ENGINEERING RESEARCH INSTITUTE UNIVERSITY OF MICHIGAN ANN ARBOR Preliminary Report ANALYTIC CONTROL EXTENSION UTILIZING A SINGLE STRIP OF OVERLAPPING PHOTOGRAPHS Eldon Schmidt Approved By Edward Young Project 1699-1 DEPARTMENT OF THE AIR FORCE WRIGHT AIR DEVELOPMENT CENTER WRIGHT-PATTERSON AIR FORCE BASE, OHIO CONTRACT NO. W33(038)ac-15318 PROJECT NO. 54-650A-460 June 1955

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN ABSTRACT This report is an outline of a possible method of control extension utilizing a single strip of overlapping photographs. This method if proved useful would hold manual operations to a minimum at the expense of computational complexity. From a practical point of view it could be useful only with automatic computation. The report only introduces the method in theory and does not represent a full investigation of this method. OBJECTIVE The objective of this project is to determine methods of photogrammetric reductions for aerial defense purposes. ~~~___~~______________ ii _____........

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN The problem of finding photogrammetric solutions in uncontrolled areas is one of the basic problems of photogrammetry. Unfortunately there is no direct solution to this problem. As is well known to photogrammetrists at least three control points are necessary for the solution of a photographic frame. This idea is tied in with the fact that it is necessary to know at least three points in a rigid body to completely determine its position. It is possible to find the ground control of an unknown point if that point appears on at least two frames that otherwise contain the necessary known control. By successive application of this idea it is possible, by using strips of overlapping frames, to find the ground control throughout the strip if that strip starts over known control. In this way a partial solution to the basic problem is achieved. Thus a completely unknown frame can be evaluated if it can be tied to known control by means of a strip of overlapping frames. This is the idea of control extension. Such a procedure is illustrated in Figure 1. Referring to this figure, points 1, 2, 6, 7, 11, and 12 are to be regarded as known control at.1.2.3.4.5 I III V VIi.6 1 7 1 fi__s *o10 II IV VI VIII.11 12'13 4 15 Figure 1 ____________________________ 1 ________________________

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN the start of the strip. The remaining points represent unknown control and are called pass points. From the known control frames I and II can be reduced. From the reduced data it is possible to determine pass point 8 as that point appears on both frame I and II. Now since 2, 7, 8 and 7, 12, 8 are known, frames III and IV can reduced. But now pass point 3 can be found from frames I and III, and pass point 13 from frames II and IV. Pass point 9 can now be found from frames III and IV, and therefore frames V and VI can be reduced determining pass points 4 and 14, etc. By this means the pass points throughout the strip can be determined. After the points have been determined it is possible to adjust their values, somewhat as each point appears on at least three frames which is an overdetermination. From a practical point of view this type of strip is not very useful. As can be seen it either requires two passes by one aircraft or a parallel flight of two aircraft. To obtain the desired overlaps would require both very accurate flying and very careful synchronization of the camera shutter mechanisms. Assuming that such a strip has been flown, there is still great difficulty in finding pass points. The reason for this can be seen by taking pass point 8 as an example. It is noted that it must appear in the corner of frames I, II, V, and VI. This imposes the necessity of finding this point in a very restricted area. In general over unknown terrain the selection of a pass point in a restricted area is a difficult problem and often proves impossible. Since failure to find a point would cut off the strip, this situation makes the use of this type of strip impractical, except over special types of terrain where the distribution of possible pass points is very dense. A less demanding type of extension can be made with the use of a strip as illustrated in Figure 2. Here points 1, 2, 3, and 4 are known and the rest are pass points. Now the physical assumption is made: that from.1'3 5'7'3 I I III IV *2 ~4 F6 *8 2 Figure 2 _________________________________ 2 _________________________________

