D5 = -2b2(nx s + nysy)
D6 = nx22 + ny2c2 + aS2 ay2c2 + px2 + py2 +
2nxaxscc + 2nyay sc 2pxnxc 2pynyc 2axscpx -
2ayScPy a2
E1 = D1 D3 + D6
E2 = 2(D4 D5)
E3 = -2(D1 -2D2 Dg)
E4 = 2(D4 + D5)
E5 = D1 + D3 + Dg
It is seen that there is a maximum of four solutions of
X (or Bc) in Eq. (2.25). Back substituting the values of Sc
into Eq. (2.23) and (2.24) yields up to four sets of
solutions for ea and da-
From Eq. (2.21), we can find certain relationships by
equating corresponding elements of the two matrices on
either side of the equation. Thus we obtain the following
equations:
-axS8a + ayCea
8bI = tan-l( ----a-Sa + (2.26)
axC8a + aySea
eb2 = tan-l( aC + ) (2.27)
-azCbbl
nzSOc + SzC~c
9b3 = tan-1( n--Sc + ) (2.28)
szS9c + n+zCc