ARLEO, CARLETINE JOY P. BSCE 2A Problem Set No. 2 MEASUREMENT OF HORIZONTAL DISTANCE 1. PACING. In walking along a 75-m course, the pacer of a field party counted 43.50, 44.00, 43.50, 43.75, 44.50, and 43.25 strides. Then 105.50, 106.00, 105.75, and 106.25 strides were counted in walking from one marker to another established along a straight and level course. Determine the distance between the two markers. 2. PACING. A student paces a 50-m length five times with the following results: 57.00, 56.75, 56.50, 58.00, and 56.25 paces. Determine how many paces he must step off in order to establish a distance of 450 meters on level ground. 3. PACING. Determine the length of a line negotiated in 208 paces by a person whose pace is 0.76 meters long. 4. DISTANCE BY SUBTENSE BAR. With the use of a 1-sec theodolite positioned at the center of a six-sided lot, the following readings were taken on a 2-m subtense bar set up at each corner: 0°26'16", 0°12'35", 0°15'05", 0°22'29", 0°30'45", and 0°09'50". Determine the distance of each corner from the instrument position. 5. DISTANCE BY SUBTENSE BAR. A 2-m long subtense bar was first set up at A and subsequently at B, and the subtended angles to the bar, as read from a theodolite positioned somewhere along the middle of line AB, were recorded as 0°24'15" and 0°20'30", respectively. Determine the length of AB G. SLOPE MEASUREMENT. A traverse line was measured in three sections: 295.85 m at slope 8°45’, 149.58 m at slope 4°29', and 373.48 m at slope 4°25’. Determine the horizontal length of the line, 7. SLOPE MEASUREMENT. A slope measurement of 545.38 m is made between points A and B. The elevation of A is 424.25 m and that of B is 459.06 m. Determine the horizontal distance between the two points. 8. MEASUREMENTS WITH TAPE. The sides of a rectangular parcel of property were measured and recorded as 249.50 m and 496.85 m. It was determined, however, that the 30-m tape used in measuring was actually 30.05 m long. Determine the correct area of the rectangle in hectares. x CHECKED 1 2 FeB 20M 9. MEASUREMENTS WITH TAPE. A 30-m steel tape when compared with a standard is actually 29.95 m long. Determine the correct length of a line measured with this tape and found to be 466.55 m. 10. LAYING OUT DISTANCES. A track and field coach wishes to lay out for his team a 200-m straightaway course. If he uses a 50-m tape known to be 50.20 m long, determine the measurements to be made so that the course will have the correct length. 11. LAYING OUT DISTANCES. It is required to lay out a building 80 m by 100 m with a 30-m long metallic tape which was found to be 0.15 m too short. Determine the correct dimensions to be used in order that the building shall have the desired measurements. 12. LAYING OUT DISTANCES. A steel tape whose nominal length is supposed to be 30 m long was found to be 30.02 m long when compared with an13. 14. 15. 16. 1 18. 19. 20. 