V17_GENERAL-SURVEYING-1_CONSTANTINO_LEARNING-RESOURCE-PACK PDF

Appendix Learning Resource Materials What is Geodetic Engineering Target Outcomes At the end of the lesson, you are expected to 1. Describe the...

Appendix Learning Resource Materials What is Geodetic Engineering Target Outcomes At the end of the lesson, you are expected to 1. Describe the different types of surveying works. 2. Explain the importance of Geodetic Engineering profession. Abstraction Geodetic Engineering Geodetic Engineering is that branch of engineering which deals with the collection and measurement of spatial data above, on, or below the surface of the earth using appropriate technologies and the scientific and methodological processing and management of these data for the production of spatial information systems, maps, plans, charts, and other documents (CHED memorandum order no. 89, S 2017). Geodetic Engineering includes the establishment of geodetic control network; collection of ground data using various evaluation and analysis of collected data to generate information accordance with the accepted standards and specifications; development of survey and mapping standards and protocols; conduct of research and development activities; development of spatial information systems; and development of business entrepreneurial skills (CHED memorandum order no. 89, S 2017). Practice of Geodetic Engineering. The practice of Geodetic Engineering is a professional and organized act of gathering physical data on the surface of the earth with the use of precision instruments. It is also the scientific and methodical processing of these data and presenting them on graphs, plans, maps, charts or documents. It shall embrace, but is not limited to, the following activities: (1) Professional Geodetic Engineering services with the use of surveying and mapping equipment such as graduated rods, measuring tapes, transits, levels, theodolites, fathometers/echosounders, electronic distance meters, global positioning systems, stereo plotters and all other instruments that are used to determine metes and bounds of lands positions of points on the surface of the earth, water depths, underwater configuration, ground elevation, gravity, isostasy, crustal movements and the size and shape of the earth, and other instruments used for construction survey, and those instruments used to guide the installation of large industrial equipment and machineries; (2) Horizontal and vertical control surveys and political boundary surveys; (3) Land surveys to determine their metes and bounds and prepare the plans thereof for titling and for other purposes; (4) Subdivision, consolidation and/or consolidation subdivision of titled properties; (5) Submission of survey plans of subdivided, consolidated and/or consolidated-subdivision titled properties to the government agencies concerned; hereafter, such plans on surveyed titled properties submitted by geodetic engineers shall not be subject to verification and approval; (6) Preparation and making of sketch, lot and location plans; (7) Conduction of engineering surveys and the technical preparation of engineering survey plans such as topographic, hydrographic, tidal, profile, cross section, construction and boundary surveys; (8) Parcellary surveys of lands traversed by infrastructure projects; and the preparation of subdivision plans; (9) Conduction of gravimetric and photogrammetric survey and the technical preparation of such survey plans;  (10) Survey and mapping works such as the preparation of geographic and/or land information systems; (11) Survey to determine and establish line and grade for the construction of buildings and other structures and its attachments; (12) Construction of as-staked and as-built surveys for infrastructures; (13) Conduction of mineral and mining surveys; (14) Installation of machineries requiring the use of precision instruments; (15) Engagement in the transfer of the knowledge and technology of geodetic engineering in any institution of learning; Geodetic Engineer Shall refer to a natural person with professional expertise in the field of surveying and the corresponding survey data presentation in the form of maps, plans, geo-spatial digital maps, etc.; either in the government service or in the private practice, and who has been issued a Certificate of Registration and Identification Card by the Professional Regulation Commission (PRC) – Board of Geodetic Engineering pursuant to Republic Act of 8560, the Geodetic Engineering Act, as amended. Surveying Is an art of determining positions, measuring distances of objects above, on, or below the surface of the earth, measurement of vertical and horizontal angles, establishment of control networks, processing these data using specialized instruments and technologies in development plans, maps, charts in accordance with the standards. Two General Type of Surveying Geodetic Surveying A type of surveying that takes into account the geoid-spheroidal shape of the earth. These are surveys done on wide extent, employ principles of geodesy, and of high precision. Plane Surveying In this type of surveying to which the geoid-spheroidal shape of the earth is disregarded and therefore considered as a plane. Classification of Land Surveys Control Surveys Shall refer to the survey conducted to determine the horizontal and vertical positions of points which will form part of geodetic network or project controls over an area which will subsequently become the basis in determining the rectangular coordinates in the area. a. Geodetic Control Surveys – shall refer to the surveys conducted covering extensive areas which take into account the curvature and the geoid-spheroidal shape of the earth at sea level for the purpose of establishing basic network of reference points, covering the first and second order controls. b. Project Control Surveys – shall refer to the surveys conducted to establish the positions of points of reference for projects with a limited geographic coverage such as a municipality, a large isolated tract of land, a gourp settlement, a barangay or group of municipalities, covering primary control (third order), secondary control (fourth order)and tertiary control. Cadastral Surveys Shall refer to the survey made to determine the metes and bounds of all parcels within an entire municipality or city for land registration and other purposes. Isolated Surveys Shall be comprised of all classes of surveys of isolated parcels of land used for agricultural, residential, commercial, resettlement, or other purposes covering areas not more than 1,500 hectares. These surveys are classified as follows: a. Group Settlement/Townsite Subdivision Surveys – shall refer to the subdivision of A and D lands of not more than 1,500 hectares into 50 parcels or more. b. Public Land Surveys – shall refer to all original surveys covering A and D lands which has not been subjected to