2. OBLICZENIA. |
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2.2. KILOMETETRACJA TRASY. |
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UWAGA!: |
obliczając wpisz dane tylko w pola koloru |
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2.1. OBLICZENIE ŁUKÓW POZIOMYCH. |
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WARIANT 1 WIERZCHOŁEK 1 |
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WARIANT 1 ŁUK 1 |
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AW1= |
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0+335,00 |
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948,00 |
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PKP1= AW1-To= |
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0+335,00 |
- |
0+161,78 |
= |
0+858,32 |
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858,32 |
a) kąt zwrotu trasy: |
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α = |
23,18 |
[°] |
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PŁ= PKP1 +L/2 = |
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0+858,32 |
+ |
0+041,29 |
= |
0+886,48 |
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886,48 |
b) prędkość projektowa: |
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vp = |
50 |
[km/h] |
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KKP1= PŁ+L/2= |
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0+886,48 |
+ |
0+041,30 |
= |
0+914,65 |
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914,65 |
c) promień łuku kołowego: |
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R = |
300 |
[m] |
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KKP2= KKP1+K= |
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0+914,65 |
+ |
0+149,56 |
= |
0+979,69 |
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979,69 |
d) szerokość jezdni: |
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B = |
2,75 |
[m] |
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KŁ= KKP2+L/2= |
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0+979,69 |
+ |
0+041,29 |
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1+007,85 |
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1007,85 |
e) pochylenie poprzeczne na prostej: |
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ip = |
2 |
[%] |
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PKP2=KŁ+L/2= |
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1+007,85 |
+ |
0+041,30 |
= |
1+036,02 |
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1036,02 |
f) pochylenie poprzeczne na łuku: |
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io = |
3 |
[%] |
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1) WARUNEK DYNAMICZNY: |
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Amin1 = |
57,87 |
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GDZIE: |
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2) WARUNEK GEOMETRYCZNY ZESPOŁU KRZYWA PRZEJŚCIOWA-ŁUK KOŁOWY KRZYWA PRZEJŚCIOWA: |
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Amax2 = |
190,82 |
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3) WARUNEK ESTETYKI (WARUNEK KĄTA τ): |
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Amin3 = |
100,00 |
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To = |
89,68 |
m |
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Amax4 = |
300,00 |
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L= |
56,33 |
m |
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4) WARUNEK ESTETYKI (WIELKOŚĆ ODSUNIĘCIA ŁUKU H): |
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H = 0,5 m |
→ |
Amin5 = |
134,08 |
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L / 2 = |
28,17 |
m |
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H = 2,5 m |
→ |
Amax6 = |
200,39 |
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K = |
65,04 |
m |
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H = 0,2 m |
→ |
Amin7 = |
106,68 |
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5) WARUNEK KONSTRUKCYJNY (STOSOWANY NA ŁUKACH Z POSZERZENIEM): |
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40 / R = 0,11m |
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NIE UWZGLĘDNIAMY |
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6) WARUNEK KONSTRUKCYJNY (KOMFORTU JAZDY): |
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Amin8 = |
32,11 |
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134,08 |
≤ A ≤ |
190,82 |
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PRZY ZAŁOŻENIU PROPORCJI 1:2:1 |
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PRZY ZAŁOŻENIU PROPORCJI 1:1:1 |
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A = |
110,17 |
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A = |
134,93 |
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PRZYJMUJEMY PARAMETR KLOTOIDY RÓWNY A = |
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130 |
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1 |
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13 |
DŁUGOŚĆ KLOTOIDY: |
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L= |
56,33 |
m |
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L = A2 / R |
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KĄT |
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τ = |
0,094 |
rad |
γ = |
0,217 |
rad |
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DŁUGOŚĆ ŁUKU: |
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K = |
65,04 |
m |
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K = R · γ |
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OBLICZENIE PARAMETRÓW KLOTOIDY. |
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1) Odcięta X końca krzywej przejściowej: |
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56,28 |
m |
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2) Rzędna Y końca krzywej przejściowej: |
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1,76 |
m |
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3) Odcięta Xs środka krzywizny od początku krzywej przejściowej: |
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Xs = X - R · sin τ = |
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28,16 |
m |
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4) Odsunięcie H łuku koła od prostego kierunku trasy: |
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H = Y - R ( 1 - cos τ ) = |
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0,441 |
m |
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5) Dłudość stycznej T0 : |
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T0 = T + Xs = |
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89,68 |
m |
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T = R · tg α/2 = |
61,53 |
m |
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6) Odległość środka łuku Z0 od punktu przecięcia się stycznych: |
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6,69 |
m |
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2 |
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WARIANT 1 WIERZCHOŁEK 2 |
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W1 W2= |
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2+280,00 |
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950,00 |
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W2= AW1 +W1W2= |
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0+335,00 |
+ |
2+280,00 |
= |
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1+898,00 |
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1898,00 |
WARIANT 1 ŁUK 2 |
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PKP3= PKP2+W1W2-To-To2= |
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1+036,02 |
+ |
2+280,00 |
- |
0+089,68 |
- |
-57145+538,50 |
= |
57147+434,83 |
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57147434,83 |
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PŁ= PKP3 +L/2 = |
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57147+434,83 |
+ |
1+856,41 |
= |
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57149+291,24 |
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57149291,24 |
a) kąt zwrotu trasy: |
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α = |
35,94 |
[°] |
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KKP3= PŁ+L/2= |
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57149+291,24 |
+ |
1+856,41 |
= |
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57151+147,64 |
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57151147,64 |
b) prędkość projektowa: |
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vp = |
50 |
[km/h] |
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KKP4= KKP3+K2= |
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57151+147,64 |
+ |
0+149,56 |
= |
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57147+485,01 |
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57147485,01 |
c) promień łuku kołowego: |
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R = |
80 |
[m] |
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KŁ= KKP2+L/2= |
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57147+485,01 |
+ |
1+856,41 |
= |
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57149+341,42 |
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57149341,42 |
d) szerokość jezdni: |
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B = |
6,0 |
[m] |
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PKP4=KŁ+L/2= |
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57149+341,42 |
+ |
1+856,41 |
= |
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57151+197,82 |
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57151197,82 |
e) pochylenie poprzeczne na prostej: |
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ip = |
2 |
[%] |
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f) pochylenie poprzeczne na łuku: |
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io = |
3 |
[%] |
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GDZIE: |
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1) WARUNEK DYNAMICZNY: |
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Amin1 = |
57,87 |
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2) WARUNEK GEOMETRYCZNY ZESPOŁU KRZYWA PRZEJŚCIOWA-ŁUK KOŁOWY KRZYWA PRZEJŚCIOWA: |
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Amax2 = |
63,36 |
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To = |
89,68 |
m |
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3) WARUNEK ESTETYKI (WARUNEK KĄTA τ): |
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Amin3 = |
26,67 |
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To2 = |
-57145538 |
m |
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Amax4 = |
80,00 |
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L= |
3712,81 |
m |
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4) WARUNEK ESTETYKI (WIELKOŚĆ ODSUNIĘCIA ŁUKU H): |
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H = 0,5 m |
→ |
Amin5 = |
49,75 |
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L / 2 = |
1856,41 |
m |
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H = 2,5 m |
→ |
Amax6 = |
74,36 |
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K2 = |
-3662,63 |
m |
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H = 0,2 m |
→ |
Amin7 = |
