$\frac{dd_{1}'}{dd_{1}} = \frac{1}{2\sqrt{{d_{1}}^{2} + e^{2} - 2d_{1}e\cos\theta}}\left( 2d_{1} - 2ecos\theta \right)\ $

$\frac{dd_{1}'}{\text{de}} = \frac{1}{2\sqrt{{d_{1}}^{2} + e^{2} - 2d_{1}e\cos\theta}}(2e - 2d_{1}cos\theta)$

$\frac{dd_{1}'}{\text{dθ}} = \frac{- 1}{2\sqrt{{d_{1}}^{2} + e^{2} - 2d_{1}e\cos\theta}}(2d_{1}esin\theta)$

md₁′= $\pm \sqrt{{(\frac{dd_{1}'}{dd_{1}})}^{2}{md_{1}}^{2} + {(\frac{dd_{1}'}{\text{de}})}^{2}\text{me}^{2} + {(\frac{dd_{1}'}{\text{dθ}})}^{2}{(\frac{\text{mθ}}{\rho})}^{2}}$

md₁^′=± 0,1m

md₂^′=± 0,2m

md₃^′=± 0,3m

6. Redukcja mimośrodowa kąta θ

θ_i = θ + K_i − K₁

θ₁= 291,2639^g

θ₂= 303,1011^g

θ₃= 332,5853^g

mθi=$\ \pm \sqrt{\text{mθ}^{2} + {mK_{i}}^{2} + {mK_{1}}^{2}} = \pm \sqrt{\text{mθ}^{2} + 2{mK_{i}}^{2}}$

mθ1= ± 3,62^c

mθ2= ± 3,62^c

mθ3= ± 3,62^c

6. Wyznaczenie kąta ε

$\varepsilon_{i} = arcsin(\frac{e}{d_{i}'}\sin\theta_{i})$

ε₁= -0,6262^g

ε₂= -1,0469^g

ε₃= -0,7064^g

$\frac{d\varepsilon_{i}}{de'} = \frac{1}{\sqrt{1 - {(\frac{e^{'}}{d_{i}'}\sin\theta_{i})}^{2}}}\frac{\sin\theta_{i}}{d_{i}'}$

$\frac{d\varepsilon_{i}}{dd_{i}'} = \frac{1}{\sqrt{1 - {(\frac{e^{'}}{d_{i}'}\sin\theta_{i})}^{2}}}e'sin\theta_{i}( - \frac{1}{{d_{1}'}^{2}})$

$\frac{d\varepsilon_{i}}{d\theta_{i}} = \frac{1}{\sqrt{1 - {(\frac{e^{'}}{d_{i}'}\sin\theta_{i})}^{2}}}\frac{e'}{d_{i}'}\cos\theta_{i}$

mε₁′= $\pm \sqrt{{(\frac{d\varepsilon_{i}}{\text{de}})}^{2}{(me\rho)}^{2} + {(\frac{d\varepsilon_{i}}{dd_{1}})}^{2}{(md_{1}\rho)}^{2} + {(\frac{d\varepsilon_{i}}{\text{dθ}})}^{2}{(m\theta)}^{2}}$

mε₁=±6,0^cc

mε₂=±10,9^cc

mε₃=±20,1^cc

6. Obliczenie kierunków z punktu centrycznego

K_i^′ = K_i + ε_i

K₁^′= 399,3738^g

K₂^′= 10,7902^g

K₃^′= 40,6150^g

mK₁^′= $\sqrt{m{K_{i}}^{2} + m{\varepsilon_{i}}^{2}}$

mK₁^′= 8,5^cc

mK₂^′= 12,4^cc

mK₃^′= 21,0^cc