private rights nor devoted to public use pursuant to the provisions of public land laws.  c. Amendment Surveys – shall refer to the surveys covering untitled/undecreed properties by changing the number of lots thereof without affecting the original technical description of the boundary. d. Private Land Surveys – shall refer to the surveys covering lands claimed or owned by an individual, a partnership, a corporation, or any other form of organization, undertaken for purposes of original or subsequent land registration. e. Government Land Surveys – shall refer to the surveys of parcels of lands administered by or belonging to the National Government or any of its branches and instrumentalities. f. Conversion Surveys – shall refer to the surveys conducted for the purpose of transforming/converting the lots covered by approved graphical cadastral surveys, cadastral mapping (Cadm) and photocadastral mapping (PCadm), into numerical or regular cadastral lots, with computation and plotting in the system of the cadastral project. g. Other Land Surveys – shall refer to the surveys made for purposes of determining the metes and bounds of parcels not included in the enumeration above and intended for a specific purpose. Mineral Land Surveys Shall refer to surveys of mining claims, quarry applications, and gravel applications, and other mineral lands within private or public lands, executed for mineral agreements, premits, licenses or for other purposes pursuant to the provision of Republic Act no. 7942, otherwise known as “The Philippine Mining Act of 1995”. Forestlands and National Parks/Protected Areas Delimitation Surveys Shall refer to the surveys conducted by the Regional Composite Survey Team (RCST) or qualified private Geodetic Engineers in order to delimit on the ground the boundaries of forestlands and national parks/protected areas from the agricultural (A and D) lands as delineated pursuant to Sections 3 and 4, Article XII of the 1987 Constitution, PD 705, the “Revised Forestry Code of the Philippines” and Republic Act No. 7586, otherwise known as the “National Integrated Protected Areas System act of 1992”. It shall also include the subclassification and zonification of said areas. Hydrographic Surveys Shall refer to surveys of lakes, oceans, reservoirs, and other water bodies. Acquiring data for determination of water areas, volumes, rate of flow and to determine the shape of the area underlying the water surface. Photogrammetric/Aerial Surveys Shall refer to surveys with the use of photographs taken from an aerial cameras or ground stations, these photographs are then interpreted to produce the desired data. Topographic Surveys Are surveys made for determining the shape of the ground, including the elevation and locations of natural and artificial features of the area. Utilization of Learning Direction: You can either submit through online or you can submit to your instructor at school. With a wide works and opportunities in the Geodetic Engineering profession, select at least three (3) specialization (surveying practice/classification of survey) you want to focused with and explain why. Make a video presentation of your answer with the following conditions: a. Contains the discussion of chosen field and give some example work. b. At least 3 mins and maximum of 5 mins. c. You are in the video most of the time. d. The video resolution should be at least 340P and the sound should be comprehensible.  Surveying Measurements and Errors Target Outcomes At the end of the lesson, you are expected to 1. Explain the differences of sources of errors and kinds of errors. 2. Determine the most probable value of a quantity from a series of measurements Abstraction Units of Measurement Linear Measurement The international unit of linear measure is the meter. The meter is the unit commonly adopted in the Philippines for surveying works. Meter is subdivided into the following units: 1 kilometer (km) – 1000 meter (m) 1 decimeter (dm) = 0.1 meter (m) 1 centimeter (cm) = 0.01 meter (m) 1 millimeter (mm) = 0.001 meter (m) 1 micrometer (µm) = 0.0000001 (m) = 10-6 m Conversion of English to Metric System 1 meter (m) = 3.281 foot (ft) 1 meter (m) = 39. 3701 inch (in) 1 meter (m) = 1.09361 yard (yd) 1 mile (mi) = 1609.34 meters (m) Example 2.0 Convert the following: a. 12 ft. to meter Solution: 1𝑚 12𝑓𝑡 ( ) = 3.657 𝑚 3.281 𝑓𝑡 b. 5 miles to kilometer Solution: 1𝑚 1,000 𝑚 5 𝑚𝑖 ( )( ) = 3.11 𝑘𝑚 1609.34 𝑚𝑖 1 𝑘𝑚 Angular Measurement The sexagisimal units of angular measurement are the degree, minute, and second. It is the most commonly used unit of angular measurement accepted in the Philippines. 1 degree = 1/360th of a circle 1 degree = 60 minutes 1 minute = 60 seconds 1 degree = 3600 seconds The angular value 20 degree 49 minute 20 seconds can be expressed as 200 49’ 20”. Errors in Survey Measurements  Errors Errors in measurement or observational error is the difference between the measured quantity and the true value. e=x-t Systematic Errors Errors due to non-calibrated instrument used in the measurement. This error will have the same sign and value per measurement with the same condition, thus, can be eliminated by applying the corrections. Random Errors A random error also called accidental errors occurs in all measurement. This error remains even after eliminating mistakes and systematics errors. Mistake Mistakes or blunders are inaccuracies in measurements which are due to carelessness and improper executions in a surveying works. For example a surveyor records a distance of a line 5 times with the following values: 123.10m, 128.15m, 123.25m, 123.20m, and 123.15m, the surveyor may have read the second measurement with a value 8 instead of 3. Accuracy and Precision Accuracy is defined as the closeness of the measured value to the true value. It is said that the measurement is more accurate if it is close to a true value, and consequently the measured quantity is not accurate if it significantly deviates from the desired value. Precision denotes the degree of consistency of the measurements made. To help us visually understand, Figure 1 shows an example of a target shooting showing the difference between accuracy and precision. https://www.geavis.si/en/2017/06/difference-between-accuracy-and-precision/ Figure 1. Accuracy and Precision Most Probable Value (x̅) Refers to a quantity which, based on available on available data, as more chances of being correct than has any other. Residual (𝜐) The residuals refers to the variations or deviations in the measurement. It can be calculated by getting the difference of the most probable value and a single measurement. 𝜐 = 𝑚𝑝𝑣 − 𝑥 Standard Deviation (σ) Also called root-mean square, it is a quantity determined to indicate the extent of deviation for a group as a whole. Standard deviation of a group of observation can be estimated as 𝛴(𝑥 − 𝑥)2 𝜎=√ 𝑛−1 𝛴𝜐2 𝜎=√ 𝑛−1 Variance (V or σ2) Is a measurement of the spread of a distribution. Variance can be expressed as  𝛴𝑣 2 2 𝑉=𝜎 = 𝑛−1 Most Probable Error Shall refer to the error for which there are equal chances of the true error being less and greater than probable error. 