A METHOD OF DETERMINING THE DEPRESSION ANGLE·
AND SWING OF A TERRESTRIAL PHOTOGRAPH
WHEN THE APPARENT HORIZON IS NOT
IMAGED ON THE PHOTOGRAPH
Everett L. Merritt
Topographic Engineer, U. S. Naval Photographic Interpretation Center
A photo-theodolite is used the horizontal ahd vertical circles indicate the
I Fbearing
and angle of depression or elevation of the camera's optical axis.
However, a sturdy camera and compact precise theodolite employed interchangeably on the same tripod is the more practical for the execution of ground
control in exploratory or reconnaissance military mapping as the photo-theodolite, being extremely limited in its use, is inadequate for standard traverse and
triangulation surveying. Since a map constructed from terrestrial photographs
requires that a control net be established either before or during the terrestrial
photography, a theodolite must precede or accompany the terrestrial photography. In establishing ground control for reconnaissance maps the equipment
will generally be "packed." The control net will usually be established simultaneously with the terrestrial photography by the same personnel. In keeping
with speed, universal application, simplicity and portability, it is more suitable
for reconnaissance surveys to have the camera and theodolite used interchangeably on the same tripod.
The choice of the latter combination necessitates the computation of the
fixed angular relationship of the camera's optical axis. R. M. Wilson solves this
problem with the aid of the Wilson photo-alidade by a method of successive
approximations that requires the horizontal and vertical coordinates of three
ground objects. l In rapid reconnaissance surveys ground control is invariably
lacking. The Canadians have found the camera and theodolite used as two
separate instruments interchangeably on the same tripod more adapted than
the photo-theodolite for reconnaissance mapping in uncultured areas. 2 The Canadian method has many practical features, but is limited in use in that it requires
that the line connecting the left and right fiducial markers coincide with the
true horizon and that the optical axis be a direction line in the horizon plane with
the attendant dependence on level vials. In the method described below the
level vials are not used and therefore may be eliminated. This fact eliminates
in turn the greatest feature of vulnerability in the terrestrial camera operating in
unfamiliar terrain great distances from a base camp. The terrestrial photomapper has the maximum freedom if he can make an exposure at the most advantageous horizontal and vertical angle without concern over the level bubbles.
An exposure can be made with the camera's optical axis pointing in any
direction without leveling, as the depression angle and swing of the photograph
are easily computed when the vertical and horizontal angles have been observed with a theodolite to two ground objects that will be sharply imaged on the
photograph. The calibrated focal length, the intersection of the opposite fiducial
markers making an angle of 90° ± 1 minute, and the principal point coinciding
with this intersection ± .03 mm is the only other data that must be known. A
mathematical explanation of the computation is given below.
1 Wilson, R. M. "Oblique photographs for the surveyor." In American Society of Photogrammetry Manual of photogrammetry. Pitman Pub!. Corp., 1944, p. 597-619.
2 Bridgland, M. P. Photographic surveying. Canada, Department of the Interior, Bulletin
no. 56, 1924.
465
"
466
PHOTOGRAMMETRIC ENGINEERING
In Figure 1, C is the point where the camera's optical axis pierces the photoplane which is tangent to a sphere whose radius is f at the point on the photograph corresponding to the plate perpendicular. VA and VB are the vertical
angles measured to two ground objects, designated A and B respectively, that
7..
IPAL LINE
FIG. 1
are imaged on the photograph. The horizon plane is the equator. A' and B' are
the points where the vertical planes defined by the vertical angles cut the
horizon plane and are therefore the initial and terminal sides of the horizontal
angle ~ubtended by A and B at L. L is the rear nodal point of the camera lens
and the center 'of the sphere. The required elements are (1) the swing angle
of C, defined by the intersection of the line between t'he projection of the zero
meridian on the photograph with the line connecting the upper and lower
fiducial markers, and (2) the depression angle at L, defined by the intersection
of the optical axis with the horizon plane in the principal plane.
Formulas: In Figure 2 photo lengths A C, B C, and AB are computed by plane
analytics.
467
DETERMINING DEPRESSION ANGLE AND SWING
+ (Ye XB)2 + (Ye -
AC = V(xe - x...)2
YA)2
(1)
BC = V(Xe -
YB)2
(2)"
AB = V(XA - XB)2
+ (YA
(3)
- YB)2.