21. invar tape during standardization. If the tape is to be used in laying out a 520 m by 850 m rectangular parking lot. determine the actual dimensions. to be laid out. CORRECTION DUE TO TEMPERATURE. A 30-m steel tape is of standard length at 20°C. If the coefficient of thermal expansion of steel is 0.0000116/1°C, determine the distance to be laid out using this tape to establish two points exactly 1235.65 m apart when the temperature is 33°C. CORRECTION DUE TO TEMPERATURE. A steel tape having a correct length at 22°C was used to measure a baseline and the recorded readings gave the total of 856.815 m. If the average temperature during the measurement was 18°C, determine the correct length of the line. CORRECTION DUE TO TENSION. A heavy 30-m tape having a cross- sectional area of 0.05 cm’ has been standardized at a tension of 5 kg. If 10 x 10° kg/cm? calculate the elongation of the tape for an increase in tension from 5.5 kg to 20 kg. CORRECTION DUE TO TENSION. A steel tape is 30.0-m long under a pull of 6.0 kg when supported throughout. It has a cross-sectional area of 0.035, cm? and is applied fully supported with a 12-kg pull to measure a line whose recorded length is 308.32 m. Determine the correct length of the line if E=2.1 X 10° kg/cm’. CORRECTION DUE TO TENSION. A 30-m steel tape weighing 1.75 kg is of standard length under a pull of 4.55 kg, supported for full length. This tape was used in measuring a line (found to be 1371.50 m) on smooth level ground under a steady pull of 8 kg. Assuming E=2.05 X 10° kg/cm? and that the unit weight of steel is 7.9 X 10° kg/cm?, determine the following: cross-sectional area of the tape, correction for increase in tension for the whole length measured, and the correct length of the measured line. CORRECTION DUE TO SAG. A 30-m steel tape weighs 1.5 kg and is supported at its end points and at the 5 and 15-meter marks. If a pull of 8 kg is applied, determine the correction due to sag between supports and for one tape le! le Pee CHECKED 12 FEB 2 CORRECTION DUE TO SAG. A 30-m steel tape weighing 0.04 kg/m is constantly supported only at its end points, and used to measure a line with a steady pull of 8.5 kg. If the measured length of the line is 2465.18 determine the correct length of the line. NORMAL TENSION. Determine the normal tension required to make a tape exactly 30.0 m between its ends when used in an unsupported mode, if the tape has a cross-sectional area of 0.045 cm? and weighs 0.90 kg. Assume that the tape is exactly 30.0 m when supported throughout its length under a standard pull of 6.0 kg, and its modulus of elasticity 2.10 X 106 kg/cm’. is NORMAL TENSION. A 30-m steel tape supported at its ends weighs 0.03 kg/m and is of standard length under a pull of 6.5 kg. If the elastic modulus of steel is 2.0 X 10° kg/cm? and its weight density is 7.9 X 10° kg/cm’, determine the tension at which the effect of sag will be eliminated by the elongation of the tape due to increased tension.22. COMBINED CORRECTIONS. A 30-m tape weighs 12.5 g/m and has a cross section of 0.022 cm. It measures correctly when supported throughout under a tension of 8.0 kg and at a temperature of 20°C. When used in the field, the tape is only supported at its ends, under a pull of 9.0 kg, and at an average temperature of 28°C. Determine the distance between the zero and 30-m marks. 