survey1 - project

8/8/2019 Survey1 - Project

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Chaining & VisionChaining & Vision

Both ObstructedBoth ObstructedProject By:Project By:

Y610022Y610022 Hem ShahHem Shah

Y610023Y610023 Nirav ShahNirav Shah Y610024Y610024 Nirav SethiyaNirav Sethiya Y610025Y610025 Grishma SiswalaGrishma Siswala Y610026Y610026 Prashant SoniPrashant Soni Y610029Y610029 Jogendar YadavJogendar Yadav Y610032Y610032 Mihir PatelMihir Patel Y710101Y710101 Arfan KhanArfan Khan Y610028Y610028 Ronak Vora.Ronak Vora.


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Chaining & VisionChaining & VisionBoth ObstructedBoth Obstructed

In this case the problem consists inIn this case the problem consists inprolonging the line beyond theprolonging the line beyond theobstacle and determining theobstacle and determining the

distance across it.distance across it.

A BUILDING is a typical example ofA BUILDING is a typical example ofthis class of obstaclethis class of obstacle


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Methods to OvercomeMethods to Overcome

Such ObstaclesSuch Obstacles..

There are different methodsThere are different methods

to overcome the obstaclesto overcome the obstacles

which are related towhich are related to

CHAINING AND VISIONCHAINING AND VISIONOBSTRUCTED.OBSTRUCTED.

They are described as follows :They are described as follows :


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Method 1Method 1 Choose points A & B onChoose points A & B on

the chain line PR.the chain line PR.

At A & B erectAt A & B erect

perpendiculars AE & BF ofperpendiculars AE & BF ofequal length.equal length.

Check the diagonals BE &Check the diagonals BE &AF which should be equalAF which should be equal

and also EF which shouldand also EF which shouldbe equal to AB.be equal to AB.

Prolong the line EF. PassProlong the line EF. Passthe obstacle and select twothe obstacle and select twopoints G & H on it.points G & H on it.

P A B C D R

E F G H


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Method 1Method 1

At G & H set outAt G & H set outperpendiculars GC & HDperpendiculars GC & HD

equal in length to AE.equal in length to AE.

The points C and D areThe points C and D areobviously on the chain lineobviously on the chain linePR and BC = FG.PR and BC = FG.

Care must be taken inCare must be taken insetting out perpendicularssetting out perpendicularsvery accurately and theirvery accurately and their

lengths are equal.lengths are equal.

P A B C D R

E F G H


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METHOD # 2METHOD # 2

Select a point B on theSelect a point B on thechain line PR and erectchain line PR and erectperpendicular BE.perpendicular BE.

Mark another point A onMark another point A onPR, such as BA = BEPR, such as BA = BE

Join AE and produce itJoin AE and produce it

to a point F.to a point F.

With the help of anWith the help of anoptical square, set outoptical square, set outFD at right angles toFD at right angles toFA, making FD = FA.FA, making FD = FA.

P A B C D R

F

GE


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METHOD 2METHOD 2 On FD mark a point G soOn FD mark a point G so

that FG = FE.that FG = FE. With D as the centre andWith D as the centre and

radius equal to BE, swing anradius equal to BE, swing an

arc.arc. Similarly with G, as theSimilarly with G, as the

centre and radius equal tocentre and radius equal toBE, swing another arc.BE, swing another arc.Cutting the first one in C.Cutting the first one in C.

C and D are then on theC and D are then on thechain line PR.chain line PR.

Measure EG. Then theMeasure EG. Then theobstructed lenght BC=EG.obstructed lenght BC=EG.

P A B C D R

F

GE


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METHOD 3METHOD 3

Select two points A andSelect two points A andB on the chain line PR.B on the chain line PR.

On AB lay down theOn AB lay down theequilateral triangle ABEequilateral triangle ABEby swinging arcs with theby swinging arcs with thetapetape

Produce AE to F.Produce AE to F.

Mark a point H on FAMark a point H on FAand construct theand construct theequilateral triangle FHKequilateral triangle FHKon FH in the same way.on FH in the same way.