Angles CLB, CLA, and BLA are designated (x, fl, and 'Y respectively as they
are the sides of spherical triangle ABC whose vertex is L and may be computed
by solid analytics.
J
-
A:
•
~~'~====~~==:::;~:::=i~~~--,------_
-,
'1",
"'-'--:" ..... .....
,
\(
-
------ ---
-~
............ ---........
--
L
FIG. 2
(4)
(5)
f
(LA)2
COS'Y
=
+ (LB)2 2(LA)· (LB)
(AB)2
(6)
468
PHOTOGRAMMETRIC ENGINEERING
and
(LA)
= V(xo
- XA)2
+ (Yo -
YA)2
+P
(7)
(LB)
= V(Xe - XB)2
+ (Ye -
YB)2
+ P.
(8)
Since C is the point where the photograph is tangent to the sphere, angle C
in the plane of the photograph is equal to the dihedral angle C in the spherical
triangle whose side are a, f3, and 'Y. In the plane of the photograph C=m+n.
tan n
tanm = - - Ye - YA
Xc - XB
= ----.
(9)
Ye - YB
FIG.
3
Then by the law of cosines in spherical trigonometry
cos'Y
=
cos ex cos f3
+ sin ex sin f3 cos C.
(10)
In sphericaf triangle AB C we have the sides a, f3, and 'Y and one included angle C.
The required elements are (j and S as shown in Figure 3. It can readily be seen
that
cos f3 cos 'Y - cos ex
cos A
cos B
(11)
sin f3 sin 'Y
=
cos 'Y cos ex - cos f3
sin'Y sin ex
.
(12)
Since A'B' is a horizontal angle in the plane of the equator and VA and VB are
vertical arcs perpendicular to the horizon plane, both AA'B' and BB'A' are
469
DETERMINING DEPRESSION ANGLE AND SWING
right spherical triangles. Then by Napier's analogies in Figure 4
cos A B ' = cos VA . cos A'B '
sin VA
=
(13)
cot F· tan A'B '
cot F = sin VA· cos A'B '
(14)
8
4
FIG.
also
cos E
cos A B . cos 'Y - cos VB
'
= ~'--------­
(15)
sin A B ' sin 'Y
x =
(F
+ E)
- A
and
cos A'e = cos {3 cos VA
+ sin {3 sin VA
cos X.
(16)
Then by the law of sines in spherical trigonometry in Figure 5 we see that
sin G
tin X
--=
sin {3
sin A'e
sinG =
sin {3' sin x
sin A'e
(17)
(18)
470
PHOTOGRAMMETRIC ENGINEERING
Since A' is a 90° angle
G' = 90° - G
and
sin 8 = sin A'C· sin G'
(19)
also
cos A'C
=
cot G' ·cot H
cot H = cos A'C· tan G'
sin H'
sin VA ·sin G
= ----sin {j
FIG. 5
(20)
(21)
.
It can be seen in Figure 5 that the angle of swing then is equal to
(H
+ H' + m) -
180°.
(22)
Having computed the depression angle 8 and the swing angle S, the horizontal
and vertical angles to other objects imaged on the photograph are computed.
However the new coordinates must be computed first. Figure 6.
In analytic geometry we have the formula for rotation of axes
x' = x cos s - y sin s
(23)
= x sin s + y cos s.
(24)
y'
And finally the horizontal angles 8 and vertical angles V are computed employing the new coordinates x' and y' and the computed 8.
Then
cos 'Y
= tan { j . - cos cP
tan V = cos 8· tan cP.
tan 8
(25)
(26)
471
DETERMINING DEPRESSION ANGLE AND SWING
The geometric proof of equations (25) and (26) is easily demonstrated.
In Figure 7,f is the focal length, L the lens, 8 the angle of depression of the
optical axis, C the plate perpendicular, VV' the horizon line, pp' the principal
parallel, HG the principal line, and HLG the principal plane. x' and y' are the
y'
x
x·
FIG. 6
new coordinate axes coinciding with the principal parallel and the principal line
referred to origin C.
x'
tan f3J= -
f
tan
1'~=
,
y'
-
f
G'G l..to LH by construction
cos
l'
=-
f
LG
cos l'
tan B = tan f3._cos q,
x
tanB = LG'
LG'
cos q, = - q, = B ±
LG
l'
(25)
472
PHOTOGRAMMETRIC ENGINEERING
f
x
LG
xfLG
x
tan 0 = - . - - = _. _ . - - = - f
LG'
f LG LG'
Le'
LG
H~=--
-----7?'------=~-r--r------:::~ L
FIG.
and
tan V
7
= cos ~ . tan
¢.
