2. Chord and Offset Simple

National University College of Civil Engineering Civil Engineering Department

SIMPLE CURVE LAYOUT USING CHORD AND OFFSET Plate .

Engr. Joselito D. Batac Professor Rafol , Tyrone john F. Basic and Higher Surveying Field Works Civ. 143 OBJECTIVE:

 to layout simple curve using chord and offset  to learn how to compute for the chord and offset of a given curve EQUIPMENT:  fiber glass tape  chalk  string

PROCEDURE:  Consider one point (PI). From PI, measure the length of your tangent line using the steel tape and name the other endpoint PC. Swing the length on the other side. From PC, swing the length of the computed chord. Mark as PI the intersection between the two swings.  Compute for the corresponding length of every station given.  In accord to that, compute the offset of each station from the chord.  Apply the computed data and assure if it forms a curve.

DATA: station 1/8 1/4 3/8 1/2 5/8 3/4 7/8

chord 2.181 4.362 6.540 8.716 10.887 13.053 15.212

offset 0.332 0.569 0.712 0.76 0.712 0.569 0.332

FORMULA: Ѳ= 10-(X)(I)

offset= Csin(Ѳ)

C= 2Rsin(I/2)

I= angles divided

COMPUTATION: R=50 I= 20 1/8:

1/4:

1/8(20)=2.5

1/4(20)=5

C1=2(50)sin(2.5/2)= 2.181

C2=2(50)sin(5/2)= 4.362

Offset1=2.181sin(8.75)=0.332 4.362sin(7.5)= 0.569

Offset2=

3/8:

1/2:

3/8(20)= 7.5

1/2(20)= 10

C3=2(50)sin(7.5/2)= 6.540 8.716

C4=22(50)sin(10/2)=

Offset3=6.540sin(6.25)= 0.712 0.76

Offset4=8.716sin(5)=

5/8:

3/4:

5/8(20)=12.5

3/4(20)=15

C5=2(50)sin(12.5/2)= 10.887 13.053

C6=2(50)sin(15/2)=

Offset5=10.887sin(3.75)= 0.712 Offset6=13.053sin(2.5)= 0.569

7/8: 7/8(20)=17.5 C7=2(50)sin(17.5)= 15.212 Offset7=15.212sin(1.25)= 0.332

ILLUSTRATION: