file pdf - La pagina della Topografia

ISTITUTO TECNICO STATALE COMMERCIALE E PER GEOMETRI
"In Memoria dei Morti per La Patria"
Viale Enrico Millo, 1 - 16043 Chiavari
Laboratorio di Topografia - G.P.S. - G.I.S
Anno scolastico 2009 - 2010 Classe: 4^ "A" L.T.C. Allievo: ...........................................
Intersezione multipla inversa: problema di Hansen.
Metodo della base fittizia.
Sommario
Calcolare le coordinate del vertice P di una poligonale. All’uopo si è fatta stazione con
un teodolite sul vertice P di coordinate incognite e sul vertice ausiliario Q scelto a
piacere ma intervisibile con P. Tramite osservazioni a due vertici trigonometrici A, e B
si sono determinate le ampiezze angolari necessarie.
Nella tabella seguente sono riportati i dati ricavati dal quaderno di campagna e le
coordinate dei trigonometrici tratte dalle schede relative.
Calcolare inoltre le coordinate del vertice aggiuntivo Q e l’azimut (PQ).
Punto A:
XA ≡ 1520050.51m
σ XA ≡ 0.05m
YA ≡ 4550160.63m
σ YA ≡ 0.05m
Stazione P:
θ PA ≡ 95.400 ⋅
σ θPA ≡ 0.002 ⋅
Stazione Q:
π
200
π
200
θ QP ≡ 118.405 ⋅
σ θQP ≡ 0.002 ⋅
π
200
π
200
Punto B:
π
rad
θ PB ≡ 164.740 ⋅
rad
σ θPB ≡ 0.002 ⋅
rad
θ QA ≡ 153.880 ⋅
rad
σ θQA ≡ 0.002 ⋅
200
π
200
π
200
π
200
XB ≡ 1520140.83m
σ XB ≡ 0.05m
YB ≡ 4550180.92m
σ YB ≡ 0.05m
rad
θ PQ ≡ 225.625 ⋅
rad
σ θPQ ≡ 0.002 ⋅
rad
θ QB ≡ 233.510 ⋅
rad
σ θQB ≡ 0.002 ⋅
π
200
π
200
π
200
π
200
rad
rad
rad
rad
Summary
Some times the determination of the coordinates for the one single point is a
prerequisite asked by the most complicated problems. For example: the
coordinates of the points are necessary for the linear transformations of the
drawings. The Hansen problem is a typical problem for the calculation of the
coordinates of the one point. The data of the problem are the coordinates of
the three A, and B trigonometric points. The measured data are the
orizzontal angles with the instrument in P and Q point. For the measurement of
the angles, in this exercise, is used the centesimal system. The data of the
trigonometric points are the typicals data in UTM coordinates.
Aggiornamento del 29/11/2009
Pagina 1
AB
A
B
1
Q
1
P
Risposte:
XP = 1520056.15 m
σ XP = 0.08 m
XQ = 1520093.39 m
σ XQ = 0.10 m
YP = 4550120.37 m
σ YP = 0.06 m
YQ = 4550107.38 m
σ YQ = 0.09 m
Φ PQ ⋅
200
π
= 121.367
grad
σ ΦPQ ⋅
200
π
= 0.183
grad
Relazione di calcolo.
