Chapter 2: Working with map references
Introduction
In Chapter 1 we learned how the position of a place on a topographic map with a grid system
could be specified in terms of local coordinates called map references or universal coordinates called full
UTM grid references.
In this chapter we will examine the use of this locational device further.
In
particular, we will explore how additional spatial information contained in the grid system can be extracted
by applying to map references some simple notions of coordinate geometry and trigonometry.
First we need to develop a few ideas about the geometry of cartesian space. If we consider two
axes, conventionally designated X and Y, we can specify the location of any point in the domain of the
graph by the coordinate address x,y. Thus, in Figure 2.1A, addresses of the points A through D are
A(1,5), B(7,9), C(4,8), and D(9,3). That is, the location of each point is uniquely given by its X-coordinate
and its corresponding Y-coordinate ('read right up': sound familiar?!). Not only do the coordinates of a
point specify its location with respect to the axes, they also define the location of all points with respect to
all others using the same system of address. Herein lies the power of coordinate geometry. For example,
we can see that point B in Figure 2.1A is 6 horizontal units to the right (east) of point A because the
difference in their X coordinates is 7 - 1 = 6. Similarly, we know that D must be 2 vertical units below
point A because the difference in their respective Y coordinates is 5 - 3 = 2.
We can also use these coordinates together with Pythagorus' theorem to determine the straightline distance between any two points on the graph. For example, in order to determine the distance
between A and C (AC), we form a right-angle triangle ACE on the hypotenuse AC, and note that, from
2
2
2
Pythagorus, (AC) = (AE) + (CE) . We know that the horizontal distance AE is given by the difference in
the respective X coordinates for A and C (4 - 1 = 3) and that the vertical distance CE is given by the
2
2
difference in the respective Y coordinates for A and C (8 - 5 = 3). Thus, we can say that, (AC) = 3 + 3
= 9 + 9 = 18 and, therefore, that AC =
calculated as CD =
2
2
2
18 = 4.24. The straight-line distance CD similarly can be
5 + 5 = 7.07, and AD =
8 2 + 2 2 = 8.25, and so on.
You will note that, although we do not need to know the coordinates for point E in order to
determine the distance AC, they are determined readily from the coordinates of C and A as E (4, 5).
We can also use coordinate geometry, together with a little algebra, to find the location of point F
in Figure 2.1A; point F is the intersection of lines AB and CD. Let's consider this problem in a more
general way by referring to any two points P1(X1,Y2) and P2 (X2,Y2), as depicted in Figure 2.1B.
We
know that the equation of the line passing through P1 and P2 must have the form Y = mX + b, where the
Y 2 − Y1
slope m =
and b is the intercept on the Y axis where X = 0.
X 2 − X1
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Chapter 2: Working with map references
2.1: Grid coordinates (see text for explanation)
Thus the equation of the line P1P2 is
⎛ Y2 − Y1 ⎞
Y = ⎜
⎟ X + b ........................................................................(2.1)
⎝ X 2 − X 1⎠
The intercept b on the Y axis, can be evaluated by substitution of the X,Y coordinates of any point on the
line through P1 and P2.
Now we can get back to the problem of locating point F in Figure 2.1A. The coordinates for this
point can be determined by the simultaneous solution of the equations for the lines AB and CD.
From equation 2.1, the equation for AB is
⎛ 9 − 5 ⎞
⎟ X + b
Y = ⎜
⎝ 7 − 1 ⎠
or
Y = 0.67 X + b
To evaluate b for A (1,5), we substitute X = 1 and Y = 5 in the equation above, yielding
5 = (0.67)(1) + b; b = 5 - 0.67 = 4.33
Thus, the equation for the line through A and B is
Y = 0.67 X + 4.33 .............................................................................(2.2)
Similarly, the equation of the line through C and D follows from equation (2.1) as
⎛ 3 − 8 ⎞
⎟ X + b
Y = ⎜
⎝ 9 − 4 ⎠
or Y = -1.0X + b or -X + b.
Evaluating b for C(4,8), again we substitute X = 4 and Y = 8 in 8 = -4 + b; b = 12
Thus, the equation for the line through C and D is
Y = -X + 12 .......................................................................................(2.3)
All we need to do now is find the one pair of XY coordinates that satisfies both equations (2.2)
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Chapter 2: Working with map references
and (2.3) since this condition defines the point of intersection. There are a variety of ways to obtain a
simultaneous solution of two linear equations such as these.
You might like to equate the two
expressions so that, Y = 0.67X + 4.33 = Y = -X + 12, or in other words, 0.67X + 4.33 = -X + 12.
7.67
Solving for X (X + 0.67X = 12 - 4.33 and therefore 1.67X = 7.67; X =
= 4.59). To obtain the
1.67
corresponding Y coordinate we simply substitute X = 4.59 in either of the two equations in order to solve
for Y. For example, in equation (2.2), Y = 0.67 (4.59) + 4.33 so that Y = 7.41.
Although it amounts to the same thing, I prefer to obtain this simultaneous solution by subtracting
the equations one from the other. Subtracting equation (2.3) from equation (2.2) yields
Y = 0.67 X + 4.33
Y =
-X + 12.00
0 = 1.67X - 7.67 so that, once again, X =
7.67
1.67
= 4.59
Just to be different, this time we substitute X = 4.59 in equation (2.3) to find the corresponding Y
coordinate: Y = -X + 12 = -(4.59) + 12 = 7.41.
Thus, the coordinates for the intersection of the line through A and B, and that through C and D,
are F(4.56, 7.41). As before, we could now easily determine the distance of this intersection point F from
any of the other points in Figure 2.1A. Now we can transfer these mathematical notions to the map.