23. COMBINED CORRECTIONS. A line was found to be 2865.35 m long when measured with a 30-m tape under a steady pull of 6.5 kg at a mean temperature of 30°C. Determine the correct length of the line if the tape used is of standard length at 20°C under a pull of 5.5 kg. Assume the cross-sectional area of tape to be 0.025 cm’, elastic modulus as 2.10 X 10° kg/cm?, and coefficient of thermal expansion to be 0.0000116/1°C. 24. MEASURING ANGLES WITH TAPE. The sides of a triangle measure 1063.55, 840.33, and 1325.05 m. three angles in the triangle. Determine the 25. OBSTRUCTED DISTANCES. In the accompanying sketch, it is required to determine the distance between points A and B which spans a wide and deep river. Lines BD and CE, which measure 385.75 m and 529.05 m, respectively, are established perpendicular to line ABC. If points D and E are lined up with A and the length of BC=210.38 m, determine the required distance. CHECKED | 2 FEB 202)Gover, dutance = 45m Strides: 43.60 , 44.00, 43.50, 43-15, 4.50 ond 43,25 otride, Grids = W550, toudd 105.98, cfd 10.25 ste Peawimey, didtance between tun marktry GOWTON. F = (450-440 + 43.60 4497S TAH SO MAE YG RF Ass Pros Te 15m) 4575 > 194s m teres Ha = (195.50) 190.00 + 195.95 How2s) /4 Ha = 105.995 x» (%e) (er) = (wos.e7s)(u1l43m) = 481.5015 2) Pacing Given’, 0° BOM TALES? $7.00, 51.95, 6.50, 58.00, ane] 6.25 paces Reauinen! umber of pocey SoUTON PF “ J ANP = 57.004SK754 Bo.6ots$ootsu25/S = sq PFs Gom/ Boa = O8DEFm/pare” X= (ANP) « PF som = (no-ot paces) (o.gs¢in/ pace) Wo.of pous = 512.1202 paves 3 mone Given: 20 paces 5 0.90 m/pnce” AeGuNMED Length of tin caunins x = (nooek mae) te) Avon fuspuuet (69g afpace) = x = Aotpoies( o%um/pove) = 158.08 m 4) arms Oy SURE BAR, CHECKED 1 2 FEB 2024 Gwent Zim, O'Du'IU" of 2! 35*, O15'08", of 2224", O'Be'4s", ond cron so" teaueans dictumer Of each commer sowoon: Dee at (OHM) 5 244 986m Dae Mh ut (235") = se, s40am = Yea (MO) = assis * a (22g) + sos. 2 hen a = 29, Sam bes Yo ck (CUBA) = gag, coat 5) Dune oF eustense DR Goon tm, 49> O*24) 15" gud D700" Negi: ag nae Site Adhation dn > Al ove (22H) = 200.2293 do = % ue (Same —) + 335, s8e4m Dao = 28%. 5743 m +325: 3¥8Am my he > dao =_oln, 11920"©) lave weanuncmaant GWEN, 146.45 m © Hs, Ha,sHm @ At’ , And B7AHEM _@ APS! eouinens Weszor yenghh of her Ve AM FAAS. ES moras) = 202. 400% da = 194, 58m cos Carag") * 149.9219 m ds = 379, 4gm cau (P25!) = 372, 770m Droran © 242.400%m + 18, (22395 4.972. 2908 Drone 2204 7) Bore EACUREMAT Gn, 5 545-98m 5 AZAMMAS 5 B59. cum FFeuustent evade dione ehieen te pints he 4$4.00n - 424.25— he s4.81m a: Le oe anos, ssxctim| ds {Coys tem “Cann d + 544. 2090m, “6) MeARIREMENES wt THRE Goon * eMtegm , Wz A0b45m , Ts Q0m Requited’ rpreck orea of rickangie in hechates. Geunen Carr > 3.0Fm— Bom + 0,05 em ( teo long) eos exer ("Ym) = o.crm (2°"/ ) CHECKED 1 2 FEB 2024, wah dimensions Chrgth) = Cy 2 0.050 (%**Yex0n) F oerwoan Cu = e.06m( 8° c00) = otisvsam Cut AtG.s5m + O82G083M =44, CFT Sanit ts ecbaroe Cy = 2K05 + oetmegaN F249, 4BE Mm A> rar enone, tha Ar3(H) Cw?) AY = (497. 0781m) (241.4159) = [r439F,6208m™ A= NSF ha. 4) IneaaeneNne wiry TARE GINA, NL Qom GL = ANAS Length = Aue: S5 m9 epuimap tare length oF ine suinon: 30m - 2A,asm= 105m (to chat) Cy = core ( mf’) = 005i( A%0+ S5M/ soy) = 0 HFT AtMte He. 65m = OFF TEM 405.7704 mSst hom: Boney + ve tony Ieee: tore Lgth ee ee Grr ee itn) = ba2om ( 2/50) = 00 Lett, L* hoom - oem 2194-200 he) LAYING OUT DISTANCES GIVEN: WE Dm LT HOM , Sum Lous m too thet) emaiten: ee Ame. Bouinow carr = Q LM Fons Cinshot) tw = ons»(fon/rom) + oaom eas x( 12'm/ son) = 050m w! = 8p 4o.4on = oom p ovsom = ro.50m W! = wate dimenciins !