[FEL RANGING OUT SURVEY LINES survey line or ‘chain lin | stations. e’, the chain or tape must be stretcheq If the length of line is Jess than the so, If however, the length of the ling te will have to be established in line The process of fixing or establishing methods of ranging: (i) Direct While measuring the length ofa straight along the line joining its two terminal io length of the chain, there will be no difficulty, in doing eds the length of the chain, some intermediate point with the two terminal points before chaining is started. such intermediate points is known as ranging. There are two ranging, (ii) Indirect ranging. (i) Direct Ranging: Direct ranging is done when the two & In such cases, ranging can either be done by eye or throu, line ranger or a theodolite. Ranging by eye: (Fig. 3.20).LetA and Bbethetwo —9¢ points at the ends of a survey line. One ranging rod ‘Surveyor is erected at the point B while the surveyor stands . Fig. 3.20 Ranging by eye. with another ranging rod at point A, holding the rod at about half metre length. The assistant then goes with another ranging rod and establishes the rod at a point approximately in the line with AB (by judgment) at a distance not greater than one chain length from A. The surveyor at A then signals the assistant to move transverse to the chain line, till he is in line with A and B, Similarly, other intermediate points can be established. The code of signals used for this purpose is given in the table below : CODE OF SIGNAL nds of the survey lines are intervisible, igh some optical instrument such asa Beaiaielicte ; the Surveyor” Rapid sweep with right hand Slow sweep with right hand Move considerably to the right Move slowly to the right | Right arm extended Right arm up and moved to the right Rapid sweep with left hand Slow sweep with left hand Left arm extended Left arm up and moved to the left Both hands above head and then brought down Both arms extended fc i Both arma ental forward horizontally and the hands Continue to move to the right Plumb the rod to the right Move considerably to the left Move slowly to the left | Continue to move to the left Plumb the rod to the left Correct Fix the rodRanging by Line Ranger Aline ranger consists of either two plane mirrors or two right angled isosceles prisms placed one above the other, as shown in Fig. 3.21. The diagonals of the two prisms are silvered so as to reflect the incidental rays. A handle with a hook is provided at the bottom to hold the instrument in hand to transfer the point on the ground with the help of plumb-bob. ‘To range a point P, two ranging rods are fixed at the ends A and B, and the surveyor at P holds the line ranger very near to the line AB (by eye judgment). The lower prism abe receives the rays from A which are reflected by the diagonal ac towards the observer. Similarly, the upper prism dbc receives the rays from B which are reflected by the diagonal bd towards the observer. Thus, the observer views the images of ranging rods at A and B, which may not be in the same vertical line as shown in Fig. 3.21 (0). Top Image prism of pole B Sex Observer Image of pole A (a) Plan (b) Pictorial view ao een () @ Fig. 3.21 Optical Line Ranger. The surveyor then moves the instrument sideways till the two images are in the same vertical line as shown in Fig. 3.21 (d). The point Pis then transferred to the ground with the help of a plumb bob. Thus, the instrument can be conveniently used for fixing intermediate points on a long line without going to either end. Also, only one person, holding the line ranger, is required in this case. Fig. 3.18 shows a combined line ranger and a prism square. Adjustment of Line Ranger: One of the mirrors or prisms is commonly made adjustable. To test the perpendicularity between the reflecting surfaces, three poles are ranged very accurately with the help of a theodolite. The line ranger is held over the middle pole. The instrument will be in perfect adjustment if the images of the two end poles appear in exact coincidence. If not, they are made to do so turning the movable prism by means of the adjusting screw.(ii) Indirect or Reciprocal Rang- ing: Indirect or Reciprocal ranging is resorted to when both the ends of the survey line are not intervisible either due to high intervening ground or due to long distance between them. In such a case, ranging is done indirectly by selecting two intermediate points M, and N,, very near to the chain Tine (by judgement) in such a way that from M,, both N,, and Bare visible (Fig. 3.22) and from N,, both M,, and A are visible. ‘Two surveyors station them- selves at M, and N, with ranging rods. The person at M, then directs the person at N, to move to a new position N, in line with Fig. 3.22 Reciprocal Ranging. M, B. The person at Nz then directs the person at M, to move to a new position M, in line with N,A. Thus, the two persons are now at M, and N; which are nearer to the chain line than the positions M, and Nj. The process is repeated till the points M and N are located in such a way that the person at M finds the person at N in line with MB, and the person at N finds the person at M in line with NA. After having established M and N, other points can be fixed by direct ranging.CHAIN TRIANGULATION Chain surveying is that type of surveying in which only linear measurements are made in the field. This type of surveying is suitable for surveys of small extent on open ground to secure "data for exact description of the boundaries of a piece of land or to take simple details. The principle of chain survey or Chain Triangulation, as is sometimes called, is to provide a skeleton or framework consisting of a number of connected triangles, as triangle is the only _simple figure that can be plotted from the lengths of its sides measured in the field. To get good ‘results in plotting, the framework should consist of triangles which are as nearly equilateral _as possible. Eg survey sTaTIONs A survey station is a prominent point on the chain line and can be either at the beginning of the chain line or at the end. Such station is known as main station. However, subsidiary or tie station can also be selected anywhere on the chain line and subsidiary or tie lines may be run through them. Asurvey station may be marked on the ground by driving pegs if the ground is soft. However, on roads and streets etc., the survey station can be marked or located by making two or preferably three tie measurements with respect to some permanent reference objects near the station. The more nearly the lines joining the peg to the reference points intersect at right angles, the more definitely will the station be fixed. A diagram of the survey lines with main stations numbered should be inserted in the beginning of the field note book. EER. survey Lines The lines joining the main survey stations are called main survey lines. The biggest of the main Survey line is called the base line and the various survey stations are plotted with reference to this. If the area to be surveyed has more than three straight boundaries, the field measurements ae fe so arranged that they can be plotted by laying down the triangles as shown in Fig. 4.1 (a) or (b).