𝛴𝜐 2 𝑒 = ±0.6745√ 𝑛−1 Example: 2.1 The following are the measurements made on the same angle: 61o46’ 15” 61o46’ 12” 61o46’ 18” 61o46’ 14” 61o46’ 15” 61o46’ 18” 61o46’ 10” 61o46’ 13” 61o46’ 16” 61o46’ 09” Determine the following: a. most probable value of the angle b. range, c. standard deviation Solution: a. Most probable value number of observation n = 10 610 46’ 15” + 610 46’ 18” + 610 46’ 15” + 610 46’ 10” + 610 46’ 16” + 610 46’ 12” + 610 46’ 14” + 610 46’ 18” + 610 46’ 13” + 610 46’ 09" 𝑀𝑃𝑉 𝑜𝑟 𝑥 = 10 𝑀𝑃𝑉 𝑜𝑟 𝑥 = 610 46’ 14” b. Range Range = Highest value – Lowest value Range = 61o46’ 18” - 61o46’ 09” Range = 09” c. Standard deviation Most probable value = x̅ observed angles = x 𝜐 = x̅ - 𝜐2 = (x̅ - x x)2 61o46’ 14” 61o46’ 15” -01” 01” 61o46’ 14” 61o46’ 18” -04” 16” 61o46’ 14” 61o46’ 15” -01” 01” 61o46’ 14” 61o46’ 10” +04” 16” 61o46’ 14” 61o46’ 16” -02” 04” 61o46’ 14” 61o46’ 12” +02” 04” 61o46’ 14” 61o46’ 14” 00” 00” 61o46’ 14” 61o46’ 18” -04” 16” 61o46’ 14” 61o46’ 13” +01” 01” 61o46’ 14” 61o46’ 09” +05” 25” 𝛴𝜐 = 84” 2 𝛴𝜐2 𝜎=√ 𝑛−1  842 𝜎=√ 10 − 1 𝜎 = ±3.1" 2.2 Five group of students were sent out to measure the same line. The distances are 362.10, 362.42, 362.25, 362.38, and 361.95 m. Determine the most probable value of the distance of the line. 𝛴𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑚𝑝𝑣 = 𝑛𝑜. 𝑡𝑖𝑚𝑒𝑠 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 362.10 + 362.42 + 362.25 + 362.38 + 361.95 𝑚𝑝𝑣 = 5 mpv = 362.22 m 2.3 The angles about a point O were observed as follows: ∠XOY = 120°30’, ∠YOZ = 85°20’, and ∠ZOX = 154°40’. What is the most probable value of the observed angles? Given: θ1 = 120°30’ θ2 = 85°20’ θ3 = 154°40’ no. of observation = 3 Error = Full rotation angle – summation of observed angles Error = 360° - (120°30'+85°20'+154° 40') Error = 30’ Correction per observation: Corr. = Error/no. of observation Corr. = 30’/3 Corr. = 10’ To get the most probable value of the angles the correction must be subtracted from the observed angles since the summation of angles exceeds the true angle or a full rotation angle. mpv θ1’ = θ1 – 10’ = 120°30’ – 10’ = 120°20’ mpv θ2’ = θ2 – 10’ = 85°20’ – 10’ = 85°10’ mpv θ3’ = θ3 – 10’ = 154°40’ – 10’ = 154° 30’ Checking: 360° = 120°20'+85°10'+154° 30' 360° = 360° Checked. 2.4 The measured angles about point O have the following observed values. ∠AOB = 15°25’, ∠BOC = 36°15’, ∠COD = 52°10’, and a single angle of ∠AOD = 103°10’. Determine the most probable value of each observation. Given: θ1 = 15°25’ θ2 = 36°15’  θ3 = 52°10’ θ4 = 103°10’ no. of observation = 4 Check for Error θ4 = θ 1 + θ 2 + θ3 103°10’ = 15°25’ + 36°15’ + 52°10’ 103°10’ = 103°50’ Error = 103°10’ - 103°50’ = -40’ (the negative sign indicates that summation of the three observe angles exceeds the single observation) Compute for correction per observe angle: Corr. = 40’/4 = 10’ Most probable value The correction will be subtracted to the three observed angles and added to the single observation. mpv θ1’ = θ1 – 10’ = 15°25’ – 10’ = 15°15’ mpv θ2’ = θ2 – 10’ = 36°15’ – 10’ = 36°05’ mpv θ3’ = θ3 – 10’ = 52°10’– 10’ = 52°00’ mpv θ4’ = θ4 + 10’ = 103°10’ + 10’ = 103°20’ Utilization of Learning Direction: Choose the best answer to the following questions. These questions will also be posted on your google classroom. You can either answer through online or you can submit your answers to your instructor. 1. What is the equivalence of 35 inches to centimeter? a. 2.92 cm b. 0.97 cm c. 0.89 cm d. 88.90 cm 2. If the distance of a line equal to 1.23 km what will be the equivalent value in inches? a. 48,425.20 in b. 42845.20 in c. 1345.14 in d. 1435.41 in 3. Determine the value of 0.65 miles in centimeter. a. 104,607.40 cm b. 106,407.04 cm c. 41,184.00 cm d. 41841.00 cm 4. A complete revolution of a circle is equal to 3600, calculate the value in seconds. a. 3,600,000” b. 1,920,060” c. 1,290,000” c. 60” 5. A measurement accuracy is said to be accurate if a. the measured value is different to a true value. b. the measured value is close to a true value. c. the measured value is somewhat near the given value. d. the measured value is far to a true value. 6. Systematic error is a cumulative error since the error will have the same sign and can be eliminated by applying the corrections. What type of error which is present in all measurement? a. error b. random error  c. mistake d. erroneous error 7. Given the following data from a leveling operation are the observed elevations of a point. 143.30 m. 143.50 m. 142.90 m. 142.95 m. 143.35 m. 143.60 m. 143.30 m. 143.45 m. 143.75 m. 143.55. 7.1 Determine the most probable value. a. 143.365 m b. 142.925 m c. 143.455 m d. 144.123 m 7.2 Determine the range. a. 0.35 m b. 0.25 m c. 0.15 m d. O.45 m 7.3 Determine the standard deviation. a. 0.27 b. 0.37 c. 0.40 d. 0.11 8. A geodetic engineer sent out 10 surveying teams to measure the same line and found to be: 100.10, 100.15, 100.43, 100.27, 100.05, 100.21, 99.98, 100.38, 99.95, and 100. 20 m. Compute for the most probable value of the line measured. a. 100.15 m b. 100.16 m c. 100.17 m c. 100.18 m 9. The angles about point 1 are observed as follows: 87°30’, 140°23’ and 132°04’. Determine the most probable value of each observed angles. a. 87°31’, 140°24’ and 132°05’ b. 87°29’, 140°22’ and 132°03’ c. 87°33’, 140°20’ and 132°04’ d. 87°28’, 140°21’ and 132°02’ 10. The measured angles about point P have the following observed values. ∠APB = 12°20’, ∠BPC = 37°35’, ∠COD = 35°55’, and a single angle of ∠AOD = 87°10’. 10.1 What is the error in the observation? a. 1°20’ b. 30’ c. 85°50’ d. 20’ 10.2 What is the Most Probable Value of ∠APB? a. 12°00’  b. 12°40’ c. 20’ d. 11°50’ 10.3 What is the Most Probable Value of ∠AOD? a. 86°50’ b. 87°30’ c. 88°20’ d. 87°10’  Horizontal Distance Measurement Target Outcomes At the end of the lesson, you are expected to 1. Enumerate different methods and instruments used in determining horizontal distances. 2. Identify procedures in taping operation. Abstraction Horizontal Distance Measurements In surveying, the distance between two points with the same elevation is called the horizontal distance. Measurements are never perfectly accurate they always contain some error, so the measured value is an approximate representation of the true value. Methods in obtaining Distance measurements. 1. Pacing 2. Taping 3. Stadia measurements 4. Electronic Distance measurements (EDM) A online meeting will be conducted discussing Horizontal Distance Measurements. Synchronous online meeting particulars: Title: Horizontal distance measurements Date: August 28,2020 Time: 09:00 AM Duration: 3 hrs. Guest Speaker/Professor: To be announced. Online meeting class rules and etiquette: 1. Show up on time. Call in is 10 minutes before the time. Avoid being late, you may disrupt the webinar by logging late. 2. Turn your video on. 3. Mute your microphone, unless it’s your turn to speak. 