i = INT | (t-ty)/dt ] * 1 
where INT is the integer truncation function 
(2) Compute the value of the independent variable at row 
i: 
t(i) = tg + (1-1)*dt 
(3) Compute the Lagrange interpolation coefficients: 
h » (t-t(1))/dt 
a4 - -0.5(h-n2) 
a9 = 7284 th 
a, = aj - h + 1 
(4) Apply the interpolation coefficients: 
v(t) = v(i)a, + v(i+l)a, tv (1*2)a4 
(5) Note: for values of the independent variable below or 
above the table, the Lagrange coefficients are 
replaced by terms which force a linear extrapolation: 
If i is less than 1 ("below" the table): 
set 1 =1 
23 = 1.0 - h 
a» h 
a, = 0.0 
® 
m NS 
"ow 
5 —o 
oo 
> 
In actual analysis, the time spent in the Lagrange 
Interpolation compared to the linear interpolation is insignificant 
compared to the computation of the trigonometric functions to fill the 
orientation matrix of the instantaneous framelet, so the overall impact 
upon computation time is negligible. The impact upon the time necessary to 
compute the tables is great, on the other hand, because only about one 
third or fewer of the points must be computed using the rigorous 
equations. Thus there is a net savings in preprocessing time, with no 
impact upon the real-time computations, yielding an overall net savings. 
SOFTWARE IMPLEMENTATION 
Photogrammetric functions required in most exploitation systems 
consist of space intersections and inverse problems. Space intersection is 
defined as either the intersection of two rays, one from each image of a 
stereo pair, in order to determine ground point horizontal and vertical 
position; or the intersection of a ray from one image with a surface 
represented by digital terrain elevation data (DTED) (or by a plane) which 
576 
m Ct Cu C 
PF?