—————— Check lines: Check lines or proof lines are the lines which are run in the field to check the accuracy of the work. The length of the check line measured in the field must agree with its length on the plan. A check line may be laid by joining the apex of the triangle to any point on the opposite side or by joining two points on any two sides of a triangle. Each triangle must have a check line. For the framework shown in Fig. 4.1 (a), the various an Fig. 4.2 (@), (0), (©) and (d) by dotted lines. In Fig. 4.1 of check lines for the framework. @ Fig. 4.4 (b) (a) (b) Fig. 4.2 rangements of the check lines are shown in (b), the dotted lines show the arrangements Tie lines: A tie line is a line which joins subsidiary or tie stations on the main line. The main object of running a tie line is to take the details of nearby objects but it also serves the purposed! acheck line. The accuracy in the location of the objects depends upon the accuracy in |: faying the tie line. A framework may have one or more tie lines depending upon the circumstances (Fig. 43) Land boundary Fig. 4.3,pee ee ie : TS LOCATING GROUND FEATURES : OFFSE An offset is the lateral distance of an object or ground feature measured from a survey line, By method of offsets, the point or object is located by measurement of a distance and angle (usually 90°) from a point on the chain line. When the angle of offset is 90°, it is called Perpendicular offset [Fig. 4.5 (a), (0)] or sometimes, simply, offset and when the angle is other than 90°, it is called an oblique offset (Fig. 4.5 (b)]. Another method of locating a point is called the method of ‘ties’ in which the distance of the point is measured from two separate points on the chain line such that the three points form, as nearly as possible an equilateral triangle [Fig. 4.5 (d)]. The method of perpendicular offsets involves less measuring on the ground. (a) Perpendicular ™NS offset (©) Perpendicular offset Fig. 4.5 Offsets should be taken in order of their chainages. In offsetting to buildings, check can be made by noting the chainages at which the directions of the walls cut the survey line, as shown by dotted lines in Fig. 4.5 (c), (d). In general, an offset should be taken wherever the outline of an object changes. In the cased a straight wall or boundary, an offset at each end is sufficient. To locate irregular boundarie, sufficient number of offsets are taken at suitable interval and at such point where the direction suddenly changes, as shown in Fig. 4.6 (a). In the case of a nallah, offsets should be taken to bot! the sides of its width, as shown in Fig. 4.6 (b). However, in the case of regular curves with constat! width, the offsets should be taken to the centre line only and the width should also be meas Taking Perpendicular Offsets: Fig. 4.7 illustrates the procedure for finding the length and position of the perpendicular offset. ‘The leader holds the zero end of the tape at the point P to be located and the follower carries the tape box and swings the tape along the chain. The length of the offset is the shortest distance from the object to the chain obtained by swinging the tape about the object as centre. Such an offset is called swing offset. The position of the offect on the chain is located by the point where the arc is tangential to the chain. Fig. 47Degree of Precision in Measuring the Offsets: Before commencing the field measurements, one should know the degree of precision to be maintained in measuring the length of the offset. This mainly depends on the scale of survey. Normally, the limit of precision in plotting is 0.25 mm. If the scale of plotting is 1 cm = 2 m, 0.25 mm on paper will correspond 2x0.25 o 10 to the nearest 5 cm. On the other hand, if the scale of plotting is 1 cm = 10 m. 0.25 mm 10x0.25 10 measured to the nearest 25 cm. However, if there is likelihood of changing the scale of plotting at a later stage, it is better always to be over-accurate than to be under-accurate. t = 0.05 m on the ground. Hence, in such a case, the offsets should be measured on paper will correspond to = 0.25 m on the ground. Hence the offset should be Long Offsets: The survey work can be accurately and expeditiously accomplished if the objects and features that are to be surveyed are near to the survey lines. The aim should always be to make the offset as small as possible. Long offset may be largely obviated by judiciously placing the main lines of the survey near the object or by running subsidiary lines from the main lines. Fig. 4.8 shows a well-proportioned subsidiary triangle abc run to locate the deep bend of the outline of the fence. The base of the triangle is on the main line and bd is Fig. 4.8 the check line.EBA INSTRUMENTS FOR SETTING OUT RIGHT ANGLES i There are several types of instruments used to set out a right angle to a chain line, the most common being (i) cross staff (ii) optical square (jii) prism square (jv) site square. ( Cross Staff; The simplest instrument used for setting out right angles is a cross staff. It consists of either a frame or box with two pairs of vertical slits and is mounted on a pole shod for fixing in the ground. The common forms of cross staff are (a) open cross staff (b) French cross staff (c) adjustable cross staff. (a) Open Cross Staff: Fig. 4.14. (@) shows an open cross staff. It is provided with two pairs of vertical slits giving two lines of sights