4. Raise your hand if you want to say something. 5. When asking questions, ask briefly. You can provide questions via chat or at Q&A window. A moderator will be designated which will read the question one at a time, and the moderator can likewise call out the participant to ask questions directly to the speaker. Question and Answer will begin after the presentation. 6. Be respectful. Pick a quiet spot and remove distractions such as background noise. 7. Stay seated 8. Don’t do other task while the presentation is ongoing. 9. Be an active participant. Respond to polling questions or comments. 10. Jot down lessons learned. Utilization of Learning 1. An evaluation form will be sent after the online meeting. 2. Activities will be given after the webinar. Answer or provide the requirement within the period given.  FIELDWORK NO. 1: PACING Target Outcomes At the end of the lesson, you are expected to 1. Calculate your own pace factor 2. Determine the unknown distance by pacing Abstraction Pacing Pacing is one of the most convenient method for rough/estimating distance of a line. It includes counting the number of steps to get your average distance of one step also called the pace factor. The following are the procedure and requirements to determine one’s pace factor. Materials to be use: o Measuring Tape (Ribbon tape, carpenters tape, or any other tape) o Marker for marking ground line (chalk, paint, or any kind) o Notebook Procedure: A. Determining Pace Factor 1. Select a straight and smooth level ground with no obstacles of at least 60 m. in distance. 2. Marked the starting line with chalk and designate it as point A. 3. Using the measuring tape, set the 0 mark on tape at point A and measure a distance of 30m. 4. Mark the end line with chalk and designate it as point B. 5. Start pacing at desired marked line A or B. Starting from heel or toe. 6. Count your number of paces until you reached the other end of the line, count the last step or pace up to nearest one fourth. 7. Walk with your normal or usual steps on a straight line. 8. Do not shout the count on your steps to avoid the confusion of other students doing the fieldwork. 9. Avoid walking side by side or simultaneously with another student taking the fieldwork, as it affects or influence your pace, and therefore not getting accurate results. 10. Continue pacing line A to B and B to A up to 10 times. 11. Calculate your mean no. of paces. Total no. of paces / No. of Trials 12. Calculate your Pace Factor. Tape distance / Mean no. of Paces Example: Field notes:  Tape distance = 30 m. NO. OF NO. OF TRIAL LINE TRIAL LINE PACE PACE 1 AB 51 6 BA 52 2 BA 51.5 7 AB 51.5 3 AB 51.25 8 BA 49.75 4 BA 49 9 AB 51 5 AB 49.5 10 BA 51.5 Computation: 𝑇𝑜𝑡𝑎𝑙 𝑛𝑜. 𝑜𝑓 𝑝𝑎𝑐𝑒𝑠 𝑇𝑟𝑖𝑎𝑙 1 + 𝑇𝑟𝑖𝑎𝑙 2+... 𝑇𝑟𝑖𝑎𝑙 10 𝑀𝑒𝑎𝑛 𝑛𝑜. 𝑜𝑓 𝑝𝑎𝑐𝑒𝑠 = = 𝑁𝑜. 𝑜𝑓 𝑇𝑟𝑖𝑎𝑙 10 51 + 51.5 + 51.25 + 49 + 49.5 + 52 + 51.5 + 49.75 + 51 + 51.5 𝑀𝑒𝑎𝑛 𝑛𝑜. 𝑜𝑓 𝑝𝑎𝑐𝑒𝑠 = 10 508 𝑀𝑒𝑎𝑛 𝑛𝑜. 𝑜𝑓 𝑝𝑎𝑐𝑒𝑠 = = 50.80 𝑝𝑎𝑐𝑒𝑠 10 𝑇𝑎𝑝𝑒 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑃𝑎𝑐𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑀𝑒𝑎𝑛 𝑛𝑜. 𝑜𝑓 𝑝𝑎𝑐𝑒𝑠 30 𝑚. 𝑃𝑎𝑐𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 = = 0.59 𝑚./𝑝𝑎𝑐𝑒 50.80 𝑝𝑎𝑐𝑒𝑠 B. Determining Unknown Distance by Pacing Since we have determined your pace factor from Part A, we can now determine any unknown distance by walking through the line. The following are the procedures for determining the unknown distance. 1. Measure another 30 m. distance from B. Marked of the tape mark the ground, designate it as point C. 2. Repeat steps 5 to 9 of Part A, but this time from A to C and C to A. 3. Calculate the mean no. of paces. Total no. of paces / No. of Trials 4. Calculate the paced distance. Pace Factor x Mean no. of Paces (the value of pace factor is as computed from Procedure A, no need to compute again). 5. Determine the relative precision of the paced distance. (Paced Distance – Tape Distance) / Tape Distance 6. The answer should be in fraction. Ex. 1/100 7. Your relative precision must be greater than 1/100. Repeat all steps from part A to part B if otherwise. Example Field notes: NO. OF NO. OF TRIAL LINE TRIAL LINE PACE PACE 1 AC 102 6 CA 102.5 2 CA 103 7 AC 101.75 3 AC 102.25 8 CA 102.75 4 CA 101.25 9 AC 102 5 AC 102.5 10 CA 101.5 Computation: 𝑇𝑜𝑡𝑎𝑙 𝑛𝑜. 𝑜𝑓 𝑝𝑎𝑐𝑒𝑠 𝑇𝑟𝑖𝑎𝑙 1 + 𝑇𝑟𝑖𝑎𝑙 2+... 𝑇𝑟𝑖𝑎𝑙 10 𝑀𝑒𝑎𝑛 𝑛𝑜. 𝑜𝑓 𝑝𝑎𝑐𝑒𝑠 = = 𝑁𝑜. 𝑜𝑓 𝑇𝑟𝑖𝑎𝑙 10 102 + 103 + 102.25 + 101.25 + 102.5 + 102.5 + 101.75 + 102.75 + 102 + 101.5 𝑀𝑒𝑎𝑛 𝑛𝑜. 𝑜𝑓 𝑝𝑎𝑐𝑒𝑠 = 10 1021.5 𝑀𝑒𝑎𝑛 𝑛𝑜. 𝑜𝑓 𝑝𝑎𝑐𝑒𝑠 = = 102.15 𝑝𝑎𝑐𝑒𝑠 10 Note: Since we have determined your pace factor from Part A, we will not need to calculate it again here. Instead we will use the pace factor value. 𝑃𝑎𝑐𝑒𝑑 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑃𝐷 ) = 𝑃𝑎𝑐𝑒𝐹𝑎𝑐𝑡𝑜𝑟 𝑥 𝑀𝑒𝑎𝑛 𝑛𝑜. 𝑜𝑓 𝑃𝑎𝑐𝑒𝑠 𝑃𝑎𝑐𝑒𝑑 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑃𝐷 ) = 102.15 𝑝𝑎𝑐𝑒𝑠 𝑥 0.59 𝑚./𝑝𝑎𝑐𝑒 𝑃𝑎𝑐𝑒𝑑 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑃𝐷 ) = 60.27 𝑚. Therefore, the unknown distance is equal to 60.27 meters. 𝑇𝑎𝑝𝑒 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 − 𝑃𝑎𝑐𝑒𝑑 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑃𝑟𝑒𝑐𝑖𝑠𝑖𝑜𝑛 = 𝑇𝑎𝑝𝑒 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 60 𝑚. −60.27 𝑚. 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑃𝑟𝑒𝑐𝑖𝑠𝑖𝑜𝑛 = 60 𝑚. 1 1 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑃𝑟𝑒𝑐𝑖𝑠𝑖𝑜𝑛 = 𝑜𝑟 223.46 200 Utilization of Learning Direction: Answer the following questions. These questions will also be posted on your google classroom. You can either answer through online or you can submit your answers to your instructor. I. Follow the procedure and determine your face factor and an unknown distance. Tabulate your data and computation. Take a picture/video of you doing this fieldwork and make a power point presentation. Question no. 1-3: What is the time, date, and weather condition during the fieldwork? Question no. 4: What is the unit of the measuring tape used? Is it graduated up to what nearest unit? Question no. 5: What are the numbers on the measuring tape indicates? Question no. 6: What is your pace factor represents?  Corrections in Taping Target Outcomes At the end of the lesson, you are expected to 1. Adjust horizontal distance measurement by applying tape corrections. 2. Calculate different corrections in taping measurements. Abstraction Taping Is simple method in distance measurements, using a tape with known graduated distance laid out to a line to determine the distance between points. Rules for applying the correction Measured distance 1. Measured distance tape too long add correction 2. Measured distance tape too short subtract correction Laying out distance 1. Laying out distance tape to long subtract correction 2. Laying out distance tape to short add correction Corrections in Taping A. Correction due to Temperature. 