Calcolo degli angoli con vertice nelle stazioni P e Q:
200
α ≡ θ PQ − θ PA
α⋅
β ≡ θ QA − θ QP
β⋅
α 1 ≡ θ PQ − θ PB
α1⋅
β 1 ≡ θ QB − θ QP
β 1⋅
π
σα ≡
σ θPQ + σ θPA
= 35.4750
grad
σβ ≡
σ θQA + σ θQP
= 60.8850
grad
σ α1 ≡
σ θPQ + σ θPB
= 115.1050
grad
σ β1 ≡
σ θQB + σ θQP
200
π
200
π
Aggiornamento del 29/11/2009
2
grad
π
200
2
= 130.2250
2
2
2
2
2
2
σα ⋅
σβ ⋅
200
π
200
σ α1 ⋅
σ β1 ⋅
π
200
π
200
π
= 0.00283
grad
= 0.00283
grad
= 0.00283
grad
= 0.00283
grad
Pagina 2
Calcolo della distanza AB:
AB ≡
(XB − XA)2 + (YB − YA)2
σ AB.XB ≡ σ XB ⋅
σ AB.YB ≡ σ YB ⋅
dX B
σ AB.XB = 0.049 m
(XB − XA)2 + (YB − YA)2
d
dX A
σ AB.XA = −0.049 m
(XB − XA)2 + (YB − YA)2
d
dYB
σ AB.YA ≡ σ YA ⋅
σ AB ≡
(XB − XA)2 + (YB − YA)2
d
σ AB.XA ≡ σ XA ⋅
AB = 92.571 m
σ AB.YB = 0.011 m
(XB − XA)2 + (YB − YA)2
d
dYA
2
2
2
σ AB.XB + σ AB.XA + σ AB.YB + σ AB.YA
σ AB.YA = −0.011 m
2
σ AB = 0.071 m
Calcolo dll'azimut Φ AB:
⎛ XB − XA ⎞
⎟
⎜Y −Y ⎟
⎝ B A⎠
Φ AB ≡ atan ⎜
Φ AB ⋅
200
= 85.9322
π
⎛ XB − XA ⎞
⎟
σ ΦAB.XB ≡ σ XB ⋅
atan ⎜
⎜
Y −Y ⎟
dX B
⎝ B A⎠
σ ΦAB.XB ⋅
⎛ XB − XA ⎞
⎟
σ ΦAB.XA ≡ σ XA ⋅
atan ⎜
⎜
Y −Y ⎟
dX A
⎝ B A⎠
σ ΦAB.XA ⋅
⎛ XB − XA ⎞
⎟
σ ΦAB.YB ≡ σ YB ⋅
atan ⎜
⎜
Y −Y ⎟
dYB
⎝ B A⎠
σ ΦAB.YB ⋅
⎛ XB − XA ⎞
⎟
σ ΦAB.YA ≡ σ YA ⋅
atan ⎜
⎜
Y −Y ⎟
dYA
⎝ B A⎠
σ ΦAB.YA ⋅
d
d
d
d
σ ΦAB ≡
2
2
2
σ ΦAB.XB + σ ΦAB.XA + σ ΦAB.YB + σ ΦAB.YA
Aggiornamento del 29/11/2009
2
200
π
200
π
200
π
200
π
grad
= 0.0075
grad
= −0.0075
grad
= −0.0335
grad
= 0.0335
grad
σ ΦAB ⋅
200
π
= 0.0486
grad
Pagina 3
Risoluzione dei 2 triangoli A'P'Q' e B'P'Q' composti da lati proporzionali della figura simile
dopo aver scelto arbitrariamente la lunghezza del lato PQ fittizia :
PQf ≡ 10.00m
sin ( β )
APf ≡ PQf ⋅
sin ( α + β )
σ APf.β ≡ σ β ⋅
d
dβ
σ APf.α ≡ σ α ⋅
APf = 10.307 m
sin ( β )
PQf ⋅
sin ( α + β )
σ APf.β = 0.002 m
sin ( β )
PQf ⋅
sin ( α + β )
σ APf.α = 0.001 m
d
dα
2
σ APf ≡
2
σ APf = 0.002 m
σ APf.β + σ APf.α
sin ( α )
AQf ≡ PQf ⋅
sin ( α + β )
σ AQf.β ≡ σ β ⋅
d
dβ
σ AQf.α ≡ σ α ⋅
σ AQf ≡
AQf = 17.334 m
sin ( β )
PQf ⋅
sin ( α + β )
σ AQf.β = 0.002 m
sin ( β )
PQf ⋅
sin ( α + β )
σ AQf.α = 0.001 m
d
dα
2
σ AQf = 0.002 m
2
σ AQf.β + σ AQf.α
γ ≡ π − (α + β)
σγ ≡
γ⋅
2
σα + σβ
2
200
σγ ⋅
π
= 34.3000
200
π
= 0.0040
grad
grad