at right angles to each other. The cross Staff is set up at a point on the line from which the right angle is to run, and is then turned until one line of sight passes through the ranging pole at the end of the survey line. The line of sight through the other two vanes will be a line at right angles to the survey Jine and a ranging rod may be established in that direction. If, however, itis to be used to take offsets, it is held vertically on the chain line at a point where the foot of the offsets is likely to occur. It is then turned so that one line of sight passes through the ranging rod fixed at the end (a) Fig, 4.14. Various Forms of Cross Staff.er pai its, it i if the point ine. Looking through the other pair of slits, it is seen if the point », which ne ibae is to be eke is bisected. If not, the cross oa is moved backwarq or forward till the line of sight also passes through the point. on French Cross Staff: Fig. 4.14 (b) shows a French cross ae Se octagonal box, Vertical sighting slits are cut in the middle o! acl a fae aa t the lines between the centres of opposite slits make angles of 45 witl each other. It jz possible, therefore, to set out angles of either 45° or 90' with this instrument. Adjustable Cross Staff; The adjustable cross staff (Fig. 4.14 (c)] consists ty te cylinders of equal diameter placed one on top of the other Both are eat with sighting slits. The upper box carries a vernier and can be rotated rel eee to the lower by a circular rack and pinion arrangement actuated by a milled headed screw, ‘The lower box is graduated to degrees and sub-divisions. It is, therefore, possible tp set out any angle with the help of this instrument. (ii) Optical Square: Optical square is somewhat more convenient and accurate instrument than the cross staff for setting out a line at right angles to another line. Fig. 4.16 (a) illustrates the principle on which it works. It consists of a circular box with three slits at E, F and G. In line with the openings E and G, a glass silvered at the top and unsilvered at the bottom, is fixed facing the opening E. Opposite to the opening F, a silvered (b) ©) 4 From P glass is fixed at A making an angle of 46° to the previous glass. A ray from @ © the ranging rod at @ passes through the lower unsilvered portion of the Fig. 4.15 Optical Square. mirror at B, and is seen directly by eye at the slit E. Another ray from the object at Pis received by the mirror at A and is reflected towards the mirror at B which reflects it towards the eye. Thus, the images of P and @ are visible at B. If both the images are in the same vertical line as shown in Fig. 4.15 (b), the line PD and QD will be at right angles to each other. Let the ray PA make an angle a with the mirror at A, ACB =45° or ZABC=180°- (45° + a) = 135°- a By law of reflection ZEBb, = ZABC = 135°—a Hence ABE = 180° — 2(135° — @) = 2a — 90° ; Also DAB = 180° — 2a From AABD, ZADB = 180° — (2a ~ 90°) — (180° — 2a) = 180° — 2a + 90° — 180° + 20 = go* Thus, if the images of P and @ lie in the same vertical line, ag shown in Fig. 4.15 (6) i f » 4, ), the line PD and QD will be at right angles to each other, : aeTo set a right angle: To set a right angle on a survey line, the instrument is held on the line with its centre on the point at which perpendicular is erected. The slits F and G are directed towards the ranging rod fixed at the end of the line. The surveyor (holding the instrument) then directs person, holding a ranging rod and stationing in a direction roughly perpendicular to the chain line, to move till the two images described above coincide. Testing the Optical Square (Fig. 4.16) (@ Hold the instrument in hand at any intermediate point C on AB, sight a pole held at Aand direct an assistant to fix a ranging rod at a, such that the images of the ranging rods at a and A coincide in the instrument. A c B (i) Turn round to face B and sight the ranging rod at a. If the image of the ranging rod at B coincides with the image of ranging rod at a, the instrument is in adjustment. (ii) If not, direct the assistant to move to a new position 6 so that both the images coincide. Mark a point don the ground mid-way between aand b. Fix a ranging rod at d. (iv) Turn the adjustable mirror till the image of the ranging rod at d as coincide with the image of the ranging rod at B. Repeat the test till correct. Fig. 4.16 (iii) Prism Square: The prism square shown in Fig. 4.17 works on the same principle as that of optical square. It is a more modern and precise instrument and is used in a similar manner. It has the merit that no adjustment is required since the angle between the reflecting surfaces (i.e. 45°) cannot vary. Fig. 4.18 shows a combined prism square as well as line ranger. Fig. 4.17. Prism Square Fig. 4.18 Combined Prism Square and Line Ranger. (iv) Site Square (Fig. 4.19): A site square, designed for setting out straight lines and offset lines at 90°, consists of a cylindrical metal case containing two telescopes set at 90° to each other, a fine setting screw near the base, a circular spirit level at the top and a knurled ring atthe base. It is used in conjunction with a datum rod screwed into the base of the instrument.| | | 1 Telescopes | 2 Clamp | 3 Tripod 4 Cylindrical case | | } 5 Fine setting screw 6 Knurled ring 7 Datum rod 8 Clamp arm | | Fig. 4.19 The Site Square. | EE BAsic PROBLEMS IN CHAINING (A) To Erect a Perpendicular to a Chain Line from a Point on it: The method of establishing perpendiculars with the tape are based on familiar geometric constructions, The following are some of the methods most commonly used. The illustrations given are for a 10 m tape. However, a 20 m tape may also be used. @ The 3-4-5 method: Let it be required to erect a perpendicular to the chain line ata point C in it [Fig. 4.20 (a)]. Establish a point Z at a distance of 3 m from C. Put thed m long) at and the 10 m end at C. The 5 m and 4 m marks are 0 form a loop of 1 m. The tape is now stretched tight by fastening the ends E and C. The point Dis thus established. Angle DCE will be 90°. One person can set out a right angle by this method. Dfim Rr D Loop 5 4m F, ° Ao A it 3m B AE Ce ran 26 eros E c (a) (b) «© Fig. 4.20 establish D. Join DC.