In Figure 8
y
tan V
=--
cos 0
=--
tan ¢
LG"
LG'
LG"
y
=--
LG'
LG'
y
Y
tan V = - - - = -- •
LG"
LG'
LG"
(26)
473
DETERMINING DEPRESSION ANGLE AND SWING
The Cl;bove formulas are adapted to machine computation.
To further illustrate the method of analytical rectification of terrestrial
photographs in' accordance with the principles discussed, a sample problem is
computed.
The horizontal angle measured between two ground objects inraged on the
photograph is 34° 34' 55". The vertical angle of each of the ground objects desig-
,.-
y
~:::"",_-------::;.r-------':'I'l"""---:~L
FIG. 8
nated A and B respectively is VA = 14° 4' 9" and VB = 35° 2' 29". The coordinates of the two phot,o images and the plate perpendicular were measured
with a comparator. The principal point is designated C. The mean of five separate x and y readings on each photo image is used in the computation.
"
A
Points
Sets
1
2
3
4
5
•
x
B
y
x
C
y
x
y
mm,
mm.
mm.
mm.
mm.
mm.
47,02
47.02
47.03
47.03
47.03
97.00
97.00
97.02
97,01
97.00
123.93
123.94
123.91
123.93
123.91
173.44
173.44
173.46
173.45
173.44
98.00
98.01
98.01
98.01
98.01
93.00
92.99
93.00
92.98
92.98
474
PHOTOGRAMMETRIC ENGINEERING
Mean values of coordinates
XA = 47.025 rom
XB = 123.924 rom
Xc = 98.02 rom
YA. = 97.006 rom
YB = 173.446 rom
Yc = 92.99 rom
The photo lengths are computed according to equations (1), (2) and (3).
Xc =
98.010
Yc =
92.990
XA. =
47.025
YA =
97.006
(yc - YA) = -
(xc - XA) = - 50.985
+ (Yo -
AC = V(Xc - XA)2
AC =
YA)2 = Y(50.985)2
+ (- 4.016)2
•
51.1429
Xo =
98.010
yo = 92.990
XB = 123.924
YB = 173.446
(xc - XB) =
(yc - YB) = 80.456
25.914
+ (Yc -
BC = V(xc - XB)2
YB)2 = Y(25.914}2
BC = 84.526
XB = 123.924
XA =
47.025
(XB - XA) =
76.899
+ (80.456)2
YB = 173.446
YA = 97.006
(YB - YA) =
AB = Y(XB - XA)2
+ (YB -
YA)2 = Y(76.899)'2
f
AB = 108.427
AC =
51.1429
BC =
84.526
AB
Eq. (4)
4.016
76.440
+ (76.440)2
= 152.4003
= 108.427
AC
51.1429
tan {j = =
= .3355787
f
152.4003
{j = 18 0 33' 02"
Eq. (5)
BC
tan a = =
Eq. (6)
cos'Y =
f
(LA)2
84.526
152.4003
+ (LB)2
= .5546314 a = 290 00' 51"
- (AB)2
2(LA) (LB)
f'
152.4003
cos {j
.9480449
= -- =
= 160.7521
but
LA
and
= 174.2713
.8744998
LA = Yj2 + (A C) 2 = yF'7(-:-I:
l ~::-=2or-:.4-:!-607':3:-C-)2°-+----'('-'-5~1."--14-=2"""3)C-::-2
Eq. (7) also
f
LB= - cos a
=
152.4003·
= Y25841.3863
= 160.7521
•
475
DETERMINING DEPRESSION ANGLE AND SWING
LB == Vr
Eq. (8)
+ (BC)2
= V(152.4003)2 + (84.526)2
= v30370.49612
174.2713
=
cos'Y
=
+ (174.2713)2 -
(160.7521)2
(108.427)2
2(160.7521) . (174.2713)
44455.309329
56028.95488
cos 'Y
= .7934346
'Y = 37 0 29' 33"
(X,
fl,
and l' are the sides of a spherical triangle whose vertex is L. Therefore:
Eq. (10)
cos 'Y = cos a cos f3
+ sin a sin f3 cos C
C is a point where the photograph is tangent to the sphere. Angle C in the
plane of the photograph is also angle C in spherical triangle AB C. In the plane
of the photograph C=m+n.