_ 100: Som by SO4em 8) Wawa our pirances GWEN: Met Bom tLe toozm img, Som by S5on RERUIEED: Acruel dioendong sownon; Core > Th -NI 2 geoamaton + conn (os long) Gas ner (Zao) + osmer Cas oes ("7am # 0.5eT ©! = gron-ooney > “ste Bm U > etom-o.serm > eat Acton dwenctou 4.4m oy Fia.uBm 13) ceca DUE ro TEMPERATURE Gwen Ne* 20m = o100POHG/ T° THT 20°C jTe> BOTO reaunrer’, pyemuce te ber lad ost founow, = LACT Te) C2 23.06 (ovomote/) (sae - 200) C2 onsum L) eyra.es m= 0.1800 ‘a CHECKED 1 2 FEB 202% 1h) crmamren DF to TereRATORE One; Mis $54. %Sm 06 = B.00001U/1"C , To2 22% , Te * 18*O epuinep: Correct length enunteh: C+ La Ltr te) co? 85¢.81Sm -BOKOm C2 @surtism (oovoorw/re) Crvre~n2te) CL g5u. 975m) © = -oaion W5)coRREETON IIE TO TENAEN cent Leyem, As 05m! Pos o3hy, Pr TNS BF DMoxto” ky fem? HetuweDs elongation of the Tape enon cs Ure -b) L Ct A gto AR C2 Alfa xto™m os Le ae Co.2s ont) (2.018% Lalo)| ccs a 2 sn Goren; mes hm] HE REG ML B08. 32m A> owosemt Pes Ging 4 n80" ig Jent rsoaeens +1 oust: Gt Ue-R) U7 308.01m +0.035am as Ut aes. s4cam Tach susie ©. + Cina-¢in) Caoetim) 4. ovezesg ato Cotto" bent 1 conneeney ou Te cero War ton PEN My MLE BI Som WLastg ts aas tet to/mt Porassig 4x10 ko/om esuwe: Aj eet we soumon: A= W/Ane + 17589 /Cransatg/en? sens Sie) + 10s Bemt Che =P.) me 55 2 9.0373. = RE Corova tem? Xausmtotbgend) * 28213, Uemee, js += (39 1.56m +0,0319m = 1371, 53i2m, Xs a Me eaten 8Ue 10 6h covery LeBom j e Ley. cag Ak Sand Sy YH Rouse: Car?2 Coe2T Coot? Gre? sown: oe en Gs = th faapt Gey (1519) Caon-tim) /a4Caug y= 0.02306 ° Cas = (819)(m) 240009)" = p.soram Car seas teae 49 70.00Dm+ Bot0m # .0%00 ou + C4.545)"(16a- 6m) faa [4hg)* = ovcrv0n, = co43%m IW) enneonon ove te Aer fails Ta. AeowiReD = UL 27 enanon’s eg MEYaxp? + Court in)™Coom)? fan Lest = o.ontam Ch sie £06 * O108.18m -O.024aM > aHy BIE Be. WoRMAL TEMA orkgin , ML? OH6bI8m _P> SRG Te vol" OHECKED 1 2 FEB Ald ream ta? four: (twyt = WAG /24 (Pw -Pe) Cony? = (0.9049 )Covorsem* Yz.s0nr0* Ka /em?)) / 29 Cen -Ore) fut Voie boJ jy nemma, Tension; Gwar: L290 5 Vowakgin = OMky R= asky Ww = 0.0005 kg /em © sorte! tglent De Fan to kg sem? Pye Reouined: Pye? Sounen: Formic: O04 WYRE he fr A= Ae v.UBtglem = oo3tem* Yo -% Dd Faxto%gleat oo Fr py seta ETT, oe 6 Vissi "ie Coady me etuag Ty re aay ae or waayy = matey Trig -@stg eS stent lly © ett ty wey = water tg 22) Compini conRECNS even: Le20m 4 W>)e.S9/m , = fokg tr? 200 jm FROKg , TH8C Regul: 1"? COUN, Frnemlgs UAL tor toe ” Cia = Re er CL CTH) = tyuxto*caun)(aete-200) As ss eg etaest0%m L' = g0m7 2.784 x10 + 6162x104 = 30.0035 m (23. ComBINEO COTES GONEW: LF Aus 35m, P= Giskg, T2IT, Te? MOC, fo = 5.5Kg A+ ocasen™, & = tome? kg /om® , C= o.coccre/ *e Ropu: Le? Bourn: tee F & = e.000e/*U( a5 35m)(aete-20C) LF 28UE. am + 0.3824 em + O.0FFBM = gaazem 2 agus. 797M = usig"6k9 (2965. 55m .025un* (2 +10" ig/em*) Euosrsn CHECKED 1 2 FEB 202t,pg MEABIRNG ANGLES VOTH TAPE NEN: WEE + (003.55 m, $40.99 my, an 1925. oF eaUIRED: ° angle o.unen c 1 Det aA vemnss ae peasetn © 4A A 3 « o SEN Aegwt ace oi" _t0o3.sema 10027" snling fer Le ¢ 1640.35 > court AZ tt -c% ab sequ*t 11063.55n¥ + (1926. 260 2 Co psmn K2Fe5e) Ma? = tA tte two? * PAGG*ED + 00.27" fy = 45, 25}) oBermereD DssomKS Gey: 99 > 385,95 , CE =529-05m , BE= 210.98m Aeing, Srviler Tear: I f —~ \ © © Remecn: oltonee, € ‘angst Ac = AB +00, AB LAU wet As, Ap toe »D Cs AD, AB +210.38m CHECKED | 2 FEB 2024 397.35 524.05m AG = Suwe.229m