𝐶𝑡 = 𝐶 (𝑇𝑜 − 𝑇𝑠 )𝑙 Where: Ct = corrections due to Temperature C = coefficient of thermal expansion (0.0000116/oC) To = temperature observed Ts = temperature standard l = nominal length of the tape L = total length of the measured line Example: 1. A steel tape with a coefficient of thermal expansion of 0.0000116/oC is known to be 30 m. long at 25o C. The tape was then used to measure a distance and found out to be 340.25 m long. During the instant observation the temperature was 35oC. Determine the following a. Temperature correction per tape length.  b. Temperature correction for the measured line c. Correct length of the tape during measurement. d. Correct length of the line. Solution: Given: C = 0.0000116/oC To = 35oC Ts = 25oC l = 50 m L = 340.25 m a. Temperature correction per tape length. 𝐶𝑡 = 𝐶 (𝑇𝑜 − 𝑇𝑠 )𝑙 𝐶𝑡 = 0.0000116/0 𝐶(35𝑜 𝐶 − 25𝑜 𝐶) 30 𝑚 𝐶𝑡 = 0.003 𝑚 tape too long b. Temperature correction for the measured line 𝐶𝑡 = 𝐶 (𝑇𝑜 − 𝑇𝑠 )𝐿 𝐶𝑡 = 0.0000116/0 𝐶(35𝑜 𝐶 − 25𝑜 𝐶) 340.25 𝑚 𝐶𝑡 = 0.039 𝑚 tape to long c. Correct length of the tape during measurement. l' = l ± Ct l’ = 30 m + 0.003 m l’ = 30.003 m d. Correct length of the line L’ = L ± Ct L’ = 3340.25 m + 0.039 m L’ = 340.289 m 2. A surveyor measured the distance between two points using a 50 m steel tape and found out to be 2125.30 m long, during the instant of observation the temperature recorded was 150C. During standardization the tape has a standard distance when the temperature is 270C with a coefficient of linear expansion of 0.0000116/0C. Compute for the following: a. Correction due to temperature of the measured line b. Correct length of the line. Given: l = 50 m L = 2125.30 m T0 = 150C Ts = 270C C = 0.0000116/0C Solution: a. Correction due to temperature of the measured line 𝐶𝑡 = 𝐶 (𝑇𝑜 − 𝑇𝑠 )𝐿 𝐶𝑡 = 0.0000116/0 𝐶(15𝑜 𝐶 − 27𝑜 𝐶) 2125.30 𝑚  𝐶𝑡 = −0.296 𝑚 tape to short b. Correct length of the line L’ = L ± Ct L’ = 2125.30 m – 0.296 m L’ = 2125.004 m 3. A surveyor was task to lay out two points that are 120 meters apart. Field conditions indicate that the temperature rise from 200C standard temperature to 270C. The coefficient of thermal expansion of the tape used is 0.0000116/0C. Compute for the distance to be laid out to achieve the required distance. Given: L = 120 m Ts = 200C T0 = 270C C = 0.0000116/0C Solution: 𝐶𝑡 = 𝐶 (𝑇𝑜 − 𝑇𝑠 )𝐿 𝐶𝑡 = 0.0000116/0 𝐶(27𝑜 𝐶 − 20𝑜 𝐶) 120 𝑚 𝐶𝑡 = 0.01 𝑚 tape to long Measurements with tape to be laid out to achieve the required distance. The correction should be subtracted since it to be laid out. L’ = L ± Ct L’ = 120 m + 0.01 m L’ = 120.01 m B. Correction due to Pull (𝑃0 − 𝑃𝑠 )𝐿 𝐶𝑝 = 𝐴𝐸 Where: Cp = Correction due to pull P0 = Pull observed Ps = Pull standard L = Nominal length of the tape or total distance A = Cross sectional area of the tape E = Modulus elasticity of the tape Example: The measured distance of points A and B was 312.45 m using a tape with a cross sectional area of 0.05 cm2 has been standardize at a tension of 5.5 kg. If the tension applied during the measurement is 7.5 kg and the modulus elasticity of the tape is 2.10x106 kg/cm2, what is the correct length of the line? Given: L = 312.45 m A = 0.05 cm2 Ps = 5.5 kg P0 = 7.5 kg E = 2.1x106 kg/cm2 Solution: First solve for correction due to pull  (𝑃0 − 𝑃𝑠 )𝐿 𝐶𝑝 = 𝐴𝐸 (7.5 𝑘𝑔 − 5.5 𝑘𝑔) 312.45 𝑚 𝐶𝑝 = 𝑘𝑔 0.05 𝑐𝑚2 (2.10 𝑥 106 ) 𝑐𝑚2 Cp = 0.006 m tape too long Apply the correction to get the corrected length of the line. L’ = L ± Cp L’ = 312.45 m + 0.006 L’ = 312.456 m C. Correction due to Sag 𝑤 2 𝐿3 𝑊2𝐿 𝐶𝑠 = or 𝐶𝑠 = 24 𝑃𝑜2 24 𝑃𝑜2 Where: w = unit weight of the tape L = length of the tape between support W = total weight of the tape Po = Pull applied or observed Example: A 50 meter weighs 1.2kg was used to measure a line and found out to be 80 m long supported only at end points, with a steady pull of 6kg. Calculate the Sag correction per support and the true distance of the line. Given: NL = 50 m L = 80 m W = 1.2 kg Po = 6 kg Solution: Correction due to sag 𝑊 2𝐿 𝐶𝑠 = 24 𝑃𝑜 2 (1.2 𝑘𝑔)2 (50 𝑚) (1.2 𝑘𝑔)2 (30 𝑚) 𝐶𝑠 = + 24 (6 𝑘𝑔)2 24 (6 𝑘𝑔)2 Cs = 0.133 m sag correction is always subtracted True distance of the line L’ = L – Cs L’ = 80 m – 0.133 m L’ = 79.867 m Combined Corrections A 50 m. tape of standard length has a weight of 0.05 kg/m, with a cross sectional area of 0.04 cm 2. With a modulus of elasticity of 2.10 x 106 kg/cm2. The tape in its standard length with a pull of 5.5 kg when supported throughout its length and a temperature of 200C. The tape was used to measure the distance between points A and B and was recorded to be 625.90 m long. At the instant of observation, the pull applied was 8 kg. with the tape supported only at its end points and the temperature observed was 180C. Coefficient of linear expansion of the tape is 0.0000116/0C. Determine the following: a. Correction due to tension. b. Correction due to sag. c. Correct distance of line AB. Given: NL = 50 m W = 0.05 kg/m A = 0.04 cm2 E = 2.10 x 106 kg/cm2 Ps = 5.5 kg Ts = 200C L = 625.90 m P0 = 8 kg T0 = 180C C = 0.0000116/0C Solution: a. Correction due to tension/pull (𝑃0 − 𝑃𝑠 )𝐿 (8 𝑘𝑔 −5.5 𝑘𝑔) 625.90 𝑚 𝐶𝑝 = 𝐶𝑝 = 𝑘𝑔 𝐴𝐸 0.05 𝑐𝑚2 (2.10 𝑥 106 ) 𝑐𝑚2 Cp = 0.016 m tape too long b. Correction due to sag 𝑤 2 𝐿3 𝐶𝑠 = 24 𝑃𝑜 2 (0.05 𝑘𝑔/𝑚)2 (50 𝑚)3 (12) (0.05 𝑘𝑔/𝑚)2 (25.90 𝑚)3 𝐶𝑠 = + 24 (8 𝑘𝑔)2 24 (8 𝑘𝑔)2 Cs = 2.441 m + 0.028 m Cs = 2.469 m (sag correction is always subtracted) c. Correct distance of line AB First compute for correction due to temperature 𝐶𝑡 = 𝐶 (𝑇𝑜 − 𝑇𝑠 )𝐿 𝐶𝑡 = 0.0000116/0 𝐶(18𝑜 𝐶 − 20𝑜 𝐶) 625.90 𝑚 Ct = -0.015 m tape too short L’ = L ± Cp - Cs ± Ct L’ = 625.90 + 0.016 m – 2.469 m – 0.015 m L’ = 623.432 m Normal Tension 0.204𝑊√𝐴𝐸 𝑃𝑁 = √𝑃𝑁 − 𝑃𝑠 Where: PN = Normal Tension, Tension/Pull to be applied to eliminate the errors due to sag and elongation of the tape.  W = Total weight of the tape A = Cross-sectional Area of the tape E = Modulus elasticity Ps = Standard pull Example: Determine the normal pull tension of 30 m tape if the unit weight of tape is 0.14 N/m. Tape is only supported at end points. The standard pull is 50 N, cross-sectional area of tape equal to 1.8mm2 and the modulus elasticity of tape is 200 x 109 Pa. Given: L = 30 m w = 0.14 N/m Ps = 50 N A = 1.8mm2 E = 200 x 109 Pa Solution: 0.204𝑊√𝐴𝐸 𝑃𝑁 = √𝑃𝑁 − 𝑃𝑠 First find the total weight of the tape W. W = unit weight of tape x length of the tape W=wxL W = 0.14 N/m x 30 m W = 4.2 N Calculate PN by assuming value (trial and error) Try 85.8 N 0.204(4.2 𝑁)√(1.8 𝑚𝑚2 (200 𝑥 109 𝑃𝑎) 85.8 𝑁 = √85.8𝑁 − 50 𝑁 85.8 N = 85.8 N Therefore PN = 85.8 N Utilization of Learning Direction: Answer the following questions. These questions will also be posted on your google classroom. You can either answer through online or you can submit your answers to your instructor. 1. A line was determined to be 2123.12 m. when measured with a 30 m steel tape supported throughout its length with a pull of 4 kg at a temperature of 350C. Tape is in standard length at 200C under a pull of 5 kg. The cross-sectional area of tape is 0.03 cm2, coefficient of thermal expansion is 0.0000116/0C, modulus elasticity of tape is 2x106 kg/cm2. 