Prolong the line FK to D,Prolong the line FK to D,

making FD = FA.making FD = FA.

P A B C D R

E

F

Q

H K


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METHOD 3METHOD 3

Choose a suitable point Q onChoose a suitable point Q onFD.FD.

Determine a second point C onDetermine a second point C onthe chain line PR forming thethe chain line PR forming the

equilateral triangle DQC onequilateral triangle DQC onDQ as the base.DQ as the base. If possible make DQ = AE.If possible make DQ = AE. The line joining the points CThe line joining the points C

and D thus determined,and D thus determined,

determine the direction of thedetermine the direction of thechain line past the obstacle.chain line past the obstacle. Then the obstructed lengthThen the obstructed length

BC = ADBC = AD ABAB CDCD= FA= FA ABAB CD.CD.

P A B C D R

E

F

Q


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METHOD 4METHOD 4 Choose two points A and BChoose two points A and B

on the chain line PR and seton the chain line PR and setout a line DBE as shown.out a line DBE as shown.

Prolong AD and AE to FProlong AD and AE to Fand G respectively makingand G respectively making

AF = n x ADAF = n x AD andandAG = n X AEAG = n X AE

Join FG and making the pointJoin FG and making the point

C on it so that F

C = n x

C on it so that F

C = n xBD.BD.

In order to locate the secondIn order to locate the secondpoint on PR, continue AFpoint on PR, continue AFand AG to H and K resp.and AG to H and K resp.

making AH = n x AD

and AKmaking AH = n x AD

and AK= n x AE.= n x AE.

H M

R

K

F

D

C

E

G

B

P

AA


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METHOD 4METHOD 4

Mark the point AH on HK soMark the point AH on HK sothat HM = n x DB.that HM = n x DB.

The points C and M are on theThe points C and M are on the

chain line.chain line. The obstructed distance BC isThe obstructed distance BC is

equal to ACequal to AC AB.AB. But by similarity of trianglesBut by similarity of triangles

ABD

and ACFAB

Dand A

CFACAC == AFAF = n= n

AB ADAB ADThus AC = n x AB and henceThus AC = n x AB and hence

BC

= ( nBC

= ( n--1

)AB.1

)AB.

H M

R

K

F

D

C

E

G

B

P


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TO DETERMINETHE DISTANCETO DETERMINETHE DISTANCE

BETWEENTWO POINTS WHENBETWEENTWO POINTS WHEN

BOTHAREINACCESSIBLEBOTHAREINACCESSIBLE

Refering to the figure let ARefering to the figure let A

and B be the twoand B be the twoinaccessible figures.inaccessible figures. It is required to determineIt is required to determine

the distance AB.the distance AB. Choose any convenient pointChoose any convenient pointC.C.

Determine the obstructedDetermine the obstructeddistances CA and CB bydistances CA and CB byany type of method.any type of method.

A D B

C

a d b


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TO DETERMINE THE DISTANCETO DETERMINE THE DISTANCE

BETWEEN TWO POINTS WHENBETWEEN TWO POINTS WHEN

BOTH ARE INACCESSIBLEBOTH ARE INACCESSIBLE

Take any convenient point aTake any convenient point aon CA and measure Ca.on CA and measure Ca.

Lay off along CB a length CBLay off along CB a length CBequal toequal to CA x CBCA x CB

CACA

Measure ab.Measure ab. Then the required distanceThen the required distanceAB =AB = ab x CAab x CA == ab x CBab x CB

CaCa CbCb

A D B

C

a d b


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IMPLEMENTATION OFIMPLEMENTATION OF

CHAININGAND VISION BOTHCHAININGAND VISION BOTH

OBSTRUCTEDOBSTRUCTED

The method Chaining And Vision Both

Obstructed is used in the constructionof:

Construction of bridges.

Construction of fly-overs.

It is very important that the placement ofcolumns and beams are correct becauseif it is not than the load distributionbetween the columns will be unequalwhich may damage the structure.

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