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Unformatted text preview: A ENGINEERlNG SURVEYiNG .0 46%;“)
Examination 1 '
Fall 2007 Name PROBLEM 1. Short questions (4 points each). ,.
. K” Kw «‘ ‘ 2 a. What is the bearing of a line with an azimuth of 297 degrees 18 minutes? Sketch
the line on a true north axis system. ‘ I" //7 , Ma" —z=n” g' a (01°41 ’>
[5”
,i4ig M 5
b. A steel tape is has a known length of 100.00 feet at 68 degrees when fully supported
and pulled at 0 lb. When used to measure a line along a level sidewalk with a pull of
just 10 pounds (also fully supported and at 68 degrees), the tape gave a measurement
of 386.81 feet. Assume the tape’s crosssectional area is 0.004 sq. in. and the elastic
modulus is 29*106 psi. What is the corrected length of the line?
CF ; {1"} ~ 93L ; (ions — Olét 330ml 3 ‘ a k {x A A A («5:340:34 gi'izvgﬁfamo 01%;) ,\xwm’//;l "’1‘. “if. " "A '1‘ c. A surveyor ran a open series of differential leveling shots through 18 setups (36
shots). The elevation of the starting benchmark is known exactly. It is estimated that the 95% error of a single shot is i0.015 feet. Estimate the 95% error in the elevation of
the ending benchmark, as measured. ' F] d. Today we know that the true bearing of the line AB is S 35 degrees 15 minutes E. In
_ 1890 the magnetic bearing of the line was reported to be S 38 degrees 45 minutes E.
What is the 1890 magnetic declination?
38"*13’— '33”)? z 3 30'
[(670 an: ! >n¢+ﬂ9ﬂ :
142;,”—
a/r"
e. BMD has a known elevation of 643.88 feet. A surveyor needed to quickly establish
the elevation of BME across a river. He set up his level near BMD and recorded a rod
reading of 6.45 feet. From the same instrument position he shot BME, 800 feet away,
and recorded a rod reading of 7.23 feet. What is the elevation of BME, ignoring earth
curvature and refraction?
' g ; (4519.33 LB,“
HI ‘— (a I I I“? “30
a; (/19 gm: : 33 ~ 32 ~—'« W's I‘M, a NMNJ
f. The measured interior angles of a 4course loop traverse (almost rectangle in shape)
were (in clockwise order); (3
[S A = 90 degrees 12 minutes 7"
B = 89 degrees 54 minutes D = 90 degrees 32 minutes W “I;
E = 89 degrees 38 minutes After adjusting for closure (equally weighted), what should be the corrected
measurement of angle Z? I?
Isa ( +1) ' ‘IVrz' +8905‘rr'+90°31'+“ 39 1 360 W / I
La 1.0, : 4
a. L, D 21.2” :: (1;.rrrt'ci'cﬁrwv's “5'6 0 C708? M91, 3* [20 8H? PROBLEM 2. A five course traverse has the following internal angles in a clockwise
direction: A = 137 degrees, B = 81 degrees, C = 98 degrees, D = 124 degrees and E =
I. 100 degrees. The bearing on line AB is N 37 degrees E. Sketch the traverse. What is i the bearing of line CD? (Hint: this traverse closes exactly.) (8 points) 1% ; N37°00'E : 570 8%: (5c :— 37% 816° 7* ‘39 L 3‘ lie” 4 if; 2 2,1?" 8; QC) > A \ r" a I CD > zis°’/s’o’ " is?” ‘9 [538’ 90 w; 0 \ _
Lg i y‘ PROBLEM 3. A level circuit was run from BMR'to BMS and then back to BMR. The
known elevation of BMR was 684.80 ft. The recorded elevation of BMS was 671.00.
Five turning points were established between BMR and BMS and three more turning
points were established between BMS and BMR. The calculated elevation of BMR was
684.96. Sketch the circuit. What should be the corrected elevation of BMS? (8 points)  BMR (ammo ~czrma‘ « 0! W 3 TH ’ . T?!
“if M 2 19.2.3589 7P .  T??
ji: ED ’ l ’ T?)
r 'Tna/
’— 0.0l (a se‘FDFSB : T 0107‘? 76,6 ft ms (945 a (07!.00 » (9,0070 awﬂfllﬁwé W‘s )l o o w
A EXTRA CREDIT A. Two inexperienced surveyors ran a level circuit from BMC to BMD and then back to
BMC. BMC has a known elevation of 679.43 ft. One turning point was established
between BMC and BMD and a second turning point was established between BMD and
BMC. Unfortunately, they forgot to record the foresight on BMD, The measurements
(in the order taken) were 3.95, 4.35, 7.37, ?.??, 4.87, 6.77, 2.14, 5.01. Set up the level
notes and compute the missing foresight and the elevation of BMD. State any
assumptions. (3 points) é] VLF] ZIZ Ckﬁvcgs ‘ I no
“0”; («meagre B. Refer to PROBLEM 1e. What is the corrected elevation onBé/IE?/(2 points)
800 ~ .3 v f/
1000) to, 0? “if C. A treasure map was recently found in some old materials at the Smithsonian
institution. The map has an X on it that is labeled as being halfway in longitude,
between London England and New Zealand and halfway in latitude between the
equator and the North Pole. About where in North America is the treasure? (1 point) . LAM“— D. Refer to PROBLEM 1f. The traverse is the foundation of an ancient temple in
Uganda. Archeologists know that course AB is oriented toward the setting sun on June
let of each year. What is the approximate bearing of course36’? (2 points) Q0
5/ suiting Sun Erameer‘m «(Uf‘ﬁﬂﬂ‘n
52mm 1 jwews ZOO“? A} B sw‘zqi {8i C? .__ (koﬁoyg 0.033
= N “042(k) \_ 2 324.8“ 5.033: 3%.9“: 5A2
A 7 7_ , , ‘1' gas a *5 0.0x: m ;: 1: 0.03 a P! ° ’ V 6..) Q" Hi: (4.3.?>g=* 69:5: 635.33
:9 33335545 ‘5 exam (was 650353—113 = Mama
13. Z39 = aeo'xL'; SHOJH be. 3go‘o’; g: ‘5’
ﬂ ‘3
" Am A 6/4 = “971: 4,“) no :qo°sz’~<t’ = “W 2% ' b A.
@ 2‘35 : FIVE: \‘M:0 ’v x v x x a = L 34.14  03430 :
\ 4‘ 9;ch = gym 22, = 2&6 o a 5:15 aura. Luqh
A 5 NC? = 539%) one 2‘ a! v 93/ka 3 gag/.0 = 0.0%
' 1° 5% saws +5 8H3 = é
39‘ 98% 6M1 M15 : 471.00 ~ 6 (0M6) —'
E 7‘ mm L+ ‘ ...
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 Fall '07
 Horowitz

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