1.1. Compute for the correction due to temperature a. 0.005 m tape too long b. 0.369 m tape too long c. 0.369 m tape too short d. 0.005 m tape too short  1.2. Calculate the error due to tension a. 0.087 m tape too long b. -0.035 m tape too short c. -0.035 m tape too long d. -0.055 m tape too short 1.3. Determine the corrected length of the line. a. 2123.454 m b. 2122.716 m c. 2123.524 m d. 2122.786 m 2. A base line was measured using a 100 m steel tape and found to be 4345.76 m long. During the observation the temperature is 170C and the pull applied is 8 kg. The total weight of the tape is 0.90 kg, modulus elasticity of the tape 2.10 x 106 kg/cm2, cross-sectional area of tape is 0.05 sq. cm. The tape is of standard length under a temperature of 200C, standard pull of 5 kg. If the coefficient of linear expansion of steel tape is 0.0000116/ 0C, Determine the following: 2.1 Error due to change in temperature a. -0.003 m b. -0.034 m c. -0.151 m d. -0.434 m 2.2 Error due to change in tension a. 0.183 m b. 0.211 m c. 0.167 m d. 0.124 m 2.3 Error due to sag a. 1.765 m b. 2.900 m c. 2.508 m d. 2.188 m 2.4 true distance of the line a. 4345.76 m b. 4348.543 m c. 4343.225 m d. 4343.279 m  Angles and Directions Target Outcomes At the end of the lesson, you are expected to 1. Discuss different methods in determining angles and directions 2. Calculate Azimuth and Bearing Abstraction Meridian An imaginary fix line of reference where angles and directions of lines are reckoned. Assumed meridian is an arbitrarily chosen meridian, it is usually used when the observer doesn’t have a compass or no known lines are around the area. True meridian or astronomic meridian is the north and south line passes through the geographical poles of the earth. Grid meridian, a meridian line that is parallel to the line of central meridian. Magnetic meridian is line that follows the magnetic line force of the earth, the meridian arrow of magnetic meridian is half arrowed and half feathered. Angles and Directions Angles and directions can be determined by means of bearings, azimuth, deflection angles, angle to the right, and interior or exterior angles. Bearing Bearing is a convenient method of determining directions, it is an acute angle (less than 90 0) either from north or south lines. Figure 4.1. Bearing Bearing of line EF = S 30° 1’ W Bearing of line CD = S 60° 10’ E Bearing of line AB = N 45° 00’ E Bearing of line GH = N 80° 51’ W Azimuth Another method of determining horizontal angles is by azimuth, the azimuth of a line is defined by the clockwise angle (0° to 359°59’59”) either from north or south lines. The figure below shows the azimuth of line PQ and line XY can be expressed as follows: AziN PQ = 125° 12’ AziS XY = 348° 29’  Figure 4.2. Azimuth Convert Azimuth to Bearing From figure 4.2 determine the Bearing from the given azimuths. Solution: a. For Line PQ Bearing of PQ = 180° - 125° 12’ Bearing of PQ = 54° 48’ Bearing should have direction. Since line PQ is in the SW quadrant, therefore the Bearing of PQ = S 54° 48’ E b. For line XY Since one complete revolution or a circle is 360°, then Bearing of XY = 360° - 348° 29’ Bearing of XY = 31° 31’ Bearing of XY = S 31° 31’ E Convert Bearing to Azimuth From figure 4.1 determine the Azimuth from north and south of the line from the given bearings a. For line EF Azimuth from north EF = 180° + 30° 1’ AziN = 210° 1’ Since azimuth is an angle in clockwise direction from south or north, therefore the value of azimuth of line EF from south is equal to its bearing. Azis = 30° 1’ b. For line CD Azin = 180° - 60° 10’ Azin = 119° 50’ AziS = 360° - 60° 10’ AziS = 299°50’ c. For line AB Azin = 45° (clockwise angle from north) AziS = 180° + 45°  AziS = 225° d. For line GH Azin = 360° - 80° 51’ Azin = 279° 09’ AziS = 180° - 80° 51’ AziS = 99° 09’ Deflection Angle The angle between the prolongation of the preceding line and the next line, recorded with R or right if clockwise rotation and L or left if counter clockwise. The deflection angles from the above figure can be expressed as: AB = 51° R BC = 97° 20’ R CD = 79° 46’ L DE = 32° 7’ R Angle to the right Angle to the right are angles measured clockwise from the preceding line to the next line. Interior Angle Determining the direction of line through the interior angle of a polygon. Calculating Included Angle Included angle between two lines is the angle less than 180°. Example: Find the included angle from the following: 1. Bearing of line AB = N 34° 18’ E Bearing of line AC = N 79° 32’ E Solution: First plot or draw the lines to determine the required angle. α1 = 79° 32’ – 34° 18’ α1 = 45° 14’ 2. AziN AB = 41° 20’ Brg AC = 57° 45’ Solution: First plot or draw the lines to determine the required angle. θ2 = 180° - (41°20’ + 57° 45) θ2 = 180° - 99° 05’ θ2 = 80° 55’ 3. Brg PX = 76° 30’ Brg PY = 64° 50’ Solution: First plot or draw the lines to determine the required angle. θ3 = 180° - 76° 30’ + 64° 50’ θ3 = 174° 20’  Utilization of Learning Direction: Answer the following questions. These questions will also be posted on your google classroom. You can either answer through online or you can submit your answers to your instructor. 1. Determine the bearing and azimuth from north and south of the following lines: 2. Compute the included angle from the following. A. BC = N 30° 01’ E and BA = N 29° 59’ W B. Azis PX = 3° 30’ and AziS PY = 210° 20’ C. AziS XY = 255° 11’ and XZ = S 88° 15’ W 3. From the figure below determine the angles between intersection of lines using protractor. Some are filled out for example (96°30’ and 180°).  Determining Angles and Directions Target Outcomes At the end of the lesson, you are expected to 1. Recognize equipment used in determining angles and directions 2. Determine angles using tape. Abstraction Instruments used in Determining Angles and Directions Angles can be measured using the using the following: 1. Compass. Has a magnetized needle suspended about its center showing the direction of magnetic north. Around the needle is graduated circle which is divided up to 30’ or 20’ (up to nearest seconds for surveyors’ compass) to read the direction of the line where quadrantal system (NE, NW, SE, and SW quadrant) is employed. Images from https://www.newbecca.com/product/4391863849 2. Tape Tapes can also be used to determine angles, by employing trigonometric functions in the measured distances. Laying out a right angle. In laying out special or right angle, you apply the basic trigonometric theory that a triangle with sides in the ratio of 3:4:5 is always a right angle. As shown in the figure, assume that line AB is known and lay out a perpendicular line from point C. The following are procedures in laying out the tape to make a perpendicular or right angle: 1. Establish line AB. Distance of line AB must be more than 5 m. 2. Designate point C as the starting point of the perpendicular line. 3. Lay out a 3 m along line AB from point C. Mark it with a stake and designate it as point D. 4. From point D, lay out a distance of 5 meters; make a loop at the end, hold the 5 m and 6 m mark over each other, to have the exact full meter mark and connect the other end of the tape to point C with a distance equal to 4 meters at the 10 m mark on the tape. 5. Mark the point E with stake. 