Determine the horizontal distance from A to B if the readings are: ROD STADIA READING POSITION UPPER LOWER 1.68 1.34 2.56 1.92 * Stadia constant 0.20m Stadia interval factor 100
Q: Fundamentals of Surveying: Please show your solution. A line with bearing, N 9° 45' E intersects…
A: The representation of lines are shown below:
Q: 1. Refer to the setup and data below. Determine the length of side AB. Stadia interval factor: 100…
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A: Given: Stadia intercept for Q = 0.320 m Stadia intercept for R = 0.210 m Angle QPR = 61°30’30”…
Q: 1. Refer to the setup and data below. Determine the length of side AB. Stadia interval factor: 100…
A: To Determine Length of side AB.
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Q: Fundamentals of Surveying: Please show your solution. A line with bearing, S 9° 45' W intersects…
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Q: To determine the horizontal distance between Q and R, a tacheometer was used and was placed at P.…
A: Given: Stadia intercept for Q = 0.320 m Stadia intercept for R = 0.210 m Angle QPR = 61°30’30”…
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- Azimuth Azimuth Fraction) Length Degrees Minutes Seconds Line 10047 86.99 100 28 AB 44 186.18 186 10 BC 259 89 0.00 CD 100.11 125.32 115 12 44 115.21 DE 323 10 323.17 EA 284.548 SUM 856.858 7. Determine the FGCC dassification of the angle measurements in the problem shown here. Interior Angles Degrees Minutes Seconds 80 28 86 10 44 100 01 215 12 4. 58 3. 18 SUM 69The observed lengths of the sides (in feet) of a five-sided closed polygon traverse are as follows: AB= 202.74, BC = 285.87, CD= 499.37, DE = 321.33, and EA = 382.78. (Note: Assume units of feet for all distances.) The preliminary azimuths obtained after balancing the angles are as follows: AB= 138° 16'27", BC = 67°02′46", CD= 354°22'24", DE = 256° 13'01", EA = 185°29/20". Compute the departures of the sides. Express your answers in feet to five significant figures and separated by commas. AB, BC, CD, DE, EA = Part B ||| ΑΣΦ ΑΣΦ AB, BC, CD, DE, EA = vec Compute the latitudes of the sides. Express your answers in feet to five significant figures and separated by commas. [ΠΙ —| ΑΣΦ | w vec ? ? ft, ft, ft, ft, ft ft, ft, ft, ft, ft30 Y 5OKN 3. 4 40° 7OKN 450 l00 KN z Fy=? ī Fx = ? %3D RF = ? %3D 20
- From the following measured interior angles of a five sided figure: Station Values of Angles (No. of Measurements W=1/#of obs A 99.000 B 95.000 C 120.00 D 115.00 E 113.00 > Preview Compute for the following: 98.424 E = NOTE: Final answer must be in decimal degrees. HINT: Sum of Interior Angles - (n-2)*180* 119.784 3 3 8 114.655 5 6 1/3 1/3 1/8 1/5 1/6 139/120 a) Probable value of Angle A. b) Probable value of Angle C. c) Probable value of Angle D. Angle corrected = Given Angle +/- discrepancy (W/EW) discrepancy = 540 - 542 discrepancyFrom the following measured interiror angles of a five sided figure, compute the following: Station Values of Angles (No. of Measurements A 110.00 98.000 108.00 120.00 105.00 BUDE с 5 NOTE: Final answer must be in decimal degrees. HINT: Sum of Interior Angles = (n-2)*180* NAO сл 6 4 2 1 a) Probable value of Angle A. b) Probable value of Angle C. °c) Probable value of Angle DWhich of the following indicates the correct value of precise closing error if e = 0.54 and lengths of sides are 92.69 m, 119.23 m, 92.64 m, 42.96 m, and 60.96 m? O a. 1/766.445 O b. 1/746.445 O c. 1/756.445 O d. 1/765.445
- Data: LINE AB BC LENGTH (m) OBSERVED BEARINGS (QBS) FORWARD | BACKWARD FORWARD BACKWARD D' D' 178.00 177.80 N 56 5 E 56 10 W 209.00 210.60 N 78 23 78 20 W CD 367.67 367.60 N 12 49 E 12 50 W DE 512.67 512.38 S 37 12 E N 37 15 W EF 781.55 781.50 S 65 54 E N 65 45 W FG 127.90 128.35 N 43 30 43 18 W N 23 32 E GA 564.44 564.20 S 23 40 W Question: 1.What is the adjusted bearing angle of Line DE, CD, EF, FG 2.What is the total length (summation of all the lengths of the lines) of the adjusted open traverse? S E S E W SZZS5:18 M LAZ Zoom Leave V ERIALS/1ST%20sem/CE%20PC%20313/Structural%20Analysis%20by%20Aslam%20.. Sign in ... TU IL TO IL El = constant E = 29,000 ksi I = 4,000 in.4 FIG. P6.16, P6.42 64 k t point A C В 9 ft 9 ft- -18 ft El = constant E = 29,000 ksi I = 500 in.4 FIG. P6.17, P6.43 6.18 through 6.22 Determine the smallest moment of inertia I required for the beam shown, so that its maximum deflection does not execed the limit of 1/360 of the span length (i.e., Amax < L/360). Use the moment-area method. C 60 kN 300 kN - m A B. .1 Unmute Start Video Share Participants More IIThank you
- 1 7:46 0.60 4G 49 KB/s 2 A surveying transcript of a closed traverse is defined by the vertices ABCDEFGHIJKL. Engr. Pot is performing Double Meridian Distance Method in order to compute the area of the traverse. Midway in his work, his nephew Ashton accidentally spilled Engr. Pots' coffee onto the notes. Engr. Pot decided to rewrite the data, only to find out that the DMD of the lines CD and DE are unreadable. He figures that he can just solve for the missing data knowing that the DMD of line DE is twice the DMD of the previous line CD. If the departure of line CD is -124.25 and that of line DE is 301.965, determine the DMD of line DE. Your answer A lot has a frontage along a road. The 2 IIQuestion 1 of 14 - /5 View Policies Current Attempt in Progress With your pencil, make a dot on the position of your best visual estimate of the centroid of the area of the circular sector. y 190 mm 30° 30° Now find the exact location of the centroid. (x, y) = ( i i How close did you come with your guess? ... IIQ2) a.) Find the intersection coordinates of lines passing from AC and BD. b.) Find the area defined by the corner points A, B, C, and D. The Lengths in the below given figure are in meters. Use Gauss Method only. 17.42 B e----165.05 37.15 55.98 Yp Хр 39.47 22.75L D ol4.90 i 13.90 A 0.00 ty