PARTITION OF LAND
SUBDIVISION OF LAND
The subdivision of land is a usual task undertaken by surveyors and engineers in property
surveys. When tracts of land are sold or inherited it is often necessary to divide it into smaller
parcels or into areas having certain proportions to each other.
There are numerous problems which may arise in the subdivision of land. The process of
subdivision may require the application of the principles of geometry and trigonometry or the
use of special techniques in computations. Other cases are performed by trial and error
methods. This usually involves an initial assumption such as the starting point or the direction to
be taken by a selected cut off line which will separate a certain area from the main parcel. Most
of the problems encountered are so common and frequently involved in the working out of
more complicated cases.
In this lesson and the succeeding lesson, four of the most common cases encountered in
the subdivision of land will be explained. Thses cases are:
1.Dividing an area into two parts by a line between two points.
2. Dividing an area by a line running through a point and in a given direction.
3.To cut off a required area by a line through a given point.
4.To cut off a required area by a line running in a given direction.
For any of the above mentioned cases, a resurvey is run, the latitudes and departures
area determined, the surveyed traverse is balanced, and the area of the entire tract is computed.
When computing the desired subdivision scheme only the adjusted latitudes and departures
area used.
47-2. DIVIDING AN AREA INTO TWO PARTS BY A LINE BETWEEN TWO POINTS.
In Figure 47-1, ABCDEF represents an irregular parcel of land to be divided into two parts (Tract
ABCD and tract DEFA) by a cut off line extending from D to A.
It is assumed that the length and direction of each course has been earlier determined,
the latitudes and departures computed and adjusted, and the area of the whole tract computed.
The solution ere is to determine the length and direction of the dividing line DA by
computations, and to calculate the area of each of the two tracts into which the parcel of land is
divided.
Either of the two tracts may be considered as a closed traverse with the length and
direction of side DA unknown.
Considering tract ABCD, the latitudes and departures of AB, BC and CD area known; hence the
latitude, departure, length and bearing of DA can be determined.
Similarly, the same quantities could also be determined by considering tract DEFA, since
the latitudes and departures of DE, EF, and FA are known. The area of either tract can then be
found by any of the methods (DMD, DOD, Coordinates, etc.) used for determining areas.
A check is obtained by actually determining the length and direction of the line DA and
comparing these with the calculated values. A further check may also be made by noting that
the sum of the areas of the two tracts, each calculated independently. Is equal to the calculated
area of the entire parcel of land.
47-3. DIVIDING AN AREA BY A LKINE RUNNING THROUGH A POINT IN A GIVEN DIRECTION.
In Figure 47-2, ABCDEF represents an irregular parcel of land to be subdivided into two parts by
a cut off line (BP) running in a given direction (Indicated by Θ) which passes through point B. Out
of the desired division, tract BCDEP is formed on one side of the dividing line and tract FABP on
the other side.
It is assumed that the length and direction of each course area known, the latitudes and
departures computed and adjusted, and the area of the whole tract computed. The solution will
require the calculation of the lengths BP and PP and the area of each of the two tracts.
Either of the two tracts may be considered as a closed traverse for which the lengths of
two sides area unknown. A trial line BF is first established. Since the latitude and departure of FA
and AB area known, from these the length and bearing of the trial line BF are computed. Then,
in the triangle BFP the lengths of the sides BP and FP area easily determined by applying the sine
law, and their latitudes and departures area also computed. The area of tract FABP is calculated
by any of the methods used for determining areas.
In the field, the length and direction of the dividing line area laid off from B, and a check
on fieldwork and computations is obtained if point P thus established lies on line EF, and if the
computed distance FP agrees with the observed distance.
The computations area further checked by determining if the algebraic sums of the
latitudes and of the departures of AB, BP, PF and FA area equal to zero.
The length of EP can be determined by subtracting the computed length of FP from the
known length of EF, also its latitude and departure area computed.
The areas of the two tracts area calculated independently based on the given and
calculated latitudes and departures of their respective courses. As a check, the sum of the areas
of the two tracts should equal the area of the entire parcel of land.
A small amount of error may result in the above computations which, however, is
negligible. The error may appear in the summation of latitudes and departures. This error is
attributed to the rounding off of values and in determining the values computed from the sine or
cosine of angles.