39,59 |
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5) WARUNEK KONSTRUKCYJNY (STOSOWANY NA ŁUKACH Z POSZERZENIEM): |
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40 / R = |
0,50 |
NIE UWZGLĘDNIAMY |
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6) WARUNEK KONSTRUKCYJNY (KOMFORTU JAZDY): |
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Amin8 = |
24,49 |
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57,87 |
≤ A ≤ |
63,36 |
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PRZY ZAŁOŻENIU PROPORCJI 1:2:1 |
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PRZY ZAŁOŻENIU PROPORCJI 1:1:1 |
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A = |
36,58 |
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A = |
44,80 |
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PRZYJMUJEMY PARAMETR KLOTOIDY RÓWNY A = |
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545 |
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14 |
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3 |
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DŁUGOŚĆ KLOTOIDY: |
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L= |
3712,81 |
m |
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L = A2 / R |
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KĄT |
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τ = |
23,205 |
rad |
γ = |
-45,783 |
rad |
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DŁUGOŚĆ ŁUKU: |
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K = |
-3662,63 |
m |
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K = R · γ |
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OBLICZENIE PARAMETRÓW KLOTOIDY. |
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1) Odcięta X końca krzywej przejściowej: |
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-57145639,4 |
m |
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2) Rzędna Y końca krzywej przejściowej: |
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17940029,69 |
m |
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3) Odcięta Xs środka krzywizny od początku krzywej przejściowej: |
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Xs = X - R · sin τ = |
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-57145564,4 |
m |
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4) Odsunięcie H łuku koła od prostego kierunku trasy: |
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H = Y - R ( 1 - cos τ ) = |
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17939921,75 |
m |
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5) Dłudość stycznej T0 : |
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T0 = T + Xs = |
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|
-57145538,5 |
m |
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T = R · tg α/2 = |
25,95 |
m |
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6) Odległość środka łuku Z0 od punktu przecięcia się stycznych: |
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|
18859948,12 |
m |
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4 |
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WARIANT 2 WIERZCHOŁEK 1 |
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AW1= |
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0+722,00 |
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722,00 |
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PKP1= AW1-To= |
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0+722,00 |
- |
0+161,78 |
= |
#DIV/0! |
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#DIV/0! |
WARIANT 2 ŁUK 1 |
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PŁ= PKP1 +L/2 = |
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#DIV/0! |
+ |
#DIV/0! |
= |
#DIV/0! |
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#DIV/0! |
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KKP1= PŁ+L/2= |
|
#DIV/0! |
+ |
#DIV/0! |
= |
#DIV/0! |
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#DIV/0! |
a) kąt zwrotu trasy: |
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α = |
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[°] |
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KKP2= KKP1+K= |
|
#DIV/0! |
+ |
0+149,56 |
= |
#DIV/0! |
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#DIV/0! |
b) prędkość projektowa: |
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vp = |
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[km/h] |
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KŁ= KKP2+L/2= |
|
#DIV/0! |
+ |
#DIV/0! |
= |
#DIV/0! |
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#DIV/0! |
c) promień łuku kołowego: |
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R = |
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[m] |
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PKP2=KŁ+L/2= |
|
#DIV/0! |
+ |
#DIV/0! |
= |
#DIV/0! |
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#DIV/0! |
d) szerokość jezdni: |
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B = |
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[m] |
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e) pochylenie poprzeczne na prostej: |
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ip = |
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[%] |
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f) pochylenie poprzeczne na łuku: |
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io = |
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[%] |
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GDZIE: |
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1) WARUNEK DYNAMICZNY: |
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Amin1 = |
0,00 |
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To = |
#DIV/0! |
m |
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2) WARUNEK GEOMETRYCZNY ZESPOŁU KRZYWA PRZEJŚCIOWA-ŁUK KOŁOWY KRZYWA PRZEJŚCIOWA: |
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Amax2 = |
0,00 |
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L= |
#DIV/0! |
m |
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3) WARUNEK ESTETYKI (WARUNEK KĄTA τ): |
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Amin3 = |
0,00 |
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L / 2 = |
#DIV/0! |
m |
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Amax4 = |
0,00 |
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K = |
#DIV/0! |
m |
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4) WARUNEK ESTETYKI (WIELKOŚĆ ODSUNIĘCIA ŁUKU H): |
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|
|
H = 0,5 m |
→ |
Amin5 = |
0,00 |
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H = 2,5 m |
→ |
Amax6 = |
0,00 |
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H = 0,2 m |
→ |
Amin7 = |
0,00 |
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5) WARUNEK KONSTRUKCYJNY (STOSOWANY NA ŁUKACH Z POSZERZENIEM): |
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|
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40 / R = |
#DIV/0! |
NIE UWZGLĘDNIAMY |
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6) WARUNEK KONSTRUKCYJNY (KOMFORTU JAZDY): |
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Amin8 = |
0,00 |
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0,00 |
≤ A ≤ |
0,00 |
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PRZY ZAŁOŻENIU PROPORCJI 1:2:1 |
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PRZY ZAŁOŻENIU PROPORCJI 1:1:1 |
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A = |
0,00 |
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A = |
0,00 |
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PRZYJMUJEMY PARAMETR KLOTOIDY RÓWNY A = |
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185 |
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7 |
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16 |
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DŁUGOŚĆ KLOTOIDY: |
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L= |
#DIV/0! |
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L = A2 / R |
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KĄT |
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τ = |
#DIV/0! |
rad |
γ = |
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DŁUGOŚĆ ŁUKU: |
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K = |
#DIV/0! |
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K = R · γ |
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OBLICZENIE PARAMETRÓW KLOTOIDY. |
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1) Odcięta X końca krzywej przejściowej: |
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#DIV/0! |
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2) Rzędna Y końca krzywej przejściowej: |
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#DIV/0! |
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3) Odcięta Xs środka krzywizny od początku krzywej przejściowej: |
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Xs = X - R · sin τ = |
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#DIV/0! |
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4) Odsunięcie H łuku koła od prostego kierunku trasy: |
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H = Y - R ( 1 - cos τ ) = |
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#DIV/0! |
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5) Dłudość stycznej T0 : |
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T0 = T + Xs = |
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#DIV/0! |
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T = R · tg α/2 = |
0,00 |
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6) Odległość środka łuku Z0 od punktu przecięcia się stycznych: |
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#DIV/0! |
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8 |
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WARIANT 2 WIERZCHOŁEK 2 |
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W1 W2= |
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0+921,00 |
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921,00 |
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W2= AW1 +W1W2= |
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0+722,00 |
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#DIV/0! |
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#DIV/0! |
WARIANT 2 ŁUK 2 |
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PKP3= PKP2+W1W2-To-To2= |
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0,0 |
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0+921,00 |
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#DIV/0! |
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#DIV/0! |
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#DIV/0! |
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#DIV/0! |
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PŁ= PKP3 +L/2 = |
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#DIV/0! |
+ |
#DIV/0! |
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#DIV/0! |
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#DIV/0! |
a) kąt zwrotu trasy: |
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α = |
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[°] |
