The stadia method In determining stadia interval factor (K) of a transit, a stadia rod was held vertically at several points along measured distances from the instrument (see accompanying figure), and PDF

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In determining stadia interval factor (K) of a transit, a stadia rod was held vertically at several points along
measured distances from the instrument (see accompanying figure), and the corresponding stadia hair readings
were observed. The distance and the observed readings were recorde...

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THE STADIA METHOD  The word stadia is the plural of stadium. It comes from the Greek word for a unit of length originally applied in measuring distances for athletic contests.  A stadia denoted 600Greek units, or 184m 93cm (606ft 9in) by present day international standards.  The term stadia is now applied to the cross hairs and rod used in making measurements, as well as to the method itself.  Stadia readings can be taken with most surveying instruments.  The equipment for stadia measurements consists of a telescope with two horizontal hairs called stadia hairs and a graduated rod called a stadia rod.  Well adapted to mapping requirements and is widey used for locating details and contour points in topographic surveys Stadia Hairs The telescope of most surveying instrument are equipped with stadia hairs in addition to the regular interval and horizontal hairs. When the stadia hairs and the cross hairs are simultaneously visible and in focus they are called stadia hairs.

Typical Cross Hair Ring (or Reticule) Principle of the Stadia The stadia method is based on the principle that in similar triangles corresponding sides are proportional. Figure 1 illustrates the principle upon which the stadia method is based.

f f :i=d : s∧d= s i Also

D=d+ (f + c ) f ¿ s+C i D=K s + C The equation D = Ks + C is employed in computing the horizontal distances from stadia intervals when sights are horizontal the stadia constant C is the distance from the center of the instrument to the principal focus. Its value is composed of the focal length of the lens (f) and the distance from the center of the instrument to the center of of the objective lens (c) K or f/i is the stadia interval factor of the instrument s is the stadia or rod intercept, and is determined in the field by observing the difference between the upper stadia hair reading and the lower stasdia hair reading

Figure1 Stadia Constants With present day surveying instrument and under ordinary conditions C may be considered 0.30m for external focusing telescope. The important advantage of the internal focusing telescopes used in a stadia work is that they are so constructed that C is either zero or small enough to be neglected. Stadia interval Factor The ratio f/i is called stadia interval factor and is designated by the letter K for any given instrument, this value remain constant and depends only on the spacing between the stadia hairs. The manufactured of the instrument can space the stadia hair with relation to the focal length so as to obtain any convenient value of K desired. The most common value of K is 100

Sample Problem: 1. In determining stadia interval factor (K) of a transit, a stadia rod was held vertically at several points along measured distances from the instrument (see accompanying figure), and the corresponding stadia hair readings were observed. The distance and the observed readings were recorded as follows: Point Distance from Stadia hair readings transit to rod Upper (m) Lower(m) a 30 0.96 0.66 b 45 1.10 0.64 c 60 1.21 0.60 d 75 1.35 0.58 e 90 1.47 0.56 f 105 1.57 0.53 g 120 1.72 0.50 Determine the stadia interval factor of the instrument. Assume that the stadia constant © is zero.

Solution:

D=Ks + C= Ks +0= Ks K= D /s 30 45 K 1= =100 K 2= =97.826 ( 0.96−0.66 ) (1.10−0.64) 60 75 K 3= =98.361 K 4= =97.403 (1.21−0.60) (1.35−0.58) 90 105 K 5= =98.901 K 6= =100.962 (1.47−0.56) (1.57−0.53) 120 K 7= =98.361 (1.72−0.50)

average K=

100 + 97.826 + 98.361+ 97.403 + 98.901 + 100.962 + 98.361 =98.83 7

Sample problem no 2. An engineer’s transit with a stadia constant of 0.30 was set up on the line between two points, A and B, and the following hair readings were observed. ROD HAIR READINGS POSITION UPPER MIDDLE LOWER Rod held at A 1.330 1.175 1.020 Rod held at B 1.972 1.854 1.736 If the stadia interval factor of the level is 99.5, determine the length of the line AB. Solution:

s=upper hair rod reading−lower hair rod readi ng s1=1.330 −1.020=0.31 s2=1.972 −1.736=0.236 m D=Ks + c D 1 =99.5 x 0.31+ 0.3=31.145 m D2 =99.5 x 0.236+ 0.3=23.782 m L=31.145+23.782=54.927 m(answer)

