BÌNH SAI ĐIỀU KIỆN (Nguyễn Quang Minh) - Lập hệ phương trình chuẩn

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12/25/2008 Nguy n Quang Minh NK + W = 0 BP−1BTK + W = 0 b1 ... r1 Ka  ωa  a1 a2 ... an 1/ p1 0  a1 0 0 b   0 1/ p 0 a2 b2 ... r2 Kb  ωb  b2 ... bn  0 1    +   = 0 2 ... ... ... ... 0 .. 0  ... ... ... ... ..   ..  0        bn ... rn Kr  ωr   r1 r2 ... r3  0 0 0 1/ pn an  aa  ar   ab   ..    p  p Ka  ωa   p    br      ab  bb ...  Kb  + ωb  = 0    p  p    p  ..   ..  ..      .. .. ..  rr   Kr  ωr    ar   br  ..     p   p     p  1
12/25/2008 NK + W = 0 BP −1 B T K + W = 0  aa   ab   ar  + ... +   K r + ω a = 0   K a +   K b  p  p  p   ab   bb   br   K a +  Kb + ... +   K r + ω b = 0  p   p  p  ....    ar  K a +  br  K b  rr  + ... +   K r + ω r = 0  p   p     p NK + W = 0 BP −1B T K + W = 0 [taa ]K a + [tab]K b + ... + [tar ]K r + ωa = 0 [taa ]K + [tbb]K + ... + [tbr ]K + ω = 0  a b r b  ....   [taa ]K a + [tbr ]K b + ... + [trr ]K r + ωr = 0  [taa ]K a + [tab]K b + [tac ]K c + ωa = 0  [taa ]K a + [tbb]K b + [tbc ]K c + ωb = 0  [taa ]K + [tbc ]K + [tcc ]K + ω = 0  a b c c 1 ([tab]K b + [tac ]K c + ωa ) Ka = − [taa ] 2
12/25/2008 [taa ]K + [tab]K + [tac ]K + ω = 0  a b c a [tab ]K + [tbb]K + [tbc ]K + ω = 0 a b c b  [tac ]K + [tbc]K + [tcc ]K + ω = 0  a b c c 1 ([tab]K + [tac ]K + ω ) K =− [taa] a b c a [tab] [tac] ω      [tab] = 0  [tbb ] − [taa ] [tab] K +  [tbc] − [taa ] [tab] K + ω −  a [taa ]       b c b  [tab] [tac ] K +  [tcc] − [tab] [tac] K ω   [tac ] = 0   [tbc ] − + ω −     a [taa ]  [taa ]  [taa]     b c c  [tbb.1]K + [tbc.1]K + ω = 0  b c b .1 [tbc.1]K + [tcc.1]K + ω = 0 b c c .1 1 ([tbc.1]K + ω ) K =− [tbb.1] b c b .1 [tbc.1] [tbc.1] K +  ω − ω [tbc.1] = 0   [tcc.1] −    b .1 [tbb.1] [tbb.1] c c .1     ω [tcc.2]K + ω =0⇒K =− c .2 [tcc.2] c c .2 c Ký hi u dòng a Ka b Ka c Kc a -1 Ka E1 ( b E1b×a b.1 -1 Kb E2 ( c E1c×a E2c×b.1 c.2 E3 -1 Kc Ka Kb Kc 3
12/25/2008 Ký hi u dòng a b Ka c Kc Ka a -1 Ka E1 ( b 1 ([tab]K + [tac ]K + ω ) K =− [taa ] a b c a Ký hi u dòng a b Ka c Kc Ka a -1 Ka E1 ( b E1b×a b.1 [taa ]K + [tab]K + [tac ]K + ω =0 a b c a [tab]K + [tbb]K + [tbc ]K + ω =0 a b c b [tac]K + [tbc ]K + [tcc]K + ω =0 a b c c 1 ([tab]K + [tac]K + ω ) K =− [taa ] a b c a [tab] [tab] K +  [tbc] − [tac] [tab] K +  ω − ω [tab] = 0   [tbb ] −      a [taa ]  [taa ] [taa ]      b c b [tbb.1]K + [tbc.1]K + ω = 0 b c b .1 4
