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CASE I: DIVIDING AN AREA INTO TWO PARTS BY A LINE BETWEEN TWO POINTS PROBLEM:

Given the following data of a tract of land, determine the area of the two parts into which the tract is divided by a line running from F to C. Calculate the length and bearing of the dividing line. Line Length Bearing AB 650 m N 15 o 30 ' E BC

600

N 75 o 40' E

CD

550

S 80o 10' E

DE

700

S 20o 10 ' W

EF

600

N 76 o 30' W

FA

499.8 5

S 70o 51' 5'' W

SOLUTION:

C

N

1. Solving for thearea of the whole traverse (tract ABCDEF ) .

Tarlac State University Course: CE 313 – Surveying 1 COLLEGE OF ENGINEERING Engr. Randy G. Policarpio Civil Engineering Department 10:46:50 PM

Prepared by: 12/3/2015

2 Page

A ABCDEF =

1 ( Double Area ) 2 ∑

1 A ABCDEF = (1,355,224.67 ) 2 A ABCDEF =677,612.34 sq . m .

2.Consider tract FABC

C

3. Solving for the Length∧Bearing of the Dividing Line C

Tarlac State University Course: CE 313 – Surveying 1 COLLEGE OF ENGINEERING Engr. Randy G. Policarpio Civil Engineering Department 10:46:50 PM

F

Prepared by: 12/3/2015

3 Page

LCF =√ ( 610.94 ) + ( 282.83 ) =673.23 m 2

θCF =tan−1

2

=24 50 ( 282.83 610.94 ) o

'

o

'

Bearing of the Dividing Line CF: S 24 50 W 4. Solving for the area of Tract FAB

C

1 A FABC= ( 580421.15 ) 2 A FABC=290,210.575 sq . m.

5.Consider tract CDEF

Tarlac State University Course: CE 313 – Surveying 1 COLLEGE OF ENGINEERING Engr. Randy G. Policarpio Civil Engineering Department 10:46:50 PM

Prepared by: 12/3/2015

4 Page

6. Solving for the Length∧Bearing of the Dividing Line FC

LFC = √( 610.94 ) + (282.83 ) =673.23 m 2

θ FC =tan −1

2

=24 50 ( 282.83 610.94 ) o

'

Bearing of the Dividing Line CF: N 24 o 50' E 7. Solving for thearea of Tract CDEF

Tarlac State University Course: CE 313 – Surveying 1 COLLEGE OF ENGINEERING Engr. Randy G. Policarpio Civil Engineering Department 10:46:50 PM

Prepared by: 12/3/2015

5 Page

1 A CDEF = ( 774,803.53 ) 2 A CDEF =387,401.765 sq .m .

Checking :

A FABC + A CDEF = A ABCDEF 290,210.575+387,401.765=677, 612.34 677, 612.34=677, 612.34 therefore OK

CASE II: DIVIDING AN AREA BY A LINE RUNNING THROUGH A POINT AND A GIVEN DIRECTION PROBLEM: Given the following data of a tract of land, find the area of each of the two parts into which the tract is divided by a line through F with a bearing of Line AB

Length 650 m

N 15 o 30 ' E

BC

600

N 75 o 40' E

CD

550

S 80o 10' E

DE

700

S 20o 10 ' W

N 35o 30 ' E .

Bearing

Tarlac State University Course: CE 313 – Surveying 1 COLLEGE OF ENGINEERING Engr. Randy G. Policarpio Civil Engineering Department 10:46:50 PM

Prepared by: 12/3/2015

N 76 o 30' W

FA

499.8 5

S 70o 51' 5'' W

6

600

Page

EF

SOLUTION:

C

P

1. Solving for thearea of the whole traverse (tract ABCDEF ) .

A ABCDEF =

1 ( Double Area ) 2 ∑

1 A ABCDEF = (1,355,224.67 ) 2 Tarlac State University Course: CE 313 – Surveying 1 COLLEGE OF ENGINEERING Engr. Randy G. Policarpio Civil Engineering Department 10:46:50 PM

Prepared by: 12/3/2015

Page

7

A ABCDEF =677,612.34 sq . m .

2.Consider tract FABC

C

3. Solving for the Length∧Bearing of the Trial Line C

F

LCF =√ ( 610.94 ) + ( 282.83 ) =673.23 m 2

θCF =tan−1

2

=24 50 ( 282.83 610.94 ) o

'

Bearing of theTrial Line CF :S 24o 50' W Tarlac State University Course: CE 313 – Surveying 1 COLLEGE OF ENGINEERING Engr. Randy G. Policarpio Civil Engineering Department 10:46:50 PM

Prepared by: 12/3/2015

8 Page

2.Consider tract CP F

f =¿ λ β

673.23 m ρ

c=¿

Tarlac State University Course: CE 313 – Surveying 1 COLLEGE OF ENGINEERING Engr. Randy G. Policarpio Civil Engineering Department 10:46:50 PM

