Handbook of Civil Engineering Calculations

in the principal plane. Then kq = cos t and Lk =/sec t—y\ sin t. From Fig. 212b, QA/qa =

LQfLq. From Fig. 3 Ia, LQILq = LNILk = (H- /z)/(/sec t-yl sin f). Setting 04 = XA, we

have aa = xa', and omitting subscripts gives X= jc '(H - /*)/(/sec f ->>' ,siw t), Eq. a. From

Fig. 3 Ia, Wg/£a = LNILk = (H- /*)/(/sec * -y\ sin f). Setting Ng = YA and omitting sub-

scripts, we get Y=y'[(H- /*)/(/sec t-ya' sin f)] cos /, Eq. b.

3. Compute the datum-plane coordinates
   First compute/sec f = 153.6 sec 2°54' = 153.8 mm. For A 9 H-h = 2360 - 180 = 2180 m,
   and/sec t -y
   r
   sin t = 153.8 - (-74.89) sin 2°54' = 157.6 mm. Then (H-h)/(fsec t -
   X sin O = 2180/0.1576 = 13,830. By Eq. a,XA = (+0.01586)(13,830) = +219.3 m. By Eq.
   b, YA = (-0.07489)(13,830) cos 2°54' --1034.4 m.
   Similarly, for B, H- h = 2360 - 130 - 2230 m and/sec t-y
   f
   sin t = 153.8 - 51.44 sin
   2°54' = 151.2 mm. Then (H-/*)/(/sec f-j/sin O = 2230/0.1512 = 14,750. By Eq. a,XB
   = (-0.06861)(14,750) = -1012.0 m. By Eq. b, YB = (+0.05144)( 14,750) cos 2°54' =
   +757.8 m.


4. Compute the required distance
   Disregarding the difference in elevation of the two points and proceeding as in the second
   previous calculation procedure, we have AX= +219.3 - (-1012.0) = + 1231.3 m, and
   A7= - 1034.4 - 757.8 = -1792.2 m. Then (AB)
   2
   = (1231.3)
   2

• (1792.2)
  2
  , or AB = 2174 m.
  Related Calculations: The X and Y coordinates found in step 3 can be verified
  by assuming that these values are correct, calculating the corresponding x' and y' coordi-
  nates, and comparing the results with the values in step 2. The procedure is as follows. In
  Fig. 3 Ia, let VA = angle NLQ. Then tan VA = NQILN= YA/(H- h). Also, angle oLq = vA-t.
  Now, x'JXA = LqILQ =/sec (VA - f)l[(H- h) sec VA]. Rearranging and omitting subscripts,
  we get x' = JLJTCOS vA/[(H- h) cos (VA -1)], Eq. c. Similarly, y'a - no + oq =/tan t +/tan
  (VA -1). Omitting the subscript gives y' =/[tan t +/tan (VA - OL Eq. d.
  As an illustration, consider point A in the present calculation procedure, which has the
  computed coordinates XA = +219.3 m and YA = -1034.4 m. Then tan VA = -1034.4/2180 =

• 0.4745. Thus, VA=- 25°23' and vA-t = -25°23' - 2°54' = -28
  0
  H'. By Eq. c, x' =
  (+219.3)(0.1536)(0.9035)/(2180)(0.8806) = +0.01585 m = + 15.85 mm. Applying Eq. d
  with t = 2°54' gives X= 153.6(0.0507 - 0.5381) = -74.86 mm. If we allow for roundoff
  effects, these values agree with those in step 1.
  The following equation, which contains the four coordinates x't y', X, and Y 9 can be
  applied to test these values for consistency:

f^1 + (y' -/tan f)^2 _ (H- K)^2 + Y^2

~7

2 =

x

2

DETERMINING ELEVATION OFA POINTBY

OVERLAPPING VERTICAL PHOTOGRAPHS

Two overlapping vertical photographs contain point P and a control point C that lies 284
m above sea level. The air base is 768 m, and the focal length is 152.6 mm. The microm-
eter readings on a parallax bar are 15.41 mm for P and 11.37 mm for C. By measuring the
displacement of the initial principal point and obtaining its micrometer reading, it was es-
tablished that the parallax of a point equals its micrometer reading plus 76.54 mm. Find
the elevation of P.