—— (iii) Third method [Fig. 4.20 (c)]: Select any point F outside the chain, preferably at 5 m distance from C. Hold the 5 m mark at F and zero mark at C, and with Fas centre draw an arc to cut the line at E. Join EF and produce it to D such that EF = FD=5 m. Thus, point D will lie at the 10 m mark of tape laid along EF with its zero end at E. Join DC. (B) To Drop a Perpendicular to a Chain Line from a Point outside it: Let it be required to drop a perpendicular to a chain line AB from a point D outside it. ( First method [Fig. 4.21 (a)]: Select any point E on the line. With D as centre and DE as radius, draw an arc to cut the chain line in F. Bisect EF at C. CD will be perpendicular to AB. (i) Second method (Fig. 4.21 (b)]: Select any point E on the line. Join ED and bisect it at F. With F as centre and EF or FD as radius, draw an arc to cut the chain line in C. CD will be perpendicular to the chain line. (iii) Third method [Fig. 4.21 (c)]:Select any point Eon the line, With Eas centre and EDas radius, draw an are to cut the chain line in F. Measure FD and FE. Obtain the point Con 2 the line by making FC= Fe . Join Cand D. CD will be perpendicular to the chain line. D D F Ess cane K ET B AE (a) (b) (©) Fig. 4.21El OBSTACLES IN CHAINING —— nn two points and give wee! surements. Obstacles to Obstacles to chaining prevent chainman from measuring directly bet rise to a set of problems in which distances are found by indirect mea’ chaining are of three kinds : (@) Obstacles to ranging (®) Obstacles to chaining (©) Obstacles to both chaining and ranging. cle, in which the ends are not ining: Thi sti ’ (a) Obstacle to Ranging But Not Chaining: This type of obsta le, in whi this obstacle, intervisible, is quite common except in flat country. There may b« ° () Both ends of the line may be visible from intermediate points on the line. : (ii) Both ends of the line may not be visible from intermediate points on the line (Fig. 4.24). Case (i): Method of reciprocal ranging of § 3.3 may be used. | | Case (ii): In Fig. 4.24, let AB be the x Tine in which A and B are not visible from intermediate point on it. Through A, draw 5 a random line AB, in any convenient 1 direction but as nearly towards B as if random ing D2 a possible. The point B, should be so chosen 4 | that (i) B, is visible from B and (ii) BB, is 2 perpendicular to the random line. Measure . s D 8 BB,, Select points C, and D, on the random Fig. 4.24 line and erect perpendicular C, C and D, D onit. AC, Make C0, = iq . BB, and DD, = a - BB,, Join C and D, and prolong. > Chaining But Not Ranging: There may be two cases of this obstacle :@ When it is Possible to chain round the obstacle, i.e. a pond, hedge etc. i) When it is not possible to chain round the obstacle, e.g. a river. Case (I): Following are the chief methods (Fig. 4.25). c D c 7 Ze 90° oo A 8 A 5 4 ae ® ) © c Fig. 4.25 Obstacles to Chaining Method (a): Select two points A and B on either side. Set out equal perpendiculars AC and BD. Measure CD; then CD = AB [Fig. 4.25 ()]. Method (b): Set out AC perpendicular to the chain line. Measure AC and BC [Fig. 4.25 (b)]. The length AB is calculated from the relation AB = VBC? - AC? . Method (c): By optical square or cross staff, find a point C which subtends 90° with A and B. Measure AC and BC [Fig. 4.25 ()]. The length AB is calculated from the relation : AB= VAC? +BC? . Method (d): Select two points C and D to both sides of A and in the same line. Measure AC, AD, BC and BD [Fig. 4.25 (d)]. Let angle BCD be equal to 0. From ABCD, BD? = BC? + CD? - 2BC x CD cos @ BC? +CD* - BD® or oa 8 2 a a) Simi _ BC? +AC?- AB? : imilarly from ABCA, cos 0= —S5o5aq aii) Equating (i) and (ii) and solving for AB we get 2 z ap= {BC AD) + (BD) XA) _ (acy AD) Method (e): Select any point E and range Cin line with AE, making AE = EC. Range D in line with BE and make BE = ED. Measure CD ; then AB = CD [Fig. 4.25 (¢)]. Method (f): Select any suitable point Z and measure AE and BE. Mark C and Don AE and AE BE such that CE = and DE= = Measure CD ; then—™ AB=n. CD. [Fig. 4.25 (A). Case (II): Fig. 4.26) . Method (a): Select point B on one side and A and C on the other side. Erect AD ang perpendiculars to AB and range B, D and E in one line. Measure AC, AD and CE [fig ,* (@)]. Ifa line DF is drawn parallel to AB, cutting CE in F perpendicularly, then triangle, an and FDE will be similar. , AB _ DF AD FE Bue FE=CE-CF=CE-AD, and DF=AC. AB__AC . _ ACxAD SB -ASp Fromwhich AB Gap Method (b) : Erect a perpendicular AC and bisect it at D. Erect perpendicular CE at Cay range E in line with BD. Measure CE [Fig. 4.26 (b)]. Then AB = CE. iB is Bi =A — i — y Fr + i Asie | i al go | i i F pi 90° i i t 1 1 : i @ 0) © Fig. 4.26 Obstacles to Chaining. Method (c): Erect a perpendicular AC at A and choose any convenient point C. With tt help of an optical square, fix a point Don the chain line in such a way that BCD is a right ane? AC [Fig. 4.26 (o)]. Measure AC and AD. Triangles ABC and DAC are similar. Hence 22 = wD AC? AC Therefore, AB = 7D Method (d): Fix point C in such a way that it subtends 90° with AB. Range D in line wi AC and make AD = AC. At D, erect a perpendicular DE to cut the line in E IFie. 4 96). TH AB=AE. | (c) Obstacles to Both Chaining and Ranging: A building is the typical se of this of obstacle. The problem lies in prolonging the line beyond the obtacle ani dstermininé distance across it. The following are some ofthe methods (Fig, 4.2"), |_— Method (a): Choose two points A and to one side and erect perpendiculars AC and BD of equal length. Join CD and prolong it past the obstacle. Choose two points Z and Fon CD and erect rpendiculars EG and FH equal to H Frat of AC (or BD). Join GH and prolong (a) it. Measure DE. Evidently, BG = DE (Fig. 4.27 @)- Method (b): Select a point A and erect a perpendicular AC of any convenient length. Select another point B on the chain line such that AB = AC. Join B and C and prolong it to any convenient point D. At D, seta right angle DE such that DE = DB. Choose another point F on DE such that DF = DC. With Fas centre and AB as radius, draw an arc. With E as centre, draw another arc of the same radius to cut the previous arc in G_ Join GE which will be in range with the chain line. Measure CF [Fig. 4.27 (0)]. Then AG Method (c): Select two points A and Bon the chain line and construct an equilateral triangle ABE by swinging arcs. Join AE and