Risoluzione del secondo triangolo B'P'Q' composto da lati proporzionali della figura simile
( )
sin β 1
BPf ≡ PQf ⋅
sin α 1 + β 1
(
σ BPf.β1 ≡ σ β1 ⋅
BPf = 26.393 m
)
( )
sin β 1
PQf ⋅
sin α 1 + β 1
dβ 1
d
Aggiornamento del 29/11/2009
(
)
σ BPf.β1 = 0.003 m
Pagina 4
(
2
σ BPf ≡
( )
sin β 1
PQf ⋅
sin α 1 + β 1
dα 1
d
σ BPf.α1 ≡ σ α1 ⋅
σ BPf.β1 + σ BPf.α1
( )
σ BPf.α1 = 0.003 m
)
2
σ BPf = 0.004 m
sin α 1
BQf ≡ PQf ⋅
sin α 1 + β 1
(
σ BQf.β1 ≡ σ β1 ⋅
BQf = 22.188 m
)
( )
sin β 1
PQf ⋅
sin α 1 + β 1
dβ 1
d
(
( )
sin β 1
PQf ⋅
sin α 1 + β 1
dα 1
d
σ BQf.α1 ≡ σ α1 ⋅
(
2
σ BQf ≡
σ BQf.β1 = 0.003 m
)
σ BQf.α1 = 0.003 m
)
σ BQf = 0.004 m
2
σ BQf.β1 + σ BQf.α1
(
)
γ 1 ≡ π − α1 + β 1
σ γ1 ≡
γ 1⋅
2
2
σ α1 + σ β1
200
π
σ γ1 ⋅
= 24.0100
200
= 0.0040
π
grad
grad
Considerando il triangolo A'P'B' calcolo il lato A'B' proporzionale ad AB:
ABf ≡
2
2
σ ABf.APf ≡ σ APf ⋅
σ ABf.BPf ≡ σ BPf⋅
σ ABf.α ≡ σ α ⋅
d
)
2
d
(
)
2
(
)
APf + BPf − 2 ⋅ APf ⋅ BPf ⋅ cos α − α 1
2
dα
2
2
dBPf
ABf = 23.470 m
APf + BPf − 2 ⋅ APf ⋅ BPf ⋅ cos α − α 1
dAPf
d
σ ABf.α1 ≡ σ α1 ⋅
σ ABf ≡
(
APf + BPf − 2 ⋅ APf ⋅ BPf ⋅ cos α − α 1
2
(
)
APf + BPf − 2 ⋅ APf ⋅ BPf ⋅ cos α − α 1
d
2
2
2
2
(
2
)
2
σ ABf.APf + σ ABf.BPf + σ ABf.α + σ ABf.α1
Aggiornamento del 29/11/2009
σ ABf.BPf = 0.004 m
σ ABf.α = 0.000 m
APf + BPf − 2 ⋅ APf ⋅ BPf ⋅ cos α − α 1
dα 1
σ ABf.APf = −0.000 m
σ ABf.α1 = −0.000 m
σ ABf = 0.004 m
Pagina 5
Considerando il triangolo A'Q'B' calcolo il lato A'B' proporzionale ad AB in un secondo modo:
2
ABfsec ≡
2
(
AQf + BQf − 2 ⋅ AQf ⋅ BQf ⋅ cos β 1 − β
σ ABfsec.AQf ≡ σ AQf ⋅
σ ABfsec.BQf ≡ σ BQf ⋅
σ ABfsec.β1 ≡ σ β1 ⋅
σ ABfsec.β ≡ σ β ⋅
d
d
2
2
2
2
(
2
2
(
AQf + BQf − 2 ⋅ AQf ⋅ BQf ⋅ cos β 1 − β
dβ
(
)
(
)
AQf + BQf − 2 ⋅ AQf ⋅ BQf ⋅ cos β 1 − β
2
σ ABfsec ≡
2
AQf + BQf − 2 ⋅ AQf ⋅ BQf ⋅ cos β 1 − β
dBQf
dβ 1
ABfsec = 23.470 m
AQf + BQf − 2 ⋅ AQf ⋅ BQf ⋅ cos β 1 − β
dAQf
d
d
2
)
2
σ ABfsec.AQf = 0.001 m
σ ABfsec.BQf = 0.003 m
)
σ ABfsec.β1 = 0.001 m
)
2
σ ABfsec.AQf + σ ABfsec.BQf + σ ABfsec.β1 + σ ABfsec.β
σ ABfsec.β = −0.001 m
2
σ ABfsec = 0.003 m
NOTA BENE: In questo secondo conto, si nota che la distanza AB risulta identica nel rispetto dei calcoli
precedenti, non così invece per lo s.q.m. che risente delle diverse grandezze applicate nelle diverse
formule utilizzate.