Problem no 3 A dumpy level with an internal focusing telescope was set up on the left bank of a river and the rod readings tabulated below were taken on a stadia rod held successively at the left and right water edges. If the stadia interval factor of the instrument is 100 determine the width of the river. Rod position

Hair Readings Upper(a) Middle(c) Lower(b) Rod held at LWE 2.189m 2.172m 2.155m Rod held at 2.277 2.173 2.069 RWE Solution:

s=upper hair rod reading −lower hair rod reading s1 = ( 2.189 −2.155 )=0.034 m s2 = ( 2.277−2.069 ) =0.208 D=Ks + c D=Ks + c Internal focusing telescope: C = 0

D 1 =100 x 0.034=3.4 m D 2=100 x 0.208=20.8 m

D=D 1 +D2 =3.4+20.8=24.2 m(answer) INCLINED STADIA SIGHTS In actual field practice, most stadia measurements are inclined because of varying topography, but the interval is still read on a vertically held rod. The inclined measurement, which is also dependent on the observed vertical angle, is reduced to horizontal and vertical components of the incline line of sight. Figure 2 illustrates an inclined line of sight for an instrument set up at point M with the rod held vertically at N. The horizontal distance between the telescope and the rod is shown as HD or OD, and the vertical distance between the telescope axis at O and the horizontal hair reading at P is VD or PD. The length of the inclined line of sight from O to P is ' ' ID = K ( a ´b ) +C ID = Kscos + C

The horizontal component of the inclined distance may be submitted as follows

HD=(ID)cosα HD=(Kscos +C)cosα 2

HD=Ks cos +Ccosα The vertical component of the inclined distance is determined

VD=( ID ) sinα VD=( Kscos + C ) sinα VD =Kscos sin + Csin

Sample problem Problem no 1 The following data were obtained by a stadia measurement; vertical angle = +18 023’, and observed stadia intercept = 2.20m. The stadia interval factor of the instrument used is 95.5 and C = 0.30. if the height of instrument is 1.62m and the rod reading is taken at 1.95m, determine the following: a. Horizontal stadia distance (HD) from the instrument set up at A to the rod held at point B. b. Vertical stadia distance (VD) from the center of the instrument to the point on the rod bisected by the horizontal cross hair. c. Inclined or slope distance (ID) from the instrument center to the point on the rod bisected by the horizontal cross hair. d. Difference in elevation (DE) between the point over which the instrument is set up and the point on which the rod was held.

S olution : Given  = +18023’ Stadia intercept, s = 2.20m K = 95.5 C = 0.30 H.I. = 1.62 RR = 1.95

a. HD=Ks cos2 +Ccosα

HD=95.5 x 2.20 x cos 2 18 ° 23 ' +0.30 cos 18 °23 '=189.488 m b. VD =Kscos sin + Csin VD =95.5 x 2.2cos 18 ° 23' sin 18 ° 23' + 0.30 sin 18 ° 23' =62.973 m c. ID = Kscos + C '

ID=95.5 x 2.2 cos 18 °23 +0.30=199.678 m Alternate solution:

HD =199.678 cos 18 °23' =189.488 m VD=199.678 sin 18 °23' =62.973m d. Difference in elevation From the figure

VD +H . I .= RR + DE 62.973 + 1.62=1.95 + DE DE=62.643 m

Problem no. 2 The upper and lower stadia hair readings on a stadia rod held at station B were observed as 3.50 and 1.00m respectively, with the use of a transit with an internal focusing telescope and having a stadia interval factor of 99.5. The height of the instrument above station A is 1.45m and the rod reading is taken at 2.25m. If the vertical angle observed is -23034’, determine the following: a. Horizontal, vertical and inclined stadia distances. b. Difference in elevation between the two stations. c. The elevation of station B, if the elevation of station A is 155.54m above mean sea level.