It is obvious that the solution of this type of problem is greatly facilitated by first plotting
the traverse before working out the necessary computations.
47-4. ILLUSTRATIVE PROBLEMS
•
DIVIDING A TRACT OF LAND INTO TWO PARTS BY A LINE BETWEEN TWO POINTS. Given
the following data of a tract of land, determine the area east of a line running F to C and
calculate the length and bearing of FC.
COURSE
AB
BC
CD
DE
EF
FA
ADJ LAT
ADJ DEP
490.71 m
587.12
-327.41
-1002.76
-122.67
375.01
4727
608.89
786.78
218.32
-1116.62
-544.64
0.00
0.00
SUMS
Solution:
a) Determining Area of tract ABCDEF.
(Area of whole traverse):
COURSE
AB
BC
CD
DE
EF
FA
DMD
47.27
703.43
2099.10
3104.20
2205.90
544.64
SUM
DOUBLE AREA
23,195.86
412,997.82
-687,266.33
-3,112,767.59
-270,597.75
204,245.45
-3,430,192.54
2 X AREA = -3,430,192.54
AREA = -1,715,096.27 sq m or
-171.5096 Hectares (Area of tract ABCDEF with negative sign disregarded)
b) Considering Tract CDEF (Area east of the dividing line).
∑lat = ∑NL + ∑SL = Lat cd + Lat de +Latef + Lat fc =0.0
-327.41 -1002.76 – 122.67 +Lat fc = 0.0
Lat fc = 1452.84 m (Latitude of line FC)
∑Dep = ∑ED + ∑WD = Dep cd +Dep de + Dep ef + Dep fd = 0.0
Depfc = 111.52 m (Departure of line FC)
Lfc = √(latfc)2 + (Dep fc)2
Lfc = √(1452.84)2 + (111.52)2
= 1457.11 m (Length of line FC)
Tab (Θfc) = Depfc/Latfc = 111.52/1452.84 = 0.07676
Θfc = 04°23’ (Bearing of line FC is N 04°23’ E)
Note: A small amount of error may result in the above computations which, however us
negligible. The error is attributed to the rounding off of values and in determining the values
computed from te sine or cosine of angles.
2. DIVIDING A TRACT OF LAND BY A LINE RUNNING IN A GIVEN DIRECTION. Given the following
data of a tract of land, find the area of each of the two parts into which the tract is divided bu a
line through A with a bearing of N 75°30’ E.
COURSE
ADJ
AB
BC
CD
DE
EF
FA
LAT
490.71 m
587.12
-327.41
-1002.76
-122.67
375.01
47.27 m
60.89
786.78
218.36
-1116.62
-544.64
0.00
0.00
SUMS
Solution:
a) Determining Area of tract ABCDEF
COURSE
DMD
AB
BC
CD
DE
EF
FA
47.27
703.43
2099.10
3104.20
2205.90
544.64
DOUBLE AREA
23,195.886
412,997.82
-687,266.33
-3,112,767.59
-270,597.75
207,245.45
2 X AREA = -3,430,192.54
AREA = -1,715,096.27 sq m or
= -171.510 Hectares (Area of tract ABCDEF with negative sign disregarded)
b) considering Tract EFA
∑lat = ∑NL + ∑SL = Lat cd + Lat de +Latef + Lat ae =0.0
-122.67 + 375.01 + Lat ae = 0.0
Lat ae = -252.34 m (Latitude of line AE)
∑Dep = ∑ED + ∑WD = Dep cd +Dep de + Dep ef + Dep ae = 0.0
Depae = 1661.26 m (Departure of line AE)
Lfc = √(latfc)2 + (Dep fc)2
Lfc = √(-252.34)2 + (1661.26)2
= 1680.32 m (Length of line AE)
Tab (Θfc) = Depfc/Latfc = 1661.26/-252.34 = 6.5834192
Θfc = 81°22’ (Bearing of line AE is N 81°22’E)
c) Analyzing Trangle APE
Tan( Θae) =12°17’ (Bearing of line PE is S 12°17’ E)
∠A = 180° - 75°(30’ +81°22’) = 23°08’
∠E = 81°22’ – 12°17’ = 69°05’
∠P = 75°30’ + 12°17’ = 87°45’
Check
∠A +∠E +∠P = 180°
23°08’ +69°05’ + 87°45’ = 180°
180° = 180° (Checks)
By Sine Law:
AP/Sin E = PE/ Sin A = AE/Sin P
AP = (AE)Sin E/Sin P = 1680.32 Sin(69°05’)/Sin (87°47’) = 1570.76 m (Length
of dividing line AP)
AP = (AE)Sin A/Sin P = 1680.32 Sin(23°08’)/Sin (87°47’) = 660.65 m
(Distance of point P from station E)
d) Considering Tract PEFA
TABULATION FOR TRACT PEFA
COURSE
LAT
DEP
LAT CORR
PE
EF
FA
AP
-645.53
-122.67
375.01
393.29
140.55
-1116.62
-544.64
1520.73
0.05
0.00
0.00
0.05
DEP
CORR
0.01
0.00
0.00
0.01
ADJ LAT
ADJ DEP
-645.58
-122.67