Calcolo della costante di proporzionalità K e del suo s.q.m.:
K≡
AB
K = 3.9443009
ABf
σ K.AB = 0.0030129
dAB ABf
σ K.ABf ≡ σ ABf ⋅
σK ≡
AB
d
σ K.AB ≡ σ AB ⋅
d
AB
dABf ABf
2
σ K.AB + σ K.ABf
Aggiornamento del 29/11/2009
2
σ K.ABf = −0.0006277
σ K = 0.0030776
Pagina 6
Calcolo dei veri valori dei lati calcolati in precedenza sulla figura simile:
AP ≡ APf ⋅ K
AP = 40.654 m
σ AP.APf ≡ σ APf ⋅
σ AP.K ≡ σ K ⋅
σ AP ≡
d
dK
d
dAPf
APf ⋅ K
σ AP.APf = 0.007 m
σ AP.K = 0.032 m
APf ⋅ K
2
σ AP.APf + σ AP.K
σ AP = 0.032 m
2
AQ ≡ AQf ⋅ K
AQ = 68.371 m
σ AQ.AQf ≡ σ AQf ⋅
σ AQ.K ≡ σ K ⋅
σ AQ ≡
d
dK
d
dAQf
AQf ⋅ K
σ AQ.K = 0.053 m
AQf ⋅ K
2
2
σ AQ.AQf + σ AQ.K
BP ≡ BPf ⋅ K
σ BP ≡
d
dK
d
dBPf
BPf ⋅ K
σ BP.BPf = 0.016 m
σ BP.K = 0.081 m
BPf ⋅ K
2
σ BP = 0.083 m
2
σ BP.BPf + σ BP.K
BQ ≡ BQf ⋅ K
BQ = 87.515 m
σ BQ.BQf ≡ σ BQf ⋅
σ BQ.K ≡ σ K ⋅
σ BQ ≡
σ AQ = 0.054 m
BP = 104.103 m
σ BP.BPf ≡ σ BPf⋅
σ BP.K ≡ σ K ⋅
σ AQ.AQf = 0.007 m
d
dK
d
dBQf
BQf ⋅ K
σ BQ.K = 0.068 m
BQf ⋅ K
2
σ BQ.BQf + σ BQ.K
Aggiornamento del 29/11/2009
σ BQ.BQf = 0.016 m
2
σ BQ = 0.070 m
Pagina 7
Calcolo dell'angolo al vertice A del triangolo AQB:
⎛ AB2 + AQ2 − BQ2 ⎞
⎜
⎟
δ ≡ acos ⎜
⎟
2 ⋅ AB ⋅ AQ
⎝
⎠
d
σ δ.AB ≡ σ AB ⋅
dAB
σ δ.AQ ≡ σ AQ ⋅
σδ ≡
⎛ AB2 + AQ2 − BQ2 ⎞
⎜
⎟
⎟
2 ⋅ AB ⋅ AQ
⎝
⎠
acos ⎜
d
dAQ
d
σ δ.BQ ≡ σ BQ ⋅
δ⋅
dBQ
⎛ AB2 + AQ2 − BQ2 ⎞
⎜
⎟
⎟
2 ⋅ AB ⋅ AQ
⎝
⎠
acos ⎜
⎛ AB2 + AQ2 − BQ2 ⎞
⎜
⎟
⎟
2 ⋅ AB ⋅ AQ
⎝
⎠
acos ⎜
2
2
2
σ δ.AB + σ δ.AQ + σ δ.BQ
200
= 70.9095
π
σ δ.AB ⋅
200
π
200
σ δ.AQ ⋅
π
200
σ δ.BQ ⋅
σδ ⋅
π
200
grad
= −0.0495
grad
= −0.0166
grad
= 0.0687
grad
= 0.0863
grad
= 25.4505
grad
π
Calcolo dell'angolo al vertice B del triangolo APB:
⎛ AB2 + BP2 − AP2 ⎞
⎜
⎟
⎟
2 ⋅ AB ⋅ BP
⎝
⎠
δ 1 ≡ acos ⎜
δ 1⋅
200
π
⎛ AB2 + BP2 − AP2 ⎞
⎜
⎟
σ δ1.AB ≡ σ AB ⋅
acos ⎜
⎟
2 ⋅ AB ⋅ BP
dAB
⎝
⎠
σ δ1.AB ⋅
⎛ AB2 + BP2 − AP2 ⎞
