Wyrównanie kątów i długości pomierzonych oraz obliczenie błędów

1. Wyrównanie stacyjne

S i	Ksi	Ki	[cc]	∑vi	∑vi^2
	1	2		1	2
286p	0.0000	0.0000	0.0000	4.7	-4.7
283	84.2232	84.2236	84.22340	6.7	-6.7
282	85.4416	85.4404	85.44100	-1.3	1.3
B1	297.9320	297.9292	297.93060	-9.3	9.3
B2	299.0291	299.0280	299.02855	-0.8	0.8
Ks	153.32518	153.32424	153.32471 153.32471	10.1	-10.1
δs=K-Ki	-0.00047	0.00047		68.67	68.67

$m_{0} = \ \pm \sqrt{\frac{\sum({V_{\text{si}})}^{2}}{(n - 1)(s - 1)}} =$ 5.86

$m_{k} = \ \pm \ \frac{m_{0}}{\sqrt{2}} =$ 4.14

1. Wyrównanie metodą par spostrzeżeń

α₁=39,5386

α₂=39,9532

$d_{1} = \propto_{1}^{\text{II}} - \propto_{1}^{I} =$ 39.5374 - 39.5398= -0,0024

$d_{2} = \propto_{2}^{\text{II}} - \propto_{2}^{I} =$ 39.9536 – 39.9529 = +0,0007

$$m_{\propto} = \pm \frac{1}{2}\sqrt{\frac{\sum{d_{i}}^{2}}{n}}$$

$m_{\propto} = \ \pm \frac{1}{2}\sqrt{\frac{{( - 24)}^{2} + 7^{2}}{2}}$ = ± 8.8 ^cc

1. Błąd standardowy dalmierza

286pp-B1 = 127.116

B1–286pp = 127.117

$\overset{\overline{}}{b_{1}}$ = 127.1165 [m]

$$m_{\overset{\overline{}}{b_{1}}} = \ \pm \left( 5mm + 0.6mm \right) = \pm 5.6mm$$

b₁ = 127.1165m ± 5.6mm

286pp-B2 = 127.038

B2–286pp = 127.035

$\overset{\overline{}}{b_{2}}$ = 127.0365 [m]

$m_{\overset{\overline{}}{b_{2}}} = \ \pm \left( 5\text{mm} + 0.6\text{mm} \right) = \pm 5.6\text{mm}$

b₂ = 127.0365m ± 5.6mm

Zestawienie współrzędnych z błędami

Xpp= 5404750.204

Ypp= 4554090.279

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