12/25/2008 Nguy n Quang Minh ϕ1 (L1 , L2 , , Ln ) = 0 ϕ (L1 , L2 , , Ln ) = 0    ϕ r (L1 , L2 , , Ln ) = 0  12/25/2008 + + = 180o + + + 11 + 14 + 17 = 360 o sin 1sin sin sin 10 sin 13 sin 16 =1 sin sin sin sin 12 sin 15 sin 18 Phương trình u ki n M i tr đo th a s cho m t phương trình u ki n, r tr đo th a ->r phương trình ñi u ki n 12/25/2008 Dùng khai tri n Taylor, gi ñ n s h ng b c nh t: Áp d ng cho phương trình u ki n, s có: đo ϕ1 (L1, L2 , , Ln ) = ϕ1 (L1 , Lño , , Lño ) + n ∂ϕ1 ∂ϕ v1 + v2 + = ∂L1 ∂L2 ño ϕi (L1, L2 , , Ln ) = ϕi (L1 , Lño , , Lño ) + n ∂ϕi ∂ϕ v1 + i v2 + = ∂L1 ∂L2 ∂ϕ1 ∂ϕ ∂ϕ ño = a1 ; = a2 , , = an ; ϕ1 L1 , Lño , , Lño = ωa n ∂L1 ∂L2 ∂Ln ( ) ∂ϕ ∂ϕ ∂ϕ ño = b1 ; = b2 , , = bn ; ϕ L1 , Lño , , Lño = ωb n ∂L1 ∂L2 ∂Ln ( ) ∂ϕ ∂ϕ ∂ϕ r ño = r1 ; r = r2 , , r = rn ; ϕ r L1 , Lño , , Lño = ωr n ∂Ln ∂L2 ∂L1 ( ) 12/25/2008 a1v1 + a2 v2 + + an + ωa = b1v1 + b2 v2 + + bn + ωb = r1v1 + r2 v2 + + rn + ωr = an   v1  ω1  bn  v2  ω2    +   =          r3  vn  ωr  BV + W = a1 a2 b b     r1 r2 12/25/2008  a1 b     r1 a2 b2 r2 a n   v1  bn  v2    +      r3   v n  =  [av ] + ω a  [bv ] + ω =  b ⇒    [rv ] + ω r =  ⇒ BV + W = ω1  ω   2 =     ω r  Φ = [ pvv] = VT PV = VT PV− 2Ka ([av] +ωa ) − 2Kb ([bv] +ωb ) − − 2Kr ([rv] +ωr ) = VT PV− 2KT (BV+ W) ∂Φ = = 2VTP − 2KTB ⇒ VTP = KTB ∂v VT = KTBP−1 Ka  K  K =  b     Kr  V = P−1BTK ⇒ BV+ W = BP−1BTK + W = NK+ W = 0; Nr×r = BP−1BT 12/25/2008  p1 0 P= 0  0 p2 0 0 0 0  0  pn  NK + W = BP−1BTK + W = a1 b 1    r1 a2 an 1/ p1   1/ p b2 bn    r2 r3  0  aa    p   ab    p     ar   p     ab  p    bb  p    br   p    a1 a2    1/ pn an 0 b1 r1 Ka  ωa  b2 r2 Kb  ωb    +   =         bn rn Kr  ωr   ar     p Ka  ωa   br      Kb  + ωb  =  p          rr   Kr  ωr      p  12/25/2008 r = n −t t Tr ño c n thi t n T ng s tr ño h1 h2 h1 + h2 + h3 = ño ño h1ño + v1 + h2 + v2 + h3 + v3 = h3 ño ño v1 + v2 + v3 + (h1ño + h2 + h3 ) = v1 + v2 + v3 +ω = h2 h1 h3 B A HA + h1 + h2 + h3 − HB = ño ño HA + h1ño + v1 + h2 + v2 + h3 + v3 − HB = ño ño v1 + v2 + v3 + (h1ño + h2 + h3 + HA − HB ) = v1 + v2 + v3 +ω = 12/25/2008 Tên Tr ño Chi u dài 1.005 2.007 -3.010 1.503 1.504 h1 h2 h3 h4 h5 Tên Tr ño 1.005 2.007 -3.010 1.503 1.504 1 h1 Chi u dài (km) h2 A h3 h4 h5 12/25/2008 r = 5−3 = h1 h2 h3 v1 + v2 + v3 + (1.005+ 2.007−3.010 = ) v1 + v2 + v3 + 0.002= r = 5−3 = h1 h2 h3 v3 + v4 + v5 + (−3.010+1.503+1.504) = v3 + v4 + v5 − 0.003= 12/25/2008 BV + W = v1  v   2 1 1 0  0.002  B= ;V = v3 ; W = − 0.003   0 1 1   v4  v5    1 0  P = 0  0 0  0 0 0.5 0 0 0 0  0  0 1  0 10 12/25/2008 NK+ W = BP−1 BTK + W = 1 0 1 1 0 0 1 10   0 0  0 0 0 0 01 01  01  00 10  0 0  K   0.002  1  +   =0 K2  − 0.003 1 1  −1  K1  4 2  0.002  4 2 K1   0.002  2 4K  + − 0.003 = ⇒ K  = −2 4 − 0.003          2  −1  K  4 2  0.002  - 0.001167  ⇒   = −  − 0.003 =  0.00133     K2  2 4   - 0.001167 1 0 01 0  - 0.001167 0 0 01 0  - 0.001167     −1 T  V = P B K = 0 01 1  =  0.000326   0.00133     0.00133 0 0 00 1  0.00133 0 0 10 1      11 ... sin 18 Phương trình u ki n M i tr ño th a s cho m t phương trình u ki n, r tr ño th a -> r phương trình ñi u ki n 12/25/2008 Dùng khai tri n Taylor, gi ñ n s h ng b c nh t: Áp d ng cho phương. .. 2  −1  K  4 2  0.002  ? ?- 0.001167  ⇒   = −  − 0.003 =  0.00133     K2  2 4   ? ?- 0.001167 1 0 01 0  ? ?- 0.001167 0 0 01 0  ? ?- 0.001167     −1 T  V = P... HB ) = v1 + v2 + v3 +ω = 12/25/2008 Tên Tr ño Chi u dài 1.005 2.007 -3 .010 1.503 1.504 h1 h2 h3 h4 h5 Tên Tr ño 1.005 2.007 -3 .010 1.503 1.504 1 h1 Chi u dài (km) h2 A h3 h4 h5 12/25/2008 r =