⎜
⎟
σ δ1.BP ≡ σ BP ⋅
acos ⎜
⎟
2 ⋅ AB ⋅ BP
dBP
⎝
⎠
σ δ1.BP⋅
⎛ AB2 + BP2 − AP2 ⎞
⎜
⎟
σ δ1.AP ≡ σ AP ⋅
acos ⎜
⎟
2 ⋅ AB ⋅ BP
dAP
⎝
⎠
σ δ1.AP ⋅
d
d
d
σ δ1 ≡
2
2
2
σ δ1.AB + σ δ1.BP + σ δ1.AP
Aggiornamento del 29/11/2009
σ δ1 ⋅
200
π
200
π
200
200
π
π
= 0.0040
grad
= −0.0264
grad
= 0.0224
grad
= 0.0349
grad
Pagina 8
Calcolo dll'azimut Φ AQ:
Φ AQ ≡ Φ AB + δ
σ ΦAQ ≡
200
Φ AQ ⋅
2
2
σ ΦAB + σ δ
= 156.8416
π
200
σ ΦAQ ⋅
grad
= 0.0991
grad
= 191.1416
grad
π
Calcolo dll'azimut Φ AP:
Φ AP ≡ Φ AQ + γ
σ ΦAP ≡
Φ AP ⋅
2
σ ΦAQ + σ γ
2
200
π
σ ΦAP ⋅
200
= 0.0991
grad
= 260.4816
grad
π
Calcolo dll'azimut Φ BP:
Φ BP ≡ Φ AB + π − δ 1
σ ΦBP ≡
Φ BP ⋅
2
2
σ ΦAB + σ δ1
200
π
σ ΦBP ⋅
200
= 0.0598
grad
= 236.4716
grad
π
Calcolo dll'azimut Φ BQ:
Φ BQ ≡ Φ BP − γ 1
σ ΦBQ ≡
Φ BQ ⋅
2
2
σ ΦBP + σ γ1
200
σ ΦBQ ⋅
π
200
π
= 0.0600
grad
Calcolo delle coodinate cartesiane di "P" e di "Q":
(
)
XP ≡ XA + AP ⋅ sin Φ AP
σ XP.XA ≡ σ XA ⋅
σ XP.AP ≡ σ AP ⋅
d
dX A
d
dAP
σ XP.ΦAP ≡ σ ΦAP ⋅
σ XP ≡
2
XP = 1520056.149 m
(XA + AP⋅ sin (ΦAP))
σ XP.XA = 0.050 m
(XA + AP⋅ sin (ΦAP))
d
dΦ AP
σ XP.AP = 0.004 m
(XA + AP⋅ sin (ΦAP))
2
2
σ XP.XA + σ XP.AP + σ XP.ΦAP
Aggiornamento del 29/11/2009
σ XP.ΦAP = −0.063 m
σ XP = 0.080 m
Pagina 9
(
)
YP ≡ YA + AP ⋅ cos Φ AP
σ YP.YA ≡ σ YA ⋅
σ YP.AP ≡ σ AP ⋅
d
dYA
d
dAP
σ YP ≡
(YA + AP⋅ cos(ΦAP))
dΦ AP
2
2
σ YP.ΦAP = −0.009 m
σ YP = 0.060 m
2
σ YP.YA + σ YP.AP + σ YP.ΦAP
(
σ XQ.XA ≡ σ XA ⋅
d
dX A
)
dAQ
σ XQ.XA = 0.050 m
(XA + AQ ⋅sin (ΦAQ))
d
σ XQ.ΦAQ ≡ σ ΦAQ ⋅
XQ = 1520093.391 m
(XA + AQ ⋅sin (ΦAQ))
d
σ XQ.AQ ≡ σ AQ ⋅
dΦ AQ
2
σ XQ.AQ = 0.034 m
(XA + AQ ⋅sin (ΦAQ))
2
σ XQ.XA + σ XQ.AQ + σ XQ.ΦAQ
(
YQ ≡ YA + AQ ⋅ cos Φ AQ
σ YQ.YA ≡ σ YA ⋅
d
dYA
σ YQ.AQ ≡ σ AQ ⋅
dAQ
2
)
dΦ AQ
σ XQ = 0.102 m
σ YQ.YA = 0.050 m
(YA + AQ ⋅cos(ΦAQ))
d
σ XQ.ΦAQ = −0.083 m
YQ = 4550107.378 m