Eq. (9)
tan m
=
tan n
=
XA -
Xc
YA - Yc
XB -
Xc
YB - Yc
50.985
= -- =
4.016
25.914
= -- =
80.456
12.695468
.3220890
m = 85 0 29' 46"
180 0
-
n
= 17 51' 11"
C
=
103 0 20' 57"
C =
76 0 39' 03"
0
= .9480449
cos {3 = .8290651
cos {3
cos a
- sin
sin {3 cos C = .0356271
a
cos 'Y
=
sin {3 =
- cos C =
sin {3 cos C =
sin
cos a = .8744998
- sin
a
a
.4850258
.3181413
.2308848
.0356271
= .7934380
'Y = 37 29' 31
a
=
0
29 00' 51"
{3
= 180 33' 02"
'Y = 37 0 29' 32"
Solving for angles A and B
Eq. (11)
cos {3 cos 'Y - cos a
cos A = - - - - - - sin f3 sin 'Y
Eq. (12)
cos a cos 'Y - cos f3
cos B = - - - - - - sin a sin 'Y
476
PHOTOGRAMMETRIC ENGINEERING
.7522144
= .8744998
cos 'Y = .7934389
cos a cos 'Y = .6934389
.8744998
- cos {3 = .9480433
.1222854
cos a cos 'Y - cos {3 = .2542211
.3181413
sin a = .4850258
cos {3 = .9480433
=
. cos {3 cos 'Y =
- cos a =
cos 'Y - cos a =
sin {3 =
sin'Y =
sin {3 sin 'Y =
cos 'Y
cos {3
cos a
.7934389
=
sin a sin 'Y =
cos B =
B =
.6086537
sin'Y
.1936379
cos A = .6315158
A
= 50° 50' 17"
.6086537
.2952127
.8611455
30° 33' 16"
cos {3 cos a - cos 'Y
cos C = - - - - - - sin {3 sin a
and
=
cos a =
cos {3 cos a =
- cos 'Y =
cos a - cos 'Y =
cos {3
cos {3
.9480433
sin {3
= .3181413
.8744Q98
sin a
= .4850258
.8290637
sin {3 sin a = .1543067
.7934389
cos C = .2308701
180° - C = 76° 39' 06"
.0356248
In right spherical triangle AA'B'
cosAB' = cos VA-cosA'B'
Eq. (13)
= cotF·tanA'B'
cotF = sin VA-cotA'B'
sin VA
Eq. (14)
VB = 35° 02' 29"
VA = 14° 04' 9"
A'B'
sin VA
cot A'B'
=
34° 34' 55"
= .2430930
=
=
cos A'B' =
cos.A B' =
AB' =
cos VA
1.4505600
= .3526210
F = 70° 54' 35"
cot F
Eq. (15)
cos E
cos AB' cos 'Y - cos· VB
= --------sin AB' s n 'Y
GOS
AB' = .7986182
cOS'Y
= .7934389
cos A B' cos 'Y = .6336547
= .8187375
cos VB = .1850828
- cos VB
cos AB' cos 'Y -
.9700030
.8233152
.7986182
37°00' 90"
DETERMINING DEPRESSION ANGLE AND SWING
477
sin AB' = .6018382
sin'Y = .6086537
sin AB' sin 'Y = .3663110
cos E = .5052614
E
= 59° 39' 05"
It is readily seen that angle x=(F+E)-A
E
= 59° 39' 05"
F =
+
F
70° 34' 35"
E = 130° 13' 40"
- A =
x
50° 50' 17"
= 79° 23' 23"
(16) Then cos A'C=cos VA ·cos l1+sin VA sin 11 cos x
cos V A = .9700030
sin VA = .2430930
cos {3 = .9480433
sin {3 = .3181413
cos VA cos {3 = .9196048
cos x = .1841277
+ sin V A sin {3 cos x = .0142400
sin V A sin {3 cos x = .0142400
cos A'C = .9338448
A'C = 20° 57' 28"
A' is a 90° angle, therefore G' = 90° - G
sin x sin {3
sinG=---sin A'C
Eq. (18)
90°
sin x = .9828970
=
89° 59' 60"
sin {3 = .3181413
G = 60° 57' 21"
sin x sin (3 = .3127001
G' = 29° 02' 39"
+ sin
A'e = .3576799
sin G = .. 8742457
G = 60° 57' 21"
sin
Eq. (19)
f)
= sin (;' sin /J 'e
sin G' = .4854413
sin
A'e = .3576799
sin
f)
= .1736326
9° 59' 59"
H = tan f)·cot A'e
Eq. (20)
cos
Eq. (21)
sin VA sin G
sin H' = - - - - sin {3
(1)
478
PHOTOGRAMMETRIC ENGINEERING
.1763220
sin VA = .2430930
cos A'e = 2.6108381
sin G = .8742457
tan 0 =
•
cos H
=
sin VA sinG = .2125230
.4603482
+ sin {3 = .3181413
H = 62° 35' 26"
sin H' = .6680145
H'
H
=
=
41° 54' 50"
62° 35' 26"
H' = 41° 54' 50"
m =
Eq. (22)
(H
85° 29' 46"
+ H' + m)
= 190° 00' 02" _
180° = 180°
S =
(2)
10° 00' 02"
Having computed the angle of depression 0, and swing S, we proceed to
compute the new coordinates x' and y' with reference to the same origin, but
with axes that coincide with the principal line along the y axis and the principal
parallel along the x axis when rotated through an angle of 10°.