Distances and intersections from map references
Figure 2.2 shows an extract from the 1:50 000 Rossland-Trail map on which four locations have
been selected for analysis:
and the
(A) the Post Office at Oasis;
map reference: 46074240
(B) BM 1360 on the bridge at Trail;
map reference: 49033785
(C) the water tower in Tadanac;
map reference: 46483888
(D) triangulation station 2365;
map reference: 48423995
Before we start it may be useful if we take a moment to consider again what distances these map
references represent. On this map with its 1 000-m grid system, distances specified to the nearest metre
would require a ten-figure map reference (two 5-figure coordinates). The first two of each coordinate
specify one of the 1 000-m grid lines and the next three specify thousandths of a grid square, or metres.
Thus, if we were to subtract such 10-figure map references, one from the other, the difference would be
some number of metres. In the case of the eight-figure map references listed above, once again the first
two numbers of the coordinates locate a particular grid
line.
The remaining two numbers in
each coordinate divide the 1 000-m grid into hundredths or tens of metres. Thus, when we subtract two
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Chapter 2: Working with map references
2.2: Extract from the 1:50 000 Rossland-Trail NTS sheet (courtesy of the Governments of British
Columbia and Canada)
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Chapter 2: Working with map references
eight-figure map references, one from the other, the difference is not metres, but rather tens of metres.
Similarly, subtraction of 6-figure map references yields tenths of a grid square or so many hundreds of
metres, and 4-figure map references would simply specify the nearest whole grid square, or kilometres.
The principles are simple enough but apparently easily forgotten, so be vigilant; it has been my
experience that almost all errors in student calculations involving map references stem from confusion
over the units involved. In this regard it is far better to reason your way through problems involving map
references rather than trying to learn by rote a set of rules.
Now that we are aware of the level of precision implied by coordinates, we are in a position to
manipulate the map references to determine distances and directions. For example, if we wanted to know
the distance between the water tower (C) and the triangulation station (D), the appropriate calculation is
once again the solution of a Pythagorean right-triangle. To reiterate, note that the map references are
eight-figure references and therefore represent the number of hundredths of the 1 000-m grid square, or
tens of metres. Furthermore, the computations will be much more manageable if we express them in
kilometres and not metres. Thus, the horizontal distance is given by the difference between the eastings
(X coordinates) of (C) and (D) and is 4842 - 4648 = 194 x10 metres = 1 940 metres = 1.94 kilometres.
The vertical (north-south) distance is similarly given by the difference between the northings (Y coordinates) of (D) and (C): 3995 - 3888 = 107 x 10 metres = 1.07 kilometres.
Therefore, the distance CD =
1.9402 + 1.0702 =
3.764 + 1.145 = 2.22 km
You should verify that this calculation is correct by making the appropriate direct measurement on
the map.
As another example, the distance between (A) and (B) similarly can be calculated, thus:
AB =
(49.030− 46.070) 2 + (42.400− 37.850) 2 =
29.464 = 5.43 km.
If we were interested in establishing where a line of sight between (A) and (B) crossed the line of
sight between (C) and (D), the derivation and simultaneous solution of the two relevant equations is as
follows:
The equation of the line through (A) and (B) can be written
⎛ 42.400− 37.850⎞
⎟ X + b
Y = ⎜
⎝ 46.070− 49.030⎠
or
Y = - 1.537X + b
For Y = 42.400 km and X = 46.070 km, b = 113.210 km, so that
Y = -1.537X + 113.210 .....................................................................(2.4)
The corresponding equation through (C) and (D) can be written:
⎛ 39.950− 38.880⎞
⎟ X + b
Y = ⎜
⎝ 48.420− 46.480⎠
29
or
Y = 0.552X + b
Chapter 2: Working with map references
For Y = 38.880 km and X = 46.480 km, b = 13.223 km, so that
Y = 0.552X + 13.223 .........................................................................(2.5)
Subtracting equation (2.4) from (2.5) yields
0 = 2.089X - 99.987, so that X = 47.864 km and by substitution in equation (2.5), Y = 39.644 km.
In other words, rounding off to our 10-metre level of accuracy, the map reference for the point of
intersection of the two lines of sight is 47863964. Again, a quick direct check on the map will verify the
accuracy of our calculation.
Calculation of intersection points of lines of sight has many applications in surveying and is the
basis of a positioning method known as resection. We will return to this topic later, in Chapter 6, when we
examine various techniques of field surveying.
Bearings from map references
No doubt it has occurred to some of you that grid bearings also can be determined from map
references in exactly the same way as the slopes of lines in the examples above. For example, the grid
bearing of the Post Office at Oasis from BM 1360 on the bridge at Trail is equal to 270o + angle ø, as
illustrated in Figure 2.3.
Angle ø can be obtained from the map
references as a tangent.
Since we are only
interested in the ratio of the side opposite to the
side adjacent angle ø, we do not have to convert
the measurements to metres. Thus we have
tan ø =
4240− 3785 455
=
= 1.5372.
4903− 4607 296
A quick reference to the natural tangent tables (or
to the tangent function on your calculator)
o
o
2.3: Grid bearings from map references
indicates that tan 1.5372 = 56.95 or 56 57'. Thus
o
o
See text for explanation
the grid bearing of (A) from (B) is 326 57' (270 +
o
56 57'). Note that the grid bearing of (B) from (A),
o
o
o
o
o
therefore, must be 146 57' (90 + 56 57' or 326 57' - 180 ).
We have touched on a number of ideas in this chapter which, if they are completely new to you,
probably will seem to be a little complicated. But most 'complicated' ideas simply are unfamiliar ideas and
you will feel much more comfortable with all of these concepts as you use them.
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