Prepared by: 12/3/2015

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1

CASE I: DIVIDING AN AREA INTO TWO PARTS BY A LINE BETWEEN TWO POINTS PROBLEM:

Given the following data of a tract of land, determine the area of the two parts into which the tract is divided by a line running from F to C. Calculate the length and bearing of the dividing line. Line Length Bearing AB 650 m N 15 o 30 ' E BC

600

N 75 o 40' E

CD

550

S 80o 10' E

DE

700

S 20o 10 ' W

EF

600

N 76 o 30' W

FA

499.8 5

S 70o 51' 5'' W

SOLUTION:

C

N

1. Solving for thearea of the whole traverse (tract ABCDEF ) .

Tarlac State University Course: CE 313 – Surveying 1 COLLEGE OF ENGINEERING Engr. Randy G. Policarpio Civil Engineering Department 10:46:50 PM

Prepared by: 12/3/2015

2 Page

A ABCDEF =

1 ( Double Area ) 2 ∑

1 A ABCDEF = (1,355,224.67 ) 2 A ABCDEF =677,612.34 sq . m .

2.Consider tract FABC

C

3. Solving for the Length∧Bearing of the Dividing Line C

Tarlac State University Course: CE 313 – Surveying 1 COLLEGE OF ENGINEERING Engr. Randy G. Policarpio Civil Engineering Department 10:46:50 PM

F

Prepared by: 12/3/2015

3 Page

LCF =√ ( 610.94 ) + ( 282.83 ) =673.23 m 2

θCF =tan−1

2

=24 50 ( 282.83 610.94 ) o

'

o

'

Bearing of the Dividing Line CF: S 24 50 W 4. Solving for the area of Tract FAB

C

1 A FABC= ( 580421.15 ) 2 A FABC=290,210.575 sq . m.

5.Consider tract CDEF

Tarlac State University Course: CE 313 – Surveying 1 COLLEGE OF ENGINEERING Engr. Randy G. Policarpio Civil Engineering Department 10:46:50 PM

Prepared by: 12/3/2015

4 Page

6. Solving for the Length∧Bearing of the Dividing Line FC

LFC = √( 610.94 ) + (282.83 ) =673.23 m 2

θ FC =tan −1

2

=24 50 ( 282.83 610.94 ) o

'

Bearing of the Dividing Line CF: N 24 o 50' E 7. Solving for thearea of Tract CDEF

Tarlac State University Course: CE 313 – Surveying 1 COLLEGE OF ENGINEERING Engr. Randy G. Policarpio Civil Engineering Department 10:46:50 PM

Prepared by: 12/3/2015

5 Page

1 A CDEF = ( 774,803.53 ) 2 A CDEF =387,401.765 sq .m .

Checking :

A FABC + A CDEF = A ABCDEF 290,210.575+387,401.765=677, 612.34 677, 612.34=677, 612.34 therefore OK

CASE II: DIVIDING AN AREA BY A LINE RUNNING THROUGH A POINT AND A GIVEN DIRECTION PROBLEM: Given the following data of a tract of land, find the area of each of the two parts into which the tract is divided by a line through F with a bearing of Line AB

Length 650 m

N 15 o 30 ' E

BC

600

N 75 o 40' E

CD

550

S 80o 10' E

DE

700

S 20o 10 ' W

N 35o 30 ' E .

Bearing

Tarlac State University Course: CE 313 – Surveying 1 COLLEGE OF ENGINEERING Engr. Randy G. Policarpio Civil Engineering Department 10:46:50 PM

Prepared by: 12/3/2015

N 76 o 30' W

FA

499.8 5

S 70o 51' 5'' W

6

600

Page

EF

SOLUTION:

C

P

1. Solving for thearea of the whole traverse (tract ABCDEF ) .

A ABCDEF =

1 ( Double Area ) 2 ∑

1 A ABCDEF = (1,355,224.67 ) 2 Tarlac State University Course: CE 313 – Surveying 1 COLLEGE OF ENGINEERING Engr. Randy G. Policarpio Civil Engineering Department 10:46:50 PM

Prepared by: 12/3/2015

Page

7

A ABCDEF =677,612.34 sq . m .

2.Consider tract FABC

C

3. Solving for the Length∧Bearing of the Trial Line C

F

LCF =√ ( 610.94 ) + ( 282.83 ) =673.23 m 2

θCF =tan−1

2

=24 50 ( 282.83 610.94 ) o

'

Bearing of theTrial Line CF :S 24o 50' W Tarlac State University Course: CE 313 – Surveying 1 COLLEGE OF ENGINEERING Engr. Randy G. Policarpio Civil Engineering Department 10:46:50 PM

Prepared by: 12/3/2015

8 Page

2.Consider tract CP F

f =¿ λ β

673.23 m ρ

c=¿

Tarlac State University Course: CE 313 – Surveying 1 COLLEGE OF ENGINEERING Engr. Randy G. Policarpio Civil Engineering Department 10:46:50 PM

Prepared by: 12/3/2015

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