σ YQ.AQ = −0.042 m
(YA + AQ ⋅cos(ΦAQ))
2
σ YQ.YA + σ YQ.AQ + σ YQ.ΦAQ
Aggiornamento del 29/11/2009
2
(YA + AQ ⋅cos(ΦAQ))
d
σ YQ.ΦAQ ≡ σ ΦAQ ⋅
σ YQ ≡
σ YP.AP = −0.032 m
(YA + AP⋅ cos(ΦAP))
XQ ≡ XA + AQ ⋅ sin Φ AQ
σ XQ ≡
σ YP.YA = 0.050 m
(YA + AP⋅ cos(ΦAP))
d
σ YP.ΦAP ≡ σ ΦAP ⋅
YP = 4550120.369 m
2
σ YQ.ΦAQ = −0.067 m
σ YQ = 0.093 m
Pagina 10
Calcolo, in un secondo modo e per verifica, delle coodinate cartesiane di "P" e di "Q":
(
)
XPsec ≡ XB + BP ⋅ sin Φ BP
σ XPsec.XB ≡ σ XB ⋅
d
dX B
d
σ XPsec.BP ≡ σ BP ⋅
dBP
σ XPsec.ΦBP ≡ σ ΦBP ⋅
σ XPsec ≡
XPsec = 1520056.149 m
(XB + BP⋅ sin (ΦBP))
σ XPsec.XB = 0.050 m
(XB + BP⋅ sin (ΦBP))
d
dΦ BP
σ XPsec.BP = −0.067 m
(XB + BP⋅ sin (ΦBP))
2
2
σ XPsec.ΦBP = −0.057 m
2
σ XPsec.XB + σ XPsec.BP + σ XPsec.ΦBP
(
)
YPsec ≡ YB + BP ⋅ cos Φ BP
σ YPsec.YB ≡ σ YB ⋅
σ YPsec.BP ≡ σ BP ⋅
d
dYB
d
dBP
σ YPsec.ΦBP ≡ σ ΦBP ⋅
σ YPsec ≡
σ XPsec = 0.101 m
YPsec = 4550120.369 m
(YB + BP⋅ cos(ΦBP))
σ YPsec.YB = 0.050 m
(YB + BP⋅ cos(ΦBP))
d
dΦ BP
σ YPsec.BP = −0.048 m
(YB + BP⋅ cos(ΦBP))
2
2
σ YPsec.YB + σ YPsec.BP + σ YPsec.ΦBP
(
σ YPsec.ΦBP = 0.080 m
2
σ YPsec = 0.106 m
)
XQsec ≡ XB + BQ ⋅ sin Φ BQ
σ XQsec.XB ≡ σ XB ⋅
d
dX B
σ XQsec.BQ ≡ σ BQ ⋅
σ XQsec ≡
(XB + BQ⋅sin (ΦBQ))
d
dBQ
σ XQsec.ΦBQ ≡ σ ΦBQ ⋅
XQsec = 1520093.391 m
σ XQsec.XB = 0.050 m
(XB + BQ⋅sin (ΦBQ))
d
dΦ BQ
2
σ XQsec.BQ = −0.038 m
(XB + BQ⋅sin (ΦBQ))
2
σ XQsec.ΦBQ = −0.069 m
2
σ XQsec.XB + σ XQsec.BQ + σ XQsec.ΦBQ
Aggiornamento del 29/11/2009
σ XQsec = 0.093 m
Pagina 11
(
)
YQsec ≡ YB + BQ ⋅ cos Φ BQ
σ YQsec.YB ≡ σ YB ⋅
(YB + BQ⋅cos(ΦBQ))
d
dYB
σ YQsec.BQ ≡ σ BQ ⋅
d
dBQ
σ YQsec.ΦBQ ≡ σ ΦBQ ⋅
σ YQsec ≡
YQsec = 4550107.378 m
σ YQsec.YB = 0.050 m
(YB + BQ⋅cos(ΦBQ))
d
dΦ BQ
σ YQsec.BQ = −0.059 m
(YB + BQ⋅cos(ΦBQ))
2
σ YQsec.ΦBQ = 0.045 m
2
2
σ YQsec = 0.089 m