Solution : K= 99.5 C = internal focusing telescope = 0

a. ID = Kscos + C ID =99.5 x ( 3.5− 1.00 )cos 23° 3 4 + 0=228. 0 m '

HD=228.2915 cos 23 °3 4 '=20 8 . 987 m VD =228.2915 sin 23 °24 ' =9 1 .160 m b. DE + H . I .= RR + VD DE=2.25 + 90.655−1.45 =91. 96 m c. ElevA = DE + Elev B Elev B =155.54 −91.96 =53.58 m Problem no. 3 A transit with stadia interval factor of 100.8 was set at C on the line between points A and B, and the following stadia readings Position of Rod Vertical angle Hair readings uppe middl lower r e Rod Held at A 15035’ 1.330 1.175 1.020 Rod Held at B -8008’ 1.972 1.854 1.736 If the stadia constant is 0.381, determine the following: a. Length of line AB b. Difference in elevation between points A and B Solution:

a. HD=Ks cos2 +Ccosα HD 1=100.80 x ( 1.330−1.020 ) ( cos 15 ° 3 5 ) +0.381cos 15 ° 35 ' =29.36 m ' 2

'

HD 2=100.80 x ( 1.972−1.736 ) ( cos 8 ° 08 ') 2 +0.381 cos 8 ° 08 =23.7 m HD = HD 1+ HD 2=29.36+23.70=53.06 m

b.

VD HD VD=HD tan α V D1 =29.36 tan 15 °3 5' =8.19 m tan α=

V D 2=23.7 tan 8 ° 0 8' =3.39 m c. From the illustration

Elev A−DE ab = ElevB DE ab=ElevA−Elev B ElevA +1.175−V D 1−V D 2−1.854=ElevB ElevA +1.175− 8.19−3.39− 1.854=ElevB ElevA− ElevB =12.26 m Therefore: DE ab= ElevA − Elev B=12.26 m

Problem no 3 Given the following set of stadia level notes. The following set of stadia level notes. The instrument used has a stadia interval factor of 100 and equipped with an internal focusing telescope. Complete the tabulation and perform the customary arithmetic check. BACKSIGHT

Inter cept, s 1.55 1.74 0.95 2.49 2.14 0.92 1.55

BMa TP1 TP2 BMb TP3 TP4 BMc

Vertical Angle α -5025’ +8015’ -4048’ -12050’ +14005’ -9041’

Rod reading , RR 1.50 1.68 1.77 2.53 1.79 1.33

FORESIGHT

Vertical Distance, VD

Intercep t,s

Vertical Angle α

Rod reading , RR

+10030’ +12008’ +7022’ -15032’ -7059’ +7032’

1.48 1.66 2.05 1.92 1.25 1.88

change ∈Elev , DE=RR BS −VD BS−RR FS +VD FS VD =Kscos sin + Csin K=100 C = INTERNAL FOCUSING TELESCOPE = 0 Vertical distance, VD Intercept,s

Vertical Angle

α BMa TP1 TP2 BM b TP3 TP4 BMc sta

VD =Kscos sin+ Csin =100*1.55*cos(-5025’)*sin(-5025’)= -14.57 =100*1.74*cos(+8015’)*sin(+8015’) = 24.71 =100*0.95*cos(-4048’)*sin(-4048’) = -7.92 =100*2.49*cos(-12050’)*sin(-12050’) = -53.93

2.49 2.14 0.92

+14005’ -9041’

=100*2.14*cos(+14005’)*sin(+14005’) = 50.51 =100*0.92*cos(-9041’)*sin(-9041’) = -15.25

Intercept,s

Vertical Angle

α BMa TP1

Vertical Distance, VD

-5025’ +8015’ -4048’ -12050’

1.55 1.74 0.95

1.76

+10030’

Elevatio n

Vertical Distance, VD 550.50

1.76 1.98 1.06 2.67 2.16 2.65

Solution:

sta

Change in Elevation,D E

Vertical Distance, VD

VD =Kscos sin+ Csin =100*1.76*cos(+10030’)*sin(+10030’)= 31.54

TP2 BM b TP3 TP4 BMc

1.98 1.06

+12008’ +7022’