375.01
393.24
140.54
-1116.62
-544.64
1520.72
SUMS
+0.10
+0.02
+0.10
+0.02
0.00
0.00
Latap = (AP) Cos (75°30’) = 1570.76 Cos (75°30’) = 393.29
Depap = (AP) Sin (75°30’) = 1570.76 Sin (75°30’) = 1520.73
Latep = (AP) Cos (12°17’) = 660.65 Cos (12°17’) = -645.53
Depep = (AP) Sin (12°17’) = 660.65 Sin (12°17’) = 140.53
COURSE
DMD
PE
EF
FA
AP
140.54
-835.54
-2496.80
-1520.72
SUMS
DOUBLE
AREA
-90,729.81
102,495.69
-936,324.97
-598,007.93
1,522,567.02
2 X AREA = -1,522,567.02
AREA = -761,283.51 sq m or of Tract PEFA with negative sign disregarded
Note: A small amount of error may result in the above computations which, however us
negligible. The error is attributed to the rounding off of values and in determining the values
computed from te sine or cosine of angles.
48-1 . TO CUT OFF A REQUIRED AREA BY LINE THROUGH A GIVEN POINT.
In Figure 48-1, ABCDEF represents an irregular parcel of land with courses of
known lengths and directions, with comuted and adjusted latitudes and
departures, and its total area already determined.
A dividing line is to pass through point F cutting a required area from
the tract. The traverse may be subdivided into tracts of equal areas or into
any desired proportional parts.
Before the length and direction of the dividing line can be determined,
a trial line BF is drawn to that corner of the traverse which. From inspection of
the sketch, will come nearest to the required line of division. In the known
except those of BF. This is a simple case of omitted bearing of BF area easily
determined. By the DMD or DPD method, the area of ABD is calculated and
the difference between this area and that required for tract FABQ is found.
In the figure it is assumed that ABD gas an area lesser than the desired
size, FQ bearing the correct position of the dividing line. Triangle BFQ
represents the difference in there two areas. In this traignle, angle FBQ may
be computed from known bearings of the trial line and course BC. Since one
side BF, one angle FBQ, and the area of BFQ area known, the length of BQ is
computed as follow:
Area BFQ = (1/2)(BF)(BQ) Sin(∠FBQ) and BQ = 2 (Area BFQ)/(BF) Sin (∠FBQ)
Where : ∠FBQ = included angle
BFQ = triangle formed by the trial line, dividing line, and side
BQ
BF = known side (trial line)
The triangle is then solved for the length of the dividing line FQ by applying
the cosine law. Also, by applying the sine law, angle BFQ is determined and
the bearing of FQ is computed. The latitudes and departures of lines FQ, BQ ,
and Qc area also computed.
As a check, the areas of the two tracts area calculated independently
based on the given and calculated latitudes and departures. The sum of the
areas of the two tracts should equal the area of the entire parcel.
In the field , the length FQ is laid off im the required direction, and a
check on fieldwork and computations is obtained if points Q thus established
falls on lines BC and if the computed distances BQ and QC agrees with the
measured distances.
Sometimes, te tract will be of such shape that a line drawn from the
fiven point in the boundary to any corner will cut off an area nowhere near
that required. Under these circumstances or when the traverse has a large
number of sides, it is advisable to first plot the traverse to scale and to
establish a trial line of subdivision.
As a rough check the planimeter may be used to advantage for
determining the area cut off by a trial line. The line may be shifted until the
area cut off agrees closely with that required.