6. Line AB must now be perpendicular to line DE and the angle DCE should be equal to 90°. Chord Bisection method  Images are from http://www.fao.org/3/r7021e/r7021e05.htm As shown in the figure, assume that line AB is known. The following are procedures in laying out the tape to make a perpendicular or right angle by chord bisection method: 1. Establish line AB. 2. Hold firmly the zero end of the tape to point C. Mark all points with stake. 3. Extend the tape until it intersects to known line AB. (you can use rope or any string as alternate) 4. Swing the tape and mark the two intersecting points. Designate them as points E and F. 5. Measure the distance of line EF, divide it by 2 to get the midpoint, and mark it as point D. 6. Line AB should be perpendicular to line CD as shown on the figure. Angle EDC and FDC should be equal to 90°. 3. Total Station A total station is a surveying instrument which is a combination of electronic theodolite and Electronic Distance Measurement (EDM) instrument. Theodolite is an equipment that electronically measure both vertical and horizontal distance. While EDM measure slope and horizontal distance of a line. Moreover, total station has an on-board computer that can collect and manage data. Image is from https://eu.sokkia.com/products/optical-instruments/ reflectorlessprism/fx-advanced-total-station-series  Utilization of Learning Direction: Follow the procedure in determining right angle, the 3:4:5 and chord bisection method. You can use any kind of tape. Take a picture/video of you doing this fieldwork and make a video presentation. You can either submit through online or you can submit your output to your instructor at school.  Magnetic Declination and its Variation Target Outcomes At the end of the lesson, you are expected to 1. Discuss what is magnetic declination variation. 2. Compute for true direction of a line. Abstraction Magnetic Declination A magnetic compass does not always point north, and has a variation over time. In surveying the horizontal angles observed from compass must be reduced or changed from magnetic bearing/azimuth to true bearing/azimuth. The difference between the horizontal angle of Magnetic North and True North is called magnetic declination θ. The declination is positive at east declination and negative when at west declination. Variation of Magnetic Declination The declination of any place is not constant, these variations may be regular or irregular. There are three classification of variations in declination, these are: 1. Diurnal or Daily variations. Change Example: 1. The magnetic Bearing of line AB is equal to N 55° 45’ E and the magnetic declination at the time of observation was 2° E. Determine the true bearing of the line AB. Solution: First plot or draw the given. Given: θ = 2° E Brg AB = N 55° 45’ E True Bearing of line AB = Magnetic Bearing AB + Magnetic Declination True Bearing of line AB = Brg AB + θ True Bearing of line AB = 55° 45’ + 2° True Bearing of line AB = 57° 45’ 2. In 1997 a surveyor surveyed a line using a compass and found the magnetic azimuth of line MN to be 310° 20’. The magnetic declination of the area is equal to -1° 5’. Calculate the true azimuth of line MN. Solution: First draw the problem. Given: θ = -1° 5’ (negative declination means it is on west side) Azi = 310° 20’ (azimuth is automatically from south if it is not indicated) β = ? (True azimuth of line MN) β = 310° 20’ – θ β = 310° 20’ – 1° 5’ β = 309° 15’ True azimuth of line MN Variation of Magnetic Declination The declination of any place is not constant, these variations may be regular or irregular. There are three classification of variations in declination, these are: Diurnal or Daily variations which where magnetic declination varies daily; Secular variations the change usually for a long time (100 to 200 years); lastly the Annual variation is the change in declination annually. Example: 1. The bearing of a line in 1986 is N 65° 15’ W, the magnetic declination was 4° 20’ W at that time. The magnetic declination today at the same place is 1° 10’ E. What would be the magnetic bearing of the line today? Solution: β = 65° 15’ + 4° 20’ + 1° 10’ β = N 70° 45’ E Utilization of Learning Direction: Answer the following questions. These questions will also be posted on your google classroom. You can either answer through online or you can submit your answers to your instructor. 1. The bearing of a line OP is found out to be N 31° 57’ E, and the magnetic declination at the instant is 3° 10’ E. What is the true bearing of line OP? 2. During an instant observation the magnetic declination of a line is -2°1’ and the azimuth of a line is 91° 30’. Find the bearing of the line. 3. At Santa Ignacia the technical description of the point Lima to point Ciam in 1790 is N 78° 30’ E, the magnetic declination was 5° 5’ E at that time. Another station, point Eyt was measured from point Ciam in 2020 and found out to be N 20° 12’ W, during the observation the magnetic declination is equal to 3° 10’ W. Compute the true bearing from point Ciam and point Eyt.  Traversing Target Outcomes At the end of the lesson, you are expected to 1. Read Bearing/Azimuth using a handheld Compass; 2. Measure distance by pacing. 3. Record and Sketch survey data. Abstraction Traverse Is a series of established points called traverse stations linked by lines called traverse lines. The traverse stations were occupied and the distances and angles of traverse lines connecting these stations were observed to determine their respective positions. Equipment for Traversing Equipment used in traversing are engineers transit and tape, theodolite and EDM, or total station. The supporting equipment are: range pole, prism, target, stadia rod, plumb bob, axe or hammer, two-way radio, and field notebook. The instruments used in marking can be: crayon, chalk, stake or hub, stick with plastic flag, spray paint or regular paint, or nail. Types of traverse Open Traverse When the traverse originates from known point and terminates to another unknown point the traverse is said to be a open traverse. Closed Traverse The traverse is said to be a closed traverse if it originates from a known point and ends at another known point. Closed loop traverse is a traverse that originates and ends at the same point and forming a closed circuit or a polygon. Methods of running traverses 1. Angle to the right traverse. A method of running of traverse by determining angles and direction of a lines in the traverse through angle to the right method. This method can be used in any type of traverse. 