=100*1.98*cos(+12008’)*sin(+12008’)= 40.69 =100*1.06*cos(+7022’)*sin(+7022’)= 13.48

2.67 2.16 2.65

-15032’ -7059’ +7032’

=100*2.67*cos(-15032’)*sin(-15032’)= -68.89 =100*2.16*cos(-7059’)*sin(-7059’)= -29.71 =100*2.65*cos(+7032’)*sin(+7032’)= 34.44

CHANGE IN ELEVATION BACKSIGHT

FORESIGHT

Change in Elevation,DE=

RR BS−VD BS−RR FS+ VDFS Interc ept,s BMa TP1 TP2 BMb TP3 TP4 BMc

1.55 1.74 0.95 2.49 2.14 0.92 1.55

Vertical Angleα -5025’ +8015’ -4048’ -12050’ +14005’ -9041’

Rod reading, RR 1.50 1.68 1.77 2.53 1.79 1.33

Vertical Distance, VD -14.57 24.71 -7.92 -53.93 50.51 -15.25

Interce pt,s

Vertical Angleα

Rod reading , RR

1.76 1.98 1.06 2.67 2.16 2.65

+10030’ +12008’ +7022’ -15032’ -7059’ +7032’

1.48 1.66 2.05 1.92 1.25 1.88

Vertical Distanc e, VD 31.54 40.69 13.48 -68.89 -29.71 34.44

1.50 - (-14.57) - 1.48 + (31.54) = 46.13 1.68 - (24.71) - 1.66+(40.69)=16.00 1.77 – (-7.92) - 2.05 + (13.48) = 21.12 2.53 –(-53.93) - 1.92 + (-68.89) = -14.35 1.79 - 50.51 -1.25 +(-29.71) = -79.68 1.33 – (-15.25) - 1.88 + (34.44) = 49.14

ELEVATION: BACKSIGHT

BMa TP1 TP2 BMb TP3 TP4 BMc SUM

Inter cept, s 1.55 1.74 0.95 2.49 2.14 0.92 1.55

Vertical Angle α -5025’ +8015’ -4048’ -12050’ +14005’ -9041’

FORESIGHT

Rod reading , RR 1.50 1.68 1.77 2.53 1.79 1.33

Vertical Distance, VD -14.57 24.71 -7.92 -53.93 50.51 -15.25

10.6

-16.45

Change in Elevation,D E

Intercep t,s

Vertical Angle α

Rod reading , RR

Vertical Distance, VD

1.76 1.98 1.06 2.67 2.16 2.65

+10030’ +12008’ +7 022’ -15032’ -7059’ +7032’

1.48 1.66 2.05 1.92 1.25 1.88 10.24

31.54 40.69 13.48 -68.89 -29.71 34.44 21.55

CHECKING:

Elev of BMa+ ∑ RR BS−∑ VD BS− ∑ RR FS+ ∑ VD FS=Elev of BMc 550.5 + 10.6 − ( 16.45 ) −10.24+( 21.55 )=588.86

46.13 16 21.12 -14.35 -79.68 49.14

Elevatio n

550.5 596.63 612.63 633.75 619.4 539.72 588.86

PROBLEM no 4 The following data were obtained by stadia observations: vertical angle = +9 025’, upper stadia hair and lower stadia hair readings are 2.352 and 0.995m respectively. The stadia interval factor is known to be 99.0 and C is 0.381m. The height of instrument above the instrument station(A) is 1.496 and rod readings is taken at 1.589m. determine the following: a. Horizontal, vertical and inclined distance. b. Elevation of the point sighted (pointB) if the elevation of point A is 776.545m. c. Difference in elevation between the two points.

SOLUTION :

a. ID = Kscos + C ID =99 x ( 2.352− 0.995 )cos 9° 2 5' + 0.381=132.91 m '

HD=132.91 cos 9 °25 =131.12m VD =132.91 sin 9 °25' =21.75 m

b. ( ElevA=776.545 ) + ( RRBS =1.496 ) + ( VD =21.75)− ( RR BS=1.496 ) =Elev B

ElevB=798.295 m c. Elev A + DE=Elev B DE= ElevB − ElevA =798.295 −776.545=21.75 m...