48-2. TO CUT OFFA REQUIRED AREA BY A LINE RUNNING IN A GIVEN
DIRECTION
In Figure 48-2. ABCDEFG represents an irregular parcel of land with courses of
known lengths and bearings, the latitudes and departures computed and
adjusted, and the total area of the tract calculated. The parcel of land is to be
divided into two parts, each of a required area, bya line MN running in a given
direction. The subdivision scheme may require the division of the whole
parcel into equal area or into any other desired proportional parts.
Sides AQ and QE area unknown. These unknown quantities area easily
determined by considering traverse EFAQ as on with two omitted quantities.
A closing line AE is established and its latitude, departure, length, and
bearing calculated.
Before the lengths of AQ and QE area determined from triangle AEQ. Since
the latitudes and departures of the traverse EFAQ area either known or can
be calculated, the area cut off by the trial line may then be determined by the
DMD or DPD method. The difference between this area and the trapezoid
MNEQ in which side QE is known. The angles Θ and Θ can be computed from the
known bearings of adjacent slides.
The altitude of the trapezoid (d) represents the distance of the dividing line from the trial line. Its
value is found by solving the above equation since it is the only unknown in the equation. The
length of the dividing line is then determined as follows:
MN = QE – NR +MP
= QE – d Tan + d Tan Θ
= QE – d (Tan – TanΘ)
In the field, the points M and N area established on lines AB and DE, at the calculated
distances from the measured value agrees with the computed value, a check is provided. A
further check is made on the computations by calculating the area of tract BCDNM and
comparing it with the required area.
PROBLEM SETS
In the succeeding pages area different problem sets which area intended to be worked out by
students in this course of study. Each problem set corresponds to a specific group of lesson
modules. They area arranged in a particular order of presentation and are properly titled to
facilitate easy cross – referencing. These problem sets area aimed at helping students attain
enough mastery f the theories and principles discussed in this book.
The problems given have varying degrees of difficulty and area designed to provide a
good balance between theory and applications. Some problems involve extension of topics
presented in the lesson modules which could not be treated because of space limitations. An
attempt should be made by the student to solve these problems immediately after a group of
lesson modules has been completely read and understood. It must not be put aside and done at
a latter time. He should only proceed to the next group of lesson modules after he is able to
solve the corresponding problem set.
A number of problems given in some sets require solutions that area relatively long. By
constant practice in solving these problems the student develops a feeling of mastery in this
course of study. He will conscientiously develop within himself a feeling of preparedness and
self-confidence in taking examinations. His solutions tend to become orderly and thoroughly
presented, and skill in the use of an electronic calculator is also developed.
It is suggested to the instructor that the students be required to submit their solutions
for each problem set after a group of lessons have been completed, preferable during a quiz or a
major unit examination. Submitted solutions should be given a corresponding grade which
should in turn form part of the final grade of the student in the course.
Where difficulty arises, the student should feel free to consult his instructor or
classmates, or preferably read back the particular lesson concerned in order to have a better
grasp and understanding of the required solution. The student should, however, be cautioned
and made to understand that he will only be depriving himself of the correct learning process by
not working out the assigned problems or by just copying the completed work of others. It is
extremely important for the student to be honest with himself.
PROBLEM SES
PS NO. 1: INTRODUCTION TO SURVEYING
PS NO. 2: MEASUREMENT OF HORIZONTAL DISTANCES
PS NO. 3: MEASUREMNT OF VERTICAL DISTANCES
PS NO. 4: LEVELING METHODS
PS NO 5: MEASUREMENT OF ANGLES AND DIRECTIONS
PS NO 6: THE ENGINEER’S TRANSIT AND THEODOLITE
PS NO 7: TRAVERSING AND TRAVERSE COMPUTATIONS
PS NO 8: OMITTED MEASUREMENTS
PS NO 9: AREA COMPUTATIONS
PS NO 10: PARTITION OF LAND
INTRODUCTION TO SUREYING
Note: Upon completion of the different lessions listed under this part of your course of study,
you will be required to solve all the problems givin in this set. The problems may be solved with
books, notes , and other references opened. You may consult your instructors, classmates, and
other whom you feel could help you better understand and solve the problems. You should,
however, refreain from copying previously prepared solutions since this will only defear the
purpose and objectives of this requirement. Althought no time limit is specified to solve each
problem, all solutions and answers must, howeer, be submitted during the scheduled
examination date for the same group of lesson modules. All solutions must b e done on short
size bond paper and write on one side only.
