When we evaluate these two integrals, the first over the rectangle of Figure 

 27-12a and the second over the rectangle of Figure 27-12b we get 





Using the result already found for H z (equation 46 and sequel) , we find that 

 to compute H s we must sum such terms as 



^3 ^3 \ / 



(2Ina + 2Ily) I —r-l—r ~ ~ = "- 

 for horizontal rectangles and such terms as 



(2Iny - 2Ila) (—**— - - 



9 IJ \ ( Z l Z 5 \ 



for vertical rectangles. 



z 

 The contour map of — shown in Figure 27-13 is used as in the 



x* + >z* 



case where H z was computed for the infinite horizontal cylinder. 



Field Component of Vertical Cylinder Measured by Airborne Magnetometer 



As before let (a,/?,y) be the direction cosines of a line in the direction 

 of which the field H s is to be computed. The z axis is taken vertically downward 

 as usual but the x and y axes are turned so that the y axis is perpendicular to 

 the direction of H s . The direction cosine (3 is therefore zero and H s may be 

 computed from 



H s = aH x + yH z 



as in the case of the horizontal cylinder. For the special case in which a = /, 

 /? = m, y = n, that is, the case where the magnetization / is in the direction of 

 the earth's field 



H s = IH X + nH z 



To find H x for the prism having the cross section shown in Figure 27-7 

 we start with the potential 



V (x, y t z) = l(C f°° \l -5-(— ) + n -$— (— ) dudvdw (56) 



J J J h du r c)w r ] 



which becomes, after carrying out the integration with respect to w, 



609 