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KKP3= PŁ+L/2= |
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#DIV/0! |
+ |
#DIV/0! |
= |
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#DIV/0! |
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#DIV/0! |
b) prędkość projektowa: |
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vp = |
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[km/h] |
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KKP4= KKP3+K2= |
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#DIV/0! |
+ |
0+149,56 |
= |
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#DIV/0! |
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#DIV/0! |
c) promień łuku kołowego: |
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R = |
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[m] |
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KŁ= KKP2+L/2= |
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#DIV/0! |
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#DIV/0! |
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#DIV/0! |
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#DIV/0! |
d) szerokość jezdni: |
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B = |
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[m] |
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PKP4=KŁ+L/2= |
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#DIV/0! |
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#DIV/0! |
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#DIV/0! |
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#DIV/0! |
e) pochylenie poprzeczne na prostej: |
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ip = |
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f) pochylenie poprzeczne na łuku: |
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io = |
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[%] |
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GDZIE: |
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1) WARUNEK DYNAMICZNY: |
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Amin1 = |
0,00 |
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2) WARUNEK GEOMETRYCZNY ZESPOŁU KRZYWA PRZEJŚCIOWA-ŁUK KOŁOWY KRZYWA PRZEJŚCIOWA: |
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Amax2 = |
0,00 |
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To = |
#DIV/0! |
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3) WARUNEK ESTETYKI (WARUNEK KĄTA τ): |
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Amin3 = |
0,00 |
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To2 = |
#DIV/0! |
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Amax4 = |
0,00 |
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L= |
#DIV/0! |
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4) WARUNEK ESTETYKI (WIELKOŚĆ ODSUNIĘCIA ŁUKU H): |
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H = 0,5 m |
→ |
Amin5 = |
0,00 |
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L / 2 = |
#DIV/0! |
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H = 2,5 m |
→ |
Amax6 = |
0,00 |
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K2 = |
#DIV/0! |
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H = 0,2 m |
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0,00 |
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5) WARUNEK KONSTRUKCYJNY (STOSOWANY NA ŁUKACH Z POSZERZENIEM): |
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40 / R = |
#DIV/0! |
NIE UWZGLĘDNIAMY |
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6) WARUNEK KONSTRUKCYJNY (KOMFORTU JAZDY): |
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Amin8 = |
0,00 |
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0,00 |
≤ A ≤ |
0,00 |
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PRZY ZAŁOŻENIU PROPORCJI 1:2:1 |
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PRZY ZAŁOŻENIU PROPORCJI 1:1:1 |
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A = |
0,00 |
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A = |
0,00 |
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PRZYJMUJEMY PARAMETR KLOTOIDY RÓWNY A = |
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185 |
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9 |
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17 |
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DŁUGOŚĆ KLOTOIDY: |
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L= |
#DIV/0! |
m |
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L = A2 / R |
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KĄT |
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τ = |
#DIV/0! |
rad |
γ = |
#DIV/0! |
rad |
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DŁUGOŚĆ ŁUKU: |
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K = |
#DIV/0! |
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K = R · γ |
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OBLICZENIE PARAMETRÓW KLOTOIDY. |
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1) Odcięta X końca krzywej przejściowej: |
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#DIV/0! |
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2) Rzędna Y końca krzywej przejściowej: |
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#DIV/0! |
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3) Odcięta Xs środka krzywizny od początku krzywej przejściowej: |
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Xs = X - R · sin τ = |
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#DIV/0! |
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4) Odsunięcie H łuku koła od prostego kierunku trasy: |
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H = Y - R ( 1 - cos τ ) = |
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#DIV/0! |
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5) Dłudość stycznej T0 : |
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T0 = T + Xs = |
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#DIV/0! |
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T = R · tg α/2 = |
0,00 |
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6) Odległość środka łuku Z0 od punktu przecięcia się stycznych: |
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#DIV/0! |
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WARIANT 1 WIERZCHOŁEK 3 |
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985 |
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W2 W3= |
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0+816,00 |
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816,00 |
WARIANT 2 ŁUK 3 |
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W3= W2 +W2W3= |
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#DIV/0! |
+ |
0+816,00 |
= |
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#DIV/0! |
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#DIV/0! |
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PKP5= PKP4+W2W3-To2-To3= |
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#DIV/0! |
+ |
#DIV/0! |
- |
#DIV/0! |
- |
#DIV/0! |
= |
#DIV/0! |
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#DIV/0! |
a) kąt zwrotu trasy: |
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α = |
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[°] |
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PŁ= PKP5+L/2 = |
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#DIV/0! |
+ |
#DIV/0! |
= |
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#DIV/0! |
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#DIV/0! |
b) prędkość projektowa: |
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vp = |
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[km/h] |
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KKP5= PŁ+L/2= |
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#DIV/0! |
+ |
#DIV/0! |
= |
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#DIV/0! |
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#DIV/0! |
c) promień łuku kołowego: |
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R = |
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[m] |
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KKP6= KKP5+K3= |
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#DIV/0! |
+ |
0+149,56 |
= |
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#DIV/0! |
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#DIV/0! |
d) szerokość jezdni: |
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B = |
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[m] |
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KŁ= KKP6+L/2= |
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#DIV/0! |
+ |
#DIV/0! |
= |
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#DIV/0! |
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#DIV/0! |
e) pochylenie poprzeczne na prostej: |
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ip = |
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[%] |
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PKP6=KŁ+L/2= |
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#DIV/0! |
+ |
#DIV/0! |
= |
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#DIV/0! |
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#DIV/0! |
f) pochylenie poprzeczne na łuku: |
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io = |
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[%] |
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B= PKP6 +W3B-To= |
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#DIV/0! |
+ |
0+985,00 |
- |
#DIV/0! |
= |
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#DIV/0! |
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#DIV/0! |
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1) WARUNEK DYNAMICZNY: |
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Amin1 = |
0,00 |
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GDZIE: |
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2) WARUNEK GEOMETRYCZNY ZESPOŁU KRZYWA PRZEJŚCIOWA-ŁUK KOŁOWY KRZYWA PRZEJŚCIOWA: |
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Amax2 = |
0,00 |
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3) WARUNEK ESTETYKI (WARUNEK KĄTA τ): |
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Amin3 = |
0,00 |
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To2 = |
#DIV/0! |
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Amax4 = |
0,00 |
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To3 = |
#DIV/0! |
m |
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4) WARUNEK ESTETYKI (WIELKOŚĆ ODSUNIĘCIA ŁUKU H): |
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H = 0,5 m |
→ |
Amin5 = |
0,00 |
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L= |
#DIV/0! |
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H = 2,5 m |
→ |
Amax6 = |
0,00 |
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L / 2 = |
#DIV/0! |
m |
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H = 0,2 m |
→ |
Amin7 = |
0,00 |
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K3 = |
#DIV/0! |
m |
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5) WARUNEK KONSTRUKCYJNY (STOSOWANY NA ŁUKACH Z POSZERZENIEM): |
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40 / R = |
#DIV/0! |