2. Deflection angle traverse. Running a traverse with by observing or determining deflection angles of the traverse lines. 3. Interior angle traverse. In this method the interior angle of lines is observed in running the traverse. This traverse is only applicable to closed loop traverse. 4. Stadia traverse. The use of optical geometry of the telescope of the instrument and stadia rod to determine the distance in running a traverse.  5. Azimuth traverse. 6. Compass traverse. The use of compass in running a traverse, it includes observing the forward and back bearing of traverse lines. Compass traverse The following are procedures and requirements in conducting a compass traverse. Materials to be used: 1 Compass 2 pcs Pole 5 Stake Procedure: 1. Select a smooth and level open ground, mark at least 5 points on the ground with stake, at least 10m apart from one point to another. See sample sketch below. 2. Denote the starting point as A and the succeeding point as B. 3. Level the compass directly above point A, until needle is moving freely. If the needle is stuck, rotate the needle lifter until it damps the needle and release it by rotating again counter clockwise. 4. Direct the Rodman to hold the pole straight over point B. 5. Observe the angle from point A to point B by rotating the compass and aligning the line of sight from the compass N arrow to the rod at point B. 6. Record the bearing of the line from either the N end needle of the compass up to nearest minutes. This angle is the Forward Bearing of line AB (Tabulate your data accordingly). 7. Make a sketch of the lines being surveyed (See sample Sketch and Tabulated Data). 8. Determine the paced distance of the line and proceed to the succeeding point. 9. Direct the Rodman to hold the range pole straight over preceding point. 10. Repeat step 3 and 5, but this time from point B to point A. 11. Record the bearing of the line from either the N end needle of the compass up to nearest minutes. This angle is the Back Bearing of line AB. 12. Repeat steps 3 to 11 until all lines were observed with forward and back bearing. Example Sketch Tabulated Data: FORWARD BACK PACED LINE REMARKS BEARING BEARING DISTANCE Staked, near AB N 600 E S 610 W 11.20 m. acacia tree BC N 260 30’ E S 260 30’ W 10.35 m. PS CD DE EF Utilization of Learning Direction: Follow the procedure and perform a compass traverse. In this fieldwork you can use the compass on your phone. Tabulate your data and computation. Take a picture/video of you doing this fieldwork and make a video presentation. You can either submit through online or you can submit your output to your instructor at school. Traverse Computation  Target Outcomes At the end of the lesson, you are expected to 1. Determine errors in traverse. Abstraction Traverse Consist of series of straight lines connecting successive points, whose distance and directions have been determined from field observations. It is a useful method of establishing horizontal control. Horizontal Control Surveys over large areas to establish the precise position (latitude and longitude) of a number of stations, which can be established in number of survey techniques. This includes traversing, triangulation, trilateration and GNSS observation. Purpose of Traverse 1. Locating or establishing precise boundaries of a property. 2. Additional horizontal control for topographic surveying. 3. To establish control for construction survey of highways, railways, and other private and public works. 4. Establishment of ground control stations for photogrammetric surveys and mapping. 5. To determine the area encompassed within the confines of a boundary. Traverse Computation Computations involving determination of: latitudes and departure of traverse lines, linear error of closure, angular error of closure, adjustment of distance and bearing of lines using compass or transit rule, and computation of position coordinates of points based on adjustment. Latitude The distance of the projection of a line onto the reference meridian or North-South line. Lines with Northerly bearings have positive (+) LAT and lines with southerly bearings are negative (-) LAT. Latitude of a line is equal to the distance multiplied by the cosine of the bearing angle, this can be expressed as: LAT = Distance*Cos(bearing/angle) Departure The distance of the projection of a line onto the reference parallel or East-West line. Lines with Easterly bearing have positive (+) DEP and lines with Westerly bearings are negative (-) DEP. Departure of a line is equal to the distance multiplied by the sine of the bearing angle, this can be expressed as: Dep = Distance*Sin(bearing/angle) Example: 1. Find the Latitude and Departure of the line having a bearing of N 35° 29’ E and distance of 201.55 m. Solution: For Latitude Lat = Dist*Cos(angle) Lat = 201.55m Cos 35° 29’ Lat = 88.35 m (positive sign for North direction) For Departure Dep = Dist*Sin(angle) Dep = 201.55m Sin 35° 29’ Dep = 181.15 m (positive sign for East direction) 2. Two station of different location were determined using a total station. The bearing of the line between stations is equal to S 55° 36’ E and a distance of 1250.80 m. Compute for the Latitude and Departure of the line. Solution: For Latitude Lat = Dist*Cos(angle) Lat = 1250.80m Cos 55° 36’ Lat = -706.66 m (negative sign for South direction) For Departure Dep = Dist*Sin(angle) Dep = 1250.80m Cos 55° 36’ Dep = 1032.05 m (positive sign for East direction) 3. A line has an azimuth of 125° 15’ with a distance of 18.10 m. What is the latitude and departure of the line? Solution: For Latitude Lat = Dist*Cos(angle) Lat = 18.10 m Cos 125° 15’ Lat = 10.45 m (opposite sign for azimuth from south, same sign for azimuth from north) For Departure Dep = Dist*Sin(angle) Dep = 18.10m Sin 125° 15’ Dep = -14.78 m (opposite sign for azimuth from south, same sign for azimuth from north) 4. Determine the Bearing and distance of a line with a Latitude of 160.87 m and Departure of -270.11 m. Solution: For Distance 𝐷𝑖𝑠𝑡 = √𝐿𝑎𝑡 2 + 𝐷𝑒𝑝2 𝐷𝑖𝑠𝑡 = √(160.87 𝑚)2 + (270.11 𝑚)2 Dist = 314.39 m For Bearing (Note: Use absolute value in computing bearings. Disregard the sign) 𝐷𝑒𝑝 𝑇𝑎𝑛 𝛽 = 𝐿𝑎𝑡  𝐷𝑒𝑝 𝛽 = 𝑇𝑎𝑛−1 { 𝐿𝑎𝑡 } 270.11 𝑚 𝛽 = 𝑇𝑎𝑛−1 { 160.87 𝑚 } β = 59° 13’ 23.34” Therefore Bearing = N 59° 13’ 23.34” W (N since the latitude is positive and W since the departure is negative) Calculator Shortcut For computing Latitude and Departure See Example no. 1 X = 164.11902153609 ; Y = 116.99294538577 Lat = X Dep = Y Recall X press for the value of Latitude Recall Y press for the value of Departure For other model of calculator, you need to recall F, just press for the value of Departure. And recall E, just press for the value of Latitude Lat = F Dep = E For computing Distance and Bearing See Example 4. 314.38601909118