produce it to any point F. On AF, choose any point H and construct an equilateral triangle FHK. Join F and K and produce it to D such that FD=FA. Choose a point G on FD and construct an equilateral triangle CDG. The direction CDis in range with the chain line [Fig. 4.27 (o)]. The length BC is given by BC=AD-AB-CD=AF-AB-CD Method (d): Select two points A and Bon the chain line and set a line CBD at any angle. Join Aand C and produce it to F'such that AF = n . AC. Similarly join A and D and produce it to G such that AG = n - AD. Join F and G and mark point E on it such that FE =n . BC. Similarly, produce AF and AG to H and K respectively such that AH = n' . AC and AK=n' . AD. Join H and K and mark J on it in such a way that HJ =n’ . CB. Join EJ, which will be in range with chain line. The obstructed distance BE is given by [Fig. 4.27 (d)] : BE=AE-AB. ButAE=n-AB BE - AB — AB = (n-1) AB. Example 4.6: To continue a survey line AB past an obstacle, a line BC 200 metres long was set out perpendicular to AB, and from C angles BCD and BCE were set out at 60° and 45° respectively. Determine the lengths which must be chained off along CD and CE in order that ED may be in Produced, Also, determine the obstructed length BE. () Fig. 4.27 Obstacles to Chaining Solution: (Fig, 4.28). From , 4 ABC is 90° From an oigaea CD = BC sec 60° = 200 x 2 = 400 m. CE, and CE = BC sec 45° = 200 x 1.4142 282.84 m. BE = BC tan 45° = 200 x 1 = 200 m. Fig. 4.28One of the primary objects of land surveying is to determine the area of the tract surveyed and to determine the quantities of earthwork. The area of land in plane surveying means the area as projected on a horizontal plane. The units of measurements of area in English units are sq. ft or acres, while in metric units, the units are sq. metres or hectares. The following table gives the relation between the two systems. TABLE 12.1 British Units of Square Measure with Metric Equivalents 258.99 ha 0.40467 ha 3,097,000 | 4,840 102,400 160 16 484 404.67 m? 1 30.25 25.29 m? 0.836 m? 929 m? 404.67 cm? Note: The standard of square measure is the Acre. TABLE 12.1(a) Metric Units of Square Measure with British Equivalents me) 1,000,000 0.3861 sq. mile 10,000 2.4710 acres 100 1,000,000 1076.4 sq. ft. 1 10,000 10.764 sq. ft. 1 0.155 sq. ft. | Note: The standard of square measure is the Are.ia GENERAL METHODS OF DETERMINING AREAS = The following are the general methods of calculating areas: 1. By computations based directly on field measurements These include: ; (a) By dividing the area into a number of triangles (®) By offsets to base line (©) By latitudes and departures: ( By double meridian distance (D.M.D. method) (ii) By double parallel distance (D.P.D. method) @) By co-ordinates. 2. By computation based on measurements scaled from a map. 3. By mechanical method: Usually by means of a planimeter. GEN AREAS COMPUTED BY SUB-DIVISION INTO TRIANGLES In this method, the area is divided into a number of p triangles, and the area of each triangle is calculated. The total area of the tract will then be equal to the sum of areas of individual triangles. Fig. 12.1 shows an area divided into several triangles. For field, ¢ work, a transit may be set up at O, and the lengths and directions of each of the lines OA, OB..... etc. may be measured. The area of each triangle can © then be computed. In addition, the sides AB, BC..... ete. can also be measured and a check may be applied by calculating the area from the three known sides ofa triangle. Thus, if two sides and one included angle of a triangle is measured, the area of the triangle is. A B given by Fig. 12.1 Area =1ab sin (td When th i ‘i . y the poe e lengths of the three sides of a triangle are measured, its.area is computed by Area= (sa G-D GT a where 8 = half perimeter = k@+o+ co). error of the figure is not important, and hence, unnecessary. The accuracy of the fi eld work, the diagonal in the field and comparing ee SU aah ch cases, may be determined by ‘ating its length to the computed length.[12:4] AREAS FROM OFFSETSTO A BASE LINE: OFFSETS AT REGULAR INTERVALS This method is suitable for long narrow strips of land. The offsets are measured from the boundary to the base line or a survey line at regular intervals. The method can also be applied toa plotted plan from which the offsets to a line can be scaled off. The area may be calculated by the following rules: 1. Mid-ordinate rule; 2. Average ordinate rule 3. Trapezoidal rule; 4. Simpson's one-third rule. 4. Mid-Ordinate Rule (Fig. 12.1) The methodis used with the assumption that the boundaries between the ex- tremities of the ordinates (or offsets) are straight lines. The base line is divided into a number of divisions and the ordinates are measured at the mid- points of each division, as illustrated in Fig. 12.2. The area is calculated by the formula Fig. 12.2 Area = A = Average ordinate x Length of base +0, +0; + n where 0}, Op ... = the ordinates at the mid-points of each division £0 = sum of the mid-ordinates; n= number of divisions L=length of base line=nd; _d = distance of each division 2. Average Ordinate Rule (Fig. 12.3) This rule also assumes that the boundaries between the extremities of the ordinates are straight lines, The offsets are measured to each of the points of the divisions of the base line. The area is given by A = Average ordinate x Length of the base —(tO.+O,\ -( alge te eae Man where Op = ordinate at one end of the base. O, = ordinate at the other end of the base divided into n equal divisions. 