12/25/2008 Ký hi u dòng a b Ka c Kc Ka a -1 Ka E1 ( b E1b×a b.1 -1 Kb E2 ( c E1c×a E2c×b.1 c.2 E3 -1 Kc Ka Kb Kc [taa ]K + [tab]K + [tac ]K + ω = 0  a b c a [tab ]K + [tbb]K + [tbc ]K + ω = 0 a b c b  [tac ]K + [tbc]K + [tcc ]K + ω = 0  a b c c 1 ([tab]K + [tac ]K + ω ) K =− [taa] a b c a [tab] [tac] ω        [tbb ] − [taa ] [tab] K +  [tbc] − [taa ] [tab] K +  ω − [taa ] [tab] = 0 a       b c b    [tbc ] − [tab ] [tac ] K +  [tcc] − [tab] [tac] K +  ω − ω [tac ] = 0       a [taa ]  [taa ]  [taa]     b c c  [tbb.1]K + [tbc.1]K + ω = 0  b c b .1 [tbc.1]K + [tcc.1]K + ω = 0 b c c .1 1 ([tbc.1]K + ω ) K =− [tbb.1] b c b .1 [tbc.1] [tbc.1] K +  ω ω  [tbc.1] = 0  [tcc.1] − −    b .1 [tbb.1] [tbb.1] c c .1     ω [tcc.2]K + ω =0⇒K =− c .2 [tcc.2] c c .2 c 5
12/25/2008 [taa ]K + [tab]K + [tac ]K + ω = 0  a b c a [tab ]K + [tbb]K + [tbc ]K + ω = 0 a b c b  [tac ]K + [tbc]K + [tcc ]K + ω = 0  a b c c 1 ([tab]K + [tac ]K + ω ) K =− [taa] a b c a [tab] [tac] ω      [tab] = 0  [tbb ] − [taa ] [tab] K +  [tbc] − [taa ] [tab] K + ω −  a [taa ]       b c b  [tab] [tac ] K +  [tcc] − [tab] [tac] K ω   [tac ] = 0   [tbc ] − + ω −     a [taa ]  [taa ]  [taa]     b c c  [tbb.1]K + [tbc.1]K + ω = 0  b c b .1 [tbc.1]K + [tcc.1]K + ω = 0 b c c .1 1 ([tbc.1]K + ω ) K =− [tbb.1] b c b .1 [tbc.1] [tbc.1] K +  ω − ω [tbc.1] = 0   [tcc.1] −    b .1 [tbb.1] [tbb.1] c c .1     ω [tcc.2]K + ω =0⇒K =− c .2 [tcc.2] c c .2 c [taa ]K + [tab]K + [tac ]K + ω = 0  a b c a [tab ]K + [tbb]K + [tbc ]K + ω = 0 a b c b  [tac ]K + [tbc]K + [tcc ]K + ω = 0  a b c c 1 ([tab]K + [tac ]K + ω ) K =− [taa] a b c a [tab] [tac] ω        [tbb ] − [taa ] [tab] K +  [tbc] − [taa ] [tab] K +  ω − [taa ] [tab] = 0 a       b c b    [tbc ] − [tab ] [tac ] K +  [tcc] − [tab] [tac] K +  ω − ω [tac ] = 0       a [taa ]  [taa ]  [taa]     b c c  [tbb.1]K + [tbc.1]K + ω = 0  b c b .1 [tbc.1]K + [tcc.1]K + ω = 0 b c c .1 1 ([tbc.1]K + ω ) K =− [tbb.1] b c b .1 [tbc.1] [tbc.1] K +  ω ω  [tbc.1] = 0  [tcc.1] − −    b .1 [tbb.1] [tbb.1] c c .1     ω [tcc.2]K + ω =0⇒K =− c .2 [tcc.2] c c .2 c 6
12/25/2008 [taa ]K + [tab]K + [tac ]K + ω = 0  a b c a [tab ]K + [tbb]K + [tbc ]K + ω = 0 a b c b  [tac ]K + [tbc]K + [tcc ]K + ω = 0  a b c c 1 ([tab]K + [tac ]K + ω ) K =− [taa] a b c a [tab] [tac] ω      [tab] = 0  [tbb ] − [taa ] [tab] K +  [tbc] − [taa ] [tab] K + ω −  a [taa ]       b c b  [tab] [tac ] K +  [tcc] − [tab] [tac] K ω   [tac ] = 0   [tbc ] − + ω −     a [taa ]  [taa ]  [taa]     b c c  [tbb.1]K + [tbc.1]K + ω = 0  b c b .1 [tbc.1]K + [tcc.1]K + ω = 0 b c c .1 1 ([tbc.1]K + ω ) K =− [tbb.1] b c b .1 [tbc.1] [tbc.1] K +  ω − ω [tbc.1] = 0   [tcc.1] −    b .1 [tbb.1] [tbb.1] c c .1     ω [tcc.2]K + ω =0⇒K =− c .2 [tcc.2] c c .2 c [taa ]K + [tab]K + [tac ]K + ω = 0  a b c a [tab ]K + [tbb]K + [tbc ]K + ω = 0 a b c b  [tac ]K + [tbc]K + [tcc ]K + ω = 0  a b c c 1 ([tab]K + [tac ]K + ω ) K =− [taa] a b c a [tab] [tac] ω        [tbb ] − [taa ] [tab] K +  [tbc] − [taa ] [tab] K +  ω − [taa ] [tab] = 0 a       b c b    [tbc ] − [tab ] [tac ] K +  [tcc] − [tab] [tac] K +  ω − ω [tac ] = 0       a [taa ]  [taa ]  [taa]     b c c  [tbb.1]K + [tbc.1]K + ω = 0  b c b .1 [tbc.1]K + [tcc.1]K + ω = 0 b c c .1 1 ([tbc.1]K + ω ) K =− [tbb.1] b c b .1 [tbc.1] [tbc.1] K +  ω ω  [tbc.1] = 0  [tcc.1] − −    b .1 [tbb.1] [tbb.1] c c .1     ω [tcc.2]K + ω =0⇒K =− c .2 [tcc.2] c c .2 c 7
12/25/2008 [taa ]K + [tab]K + [tac ]K + ω = 0  a b c a [tab ]K + [tbb]K + [tbc ]K + ω = 0 a b c b  [tac ]K + [tbc]K + [tcc ]K + ω = 0  a b c c 1 ([tab]K + [tac ]K + ω ) K =− [taa] a b c a [tab] [tac] ω      [tab] = 0  [tbb ] − [taa ] [tab] K +  [tbc] − [taa ] [tab] K + ω −  a [taa ]       b c b  [tab] [tac ] K +  [tcc] − [tab] [tac] K ω   [tac ] = 0   [tbc ] − + ω −     a [taa ]  [taa ]  [taa]     b c c  [tbb.1]K + [tbc.1]K + ω = 0  b c b .1 [tbc.1]K + [tcc.1]K + ω = 0 b c c .1 1 ([tbc.1]K + ω ) K =− [tbb.1] b c b .1 [tbc.1] [tbc.1] K +  ω − ω [tbc.1] = 0   [tcc.1] −    b .1 [tbb.1] [tbb.1] c c .1     ω [tcc.2]K + ω =0⇒K =− c .2 [tcc.2] c c .2 c Gi i h phương trình chu n s liên h như sau 3K a + K b + 2 K c + 0.5 = 0   K a + 3K b + 2 K c + 0 = 0  2K + 2K + 4K − 1 = 0 a b c 8
12/25/2008 Ký hi u dòng a Ka b Ka c Kc a 3 1 2 0.5 Ka E1 ( -1 -0.33333 -0.66667 -0.16667 b 3 2 0 E1b×a -0.33333 -0.66667 -0.16667 b.1 2.666667 1.333333 -0.16667 Kb E2 ( -1 -0.5 0.0625 c 4 -1 E1c×a -1.33333 -0.33333 E2c×b.1 -0.66667 0.083333 c.2 2 -1.25 E3 -1 0.625 Kc Ka=-0.5 Kb=-0.25 Kc=0.625 ðánh giá ñ chính xác: - Xác ñ nh sai s trung phương m02 Px = mx2 1 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 ñánh giá ñ chính xác 9
12/25/2008 ðánh giá ñ chính xác: - Xác ñ nh sai s trung phương α A 2 m ,m = ? D X Y 1 S 3α S E α12 B C mS = ? S ðánh giá ñ chính xác: - Xác ñ nh sai s trung phương m02 Px = mx2 1 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 ñánh giá ñ chính xác 10