NIE UWZGLĘDNIAMY |
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6) WARUNEK KONSTRUKCYJNY (KOMFORTU JAZDY): |
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Amin8 = |
0,00 |
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0,00 |
≤ A ≤ |
0,00 |
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PRZY ZAŁOŻENIU PROPORCJI 1:2:1 |
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PRZY ZAŁOŻENIU PROPORCJI 1:1:1 |
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A = |
0,00 |
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A = |
0,00 |
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PRZYJMUJEMY PARAMETR KLOTOIDY RÓWNY A = |
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185 |
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11 |
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18 |
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DŁUGOŚĆ KLOTOIDY: |
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L= |
#DIV/0! |
m |
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L = A2 / R |
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KĄT |
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τ = |
#DIV/0! |
rad |
γ = |
#DIV/0! |
rad |
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DŁUGOŚĆ ŁUKU: |
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K = |
#DIV/0! |
m |
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K = R · γ |
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OBLICZENIE PARAMETRÓW KLOTOIDY. |
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1) Odcięta X końca krzywej przejściowej: |
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#DIV/0! |
m |
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2) Rzędna Y końca krzywej przejściowej: |
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#DIV/0! |
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3) Odcięta Xs środka krzywizny od początku krzywej przejściowej: |
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Xs = X - R · sin τ = |
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#DIV/0! |
m |
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4) Odsunięcie H łuku koła od prostego kierunku trasy: |
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H = Y - R ( 1 - cos τ ) = |
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#DIV/0! |
m |
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5) Dłudość stycznej T0 : |
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T0 = T + Xs = |
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#DIV/0! |
m |
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T = R · tg α/2 = |
0,00 |
m |
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6) Odległość środka łuku Z0 od punktu przecięcia się stycznych: |
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#DIV/0! |
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12 |
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CO DO DALSZEJ CZĘŚCI TO ZNACZY ODTĄD W DÓŁ NIE JESTEM PEWIEN CZY ARKUSZ POLICZY SAMODZIELNIE :) |
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2.3. TABELA ZAŁOMÓW. |
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2.3.1. WARIANT I. |
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50 |
|
|
|
|
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|
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|
|
PIKIETAŻ |
ODLEGŁÓŚĆ |
POCHYLENIE [%] |
RÓŻNICA WYSOKOŚCI Δh [m] |
RZĘDNA ZAŁOMU |
|
|
|
|
|
|
|
51,25 |
|
|
|
|
|
|
|
|
|
|
|
|
WZNIESIENIE |
SPADEK |
|
|
|
|
|
|
|
52,5 |
|
|
|
|
|
|
|
|
|
|
|
|
0+000,00 |
- |
- |
- |
- |
57,50 |
|
|
|
|
|
0 |
|
53,75 |
|
|
|
|
|
|
|
|
|
|
|
|
0+354,00 |
354,00 |
- |
0,34 |
1,2 |
56,30 |
|
|
|
|
|
354 |
|
55 |
|
|
|
|
|
|
|
|
|
|
|
|
0+600,00 |
246,00 |
1,50 |
- |
3,7 |
60,00 |
|
|
|
|
|
600 |
|
56,25 |
|
|
|
|
|
|
|
|
|
|
|
|
1+020,00 |
420,00 |
- |
0,48 |
2 |
58,00 |
|
|
|
|
|
1020 |
|
57,5 |
|
|
|
|
|
|
|
|
|
|
|
|
1+250,00 |
230,00 |
0,30 |
- |
0,700000000000003 |
58,70 |
|
|
|
|
|
1250 |
|
58,75 |
|
|
|
|
|
|
|
|
|
|
|
|
2+360,00 |
1110,00 |
0,15 |
- |
1,7 |
60,40 |
|
|
|
|
|
2360 |
|
60 |
|
|
|
|
|
|
|
|
|
|
|
|
2+925,00 |
565,00 |
- |
0,29 |
1,65 |
58,75 |
|
|
|
|
|
2925 |
|
61,25 |
|
|
|
|
|
|
|
|
|
|
|
|
3+102,74 |
177,74 |
- |
0,56 |
1 |
57,75 |
|
|
|
|
|
3102,74 |
|
62,5 |
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
|
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|
63,75 |
|
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2.3.2. WARIANT II. |
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|
65 |
|
|
|
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|
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|
|
|
|
|
|
|
|
66,25 |
|
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|
|
PIKIETAŻ |
ODLEGŁÓŚĆ |
POCHYLENIE [%] |
RÓŻNICA WYSOKOŚCI Δh [m] |
RZĘDNA ZAŁOMU |
|
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|
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|
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|
|
|
|
|
|
|
|
|
WZNIESIENIE |
SPADEK |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
0+000,00 |
- |
- |
- |
- |
57,50 |
|
|
|
|
|
0 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
0+440,00 |
440,00 |
- |
0,57 |
2,5 |
55,00 |
|
|
|
|
|
440 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
0+970,00 |
530,00 |
- |
0,38 |
2 |
53,00 |
|
|
|
|
|
970 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1+430,00 |
460,00 |
1,26 |
- |
5,8 |
58,80 |
|
|
|
|
|
1430 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1+680,00 |
250,00 |
- |
0,76 |
1,9 |
56,90 |
|
|
|
|
|
1680 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1+910,00 |
230,00 |
0,13 |
- |
0,300000000000004 |
57,20 |
|
|
|
|
|
1910 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
2+700,00 |
790,00 |
0,51 |
- |
4 |
61,20 |
|
|
|
|
|
2700 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
3+338,38 |
638,38 |
- |
0,54 |
3,45 |
57,75 |
|
|
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|
|
3338,38 |
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19 |
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2.4. OBLICZENIA ŁUKÓW PIONOWYCH. |
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2.4.1. WARIANT I. |
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|
PIKIETA |
78 |
i1 |
i2 |
i |
R |
t |
f |
|
|
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|
|
0+354,00 |
8 |
0,34% |
1,50% |
1,84% |
3000 |
27,6 |
0,127 |
|
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|
|
0+600,00 |
7 |
1,50% |
0,48% |
1,98% |
3000 |
29,7 |
0,147 |
|
|
|
|
|
|
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|
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|
|
|
|
|
|
1+020,00 |
8 |
0,48% |
0,30% |
0,78% |
3000 |
11,7 |
0,023 |
|
|
|
|
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|
|
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|
|
2+360,00 |
7 |
0,15% |
0,29% |
0,44% |
3000 |
6,6 |
0,007 |
|
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2.4.2. WARIANT Ii. |
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|
PIKIETA |
78 |
i1 |
i2 |
i |
R |
t |
f |
|
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|
|
0+970,00 |
7 |
0,38% |
1,26% |
1,64% |
3000 |
24,6 |
0,101 |
|
|
|
|
|
|
|
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|
|
|
|
|
|
|
|
|
1+430,00 |
8 |
1,26% |
0,76% |
2,02% |
3000 |
30,3 |
0,153 |
|
|
|
|
|
|
|
|
|
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|
|
|
|
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|
|
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|
|
1+680,00 |
7 |
0,76% |
0,13% |
0,89% |
3000 |
13,35 |
0,030 |
|
|
|
|
|
|
|
|
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|
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|
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|
|
2+700,00 |
8 |
0,51% |
0,54% |
1,05% |
3000 |
15,75 |
0,041 |
|
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20 |
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2.5. DŁUGOŚĆ WIRTUALNA. |
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|
Lsp=Σl+Σiw/f x lw-Σis/f x ls |
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|
f = |
0,018 |
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2.5.1. WARIANT I. |
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OBLICZENIA W KIERUNKU ZGODNYM Z PIKIETAŻEM TRASY A-B |
|
OBLICZENIA W KIERUNKU PRZECIWNYM Z PIKIETAŻEM TRASY B-A |
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|
ODCINEK |
POCHYLENIE PODŁUŻNE [%] |
DŁUGOŚĆ ODCINKÓW WZNIESIEŃ ORAZ SPADKÓW NIESZKODLIWYCH [m] |
|
|
|
ODCINEK |
POCHYLENIE PODŁUŻNE [%] |
DŁUGOŚĆ ODCINKÓW WZNIESIEŃ ORAZ SPADKÓW NIESZKODLIWYCH [m] |
|
|
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|
ODCINEK 1 |
-0,34 |
354,00 |
- |
66,87 |
|
ODCINEK 1 |
0,34 |
354,00 |
66,87 |
- |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
ODCINEK 2 |
1,50 |
246,00 |
205,00 |
- |
|
ODCINEK 2 |
-1,50 |
246,00 |
- |
205,00 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
ODCINEK 3 |
-0,48 |
420,00 |
- |
112,00 |
|
ODCINEK 3 |
0,48 |
420,00 |
112,00 |
- |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
ODCINEK 4 |
0,30 |
230,00 |
38,33 |
- |
|
ODCINEK 4 |
-0,30 |
230,00 |
- |
38,33 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
ODCINEK 5 |
0,15 |
1110,00 |
92,50 |
- |
|
ODCINEK 5 |
-0,15 |
1110,00 |
- |
92,50 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
ODCINEK 6 |
-0,29 |
565,00 |
- |
91,03 |
|
ODCINEK 6 |
0,29 |
565,00 |
91,03 |
- |
|
|
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|
ODCINEK 7 |
-0,56 |
177,74 |
- |
55,30 |
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ODCINEK 7 |
0,56 |
177,74 |
55,30 |
- |
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SUMA = |
3102,74 |
335,83 |
325,19 |
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SUMA = |
3102,74 |
325,19 |
335,83 |
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Lsp= |
3113,38 |
m |
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Lsp= |
3092,10 |
m |
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DŁUGOŚĆ WIRTUALNA POTRZEBNA DO OCENY WARIANTU I WYNOSI: |
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L = |
3102,74 |
m |
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21 |
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Lsp=Σl+Σiw/f x lw-Σis/f x ls |
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f = |
0,018 |
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2.5.2. WARIANT II. |
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OBLICZENIA W KIERUNKU ZGODNYM Z PIKIETAŻEM TRASY A-B |
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OBLICZENIA W KIERUNKU PRZECIWNYM Z PIKIETAŻEM TRASY B-A |
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ODCINEK |
POCHYLENIE PODŁUŻNE [%] |
DŁUGOŚĆ ODCINKÓW WZNIESIEŃ ORAZ SPADKÓW NIESZKODLIWYCH [m] |
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ODCINEK |
POCHYLENIE PODŁUŻNE [%] |
DŁUGOŚĆ ODCINKÓW WZNIESIEŃ ORAZ SPADKÓW NIESZKODLIWYCH [m] |
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ODCINEK 1 |
-0,57 |
440,00 |
- |
139,33 |
|
ODCINEK 1 |
0,57 |
440,00 |
139,33 |
- |
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ODCINEK 2 |
-0,38 |
530,00 |
- |
111,89 |
|
ODCINEK 2 |
0,38 |
530,00 |
111,89 |
- |
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ODCINEK 3 |
1,26 |
460,00 |
322,00 |
- |
|