σ YQsec.YB + σ YQsec.BQ + σ YQsec.ΦBQ
Calcolo dll'azimut Φ PQ:
⎛ XQ − XP ⎞
⎟
⎜Y −Y ⎟
⎝ Q P⎠
Φ PQ ≡ π + atan ⎜
σ ΦPQ.XQ ≡ σ XQ ⋅
σ ΦPQ.XP ≡ σ XP ⋅
σ ΦPQ ≡
dX Q
d
dX P
σ ΦPQ.YQ ≡ σ YQ ⋅
σ ΦPQ.YP ≡ σ YP ⋅
d
atan
atan
σ ΦPQ.YP ⋅
2
σ ΦPQ.XQ + σ ΦPQ.XP + σ ΦPQ.YQ + σ ΦPQ.YP
200
π
200
π
σ ΦPQ.YQ ⋅
⎛ XQ − XP ⎞
⎜
⎟
⎜Y −Y ⎟
⎝ Q P⎠
2
= 121.3666
π
σ ΦPQ.XP ⋅
⎛ XQ − XP ⎞
⎜
⎟
⎜Y −Y ⎟
⎝ Q P⎠
2
200
σ ΦPQ.XQ ⋅
⎛ XQ − XP ⎞
⎟
⎜Y −Y ⎟
⎝ Q P⎠
dYQ
dYP
⎛ XQ − XP ⎞
⎟
⎜Y −Y ⎟
⎝ Q P⎠
atan ⎜
atan ⎜
d
d
Φ PQ ⋅
200
π
200
π
grad
= −0.0545
grad
= 0.0427
grad
= −0.1422
grad
= 0.0915
2
σ ΦPQ ⋅
grad
200
π
= 0.1827
grad
Verifica del calcolo dell'azimut Φ PQ:
Φ PQsec ≡ π + atan
⎛ XQsec − XPsec ⎞
⎜
⎟
⎜Y
⎟
−
Y
⎝ Qsec Psec ⎠
σ ΦPQsec.XQsec ≡ σ XQsec ⋅
Aggiornamento del 29/11/2009
d
dXQsec
Φ PQsec ⋅
atan
⎛ XQsec − XPsec ⎞
⎜
⎟
⎜Y
⎟
−
Y
⎝ Qsec Psec ⎠
200
π
= 121.3666
σ ΦPQsec.XQsec ⋅
200
π
= −0.0497
grad
grad
Pagina 12
σ ΦPQsec.XPsec ≡ σ XPsec ⋅
d
dXPsec
⎛ XQsec − XPsec ⎞
⎟
⎜Y
⎟
−
Y
⎝ Qsec Psec ⎠
atan ⎜
σ ΦPQsec.XPsec ⋅
⎛ XQsec − XPsec ⎞
⎟
σ ΦPQsec.YQsec ≡ σ YQsec ⋅
atan ⎜
⎜
⎟
Y
−Y
dYQsec
⎝ Qsec Psec ⎠
d
σ ΦPQsec.YPsec ≡ σ YPsec ⋅
σ ΦPQsec ≡
d
dYPsec
⎛ XQsec − XPsec ⎞
⎟
⎜Y
⎟
−
Y
⎝ Qsec Psec ⎠
atan ⎜
2
200
π
σ ΦPQsec.YQsec ⋅
σ ΦPQsec.YPsec ⋅
2
200
π
200
π
= 0.0539
grad
= −0.1360
grad
= 0.1609
2
grad
2
σ ΦPQsec.XQsec + σ ΦPQsec.XPsec + σ ΦPQsec.YQsec + σ ΦPQsec.YPsec
σ ΦPQsec ⋅
200
π
= 0.2231
grad
NOTA BENE: In questo secondo conto, si nota che l'azimut (PQ) risulta identico nel rispetto dei calcoli
precedenti, non così invece per lo s.q.m. che risente delle diverse grandezze applicate nelle diverse formule
utilizzate. Inoltre visto lo scarto quadratico medio della sua determinazione, non se ne può non tenere
conto nella sua eventuale applicazione pratica.
Aggiornamento del 29/11/2009
Pagina 13