 𝜂𝜂 and the angle at 𝐶𝐶 in the
triangle 𝐴𝐴𝐴𝐴𝐴𝐴 is 𝜉𝜉. We can compute the WCB𝐴𝐴𝐴𝐴 and the WCB𝐶𝐶𝐶𝐶

WCB𝐴𝐴𝐴𝐴 = WCB𝐴𝐴𝐴𝐴 − 𝜂𝜂
WCB𝐶𝐶𝐶𝐶 = WCB𝐶𝐶𝐶𝐶 + 𝜉𝜉
and using the control points 𝐴𝐴 and 𝐶𝐶, we can find the coordinates of 𝑆𝑆. As 𝑆𝑆 is a known point now, we can
compute the WCB𝐵𝐵𝐵𝐵 which is equal to WCB𝐵𝐵𝐵𝐵 as 𝑆𝑆 and 𝑃𝑃 are on the same line. The WCB𝐴𝐴𝐴𝐴 and WCB𝐶𝐶𝐶𝐶 can be
found:
WCB𝐴𝐴𝐴𝐴 = WCB𝐵𝐵𝐵𝐵 − 𝜉𝜉
WCB𝐶𝐶𝐶𝐶 = WCB𝐵𝐵𝐵𝐵 + 𝜂𝜂

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 Figure 10. Computing the resection using Collins’ method.

The coordinates of 𝑃𝑃 can now be found as the intersection of any combination of the lines 𝐴𝐴𝐴𝐴, 𝐵𝐵𝐵𝐵 and 𝐶𝐶𝐶𝐶 and
therefore we can check our computation.