by James Thomas Lee, Jr. 12/25/97 Copyrighted 1995 by James Thomas Lee, Jr. Copyright Number: XXx xxx-xxx
The technique used was multiple linear regression, where the dependent variable, latitude of the estimated point (yi), was estimated as a function of the longitude of the estimated point (x1i), the time of the point (x2i), and confidence radius of the point (x3i). The equation to be solved was as follows:
f = SUM (y(i) - m1x1(i) - m2x2(i) - m3x3(i) - b)**2
Taking the partial with respect to m1, m2, m3, and b yields the following expressions:
Partial(f)
1. ---------- = 2 SUM (y(i) - m1x1(i) - m2x2(i) - m3x3(i) - b) (-x1(i)) = 0
Partial(m1)
which becomes,
SUM(x1(i)y(i)) = m1SUM(x1(i)**2) + m2SUM(x1(i)x2(i)) + m3SUM(x1(i)x3(i)) + bSUM(x1(i))
-----------------------------------------------------------------------------------------------
Partial(f)
2. ---------- = 2 SUM (y(i) - m1x1(i) - m2x2(i) - m3x3(i) - b) (-x2(i)) = 0
Partial(m2)
which becomes,
SUM(x2(i)y(i)) = m1SUM(x1(i)x2(i)) + m2SUM(x2(i)**2) + m3SUM(x2(i)x3(i)) + bSUM(x2(i))
-----------------------------------------------------------------------------------------------
Partial(f)
3. ---------- = 2 SUM (y(i) - m1x1(i) - m2x2(i) - m3x3(i) - b) (-x3(i)) = 0
Partial(m3)
which becomes,
SUM(x3(i)y(i)) = m1SUM(x1(i)x3(i)) + m2SUM(x2(i)x3(i)) + m3SUM(x3(i)**2) + bSUM(x3(i))
-----------------------------------------------------------------------------------------------
Partial(f)
4. ---------- = 2 SUM (y(i) - m1x1(i) - m2x2(i) - m3x3(i) - b) (-1) = 0
Partial(b)
which becomes,
SUM(y(i)) = m1SUM(x1(i)) + m2SUM(x2(i)) + m3SUM(x3(i)) + nb
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Solving the 4x4 matrix required the following special lettered notation:
A = SUM(x1(i)**2)
B = SUM(x1(i)x2(i))
C = SUM(x1(i)x3(i))
D = SUM(x1(i))
E = SUM(x2(i)**2)
F = SUM(x2(i)x3(i))
G = SUM(x2(i))
H = SUM(x3(i)**2)
J = SUM(x3(i))
K = n
L = SUM(x1(i)y(i))
M = SUM(x2(i)y(i))
N = SUM(x3(i)y(i))
P = SUM(y(i))
| | | | | |
| A B C D | | m1 | | L |
| | | | | |
| B E F G | | m2 | | M |
| | | | | |
| C F H J | | m3 | = | N |
| | | | | |
| D G J K | | b | | P |
| | | | | |
denominator (den) = A[E(HK-J**2) - F(FK-GJ) + G(FJ-GH)]
-B[B(HK-J**2) - F(CK-DJ) + G(CJ-DH)]
+C[B(FK-GJ) - E(CK-DJ) + G(CG-DF)]
-D[B(FJ-GH) - E(CJ-DH) + F(CG-DF)]
-----------------------------------------------------------------------------------------------
m1 = (LE-BM)(HK-J**2) + (CM-FL)(FK-GJ) + (GL-DM)(FJ-GH) +
(JN-HP)(DE-BG) + (GN-FP)(CG-DF) + (BF-CE)(KN-JP)
-------------------------------------------------------
den
m2 = (AM-BL)(HK-J**2) + (FL-CM)(CK-DJ) + (DM-GL)(CJ-DH) +
(HP-JN)(BD-AG) + (KN-JP)(BC-AF) + (CG-DF)(CP-DN)
-------------------------------------------------------
den
m3 = (BL-AM)(FK-GJ) + (BM-EL)(CK-DJ) + (GL-DM)(CG-DF) +
(FP-GN)(AG-BD) + (KN-JP)(AE-B**2) + (CP-DN)(DE-BG)
-------------------------------------------------------
den
m4 = (AM-BL)(FJ-GH) + (EL-BM)(CJ-DH) + (CM-FL)(CG-DF) +
(HP-JN)(AE-B**2) + (GN-FP)(AF-BC) + (CE-BF)(DN-CP)
-------------------------------------------------------
den
Appendix H. SDC Letter of Appreciation
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