0, Oz ... = ordinates at the end of each division 3. Trapezoidal Rule (Fig. 12.3) This rule is based on the assumption that the figures are trapezoids. The 0, Tule is more accurate than the previ- cus tworules which are approximate 7] > 3 4 versions of the trapezoidal rule. «de —a. a Referring to Fig. 12.3, the area of }§ 4 L = ns _—_________ the first trapezoid is given by Fig. 12.3 40 n 1 = (0,404 Oy+ nue O,)d=dE0 (12.3)A, +O 4 0, +0, Similarly, the area of the second trapezoid is given by A, = d Area of the last trapezoid (nth) is given by Hence the total area of the figure is given by Oy +O; AF O, ac nai * bn 2 aaa dat AO, AHA + Ag to. Op + On or A= ( 2 2 +O, +02 Hover 40,4 129 Equation (12.5) gives the trapezoidal rule which may be expressed as below: ‘Add the average of the end offsets to the sum of the intermediate offsets. Multip the total sum thus obtained by the common distance between the ordinates to get - required area. 4, Simpson’s One-Third Rule This rule assumes that the short lengths of boundary between the ordinates are parabolic arcs. This method is more useful when the boundary line departs considerably from the straight line. Thus, in Fig. 12.4, the area between the line AB and the curve DFC may be considered to be equal to the area of the trapezoid ABCD plus the area of the segment between the parabolic arc DFC and the corresponding chord DC. Let Op, 0;, O2=any three cons: taken at regular interval of d. ‘Through F, draw a line EG paral «a > ——_ Fig. 12.4 lel to the chord DG to cut the ordinates in ecutive ordinates EandG. pcp = 22+% .24 2) 2 ‘To calculate the area of the segment of the curve, we will utilize the property that area of a segment (such as DFC) is equal to two-third the area of the enclosing Pé (such as CDEG). 22 ag +O, 0 DFC=3 eax anys 2{(0,- z ) ae} irst two intervals. Thus, Area of trapezoid of the parabel® arallelogt@™ ‘Thus, area of segment ‘Adding (1) and (2), we get the required area (Ay, ») of fi 0 +0, 2 Op +O; d Ay 2=— 9 x 24+ 2 f(0, 23%) 2d} 5 042 100 ) 0)Similarly, the area of next two intervals (dg, 4) is given by d As, 4 = 3 (Oz + 403 + 0,) w(4) Area of the last two intervals (A,,_,, A,) is given by d On-1n= 3 On-2+ 40, _, + 0,) +5) Adding all these to get the total area (A), we get d or A= 'g [y+ O,) + 4(O, + Og + ...... + On.4) + 2(Op + Oy + sa... Ong)} (12.6) It is clear that the rule is applicable only when the number of divisions of the area is even i.e., the total number of ordinates is odd. If there is an odd number of divisions (resulting in even number of ordinates), the area of the last division must be calculated separately, and added to equation 12.6. Simpson's one-third rule may be stated as follows: The area is equal to the sum of the two end ordinates plus four times the sum of the even intermediate ordinates + twice the sum of the odd intermediate ordinates, the whole multiplied by one-third the common interval between them. Comparison of Rules: The results obtained by the use of Simpson's rule are in all cases the more accurate. The results obtained by using ‘Simpson's rule are greater or smaller than those obtained by using the trapezoidal rule according as the curve of the boundary is concave or convex towards the base line. In dealing with irregularly shaped figures, the degree of precision of either method can be increased by increasing the number of ordinates. Example 12.1: The following perpendicular offsets were taken at 10 metres intervals from a survey line to an irregular boundary line: 3.25, 5.60, 4.20, 6.65, 8.75, 6.20, 3.25, 4.20, 5.65 Calculate the area enclosed between the survey line, the irregular boundary line, and the first and last offsets, by the application of (a) average ordinate rule, (b) trapezoidal rule, and (c) Simpson’s rule. Solution: (@) By average ordinate rule : L From equation 12.4 (a), we have A= Wee n+1=number of ordinates = 8+1=9 umber of divisions = ~ L= Length of base = 10 x 8 = 80 m EO = 3.25 + 5.60 + 4.20 + 6.65 + 8.75 + 6.20 + 3.25 + 4.20 + 5.65 = 47.75 m Here A = x 47.75 = 424.44 sq. metres = 4.2444 ares.(b) By trapezoidal rule ™~ From Eq.12.5, A= (Ap+0, $Og teen +O,}4 OptOn _ 3:25+5.65 = 4 45 Here a=10.m; 2 Oy + Op snes Og_y = 5.60 + 4.20 + 6.65 + 8.75 + 6.20 + 3.25 + 4.20 = 38.95, TENG scenes ™ = (4,45 + 38.85) 10 = 433 sq. metres = 4.33 ares (©) By Simpson’s rule From Eq. 12.6, A= ¢ [(Op + On) + 4(Oy + Oy. osoees + On 1) + 2(On + Ons Here d=10m; 0+ 0, = 3.25 + 5.65 = 8.9 m 4 (Oy + 05 + sssoss On-1) = 4 (6.60 + 6.65 + 6.20 + 4.20) = 90.60 2 (0, + O,+ 2 (4.20 + 8.75 + 3.25) = 32.40 = * (8.9 + 90.60 + 32.40) = 489.67 sq. metres = 4.3967 ne Example 12.2: A series of offsets were taken from a chain line to a curved bo intervals of 15 metres in the following order. 0, 2.65, 3.80, 3.75, 4.65, 3.60, 4.95, 5.85 m. Compute the area between the chain line, the curved boundary and the end (a) average ordinate rule, (b) trapezoidal rule, and (c) Simpson’s rule. Solution: (@ By average ordinate rule undary line t Offsets by L From Eq. 12.4 (a), we have A= —_£O n+1 Hence L=nd=7x15= 105m 20 =0 + 2.65 + 3.80 + 3.75 + 4.65 + 3.60 + 4.95 + 5.85 = 29.25 105 A= 3 * 29.25 = 383.91 Sq. m = 3.8391 ares. (6) By trapezoidal rule From equation 12.5 A= (% $08 0.460, 4 Ons Ja =igm. OotO, 045.85 Here Go 1B me = OBS 0 905m Oy + Op + oO it .65 + 3.80 + 3.75 + 4.65 + 3.60 + 4.95 = 23.40 * A= (2.925 + 23.40) 15 = 394. = . (©) By Simpson’s rule -87 sq. m = 3.9487 ares. . d Bi tion 12, a rom equation 12.6, A= °1(0,+0,) +4 (y+ 5 + 1. Ogg) +2094 Ot ve Del d 15 Here, a= prom.$$ vill bo soon that tho Si a rulo i i i wil on ee rulo is not directly applicable here since the number of ordina aa rao, and the es Pewee tho first and seventh offacts may be calculated vgimps and tl closod betwoon th by A pn tho sev ° ws rraporoidal male enth and last offsets may be found by (Oy + O,) 4(O, + Oy. 2(O,, + O44 ‘Thus, 8.60) = 40 16.90 A! = 5(4.95 + 40 + 16,90) = 309.25 5q. m. ‘Aron of the last trapezoid = (4.95 + 5.85) 3 = 81.0 sq. m. ‘Total aren = 309,25 + 81.0 = 390.25 sq. m = 3.9025 ares. Ea OFFSETS AT IRREGULAR INTERVALS (a) First Method (Fig. 12.5): In this method, the roa of ench trapezoid is calculated separately and then added together to calculate the total, area, Thus, from Fig. 12.5, d a= 4 @,+0)+ 2 0,409 : e— 4) — de —d, Pe — 1 Fig. 12.5 iB + % (03+0)) (12.7) (b) Second Method: By method of co-ordinates: See § 12.7 3: The following perpendicular offsets were taken from a chain line to an irregular Example 12.. boundary: Chainage 0 10 25 42 60 7m Offset 15.5 962 31.8 256 29.0 315 Calculate the area between the chain line, the boundary and the end offsets. Solution: Area of first trapezoid = A, = (15.5 + 26.2) = 208.5 m? 5-10 a ; Area of second trapesaid = Ap = +g" (26.2 + 81.8) = 435 m 42-25 Area of third trapezoid = 43 = —g__ (31.8 + 25.6) = 487.9 m? 60-42 Area of fourth trapezoid = A, = ——g— (25.6 + 29.0) = 491.4 mn? 75-60 Area of fifth trapezoid = 45= 9 =A, + Aa tha At As (29.0 + 31.5) = 453.7 m? Total area == 208.5 + 435 + 487.9 + 491.4 + 453.7 = 2076.5 m? = 20.765 ares, Example 12.4: The following perpendicular offsets were taken from a chain line to a hedge: Chainage(m) 0 15 30 45 60 70 SOE 100% 120) 120 Offsets(m) 7.60 85 10.7 128 106 95 83 79 64 44 Calculate the area between the survey line, the hedge and the end offsets by (a) Trapezoid rule (b) Simpson’s rule. Solution: (a) By Trapezoidal rule: The interval is constant from first offset to 5th offset. There js another interval between the 5th and 7th offset and a third interval between ‘7th offset and 10th offset. The total area A can, therefore, be divided into three sections. A= A, +Ay +g where A, = area of first section; ‘Ay = area of second section As ea of third section; d, = interval for first section = 15 m d, = interval for second section = 10 m; dg = interval for third section = 20m 7.60+10.6 Now A= (pS ssse107+128) 15 = 616.3 m? 10.6+8.3 A, pests +95] 10 = 189.5 m? 8.3+4. 