12/25/2008 Nguy n Quang Minh 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 NK + W = BP −1 B T K + W =  aa   ab    K a +   K b  p  p    ab   bb   K a +  Kb  p   p     ar  K a +  br  K b  p  p       ar  + +   K r + ω a =  p  br  + +   K r + ω b =  p  rr  + +   K r + ω r =  p NK + W = BP −1B T K + W = [taa ]K a + [tab]K b + + [tar ]K r + ωa = [taa ]K + [tbb]K + + [tbr ]K + ω =  a b r b    [taa ]K a + [tbr ]K b + + [trr ]K r + ωr =  [taa ]K a + [tab]K b + [tac ]K c + ωa =  [taa ]K a + [tbb]K b + [tbc ]K c + ωb =  [taa ]K + [tbc ]K + [tcc ]K + ω = a b c c  Ka = − ([tab]K b + [tac ]K c + ωa ) [taa ] 12/25/2008 [taa ]K + [tab]K + [tac ]K + ω =  [tab ]K + [tbb]K + [tbc ]K + ω =  [tac ]K + [tbc]K + [tcc ]K + ω =  a b c a a b c b a b c c ([tab]K + [tac ]K + ω ) [taa]  [tab] [tac]     [tbb ] − [taa ] [tab] K +  [tbc] − [taa ] [tab] K      [tab] [tac ] K +  [tcc] − [tab] [tac] K    [tbc ] −     [taa ]  [taa ]    [tbb.1]K + [tbc.1]K + ω =  [tbc.1]K + [tcc.1]K + ω = K =− a b c a c b c b K =− b c c b a c b b ω  [tab] = + ω −  [taa ]   ω  [tac ] = + ω −  [taa]   a b c 1 ([tbc.1]K + ω [tbb.1] c b ) [tbc.1] [tbc.1] K +  ω − ω [tbc.1] =   [tcc.1] −    [tbb.1] [tbb.1]     b c [tcc.2]K + ω c Ký hi u dòng a E1 ( c =0⇒K =− c a Ka c ω c [tcc.2] b Ka c Kc -1 Ka b E1b×a b.1 -1 E2 ( Kb c E1c×a E2c×b.1 c.2 E3 -1 Ka Kb Kc Kc 12/25/2008 Ký hi u dòng a Ka b Ka c Kc a -1 E1 ( Ka b K =− a Ký hi u dòng a Ka [taa ] ([tab]K + [tac ]K + ω ) b c b Ka a c Kc a -1 E1 ( Ka b E1b×a b.1 [taa ]K + [tab]K + [tac ]K + ω [tab]K + [tbb]K + [tbc ]K + ω [tac]K + [tbc ]K + [tcc]K + ω a b c a a b c b a b c c =0 =0 =0 ([tab]K + [tac]K + ω K =− [taa ] a b c a ) [tab] [tab] K +  [tbc] − [tac] [tab] K +  ω − ω [tab] =   [tbb ] −      [taa ] [taa ]  [taa ]      [tbb.1]K + [tbc.1]K + ω = a b b c c b b 12/25/2008 Ký hi u dòng a Ka b c Ka Kc a -1 E1 ( Ka b E1b×a b.1 -1 E2 ( Kb c E1c×a E2c×b.1 c.2 E3 -1 Ka Kb Kc Kc [taa ]K + [tab]K + [tac ]K + ω =  [tab ]K + [tbb]K + [tbc ]K + ω =  [tac ]K + [tbc]K + [tcc ]K + ω =  ([tab]K + [tac ]K + ω ) K =− [taa] a b c a a b c b a b c c a b c a  ω [tab] [tac]       [tbb ] − [taa ] [tab] K +  [tbc] − [taa ] [tab] K +  ω − [taa ] [tab] =          [tbc ] − [tab ] [tac ] K +  [tcc] − [tab] [tac] K +  ω − ω [tac ] =        [taa ]  [taa]  [taa ]     [tbb.1]K + [tbc.1]K + ω =  [tbc.1]K + [tcc.1]K + ω = a b c b b c c a b c b b c c 1 ([tbc.1]K + ω ) [tbb.1] [tbc.1] [tbc.1] K +  ω   [tcc.1] −   [tbb.1]    K =− b c b c [tcc.2]K + ω c c =0⇒K =− c c ω − ω b [tbb.1] [tbc.1] =   c [tcc.2] 12/25/2008 [taa ]K + [tab]K + [tac ]K + ω =  [tab ]K + [tbb]K + [tbc ]K + ω =  [tac ]K + [tbc]K + [tcc ]K + ω =  a b c a a b c b a b c c ([tab]K + [tac ]K + ω ) [taa]  [tab] [tac]     [tbb ] − [taa ] [tab] K +  [tbc] − [taa ] [tab] K      [tab] [tac ] K +  [tcc] − [tab] [tac] K    [tbc ] −     [taa ]  [taa ]    [tbb.1]K + [tbc.1]K + ω =  [tbc.1]K + [tcc.1]K + ω = K =− a b c a c b c b K =− b c c b a c b b ω  [tab] = + ω −  [taa ]   ω  [tac ] = + ω −  [taa]   a b c 1 ([tbc.1]K + ω [tbb.1] c b ) [tbc.1] [tbc.1] K +  ω − ω [tbc.1] =   [tcc.1] −    [tbb.1] [tbb.1]     b c [tcc.2]K + ω c c c =0⇒K =− c ω c [tcc.2] [taa ]K + [tab]K + [tac ]K + ω =  [tab ]K + [tbb]K + [tbc ]K + ω =  [tac ]K + [tbc]K + [tcc ]K + ω =  ([tab]K + [tac ]K + ω ) K =− [taa] a b c a a b c b a b c c a b c a  ω [tab] [tac]       [tbb ] − [taa ] [tab] K +  [tbc] − [taa ] [tab] K +  ω − [taa ] [tab] =          [tbc ] − [tab ] [tac ] K +  [tcc] − [tab] [tac] K +  ω − ω [tac ] =        [taa ]  [taa]  [taa ]     [tbb.1]K + [tbc.1]K + ω =  [tbc.1]K + [tcc.1]K + ω = a b c b b c c a b c b b c c 1 ([tbc.1]K + ω ) [tbb.1] [tbc.1] [tbc.1] K +  ω   [tcc.1] −   [tbb.1]    K =− b c b c [tcc.2]K + ω c c =0⇒K =− c c ω − ω b [tbb.1] [tbc.1] =   c [tcc.2] 12/25/2008 [taa ]K + [tab]K + [tac ]K + ω =  [tab ]K + [tbb]K + [tbc ]K + ω =  [tac ]K + [tbc]K + [tcc ]K + ω =  a b c a a b c b a b c c ([tab]K + [tac ]K + ω ) [taa]  [tab] [tac]     [tbb ] − [taa ] [tab] K +  [tbc] − [taa ] [tab] K      [tab] [tac ] K +  [tcc] − [tab] [tac] K    [tbc ] −     [taa ]  [taa ]    [tbb.1]K + [tbc.1]K + ω =  [tbc.1]K + [tcc.1]K + ω = K =− a b c a c b c b K =− b c c b a c b b ω  [tab] = + ω −  [taa ]   ω  [tac ] = + ω −  [taa]   a b c 1 ([tbc.1]K + ω [tbb.1] c b ) [tbc.1] [tbc.1] K +  ω − ω [tbc.1] =   [tcc.1] −    [tbb.1] [tbb.1]     b c [tcc.2]K + ω c c c =0⇒K =− c ω c [tcc.2] [taa ]K + [tab]K + [tac ]K + ω =  [tab ]K + [tbb]K + [tbc ]K + ω =  [tac ]K + [tbc]K + [tcc ]K + ω =  ([tab]K + [tac ]K + ω ) K =− [taa] a b c a a b c b a b c c a b c a  ω [tab] [tac]       [tbb ] − [taa ] [tab] K +  [tbc] − [taa ] [tab] K +  ω − [taa ] [tab] =          [tbc ] − [tab ] [tac ] K +  [tcc] − [tab] [tac] K +  ω − ω [tac ] =        [taa ]  [taa]  [taa ]     [tbb.1]K + [tbc.1]K + ω =  [tbc.1]K + [tcc.1]K + ω = a b c b b c c a b c b b c c 1 ([tbc.1]K + ω ) [tbb.1] [tbc.1] [tbc.1] K +  ω   [tcc.1] −   [tbb.1]    K =− b c b c [tcc.2]K + ω c c =0⇒K =− c c ω − ω b [tbb.1] [tbc.1] =   c [tcc.2] 12/25/2008 [taa ]K + [tab]K + [tac ]K + ω =  [tab ]K + [tbb]K + [tbc ]K + ω =  [tac ]K + [tbc]K + [tcc ]K + ω =  a b c a a b c b a b c c ([tab]K + [tac ]K + ω ) [taa]  [tab] [tac]     [tbb ] − [taa ] [tab] K +  [tbc] − [taa ] [tab] K      [tab] [tac ] K +  [tcc] − [tab] [tac] K    [tbc ] −     [taa ]  [taa ]    [tbb.1]K + [tbc.1]K + ω =  [tbc.1]K + [tcc.1]K + ω = K =− a b c a c b c b K =− b c c b a c b b ω  [tab] = + ω −  [taa ]   ω  [tac ] = + ω −  [taa]   a b c 1 ([tbc.1]K + ω [tbb.1] c b ) [tbc.1] [tbc.1] K +  ω − ω [tbc.1] =   [tcc.1] −    [tbb.1] [tbb.1]     b c [tcc.2]K + ω c c =0⇒K =− c c ω c [tcc.2] Gi i h phương trình chu n s liên h sau 3K a + K b + K c + 0.5 =   K a + 3K b + K c + =  2K + 2K + 4K − = b c  a 12/25/2008 Ký hi u dòng a a Ka b Ka c Kc b E1b×a b.1 0.5 -0.33333 -0.33333 2.666667 -1 -0.66667 -0.66667 1.333333 -0.16667 -0.16667 -0.16667 -1 E1 ( -0.5 -1.33333 -0.66667 -1 Kc=0.625 0.0625 -1 -0.33333 0.083333 -1.25 0.625 E2 ( c E1c×a E2c×b.1 c.2 E3 Ka=-0.5 Kb=-0.25 Ka Kb Kc ðánh giá ñ xác: - Xác đ nh sai s trung phương Px = m02 mx2 mx = m0 Px m0 - Sai s trung phương tr ng s ñơn v P – Tr ng s c a hàm ñ i lư ng c n đánh giá đ xác 12/25/2008 ðánh giá đ xác: - Xác đ nh sai s trung phương A α m ,m = ? X Y B S S E α12 D α C mS = ? S ðánh giá ñ xác: - Xác đ nh sai s trung phương Px = m02 mx2 mx = m0 Px m0 - Sai s trung phương tr ng s ñơn v P – Tr ng s c a hàm ñ i lư ng c n đánh giá đ xác 10 12/25/2008 Sai s trung phương tr ng s ñơn v xác đ nh b ng cơng th c m0 = [ pvv] n−t [ pvv] = ∑ p v v n i i i [ pvv] = −∑ k ω r i [ pvv] = V i T PV ; NK + W = ⇒ NK = − W V T = K T BP −1 V = P −1B T K ⇒ [ pvv ] = K T BP −1PP −1B T K = K T BP −1B T K = −K T W = − ∑ kiωi r − [ pvv ] = − ω12 − ω 22 [taa ] [tbb.1] Tr ng s ñ o: − − ω r2.r −1 [trr ] r −1 PF F = F ( L1 , L , , L n ) = F ( L1 , L , , L n ) + fi = ∂F ∂F ∂F v + + v1 + ∂ L1 ∂L2 ∂Ln ∂F ∂ Li  f1  f  f =       fn  [taf ]2 − [tbf 1]2 − − [trf r −1 ]2 = [tff ] − [trr r −1 ] [taa ] [tbb 1] PF [taf ][tas ] − [tbf 1][tbs 1] − − [trf r −1 ][trs r −1 ] = [tfs r ] = [tfs ] − PF [tbb 1] [trr r −1 ] [taa ] s i = a i + bi + + ri + f i 11 12/25/2008 Ký hi u dòng a a Ka b Kb c Kc [tff] -1 E1 ( b