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN one frame to the next the aircraft does not make any large changes in angular altitude. In fact it is assumed that no large error will be made if the tilt of one frame is taken to be the same as the next one. In this way a provisional determination of points 5 and 6 is obtained by letting the tilt of frame II be the same as the tilt of frame I, which is known. By using the provisional values for points 5 and 6 frame II can then be reduced, deriving better values for points 5 and 6, It can be shown that if the assumed tilt does not differ by more than a few minutes from the correct value this procedure will converge to correct values. In this way it is possible to reduce frame II and find pass points 5 and 6. This idea can then be extended to frame III, etc. This type of strip is more practical as it requires only one flight by one aircraft. In addition the areas for selection of passpoints are less restricted as each point is required to appear in the corner of only two frames. The purpose of this report is to illustrate the latter procedure from an analytical viewpoint. Analytical procedures for this purpose have already been developed (reference 1). The procedure to be given here differs from those of other investigators and should be regarded as preliminary, The investigation was not completed during the writing of this report; therefore, this material will merely be presented without any attempt to evaluate its usefulness in comparison with other methods. As this procedure involves a great deal of computation, it can be said that it is completely out of the range of hand computation and is given with the understanding that automatic computation will be used. The procedure is as follows. Before proceeding the following notation is introduced: x and y refer to photographic coordinates, X and Y refer to horizontal ground control, H is the altitude of the camera over the datum plane, and h the relief from the datum plane. Now from frame I with the data from known points 1, 2, 3, and 4 the frame is reduced. This gives the photographic nadir point (xn, yn) and the space coordinates (XN, YN, H). The azimuth determination is not needed here. From this data the frame is rectified by the following: x ffy' + fxE, y XnY(f2 - ff') x. + Yen xa +n (1) x' =n and (1) xnx + YnY + f2 5

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN ff'xn + f2y2 xn(f2 - f ) y - -- n + x nn 2 ynf2 + 22n y, = x...Yn.. (2) Xnx + yny + f2 where f = (x2 + yR + f2)1/2 and f is the focal length. Also (x', y') are the rectified coordinates of the point (x, y). Next the known control points are projected into the datum plane by X' - (X - XN) and (3) H-h yt = H (Y - YN) (4) These equations are well known from geometry. The (x', y') system is now equivalent to the (X', Y' ) system, except for scale and rotation. We therefore have the equivalent relation X' = Mx' + Ny' and (5) Yt = My - Nx'. (6) In order to mean out the photographic errors as much as possible Equations (5) and (6) are solved by least squares for M and N. It should be noted that one set of such equations is obtained from each point. In this case four sets or eight equations are obtained. Now since (xg, yg) and (x4, y&) are known on frame I, (X0, Yg) and (X6, Y6) and also the equation of the line through (XN, YN) and (X0, Y6) can be found. This is X - XN X ( x=. ^ - M(7 ) Y - YN Y6 Similarly, point 6 is obtained. It is important to notice that this line, Equation (7), also passes _______________________________ 14 ___________________________ —-_

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN through (X5, Y5). The reason for this is that the relief displacement in the datum plane is radial to the ground nadir point. We now proceed to frame II. First, the same photographic nadir point (same tilt) as in frame I is assumed. With the control data for known points 5 and 4 we can solve for XN, YN and H. Then Equations (5) and (6) can be solved as before using only the data from points 3 and 4. This permits setting up Equation (7) for points 5 and 6 as before. By intersecting these lines with the equivalent lines from frame I, values for pass points 5 and 6 are obtained. Using this data, frame II is reduced and the procedure repeated to find new values for pass points 5 and 6. This iterative operation is repeated until the values for 5 and 6 are unchanged. Thus frame II is reduced and pass points 5 and 6 determined, solving the problem. Frame III is then treated in the same way finding pass points 7 and 8, etc. We are best able to illustrate the foregoing by the use of a numerical example. In order to show that the method is accurate with respect to method fictitious data will be used. For comparison, the correct values will be given first. Frame I Frame II Pass Points (X, Y, h) Xn =3 Xn = 3.2 Yn = 5 Yn = 4.8 (15,000;15,000;150) XN = 0 XN = 15,000 YN = 0 YN = 0 (15,000;-15,000;-200) H = 20,000 H = 20,100 The initial data are photographic coordinates f = 150. FRAME I Point x _____ 1 -111.2842256 119.3248501 2 -104.5369838 -102.5751844 3 3.0387243 121.2812816 4 2.9637106 -103.9697621 5 121.2624572 123.3044678 6 113.3627081 -105 4019386 _______~______________________ 5 —---------

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN FRAME II Point x y 3 -110.4213621 118.4536728 4 -103.7012506 -102.1516296 5 3.2394599 120.5070268 6 3.1630277 -103.6126954 GROUND CONTROL Point X Y h 1 -15,000 15,000 100 2 -15,000 -15,000 -150 5 0 15,000 125 4 0 -15,000 -175 First frame I is reduced by standard methods getting n = 3 Yn = 5 XN = 0 YN = 0 H = 20,000 Equations (1) and (2) become x' = 22,512.49528 x - 7*49717 Y - 67,500 and 3 x + 5 y + 22,500 y, = 22,504.49850 y - 7.49717 x - 112,500 3 x + 5 y + 22,500 The rectified coordinates for frame I become Point x' Y 1 -113.0653266 113.0655266 2 -111.6625310 -111.6625510 5 0 113.2075472 4 0 -111.5241636 5 1135.53501259 113.3501259 6 111.5861586 -111.5861586 6