ODCINEK 3 |
-1,26 |
460,00 |
- |
322,00 |
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ODCINEK 4 |
-0,76 |
250,00 |
- |
105,56 |
|
ODCINEK 4 |
0,76 |
250,00 |
105,56 |
- |
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ODCINEK 5 |
0,13 |
230,00 |
16,61 |
- |
|
ODCINEK 5 |
-0,13 |
230,00 |
- |
16,61 |
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ODCINEK 6 |
0,51 |
790,00 |
223,83 |
- |
|
ODCINEK 6 |
-0,51 |
790,00 |
- |
223,83 |
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ODCINEK 7 |
-0,54 |
638,38 |
- |
191,51 |
|
ODCINEK 7 |
0,54 |
638,38 |
191,51 |
- |
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SUMA = |
3338,38 |
562,44 |
548,29 |
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SUMA = |
3338,38 |
548,29 |
562,44 |
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Lsp= |
3352,53 |
m |
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Lsp= |
3324,23 |
m |
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DŁUGOŚĆ WIRTUALNA POTRZEBNA DO OCENY WARIANTU I WYNOSI: |
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L = |
3338,38 |
m |
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22 |
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2. OBLICZENIA. |
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2.2. KILOMETETRACJA TRASY. |
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2.1. OBLICZENIE ŁUKÓW POZIOMYCH. |
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WARIANT 1 WIERZCHOŁEK 1 |
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WARIANT 1 ŁUK 1 |
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AW1= |
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2+373,00 |
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948,00 |
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PKP1= AW1-To= |
|
2+373,00 |
- |
0+161,78 |
= |
0+606,17 |
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|
606,17 |
a) kąt zwrotu trasy: |
|
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α = |
61 |
[°] |
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PŁ= PKP1 +L/2 = |
|
0+606,17 |
+ |
0+041,29 |
= |
0+712,63 |
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|
712,63 |
b) prędkość projektowa: |
|
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vp = |
40 |
[km/h] |
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KKP1= PŁ+L/2= |
|
0+712,63 |
+ |
0+041,30 |
= |
0+819,10 |
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|
819,10 |
c) promień łuku kołowego: |
|
|
R = |
400 |
[m] |
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KKP2= KKP1+K= |
|
0+819,10 |
+ |
0+149,56 |
= |
1+032,03 |
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|
1032,03 |
d) szerokość jezdni: |
|
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B = |
2,75 |
[m] |
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KŁ= KKP2+L/2= |
|
1+032,03 |
+ |
0+041,29 |
= |
1+138,49 |
|
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|
1138,49 |
e) pochylenie poprzeczne na prostej: |
|
|
ip = |
2 |
[%] |
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PKP2=KŁ+L/2= |
|
1+138,49 |
+ |
0+041,30 |
= |
1+244,96 |
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|
1244,96 |
f) pochylenie poprzeczne na łuku: |
|
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io = |
3 |
[%] |
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1) WARUNEK DYNAMICZNY: |
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Amin1 = |
41,41 |
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GDZIE: |
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2) WARUNEK GEOMETRYCZNY ZESPOŁU KRZYWA PRZEJŚCIOWA-ŁUK KOŁOWY KRZYWA PRZEJŚCIOWA: |
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Amax2 = |
412,73 |
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3) WARUNEK ESTETYKI (WARUNEK KĄTA τ): |
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Amin3 = |
133,33 |
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To = |
341,83 |
m |
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Amax4 = |
400,00 |
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L= |
212,93 |
m |
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4) WARUNEK ESTETYKI (WIELKOŚĆ ODSUNIĘCIA ŁUKU H): |
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|
|
H = 0,5 m |
→ |
Amin5 = |
166,36 |
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L / 2 = |
106,46 |
m |
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H = 2,5 m |
→ |
Amax6 = |
248,65 |
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K = |
212,93 |
m |
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|
H = 0,2 m |
→ |
Amin7 = |
132,38 |
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5) WARUNEK KONSTRUKCYJNY (STOSOWANY NA ŁUKACH Z POSZERZENIEM): |
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40 / R = 0,11m |
|
NIE UWZGLĘDNIAMY |
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6) WARUNEK KONSTRUKCYJNY (KOMFORTU JAZDY): |
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Amin8 = |
37,08 |
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166,36 |
≤ A ≤ |
248,65 |
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PRZY ZAŁOŻENIU PROPORCJI 1:2:1 |
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PRZY ZAŁOŻENIU PROPORCJI 1:1:1 |
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A = |
238,29 |
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A = |
291,84 |
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PRZYJMUJEMY PARAMETR KLOTOIDY RÓWNY A = |
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291,84 |
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1 |
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13 |
DŁUGOŚĆ KLOTOIDY: |
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L= |
212,93 |
m |
|
L = A2 / R |
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KĄT |
|
|
τ = |
0,266 |
rad |
γ = |
0,532 |
rad |
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DŁUGOŚĆ ŁUKU: |
|
|
K = |
212,93 |
m |
|
K = R · γ |
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OBLICZENIE PARAMETRÓW KLOTOIDY. |
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1) Odcięta X końca krzywej przejściowej: |
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211,42 |
m |
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2) Rzędna Y końca krzywej przejściowej: |
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18,80 |
m |
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3) Odcięta Xs środka krzywizny od początku krzywej przejściowej: |
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Xs = X - R · sin τ = |
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106,21 |
m |
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4) Odsunięcie H łuku koła od prostego kierunku trasy: |
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H = Y - R ( 1 - cos τ ) = |
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4,719 |
m |
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5) Dłudość stycznej T0 : |
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T0 = T + Xs = |
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341,83 |
m |
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T = R · tg α/2 = |
235,62 |
m |
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6) Odległość środka łuku Z0 od punktu przecięcia się stycznych: |
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69,71 |
m |
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2 |
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WARIANT 1 WIERZCHOŁEK 2 |
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W1 W2= |
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2+280,00 |
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950,00 |
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W2= AW1 +W1W2= |
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2+373,00 |
+ |
2+280,00 |
= |
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1+898,00 |
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1898,00 |
WARIANT 1 ŁUK 2 |
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PKP3= PKP2+W1W2-To-To2= |
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1+244,96 |
+ |
2+280,00 |
- |
0+341,83 |
- |
0+193,90 |
= |
1+659,23 |
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1659,23 |
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PŁ= PKP3 +L/2 = |
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1+659,23 |
+ |
0+054,00 |
= |
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1+713,23 |
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1713,23 |
a) kąt zwrotu trasy: |
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α = |
50,02 |
[°] |
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KKP3= PŁ+L/2= |
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1+713,23 |
+ |
0+054,00 |
= |
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1+767,23 |
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1767,23 |
b) prędkość projektowa: |
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vp = |
50 |
[km/h] |
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KKP4= KKP3+K2= |
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1+767,23 |
+ |
0+149,56 |
= |
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1+921,13 |
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1921,13 |
c) promień łuku kołowego: |
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R = |
300 |
[m] |
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KŁ= KKP2+L/2= |
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1+921,13 |
+ |
0+054,00 |
= |
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1+975,13 |
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1975,13 |
d) szerokość jezdni: |
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B = |
2,8 |
[m] |
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PKP4=KŁ+L/2= |
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1+975,13 |
+ |
0+054,00 |
= |
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2+029,13 |
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2029,13 |
e) pochylenie poprzeczne na prostej: |
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ip = |
2 |
[%] |
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f) pochylenie poprzeczne na łuku: |
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io = |
3 |
[%] |
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GDZIE: |
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1) WARUNEK DYNAMICZNY: |
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Amin1 = |
57,87 |
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2) WARUNEK GEOMETRYCZNY ZESPOŁU KRZYWA PRZEJŚCIOWA-ŁUK KOŁOWY KRZYWA PRZEJŚCIOWA: |
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Amax2 = |
280,31 |
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To = |
341,83 |
m |
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3) WARUNEK ESTETYKI (WARUNEK KĄTA τ): |
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Amin3 = |
100,00 |
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To2 = |
194 |
m |
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Amax4 = |
300,00 |
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L= |
108,00 |
m |
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4) WARUNEK ESTETYKI (WIELKOŚĆ ODSUNIĘCIA ŁUKU H): |
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H = 0,5 m |