1. LINEAR MEASUREMENTS. The measured length of airport runways in five major cities in the
Philippines are: 125.00, 1375.50, 1410.75, 1550.25, abd 1750.00 neters. Determine the
equivalent length of each runway in kilimeters, decimeters, and centimeters. Tabulate values
accordingly.
2. AREA MEASUREMENTS. Given the dimenstions of the following tracts of land:
a) 108.75 m by 76.82 m
b) 940.05 m by 1296.73 m
c) 13.36 m by 50.05m
d) 1258.30 m by 624.03 m
e) 8476.55 m by 121.79 m
Determine the area of each tract in square meters, square kilometers, areas, and hectares.
3. AREA MEASUREMENTS. Given the area and width of the following rectangular - shaped pieces
of property:
a) 2.575 ha and 195.42 m
b) 125.42 sq m and 545.0 cm
c) 0.85 sq kmn and 925.09,
d) 50.0 ares and 100.0 m
e) 42545.19 sq m and 346.72 m
Determine the length of each property in meters
4. VOLUME MEASUREMENTS. Following area dimension, length, width and depth of five
excavated borrow of a highway project:
a) 133.26 m, 35.48 m, abd 18.60 ,
b) 50.05m, 39.25 m, and 7.14 m
c) 243.55 m, 76.18m, and 26.66 m
d) 42.055 m, 8.605 m, and 12.332 m
e) 9.5 m, 6.3m, and 4.9m
Determine the volume of each pit in cubic meters.
5. VOLUME MEASUREMENTS. Given the approximate flat area and depth of excavation of the
following borrow pits:
a)3750.0 sq m and 758.0 cm
b) 0.035 sq km and 180.0 m
c) 15.6 ares and 495.0 m
d) 9.250 ha and 250.0 m
e_ 6750 sq n and 195.0 m
Determine the volume of earth removed from each pit in cubic meter.
6. SIGNIFICANT FIGURES. Given the following numbers: 45.63 ,5.700, 4010, 0.00037, 0.000940,
6.0090, 7.00, 9.5 x 198, 4.00 x 107, 2.604 x 1018 and 3.00 x 10 -6. For each number. identify the
significant figures and state the number of significant figures. Tabulate alues accordingly.
7. ROUNDING OFF NUMBERS. Given the following numbers: 0.333333, 0.412342 , 50.2155,
48.5455, and 16.3545. Round off each number to three, four and five figures. Tabulate alues
accordingly.
8. MOST PROBABLE ALUE. The three angles of a triangle were measured with the following
results: A=42° 05', B=115 °38' and C=22°08'. Determine the most preobable value of each angle.
9. MOST PROBABLE VALUES. The angles about a point have the following observed values:
87°07'50'', B=118°44'15'', C=80°54'35'', and D=59°45'50''. Determine the most probable value of
the three angles.
10. MOST PROBABLE VALUE. The interior angles of a quadrillateral were observed to be :
A=100°35'40'', B=118°44'15''. and C=80°54'35'', and D = 59°45'50''. Determine the most
probable value of each of these angles.
11. PROBABLE ERROR. A quantity was measured ten times with the following results: 340630,
34.626, 34.634, 34.628, 34.629, 34.626, 34.627, 34.633, 34.625, and 36.624 meters. Determine
the probable error of the mean and the relatie precision of the mean.
12. PROBABLE ERROR. A surveying instructor sent all the 40 students in his class out to measurea
distrance between two points marked on a runway. The students working in groups of four came
up with 10 different measurements as follows : 920.45, 921.05, 921.65, 920.25, 920.15, 921.85,
921.95, 920.45, 921.15, abd 921.35 meters. Assuming these values area equally reliable and that
ariations result only from accidental errors, determine the relative precision of a single
measurement and the relative precision of the mean.
13. PROBABLE ERROR. The following values were determined im a series of rod readings made
under identical conditrions" 3.312. 3.307, 3.304, 3.306, 3.309, 3.301, 3.311, 3.308, 3.312, 3.306,
and 3.313 meters.