Eq. (23)
x' = x cos s - y sin s
Eq. (24)
Y' = x sin s
XA
YA
+ Y cos s
= - 50.985
= - 4.016
Angle S = 10°
= - 50.985
.9848078
cos S =
cos S = - 50.21843
cos S =
.9848078
sin S =
.1736482
YA
XA
-
XA
-
XA
YA
=-
sin S =
XA
sin S = -
sinS =
.1736482
sin'S =
.69737
cos S = - 50.21043
YA sin
XA
= - 4.016
S
=
.69737
XA'
=
49.51306
50.985
-
YA
cos S
.1736482
8.85345
-
YA
cos S
XA
sin S =
8.85345
YA
cos S = -
3.95499
YA'
= -
12.80844
XA'
= -
49.51306
YA'
= - 12.80844
= - 4.016
=
.9848078
= - 3.95499
479
DETERMINING DEPRESSION ANGLE AND SWING
XB
=
+ 25.914
YB = - 80.456
=
cos S =
XB cos S =
XB
+ 25.914
YB = - 80.456
Y sin S =
25.52031
XB
.1736482
sin S =
.9848078
+
13.97104
cos S = 25.52031
YB sin S = 13.97104
XB'
sin S
= 25.914
= .1736482
sin S
=
XB
XB
= 39.49135
YB = - 80.456
cos S =
.9848078
- YB cos S = - 79.23370
4.49992
XB
sin S =
4.49992
YB cos S = - 79.23370
YB' = - 74.73378
= + 39.49135
YB' = - 74.73378
= - 49.51306
YA':= - 12.80844
XA'
XB'
With the new coordinates the horizontal and vertical angles can now be computed. To further demonstrate the validity of the method employed, the horizontal and vertical angles of the actual field angles, used in the computation of
the angle of depression (J, and angle of swing S, will be computed by equations
(25) and (26).
x'
z'
y'
A
- 49.51306
- 12.80844
00.00
B
+ 39.49135
- 74.73378
00.00
C
00.00
00.00
152.4003
x'
y'
y'
tan'Y = -
A
-.49.51306
- 12.80844
.0840447
B
+ 39.49135
- 74.73378
.4903782
'Y
f
40 48' 15"
260
x'
7' 20"
cf>=(J±'Y
140 48' 15"
360
cos 'Y
tan 0 = ta.n (3._cos cf>
cos 'Y
cos cf>
tan (3 = -
A
.9964868
.9668058
.3248882
.3348623
B
.8978569
.8077613
.2591291
.2880317
f
OA = 180 30' 50"
OB = 160 4' 5"
o=
340 34' 55"
7' 20"
480
PHOTOGRAMMETRIC ENGINEERING
tan V
0
cos 0
tan 4>
A
18 30' SO"
.9482467
.2642892
.2506114
B
16 0
.9609202
.7298068 .
.7012861
0
4' 5"
VA
=
0
14 4,9"
VB
=
=
cos 0 . tan 4>
0
35 2' 29"
The U. S. Naval Photographic Interpretation Center plans to conduct field
tests in the near future to determine the most practical equipment and methods
of terrestrial photogrammetry for exploratory surveys.
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