3= (P54 +79+64)20 = 413 m? 4= 616.5 + 189.5 + 413 = 1219 m? = 12.19 ares. (b) By Simpson's Rule: The first section and the second section have odd number of ordinates, and therefore, Simpson’s rule is directly applicable. The third section has 4 ordinates (even number); the rule is applicable for the first three ordinates only: 15 41 = —y [(7.60 + 10.6) + 4 (8.5 + 12.8) + 2 (10.7)] = 624 m® 10 a= [10.6 + 8.3) + 4 (9.5)] = 189.7 m? 20 20 4s = F(B.3 + 6.4) +4 (7.9)] + J (6.4 + 4.4) = 308.6 + 108 = 416.6 m? A= 624 + 189.7 + 416.6 = 1230.8 m* = 12.303 ares.GENERAL: ACCESSORIES Plane tabling is a graphical method of survey in which the field observations and plotting procee simultaneously. It is means of making a manuscript map in the field while the ground canbe seen by the topographer and without intermediate steps of recording and transcribing field notes. It can be used to tie topography by existing control and to carry its own control systems by triangulation or traverse and by- lines of levels. Instruments used The following instuments are used in plane table survey: 1. The plane table with levelling head having arrangements for (a) levelling, (b) rotation about vertical axis, and (c) clamping in any required position. 2. Alidade for sighting 3. Plumbing fork and plumb bob. 4, Spirit level. 5. Compass. 6. Drawing paper with a rainproof cover, 1. The Plane Table (Fig. 11.1): Three distinct types of tables (board and tripod) having devices for levelling the plane table and controlling its orientation are in common use: ( the Traverse Table, (ii) the Johnson Table and (iii) the Coast Survey Table. Johnson Table (Fig. 112) This consists of a drawing board usually 45 x 60 om Creral create head consists of a ball-and-socket joint end a vertical -pindie with © thurmb serews on the underside, The ball-and-socket joints senor s by the upper thu serew. When the upper serew is free, the table may he tilted about the ball-and-so* may be . xis an’ thus be oriented, 'y be rotated about the vertical a The Coast Survey Table: This table is superi is genet y Tt 'Perior to the aby dis gen? used for work of high precision. The levelling of tae na is done very accuatel’ ™ 264 allyif te x of cea nce foot screws. The table can be turned about the vertical axis and can e ‘ction very accurately with the help of a clamp and tangent screw. g, Alidade:A plane table alidade is a straight edge with some form of sighting device. Two Pe . Plumbing Fork: The plumbing fork (Fig. 11.5), types are used: () plain alidade and (i) telescopic alidade. Plain Alidade: Fig, 11.3 shows the simple form and used for ordinary work. It generally consist of a metal or wooden rule with two vanes at the ends. The two vanes or sights are hinged to fold down on the rule when the alidade is not in use. One of the vanes is provided with a narrow slit while the other is open and carries a hair or thin wire. Both the slits thus provide a definite line of sight which can be made to pass through the object to be sighted. The alidade can be rotated about the point representing the instrument station on the sheet so that the line of sight passes through the object to be sighted. A line is then drawn against the working edge (known as the fiducial edge) of the alidade. Itis essential to have the vanes perpendicular be the surface of the sheet. The alidade is not very much suitable on hilly area since the inclination of the line of sight is limited. A string joining the tops of the two vanes is sometimes provided to use it when sights of considerable inclination have to be taken. Telescopic Alidade: (Fig. 11.4) The telescopic alidade is used when it is required to take inlined sights. Also the accuracy and range of sights are increased by its use. It essentially consists of a smalll telescope with a level tube and graduated are mounted on horizontal axis. The horizontal axis rests on a A-frame fitted with verniers fixed in position in the same manner as that in a transit, All the parts are finally supported on a heavy rule, one side of which is used as the working edge along which line may be drawn. The inclination of the line of sight can be read on the vertical circle. The horizontal distance between the instrument and the point sighted can be computed by taking stadia readings on the staff kept at the point. The elevation of the point can also be computed by using usual tacheometric relations. Sometimes, to facilitate calculation work, a Beaman stadia are may be provided as an extra. Thus, the observer can very quickly and easily obtain the true horizontal distance from the plane table to a levelling staff placed at the point and the difference in elevation between them. The same geometric principle apply to the alidade as to the transit, but the adjustments are somewhat modified in accordance with the lower degree of accuracy required. used in large scale work, is meant for centring Point the table over the point or station occupied by the plane table when the plotted position of that point is already known on the sheet. Also, in the beginning of the work, itis meant for transferring the ground point on to the sheet so that the plotted point and the ground station are in the same vertical line. The fork consists of a hair pin-shaped light metal frame having arms of equal length, in which a plumb-bob is suspended from the end of the lower arm. The fitting can be placed with the upper arm lying on the top of the table and the lower . arm