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN Equations (3) and (4) give Point X' Y' 1 -15,075.37689 15,075.37689 2 -14,888.33747 -14,888.33747 3 0 15,094.33962 4 0 -14,869.88847 Since we have exact data we will not go to least square solution in Equations (5) and (6) and will merely solve with respect to point 1. We get -15,075.37689 = - 113.0653266 (M-N) and 15,075.37689 113.0653266 (M-N) Solving this we get M = 13355.5555533333333 and N = 0. This gives for points 5 and 6 Point X' Y' 5 15,113,35012 15,113.35012 6 14,851.48514 -14,851.48514 Equation (7) becomes point 5, X = Y and point 6, X = -Y. We now go to frame II. Since the same tilt is assumed, Equations (1) and (2) are the same as in frame I. We get at once Point x__' Y' 3 -112.2203761 112.2124619 4 -110.7704512 -111.1780214 5 0.1957777 112.4695994 6 0.2039180 -111.1467480 Solving for XN, YN and H we get _______________________________________ 7

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN XN 14,972.89747, YN = 57.67567, and H = 20,119.63835. Now from Equations (5) and (4) we get Point X' Y' 3 -15,066.50317 15,035.73890 4 -14,843.78667 -14,927.83383 Equations (5) and (6) now become Point 3 - 15,066.50317 = - 112.2203761M + 112.2124619N and 15,035.73890 = 112.2124619M + 112.2203761N; Point 4 - 14,843.78667 = - 110.7704512M - 111.1780214N and -14,927.83383 = - 111.1780214M + 110.7704512N. Solving by least squares we get M = 134.1317419 and N = -0.1323518. We get for points 5 and 6 Point X' Y' 5 11.37445 15,085.76919 6 42.06235 -14,908.27993 Equation (7) then becomes Point 5: 15,085.76919X - 11.37445Y = 225,877,019.3 and Point 6: 14,908.27993X + 42.06235Y = 223,222,572.8. Intersecting these lines with the equivalent lines from frame I we get Point X Y 5 14,984.15181 14,984.15181 6 15,015.42484 -15,015.42484 _____________________________________ 8 _______

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN Returning to Equations (3) and (4) we can now solve for h using frame I data. We get h5 = 170.97243 and he = -220.772135 These data should now be used to reduce frame II and repeat the process. As this is much too lengthy to give here we will state that the results will converge to the exact values. In order to indicate this convergence Figure 3 shows the values of X5 through the iteration. As can be seen from Figure 3 the closure is very slow. Due to the extreme length of the computation only one practical problem has been worked. This problem was selected from a flight over a controlled area. In this way by selecting known points as "pass points," an accuracy check could be obtained under actual conditions. This flight was at approximately 10,000 feet and the errors below are proportional to this altitude or double at 20,000 feet and half at 5,000 feet, etc. The results shoved a mean discrepancy in horizontal control of the pass points of 0.88 foot. The mean error in relief was 2.94 feet. A very rougt estimate of error under these conditions would be for the ninth frame (n-l)1/2 0.88 foot in horizontal and (n-1)l/2 2.94 feet in relief. This is merely applying the law of error propagation. In brief conclusion it may be stated that this method of control extension is exact in theory and does not show significant errors when normal photogrammetric errors are introduced. The practical usefulness of this method has not been established and will have to await further investigation. There are several obvious lines of investigation that should be tried. First, however, the method should be placed on an automatic computer. Some of the topics for investigation could be: 1. Find if possible any short cuts in the convergence process. It may be possible to find differential corrections that will greatly shorten the convergence. 2. Find how much the tilt assumption on frame II can be in error without causing serious difficulties. 35 Check on the error propagation along the strip. 4. Check as long a strip as possible over a controlled area to see how well the final points check. ________________________ 9 ________________________

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ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN 5. Find out how much the amount of relief effects the errors in pass points. 6. See if any practical method of adjustment is possible due to the overdetermination caused by each point appearing on three frames. BIBLIOGRAPHY 1. Church, E., "Two New Analytical Methods of Space Resection in Aerial Photogrammetry", Syracuse University, Syracuse, New York, April, 1941. 2. Schmidt, E., "Some Methods in Photogrammetric Reduction for Exterior Orientation", Univ, of Mich., Ann Arbor, Project 1699, October, 1954. 3. Rowley, J. C., and Schmidt, E., "A Rectification Transformation for Tilted Photographs', Univ. of Mich., Ann Arbor, Project 1699, 1951.