→ |
Amin5 = |
134,08 |
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L / 2 = |
54,00 |
m |
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H = 2,5 m |
→ |
Amax6 = |
200,39 |
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K2 = |
153,90 |
m |
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H = 0,2 m |
→ |
Amin7 = |
106,68 |
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5) WARUNEK KONSTRUKCYJNY (STOSOWANY NA ŁUKACH Z POSZERZENIEM): |
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40 / R = |
0,13 |
NIE UWZGLĘDNIAMY |
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6) WARUNEK KONSTRUKCYJNY (KOMFORTU JAZDY): |
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Amin8 = |
32,11 |
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134,08 |
≤ A ≤ |
200,39 |
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PRZY ZAŁOŻENIU PROPORCJI 1:2:1 |
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PRZY ZAŁOŻENIU PROPORCJI 1:1:1 |
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A = |
161,83 |
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A = |
198,21 |
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PRZYJMUJEMY PARAMETR KLOTOIDY RÓWNY A = |
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180 |
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14 |
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3 |
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DŁUGOŚĆ KLOTOIDY: |
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L= |
108,00 |
m |
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L = A2 / R |
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KĄT |
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τ = |
0,180 |
rad |
γ = |
0,513 |
rad |
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DŁUGOŚĆ ŁUKU: |
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K = |
153,90 |
m |
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K = R · γ |
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OBLICZENIE PARAMETRÓW KLOTOIDY. |
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1) Odcięta X końca krzywej przejściowej: |
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107,7 |
m |
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2) Rzędna Y końca krzywej przejściowej: |
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6,47 |
m |
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3) Odcięta Xs środka krzywizny od początku krzywej przejściowej: |
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Xs = X - R · sin τ = |
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53,9 |
m |
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4) Odsunięcie H łuku koła od prostego kierunku trasy: |
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H = Y - R ( 1 - cos τ ) = |
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1,62 |
m |
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5) Dłudość stycznej T0 : |
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T0 = T + Xs = |
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193,9 |
m |
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T = R · tg α/2 = |
139,96 |
m |
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6) Odległość środka łuku Z0 od punktu przecięcia się stycznych: |
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32,83 |
m |
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4 |
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WARIANT 2 WIERZCHOŁEK 1 |
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AW1= |
|
0+722,00 |
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722,00 |
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PKP1= AW1-To= |
|
0+722,00 |
- |
0+161,78 |
= |
0+527,91 |
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527,91 |
WARIANT 2 ŁUK 1 |
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PŁ= PKP1 +L/2 = |
|
0+527,91 |
+ |
0+057,04 |
= |
0+581,91 |
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581,91 |
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KKP1= PŁ+L/2= |
|
0+581,91 |
+ |
0+057,04 |
= |
0+635,91 |
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|
635,91 |
a) kąt zwrotu trasy: |
|
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α = |
50,08 |
[°] |
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KKP2= KKP1+K= |
|
0+635,91 |
+ |
0+149,56 |
= |
0+790,13 |
|
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|
790,13 |
b) prędkość projektowa: |
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vp = |
50 |
[km/h] |
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KŁ= KKP2+L/2= |
|
0+790,13 |
+ |
0+057,04 |
= |
0+844,13 |
|
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|
844,13 |
c) promień łuku kołowego: |
|
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R = |
300 |
[m] |
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PKP2=KŁ+L/2= |
|
0+844,13 |
+ |
0+057,04 |
= |
0+898,13 |
|
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|
898,13 |
d) szerokość jezdni: |
|
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B = |
2,75 |
[m] |
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e) pochylenie poprzeczne na prostej: |
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ip = |
2 |
[%] |
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f) pochylenie poprzeczne na łuku: |
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io = |
3 |
[%] |
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GDZIE: |
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1) WARUNEK DYNAMICZNY: |
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Amin1 = |
57,87 |
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To = |
194,09 |
m |
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2) WARUNEK GEOMETRYCZNY ZESPOŁU KRZYWA PRZEJŚCIOWA-ŁUK KOŁOWY KRZYWA PRZEJŚCIOWA: |
|
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Amax2 = |
280,47 |
|
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L= |
108,00 |
m |
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3) WARUNEK ESTETYKI (WARUNEK KĄTA τ): |
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Amin3 = |
100,00 |
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L / 2 = |
54,00 |
m |
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Amax4 = |
300,00 |
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K = |
154,22 |
m |
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4) WARUNEK ESTETYKI (WIELKOŚĆ ODSUNIĘCIA ŁUKU H): |
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|
|
H = 0,5 m |
→ |
Amin5 = |
134,08 |
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H = 2,5 m |
→ |
Amax6 = |
200,39 |
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H = 0,2 m |
→ |
Amin7 = |
106,68 |
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5) WARUNEK KONSTRUKCYJNY (STOSOWANY NA ŁUKACH Z POSZERZENIEM): |
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40 / R = |
0,13 |
NIE UWZGLĘDNIAMY |
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6) WARUNEK KONSTRUKCYJNY (KOMFORTU JAZDY): |
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Amin8 = |
32,11 |
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134,08 |
≤ A ≤ |
200,39 |
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PRZY ZAŁOŻENIU PROPORCJI 1:2:1 |
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PRZY ZAŁOŻENIU PROPORCJI 1:1:1 |
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A = |
161,93 |
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A = |
198,32 |
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PRZYJMUJEMY PARAMETR KLOTOIDY RÓWNY A = |
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180 |
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16 |
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5 |
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DŁUGOŚĆ KLOTOIDY: |
|
|
L= |
108,00 |
m |
|
L = A2 / R |
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KĄT |
|
|
τ = |
0,180 |
rad |
γ = |
0,514 |
rad |
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DŁUGOŚĆ ŁUKU: |
|
|
K = |
154,22 |
m |
|
K = R · γ |
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OBLICZENIE PARAMETRÓW KLOTOIDY. |
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1) Odcięta X końca krzywej przejściowej: |
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107,65 |
m |
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2) Rzędna Y końca krzywej przejściowej: |
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6,47 |
m |
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3) Odcięta Xs środka krzywizny od początku krzywej przejściowej: |
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Xs = X - R · sin τ = |
|
|
53,94 |
m |
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4) Odsunięcie H łuku koła od prostego kierunku trasy: |
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H = Y - R ( 1 - cos τ ) = |
|
|
1,619 |
m |
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5) Dłudość stycznej T0 : |
|
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|
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|
|
T0 = T + Xs = |
|
|
194,09 |
m |
|
|
|
T = R · tg α/2 = |
140,15 |
m |
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6) Odległość środka łuku Z0 od punktu przecięcia się stycznych: |
|
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|
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32,91 |
m |
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6 |
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|
WARIANT 2 WIERZCHOŁEK 2 |
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|
W1 W2= |
|
0+921,00 |
|
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|
921,00 |
|
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|
|
W2= AW1 +W1W2= |
|
0+722,00 |
+ |
0+921,00 |
= |
|
|
|
|
1+448,91 |
|
|
1448,91 |
WARIANT 2 ŁUK 2 |
|
|
|
|
|
|
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|
|
PKP3= PKP2+W1W2-To-To2= |
|
0,0 |
+ |
0+921,00 |
- |
0+194,09 |
- |
0+214,64 |
= |
1+410,40 |
|
|
1410,40 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
PŁ= PKP3 +L/2 = |
|
1+410,40 |
+ |
0+057,04 |
= |
|
|
|
|
1+467,44 |
|
|
1467,44 |
a) kąt zwrotu trasy: |
|
|
α = |
55,45 |
[°] |
|
|
|
|
|
|
|
|
KKP3= PŁ+L/2= |
|
1+467,44 |
+ |
0+057,04 |
= |
|
|
|
|
1+524,48 |
|
|
1524,48 |
b) prędkość projektowa: |
|
|
vp = |
50 |
[km/h] |
|
|
|
|
|
|
|
|
KKP4= KKP3+K2= |
|
1+524,48 |
+ |
0+149,56 |
= |
|
|
|
|
1+700,73 |
|
|
1700,73 |
c) promień łuku kołowego: |
|
|
R = |
300 |
[m] |
|
|
|
|
|
|
|
|
KŁ= KKP2+L/2= |
|
1+700,73 |
+ |
0+057,04 |
= |
|
|
|
|
1+757,77 |
|
|
1757,77 |
d) szerokość jezdni: |
|
|
B = |
2,75 |
[m] |
|
|
|
|
|
|
|
|
PKP4=KŁ+L/2= |
|
1+757,77 |
+ |
0+057,04 |
= |
|
|
|
|
1+814,82 |
|
|
1814,82 |
e) pochylenie poprzeczne na prostej: |
|
|
ip = |
2 |
[%] |
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f) pochylenie poprzeczne na łuku: |
|
|
io = |
3 |
[%] |
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|
GDZIE: |
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|
|
1) WARUNEK DYNAMICZNY: |
|
|
|
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|
|
|
Amin1 = |
57,87 |
|
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2) WARUNEK GEOMETRYCZNY ZESPOŁU KRZYWA PRZEJŚCIOWA-ŁUK KOŁOWY KRZYWA PRZEJŚCIOWA: |
|
|
|
|
|
|
|
