Решение:

І этап: Расчет поправки МК по заходу/восходу солнца и ПУα со входом

в карту аналитическим счислением:

₊ T = 18.50 ˍ TBō = ₊ CC = ₊ TC =

𝓝 = CBō = ΔC = α =

____ ________________ ________________________ ____ ________________ _____________________

GMT = ΔC = TC = LTC =

Dec* =

ˍ R₂ = Δt = t₂ – t₁ = ___________________ ₊ φ_R =

R₁ = _______ ½ Δφ = _____________________________________________________________

͓ ΔR = Sc = x Δt = ___________________ φ_m =

k_L =

____________________

S_L = Δφ′ = S_T ∙ cos LTC = ________________________ = __________

Sc =

_ __________________________________

S_T = Δλ′ = (S_T ∙ sin LTC)/cosφ_m = ___________________ = __________

φ_R = λ_R =

Δφ′ = _________________ Δλ′ = ____________________

φ₁ = λ₁ =

ІІ этап: Вход в МНК. Определение поправки ГК и ОМС по высоте

Полярной звезды:

₊ T = POLARIS: ₊ RS_º =

𝓝 = i+s = Ориентир 1 ____________

___________________________________________ _ _____________________________________________________

GMT = ₊ h′ = ₊ GCB =

Δh_d = ΔGC =_________

______________________________________

₊ GHA^γ = ₊ h′′ = TB =

ΔGHA = Δht =

________________________________________

_____________________________________________________________

GHA^γ_T = ₊ H_o =

λ = ________________ ₊ α_o = Ориентир 2 ____________

LHA^γ = ₊ α₁ = ₊ GCB =

α ₂ = ΔGC = _________

________________________________________________________ _

ˍ TB* = Σ = TB =

GCB* = - 1°

________________________________ _________________________________________

ΔGC = φ_o =______________ λ_o = ______________

С = ______________

ІІІ этап: Аналитическое и графическое счисление. Решение прямой и

обратной задачи с учетом дрейфа и течения. ОМС

комбинированными способами:

₊ GCC = ₊ TC = ˍ AT = ₊ α =

ΔGC = α = LTC = β =

_____________________ ____________________________ ____________________________ _______________________________

TC = LTC = β = С =

_{┴ ориентира} _________________

₊ TC = T_┴ = T₂ + (60 / V) ∙ S_L = __________________________

_┴ = R_┴ = R + (S_L / k_L) = ____________________________

__________________________

TB_┴ =

S_L =

φ_o = __________________ λ_o = __________________ С = _________________

ˍ AT = ˍ LTC = ˍ TC = ₊ α =

LTC = α = ΔGC = β =

______________ ___________ __________________________ ________________________ ----------------------------------------

β = TC = GCC = С =

ˍ R₄ =

R₃ =

___ _____________ _________

͓ ΔR = ₊ GCB = ₊ GCC = ₊ TC =

k_L = ΔGC = ΔGC = _______ α = _________

S_L = TB = TС = LTC =

φ_o = _________________ λ_o = _______________ С = _______________

ˍ R₄ =

R₃ = ________

͓ ΔR =

k_L = ________ Δt = t₄ – t₃ = ______________________________

S_L = S_c = (Vт / 60) ∙ Δt = _______________________________

S_c = ________

S_T =

Ориентир 1_____________________ Ориентир 2_________________________

₊ GCB = ₊ GCB =

ΔGC = __________ ΔGC = __________

TB = TB =

φ_o = λ_o = С =

Stopped the engine. The drift.