below it, the table being centred when the Fig. 11.5 plumb-bob hangs freely over the ground mark and the pointed end of the upper arm coincides with the equivalent point on the plan.4. Spirit Level: A small spirit level may be used for ascertaining if the table is pr level. The level may be either of the tubular variety or of the circular type, eqhr Py with a flat base so that it can be laid on the table and is truly level when the bubeely central. The table is levelled by placing the level on the board in two positions af angles and getting the bubble central in both positions. Tieht Compass: The compass is used for orienting the plane table to magnetic north compass used with a plane table is a trough compass in which the longer sides of e trough are parallel and flat so that either side can be used as a ruler or laid ay. coincide with a straight line drawn on the paper. Wty Drawing Paper: The drawing paper used for plane tabling must be of supe: quality so that it may have minimum effect of changes in the humidity fo ” aoe - The plan is drawn by the out-door surveyor himself while the country is before his eyes, and therefore, there is no possibility of omitting the necessary measurements. The surveyor can compare plotted work with the actual features of the area. Since the area is in view, contour and irregular objects may be represented accurately. . Direct measurements may be almost entirely dispensed with, as the linear and angular dimensions are both to be obtained by graphial means. Notes of measurements are seldom required and the possibility of mistakes in booking is eliminated. It is particularly useful in magnetic areas where compass may not be used. . It is simple and hence cheaper than the theodolite or any other type of survey.8. 9. It is most suitable for small scale maps. No great skill is required to produce a satisfactory map and the work may be entrusted to a subordinate. Disadvantages he oR Since notes of measurements are not recorded, it is a great inconvenience if the map is required to be reproduced to some different scale. . The plane tabling is not intended for very accurate work. It is essentially a tropical instrument. It is most inconvenient in rainy season and in wet climate. Due to heavyness, it is inconvenient to transport. Since there are so many accessories, there is every likelihood of these being lost.El METHODS (SYSTEMS) OF PLANE TABLING Methods of plane tabling can be divided into four distinct heads: 1, Radiation. 2. Intersection. 3. Traversing. 4. Resection. The first two methods are generally employed for locating the details while the other too methods are used for locating the plane table stations. Radiation In this method, a ray is drawn from the instrument station towards the point, the distance is measured between the instrument station and that point, and the point is located by plotting to some scale the distance so measured. Evidently, the method is more suitable when the distances are small (within a tape length) and one single instrument can control the points to be detailed. The method has a wider scope if the distances are obtained tacheometrically with the help of telescopic alidade (See chapter 22). ‘The following steps are necessary to locate the points from an instrument station T (Fig. 11.8): 1. Set the table at T, level it and transfer Fig. 11.8 Radiation : the point on to the sheet by means of plumbing fork, thus getting point t representins T. Clamp the table.— . . 2. Keep the alidede touching and sight toA. Dr imil " t di i : Se eeieret 1 diferent points B,C, D, Bete and draw the corresponding ays. A pin may Te, Te TD a may be kept touching the pin while sighting the points. Se eae Ge Coren cine Taree étc., in the field and plot their distances to some scale come ponding rays, thus getting a, b, c, d, e ete. Join these if needed. ‘aw the ray along the fiducial edge of the alidade. 5 INTERSECTION (GRAPHIC TRIANGULATION) Intersection is resorted to when the distance between the point and the instrument station is either too large or cannot be measured accurately due to some field conditions. The location of an object is determined by sighting at the object from two plane table stations (previously plotted) and drawing the rays. The intersection of these rays will give the position of the object. Itis therefore very essential to have at least two instrument stations to locate any point. The distance between the two instrument stations is measured and plotted on the sheet to some scale. ‘The line joining the two instrument stations is known as the base line. No linear measurement other than that of the base line is made. The point of intersection of the two rays forms the vertex of a triangle having the two rays as two sides and the base line as the third line of the triangle. Due to this reason, intersection is also sometimes known as graphic triangulation. Procedure (Fig. 11.9): The following is the procedure to locate the points by the method of intersection: 1. Set the table at A, levelitandtransfer & the point A on to the sheet by way of | plumbing fork. Clamp the table. \ 2. With the help of the trough compass, mark the north direction on the sheet. 3. Pivoting the alidade about a, sight it to B. Measure AB and plot it along the ray to get b. The base line ab is thus drawn. 4, Pivoting the alidade about a, sight the details C, D, E etc., and draw corresponding rays. 5. Shift the table at B and set it there. Orient the table roughly by compass | 4\ b , and finally by backsighting A. x — 6. Pivoting the alidade about 6, sight ee the details C, D, E etc. and draw the e of the alidade to intersect with the previously drawn corresponding rays along the edg rays inc, d, e etc. The positions of the points are thus mapped by way of intersection. ‘The method of intersection is mainly used for mapping details. If this is to be used for locating lane table station, the point should be got by way of point which will be used as subsequent p u intersection of at least three or more rays. Triangles should be well conditioned and the angle in such cases. Graphic triangulation of intersection of the rays should not be less than 45° can also proceed without preliminary measurement of the base, as the length of the base line influences only the scale of plotting.