Amax2 = |
295,13 |
|
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|
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To = |
194,09 |
m |
|
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3) WARUNEK ESTETYKI (WARUNEK KĄTA τ): |
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|
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Amin3 = |
100,00 |
|
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To2 = |
214,64 |
m |
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Amax4 = |
300,00 |
|
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|
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L= |
114,08 |
m |
|
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|
|
4) WARUNEK ESTETYKI (WIELKOŚĆ ODSUNIĘCIA ŁUKU H): |
|
|
|
|
|
H = 0,5 m |
→ |
Amin5 = |
134,08 |
|
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|
|
L / 2 = |
57,04 |
m |
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|
|
H = 2,5 m |
→ |
Amax6 = |
200,39 |
|
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|
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K2 = |
176,25 |
m |
|
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|
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H = 0,2 m |
→ |
Amin7 = |
106,68 |
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5) WARUNEK KONSTRUKCYJNY (STOSOWANY NA ŁUKACH Z POSZERZENIEM): |
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|
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40 / R = |
0,13 |
NIE UWZGLĘDNIAMY |
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6) WARUNEK KONSTRUKCYJNY (KOMFORTU JAZDY): |
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Amin8 = |
32,11 |
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134,08 |
≤ A ≤ |
200,39 |
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PRZY ZAŁOŻENIU PROPORCJI 1:2:1 |
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PRZY ZAŁOŻENIU PROPORCJI 1:1:1 |
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A = |
170,39 |
|
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A = |
208,69 |
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|
PRZYJMUJEMY PARAMETR KLOTOIDY RÓWNY A = |
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|
185 |
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7 |
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|
17 |
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DŁUGOŚĆ KLOTOIDY: |
|
|
L= |
114,08 |
m |
|
L = A2 / R |
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KĄT |
|
|
τ = |
0,190 |
rad |
γ = |
0,588 |
rad |
|
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|
DŁUGOŚĆ ŁUKU: |
|
|
K = |
176,25 |
m |
|
K = R · γ |
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|
OBLICZENIE PARAMETRÓW KLOTOIDY. |
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1) Odcięta X końca krzywej przejściowej: |
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|
113,67 |
m |
|
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2) Rzędna Y końca krzywej przejściowej: |
|
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|
7,21 |
m |
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3) Odcięta Xs środka krzywizny od początku krzywej przejściowej: |
|
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|
Xs = X - R · sin τ = |
|
|
56,97 |
m |
|
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|
4) Odsunięcie H łuku koła od prostego kierunku trasy: |
|
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|
H = Y - R ( 1 - cos τ ) = |
|
|
1,807 |
m |
|
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|
5) Dłudość stycznej T0 : |
|
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|
T0 = T + Xs = |
|
|
214,64 |
m |
|
|
|
T = R · tg α/2 = |
157,67 |
m |
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|
|
6) Odległość środka łuku Z0 od punktu przecięcia się stycznych: |
|
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|
|
40,95 |
m |
|
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8 |
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|
WARIANT 1 WIERZCHOŁEK 3 |
|
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|
985 |
|
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|
|
W2 W3= |
|
0+816,00 |
|
|
|
|
|
|
|
|
|
|
816,00 |
WARIANT 2 ŁUK 3 |
|
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|
|
W3= W2 +W2W3= |
|
1+448,91 |
+ |
0+816,00 |
= |
|
|
|
|
2+264,91 |
|
|
2264,91 |
|
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|
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|
|
PKP5= PKP4+W2W3-To2-To3= |
|
1+814,82 |
+ |
2+264,91 |
- |
0+214,64 |
- |
#DIV/0! |
= |
#DIV/0! |
|
|
#DIV/0! |
a) kąt zwrotu trasy: |
|
|
α = |
|
[°] |
|
|
|
|
|
|
|
|
PŁ= PKP5+L/2 = |
|
#DIV/0! |
+ |
#DIV/0! |
= |
|
|
|
|
#DIV/0! |
|
|
#DIV/0! |
b) prędkość projektowa: |
|
|
vp = |
|
[km/h] |
|
|
|
|
|
|
|
|
KKP5= PŁ+L/2= |
|
#DIV/0! |
+ |
#DIV/0! |
= |
|
|
|
|
#DIV/0! |
|
|
#DIV/0! |
c) promień łuku kołowego: |
|
|
R = |
|
[m] |
|
|
|
|
|
|
|
|
KKP6= KKP5+K3= |
|
#DIV/0! |
+ |
0+149,56 |
= |
|
|
|
|
#DIV/0! |
|
|
#DIV/0! |
d) szerokość jezdni: |
|
|
B = |
|
[m] |
|
|
|
|
|
|
|
|
KŁ= KKP6+L/2= |
|
#DIV/0! |
+ |
#DIV/0! |
= |
|
|
|
|
#DIV/0! |
|
|
#DIV/0! |
e) pochylenie poprzeczne na prostej: |
|
|
ip = |
|
[%] |
|
|
|
|
|
|
|
|
PKP6=KŁ+L/2= |
|
#DIV/0! |
+ |
#DIV/0! |
= |
|
|
|
|
#DIV/0! |
|
|
#DIV/0! |
f) pochylenie poprzeczne na łuku: |
|
|
io = |
|
[%] |
|
|
|
|
|
|
|
|
B= PKP6 +W3B-To= |
|
#DIV/0! |
+ |
0+985,00 |
- |
0+194,09 |
= |
|
|
#DIV/0! |
|
|
#DIV/0! |
|
|
|
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|
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|
|
|
|
|
|
|
|
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|
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|
|
|
|
|
|
|
|
|
|
|
1) WARUNEK DYNAMICZNY: |
|
|
|
|
|
|
|
Amin1 = |
0,00 |
|
|
|
|
GDZIE: |
|
|
|
|
|
|
|
|
|
|
|
|
|
2) WARUNEK GEOMETRYCZNY ZESPOŁU KRZYWA PRZEJŚCIOWA-ŁUK KOŁOWY KRZYWA PRZEJŚCIOWA: |
|
|
|
|
|
|
|
Amax2 = |
0,00 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
3) WARUNEK ESTETYKI (WARUNEK KĄTA τ): |
|
|
|
|
|
|
|
Amin3 = |
0,00 |
|
|
|
|
To2 = |
214,64 |
m |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Amax4 = |
0,00 |
|
|
|
|
To3 = |
#DIV/0! |
m |
|
|
|
|
|
|
|
|
|
|
|
4) WARUNEK ESTETYKI (WIELKOŚĆ ODSUNIĘCIA ŁUKU H): |
|
|
|
|
|
H = 0,5 m |
→ |
Amin5 = |
0,00 |
|
|
|
|
L= |
#DIV/0! |
m |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
H = 2,5 m |
→ |
Amax6 = |
0,00 |
|
|
|
|
L / 2 = |
#DIV/0! |
m |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
H = 0,2 m |
→ |
Amin7 = |
0,00 |
|
|
|
|
K3 = |
#DIV/0! |
m |
|
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5) WARUNEK KONSTRUKCYJNY (STOSOWANY NA ŁUKACH Z POSZERZENIEM): |
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40 / R = |
#DIV/0! |
NIE UWZGLĘDNIAMY |
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6) WARUNEK KONSTRUKCYJNY (KOMFORTU JAZDY): |
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Amin8 = |
0,00 |
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0,00 |
≤ A ≤ |
0,00 |
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PRZY ZAŁOŻENIU PROPORCJI 1:2:1 |
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PRZY ZAŁOŻENIU PROPORCJI 1:1:1 |
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A = |
0,00 |
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A = |
0,00 |
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PRZYJMUJEMY PARAMETR KLOTOIDY RÓWNY A = |
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185 |
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11 |
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DŁUGOŚĆ KLOTOIDY: |
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L= |
#DIV/0! |
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L = A2 / R |
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KĄT |
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τ = |
#DIV/0! |
rad |
γ = |
#DIV/0! |
rad |
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DŁUGOŚĆ ŁUKU: |
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K = |
#DIV/0! |
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K = R · γ |
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OBLICZENIE PARAMETRÓW KLOTOIDY. |
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1) Odcięta X końca krzywej przejściowej: |
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#DIV/0! |
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2) Rzędna Y końca krzywej przejściowej: |
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#DIV/0! |
m |
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3) Odcięta Xs środka krzywizny od początku krzywej przejściowej: |
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Xs = X - R · sin τ = |
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#DIV/0! |
m |
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4) Odsunięcie H łuku koła od prostego kierunku trasy: |
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H = Y - R ( 1 - cos τ ) = |
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#DIV/0! |
m |
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5) Dłudość stycznej T0 : |
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T0 = T + Xs = |
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#DIV/0! |
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T = R · tg α/2 = |
0,00 |
m |
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6) Odległość środka łuku Z0 od punktu przecięcia się stycznych: |
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#DIV/0! |
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CO DO DALSZEJ CZĘŚCI TO ZNACZY ODTĄD W DÓŁ NIE JESTEM PEWIEN CZY ARKUSZ POLICZY SAMODZIELNIE :) |
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2.3. TABELA ZAŁOMÓW. |
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2.3.1. WARIANT I. |
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50 |
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PIKIETAŻ |
ODLEGŁÓŚĆ |
POCHYLENIE [%] |
RÓŻNICA WYSOKOŚCI Δh [m] |
RZĘDNA ZAŁOMU |
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51,25 |
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WZNIESIENIE |
SPADEK |
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52,5 |
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0+000,00 |
- |
- |
- |
- |
57,50 |
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0 |
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53,75 |
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0+354,00 |
354,00 |
- |
0,34 |
1,2 |
56,30 |
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354 |
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55 |
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0+600,00 |
246,00 |
1,50 |
- |
3,7 |
60,00 |
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600 |
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56,25 |
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1+020,00 |
420,00 |
- |
0,48 |
2 |
58,00 |
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1020 |
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57,5 |
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1+250,00 |
230,00 |
0,30 |
- |
0,700000000000003 |
58,70 |
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1250 |
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58,75 |
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2+360,00 |