E1b×a b.1 -1 E2 ( c E1c×a E2c×b.1 c.2 E3 -1 Ka Kb Kc [tff] Tính s phương trình u ki n L p phương trình u ki n s hi u ch nh L p b ng h phương trình chu n s liên h Gi i h phương trình chu n s liên h ðánh giá đ xác 12 12/25/2008 Cho lư i kh ng ch ñ cao H B = 16.190 H A = 10.004 A h1 = 1.02m h4 = −0.15m h3 = 5.32m h5 = 3.85m h2 = 0.92m D B C H D = 10.105 H C = 20.185 Cho lư i kh ng ch ñ cao H B = 16.190 H A = 10.004 A h4 = −0.15m h1 = 1.02m h3 = 5.32m h2 = 0.92m D B h5 = 3.85m C H D = 10.105 r = n−t H C = 20.185 n=5 t=2 13 12/25/2008 H A + h1 − h2 − H D = v1 − v2 + ωa = 0; ωa = h1ño − h2ño + H A − H D = 1.02 − 0.92 + 10.004 − 10.105 = −1mm v1 − v2 − = H B − h4 + h5 − HC = − v4 + v5 + ωb = 0; ωb = −h4ño + h5ño + H B − HC = 0.15 + 3.85 + 16.190 − 20.185 = +5mm − v4 + v5 + = H A + h1 + h3 + h4 − H B = v1 + v3 + v4 + ωc = 0; ωc = h1ño + h3ño + h4ño + H A − H B = 1.02 + 5.32 − 0.15 + 10.004 − 16.190 = 4mm STT a b c f s v1 1 v2 -1 0 -1 v3 0 1 v4 -1 0 v5 0 ω -1 ðánh giá đ xác đ cao H1 14 12/25/2008 Ka Kc ω [f.] [.s] -1 -1 1 -1 4 Ký hi u dòng Kb b Ka a Ka c Kc [t.f] [t.s] -1 -1 -0.5 0.5 -0.5 -2 b -1 E1b×a 0 0 b.1 -1 -1 0.5 -2.5 -0.5 a E1 ( E2 ( 4 E1c×a -0.5 0.5 -0.5 -2 E2c×b.1 -0.5 2.5 0.5 c.2 0.5 2.5 E3 -1 -3.5 -0.25 -0.83333 c 2.25 -4.25 -3.5 15 12/25/2008 STT a b c f v v1 1 -1.25 v2 -1 0 -2.25 v3 0 -3.5 v4 -1 0.75 v5 0 -4.25 ω -1 K 2.25 -4.25 -3.5 [pvv]=37.5 [pvv]=37.5 Sai s trung phương ðánh giá đ xác đ cao H1 [taf ] − [tbf 1] − − [trf r−1 ] = [tff ] − [taa] [tbb.1] [trr.r−1 ] PF 2 = − 0.5×1 − × − 0.5× 0.25 = − 0.5 − − 0.125= 0.325 PF [taf ][tas] − [tbf 1][tbs.1] − − [trf r−1 ][trs.r−1 ] = [tfs.r ] = [tfs] − [taa] PF [tbb.1] [trr.r−1 ] = − 0.5 × − − 0.25× 2.5 = − − − 0.625= 0.325 mH = ±3.535mm 0.325= ±3.535mm 0.57 = ±2.016mm × × 16 ... -2 .5 -0 .5 a E1 ( E2 ( 4 E1c×a -0 .5 0.5 -0 .5 -2 E2c×b.1 -0 .5 2.5 0.5 c.2 0.5 2.5 E3 -1 -3 .5 -0 .25 -0 .83333 c 2.25 -4 .25 -3 .5 15 12/25/2008 STT a b c f v v1 1 -1 .25 v2 -1 0 -2 .25 v3 0 -3 .5 v4 -1 ... 2.666667 -1 -0 .66667 -0 .66667 1.333333 -0 .16667 -0 .16667 -0 .16667 -1 E1 ( -0 .5 -1 .33333 -0 .66667 -1 Kc=0.625 0.0625 -1 -0 .33333 0.083333 -1 .25 0.625 E2 ( c E1c×a E2c×b.1 c.2 E3 Ka =-0 .5 Kb =-0 .25... -1 0 -1 v3 0 1 v4 -1 0 v5 0 ω -1 ðánh giá đ xác đ cao H1 14 12/25/2008 Ka Kc ω [f.] [.s] -1 -1 1 -1 4 Ký hi u dòng Kb b Ka a Ka c Kc [t.f] [t.s] -1 -1 -0 .5 0.5 -0 .5 -2 b -1 E1b×a 0 0 b.1 -1 -1