Beispielberechnung Freier Polygonzug (Local)

Gegeben

Punktnummer	Y	X
1	0,000	0,000

Gemessen

Brechungswinkel, gon	Strecke, m
β_2 = 186,2397	s_1-2 = 53,541
β_3 = 175,5158	s_2-3 = 60,038
β_4 = 180,2461	s_3-4 = 55,994
	s_4-5 = 49,296

1. Punkt 1 ist Anfang lokalen Koordinatensystem, mit Koordinaten (0,000; 0,000), um negative Koordinaten zu vermeiden kann man z.B. (1000,000; 1000,000) als Anfang nehmen.

2. Die Verlängerung Strecke 1-2 ist lokale x-Achse.

3. Der Anfangsrichtungwinkel beträgt t1-2=0,000gon

4. Koordinaten des Punktes 2 betragen (y2 = 0,000; x = Strecke1-2), (0,000; 53,541)

5. Berechnung der Richtungswinkeln.

\(t_{1}= 0,0000_{gon}\) \(t_{2}= t_{1}+\beta_{1}\pm 200_{gon}= 0,000+186,2397+200=386,2397_{gon}\) \(t_{3}= t_{2}+\beta_{2}\pm 200_{gon}=386,2397+175,5158-200=361,7555_{gon}\) \(t_{4}= t_{3}+\beta_{3}\pm 200_{gon}=361,7555+180,2461-200=342,0016_{gon}\)

6. Berechnung der Koordinatenunterschiede.

\(\Delta y_{1}=s_{1} \cdot \sin t_{1}= 53,541 \cdot \sin(0,000)= 0,000m\) \(\Delta y_{2}=s_{2} \cdot \sin t_{2}= 60,038 \cdot \sin(386,2397)= -12,876m\) \(\Delta y_{3}=s_{3} \cdot \sin t_{3}= 55,994 \cdot \sin(361,7555)= -31,651m\) \(\Delta y_{4}=s_{4} \cdot \sin t_{4}= 49,296 \cdot \sin(342,0016)= -38,951m\)

\(\Delta x_{1}=s_{1} \cdot \cos t_{1}= 53,541 \cdot \cos(0,000)= +53,541m\) \(\Delta x_{2}=s_{2} \cdot \cos t_{2}= 60,038 \cdot \cos(386,2397)= +58,641m\) \(\Delta x_{3}=s_{3} \cdot \cos t_{3}= 55,994 \cdot \cos(361,7555)= +46,190m\) \(\Delta x_{4}=s_{4} \cdot \cos t_{4}= 49,296 \cdot \cos(342,0016)= +30,215m\)

7.Kontrolle der berechneten Koordinateunteschiede.

\(\sqrt2\cdot s_{1} \cdot \sin(t_{1}+50_{gon})=\sqrt2 \cdot 53,541 \cdot \sin(0+50_{gon})=+56,492\) \(\Delta x_{1}+\Delta y_{1} = 0,000+53,541=+53,541\)

\(\sqrt2\cdot s_{2} \cdot \sin(t_{2}+50_{gon})=\sqrt2 \cdot 60,038 \cdot \sin(386,2397+50_{gon})=+45,765\) \(\Delta x_{2}+\Delta y_{2} = -12,876+58,641=+45,765\)

\(\sqrt2\cdot s_{3} \cdot \sin(t_{3}+50_{gon})=\sqrt2 \cdot 55,994 \cdot \sin(361,7555+50_{gon})=+14,539\) \(\Delta x_{3}+\Delta y_{3} = -31,651+46,190=+14,539\)

\(\sqrt2\cdot s_{4} \cdot \sin(t_{4}+50_{gon})=\sqrt2 \cdot 49,296 \cdot \sin(342,0016+50_{gon})=-8,736\) \(\Delta x_{4}+\Delta y_{4} = -38,951+30,215=-8,736\)

8. Berechnung der Koordinaten.

\( y_{1}= y_{A} + \Delta y_{1} = +0,000+0,000=+0,000m\) \( y_{2}= y_{1} + \Delta y_{2} = +0,000-12,876=-12,876m\) \( y_{3}= y_{2} + \Delta y_{3} = -12,876-31,651=-44,527m\) \( y_{4}= y_{3} + \Delta y_{4} = -44,527-38,951=-83,478m\)

\( x_{1}= x_{A} + \Delta x_{1} = +0,000+53,541=+53,541m\) \( x_{2}= x_{1} + \Delta x_{2} = +53,541+58,641=+112,182m\) \( x_{3}= x_{2} + \Delta x_{3} = 112,182+46,190=158,372m\) \( x_{4}= x_{3} + \Delta x_{4} = 158,372+30,215=188,587m\)

Und so siht Berechnung im Programm aus.