12/25/2008 Sai s trung phương tr ng s ñơn v xác ñ nh b ng công th c [ pvv] m0 = n−t [ pvv] = ∑ p v v n i i i 1 [ pvv] = −∑ k ω r i i 1 [ pvv] = V PV ; NK + W = 0 ⇒ NK = − W T 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 1 ω12 ω 22 ω r2.r −1 − [ pvv ] = − − − ... − [taa ] [tbb.1] [trr. ] r −1 1 Tr ng s ñ o: PF ∂F ∂F ∂F F = F ( L1 , L 2 ,..., L n ) = F 0 ( L1 , L 2 ,..., L n ) + v1 + v 2 + ... + vn ∂ L1 ∂L2 ∂Ln ∂F fi = ∂ Li  f1  f  f =  2    fn  [taf ]2 − [tbf .1]2 − ... − [trf .r −1 ]2 1 = [tff ] − [taa ] [tbb .1] [trr .r −1 ] PF [taf ][tas ] − [tbf .1][tbs .1] − ... − [trf .r −1 ][trs .r −1 ] 1 = [tfs .r ] = [tfs ] − [taa ] [tbb .1] [trr .r −1 ] PF s i = a i + bi + ... + ri + f i 11
12/25/2008 Ký hi u dòng a Ka b Kb c Kc [tff] a -1 E1 ( b E1b×a b.1 -1 E2 ( c E1c×a E2c×b.1 c.2 E3 -1 [tff] Ka Kb Kc Tính s phương trình ñi u ki n L p các phương trình ñi 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 ñ chính xác 12
12/25/2008 Cho lư i kh ng ch ñ cao H B = 16.190 B H A = 10.004 A h1 = 1.02m h4 = −0.15m h3 = 5.32m 1 2 h5 = 3.85m h2 = 0.92m C D H D = 10.105 H C = 20.185 Cho lư i kh ng ch ñ cao H B = 16.190 B H A = 10.004 h4 = −0.15m A h1 = 1.02m h3 = 5.32m 1 2 h2 = 0.92m h5 = 3.85m C D H D = 10.105 H C = 20.185 r = n−t n=5 t=2 13
12/25/2008 H A + h1 − h2 − H D = 0 v1 − v2 + ωa = 0; ωa = h1ño − h2ño + H A − H D = 1.02 − 0.92 + 10.004 − 10.105 = −1mm v1 − v2 − 1 = 0 H B − h4 + h5 − HC = 0 − v4 + v5 + ωb = 0; ωb = −h4ño + h5ño + H B − HC = 0.15 + 3.85 + 16.190 − 20.185 = +5mm − v4 + v5 + 5 = 0 H A + h1 + h3 + h4 − H B = 0 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 v1 v2 v3 v4 v5 ω a 1 -1 0 0 0 -1 b 0 0 0 -1 1 5 c 1 0 1 1 0 4 f 1 0 0 0 0 s 3 -1 1 0 1 ðánh giá ñ chính xác ñ cao H1 14
12/25/2008 Ka Kb Kc [.s] [f.] ω 2 0 1 -1 1 4 0 2 -1 5 0 1 1 -1 3 4 1 4 1 3 Ký hi u dòng a b c Kc [t.f] [t.s] Ka Ka 2 0 1 -1 1 4 a -1 0 -0.5 0.5 -0.5 -2 E1 ( 2 -1 5 0 1 b 0 0 0 0 0 E1b×a 2 -1 5 0 1 b.1 -1 0.5 -2.5 0 -0.5 E2 ( 3 4 1 4 c -0.5 0.5 -0.5 -2 E1c×a -0.5 2.5 0 0.5 E2c×b.1 2 7 0.5 2.5 c.2 -1 -3.5 -0.25 -0.83333 E3 2.25 -4.25 -3.5 15
12/25/2008 STT v1 v2 v3 v4 v5 ω K a 1 -1 0 0 0 -1 2.25 b 0 0 0 -1 1 5 -4.25 c 1 0 1 1 0 4 -3.5 f 1 0 0 0 0 [pvv]=37.5 [pvv]=37.5 v -1.25 -2.25 -3.5 0.75 -4.25 Sai s trung phương ðánh giá ñ chính xác ñ cao H1 [taf ] − [tbf .1] − ...− [trf .r−1 ] 2 2 2 1 = [tff ] − [taa] [tbb.1] [trr.r−1 ] PF 1 = 1 − 0.5×1 − 0 × 0 − 0.5× 0.25 = 1 − 0.5 − 0 − 0.125= 0.325 PF [taf ][tas] − [tbf .1][tbs.1] − ...− [trf .r−1 ][trs.r−1 ] 1 = [tfs.r ] = [tfs] − [taa] [tbb.1] [trr.r−1 ] PF = 3 − 0.5 × 4 − 0 − 0.25× 2.5 = 3 − 2 − 0 − 0.625= 0.325 mH = ±3.535mm 0.325= ±3.535mm 0.57 = ±2.016mm × × 1 16