1110,00 |
0,15 |
- |
1,7 |
60,40 |
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2360 |
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60 |
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2+925,00 |
565,00 |
- |
0,29 |
1,65 |
58,75 |
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2925 |
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61,25 |
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3+102,74 |
177,74 |
- |
0,56 |
1 |
57,75 |
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3102,74 |
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62,5 |
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63,75 |
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2.3.2. WARIANT II. |
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65 |
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66,25 |
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PIKIETAŻ |
ODLEGŁÓŚĆ |
POCHYLENIE [%] |
RÓŻNICA WYSOKOŚCI Δh [m] |
RZĘDNA ZAŁOMU |
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WZNIESIENIE |
SPADEK |
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0+000,00 |
- |
- |
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57,50 |
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0 |
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0+440,00 |
440,00 |
- |
0,57 |
2,5 |
55,00 |
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440 |
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0+970,00 |
530,00 |
- |
0,38 |
2 |
53,00 |
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970 |
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1+430,00 |
460,00 |
1,26 |
- |
5,8 |
58,80 |
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1430 |
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1+680,00 |
250,00 |
- |
0,76 |
1,9 |
56,90 |
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1680 |
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1+910,00 |
230,00 |
0,13 |
- |
0,300000000000004 |
57,20 |
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1910 |
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2+700,00 |
790,00 |
0,51 |
- |
4 |
61,20 |
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2700 |
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3+338,38 |
638,38 |
- |
0,54 |
3,45 |
57,75 |
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3338,38 |
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2.4. OBLICZENIA ŁUKÓW PIONOWYCH. |
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2.4.1. WARIANT I. |
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|
PIKIETA |
78 |
i1 |
i2 |
i |
R |
t |
f |
|
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|
0+354,00 |
8 |
0,34% |
1,50% |
1,84% |
3000 |
27,6 |
0,127 |
|
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|
0+600,00 |
7 |
1,50% |
0,48% |
1,98% |
3000 |
29,7 |
0,147 |
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|
1+020,00 |
8 |
0,48% |
0,30% |
0,78% |
3000 |
11,7 |
0,023 |
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|
2+360,00 |
7 |
0,15% |
0,29% |
0,44% |
3000 |
6,6 |
0,007 |
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2.4.2. WARIANT Ii. |
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|
PIKIETA |
78 |
i1 |
i2 |
i |
R |
t |
f |
|
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|
0+970,00 |
7 |
0,38% |
1,26% |
1,64% |
3000 |
24,6 |
0,101 |
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1+430,00 |
8 |
1,26% |
0,76% |
2,02% |
3000 |
30,3 |
0,153 |
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1+680,00 |
7 |
0,76% |
0,13% |
0,89% |
3000 |
13,35 |
0,030 |
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2+700,00 |
8 |
0,51% |
0,54% |
1,05% |
3000 |
15,75 |
0,041 |
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20 |
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2.5. DŁUGOŚĆ WIRTUALNA. |
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Lsp=Σl+Σiw/f x lw-Σis/f x ls |
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f = |
0,018 |
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2.5.1. WARIANT I. |
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OBLICZENIA W KIERUNKU ZGODNYM Z PIKIETAŻEM TRASY A-B |
|
OBLICZENIA W KIERUNKU PRZECIWNYM Z PIKIETAŻEM TRASY B-A |
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ODCINEK |
POCHYLENIE PODŁUŻNE [%] |
DŁUGOŚĆ ODCINKÓW WZNIESIEŃ ORAZ SPADKÓW NIESZKODLIWYCH [m] |
|
|
|
ODCINEK |
POCHYLENIE PODŁUŻNE [%] |
DŁUGOŚĆ ODCINKÓW WZNIESIEŃ ORAZ SPADKÓW NIESZKODLIWYCH [m] |
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ODCINEK 1 |
-0,34 |
354,00 |
- |
66,87 |
|
ODCINEK 1 |
0,34 |
354,00 |
66,87 |
- |
|
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|
ODCINEK 2 |
1,50 |
246,00 |
205,00 |
- |
|
ODCINEK 2 |
-1,50 |
246,00 |
- |
205,00 |
|
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|
ODCINEK 3 |
-0,48 |
420,00 |
- |
112,00 |
|
ODCINEK 3 |
0,48 |
420,00 |
112,00 |
- |
|
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ODCINEK 4 |
0,30 |
230,00 |
38,33 |
- |
|
ODCINEK 4 |
-0,30 |
230,00 |
- |
38,33 |
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ODCINEK 5 |
0,15 |
1110,00 |
92,50 |
- |
|
ODCINEK 5 |
-0,15 |
1110,00 |
- |
92,50 |
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|
ODCINEK 6 |
-0,29 |
565,00 |
- |
91,03 |
|
ODCINEK 6 |
0,29 |
565,00 |
91,03 |
- |
|
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ODCINEK 7 |
-0,56 |
177,74 |
- |
55,30 |
|
ODCINEK 7 |
0,56 |
177,74 |
55,30 |
- |
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SUMA = |
3102,74 |
335,83 |
325,19 |
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SUMA = |
3102,74 |
325,19 |
335,83 |
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Lsp= |
3113,38 |
m |
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Lsp= |
3092,10 |
m |
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|
DŁUGOŚĆ WIRTUALNA POTRZEBNA DO OCENY WARIANTU I WYNOSI: |
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L = |
3102,74 |
m |
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21 |
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Lsp=Σl+Σiw/f x lw-Σis/f x ls |
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f = |
0,018 |
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2.5.2. WARIANT II. |
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|
OBLICZENIA W KIERUNKU ZGODNYM Z PIKIETAŻEM TRASY A-B |
|
OBLICZENIA W KIERUNKU PRZECIWNYM Z PIKIETAŻEM TRASY B-A |
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|
ODCINEK |
POCHYLENIE PODŁUŻNE [%] |
DŁUGOŚĆ ODCINKÓW WZNIESIEŃ ORAZ SPADKÓW NIESZKODLIWYCH [m] |
|
|
|
ODCINEK |
POCHYLENIE PODŁUŻNE [%] |
DŁUGOŚĆ ODCINKÓW WZNIESIEŃ ORAZ SPADKÓW NIESZKODLIWYCH [m] |
|
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|
ODCINEK 1 |
-0,57 |
440,00 |
- |
139,33 |
|
ODCINEK 1 |
0,57 |
440,00 |
139,33 |
- |
|
|
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|
|
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|
|
|
ODCINEK 2 |
-0,38 |
530,00 |
- |
111,89 |
|
ODCINEK 2 |
0,38 |
530,00 |
111,89 |
- |
|
|
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|
|
|
ODCINEK 3 |
1,26 |
460,00 |
322,00 |
- |
|
ODCINEK 3 |
-1,26 |
460,00 |
- |
322,00 |
|
|
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|
|
|
|
ODCINEK 4 |
-0,76 |
250,00 |
- |
105,56 |
|
ODCINEK 4 |
0,76 |
250,00 |
105,56 |
- |
|
|
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|
ODCINEK 5 |
0,13 |
230,00 |
16,61 |
- |
|
ODCINEK 5 |
-0,13 |
230,00 |
- |
16,61 |
|
|
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|
|
ODCINEK 6 |
0,51 |
790,00 |
223,83 |
- |
|
ODCINEK 6 |
-0,51 |
790,00 |
- |
223,83 |
|
|
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|
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|
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|
|
|
|
|
|
ODCINEK 7 |
-0,54 |
638,38 |
- |
191,51 |
|
ODCINEK 7 |
0,54 |
638,38 |
191,51 |
- |
|
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|
|
SUMA = |
3338,38 |
562,44 |
548,29 |
|
|
|
SUMA = |
3338,38 |
548,29 |
562,44 |
|
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|
Lsp= |
3352,53 |
m |
|
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|
Lsp= |
3324,23 |
m |
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|
DŁUGOŚĆ WIRTUALNA POTRZEBNA DO OCENY WARIANTU I WYNOSI: |
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L = |
3338,38 |
m |
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22 |
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2. OBLICZENIA. |
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2.2. PUNKTY GŁÓWNE TRASY (PIKIETAŻ KILOMETRACJI) |
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2.1. OBLICZENIE ŁUKÓW POZIOMYCH. |
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WARIANT 1 WIERZCHOŁEK 1 |
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WARIANT 1 ŁUK 1 |
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AW1= |
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0+678,00 |
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678,00 |
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PKP1= AW1-To= |
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0+678,00 |
- |
0+089,68 |
= |
0+588,32 |
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588,32 |
a) kąt zwrotu trasy: |
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α = |
23,18 |
[°] |
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PŁ= PKP1 +L/2 = |
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0+588,32 |
+ |
0+041,29 |
= |
0+616,48 |
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616,48 |
b) prędkość projektowa: |
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vp = |
50 |
[km/h] |
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KKP1= PŁ+L/2= |
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0+616,48 |
+ |
0+041,30 |
= |
0+644,65 |
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644,65 |
c) promień łuku kołowego: |
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R = |
300 |
[m] |
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KKP2= KKP1+K= |
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0+644,65 |
+ |
0+149,56 |
= |
0+709,69 |
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709,69 |
d) szerokość jezdni: |
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B = |
2,75 |
[m] |
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KŁ= KKP2+L/2= |
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0+709,69 |
+ |
0+041,29 |
= |
0+737,85 |
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737,85 |
e) pochylenie poprzeczne na prostej: |
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ip = |
2 |
[%] |
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PKP2=KŁ+L/2= |
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0+737,85 |
+ |
0+041,30 |
= |
0+766,02 |
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766,02 |
f) pochylenie poprzeczne na łuku: |
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io = |
3 |
[%] |
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1) WARUNEK DYNAMICZNY: |
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Amin1 = |
57,87 |
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GDZIE: |
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2) WARUNEK GEOMETRYCZNY ZESPOŁU KRZYWA PRZEJŚCIOWA-ŁUK KOŁOWY KRZYWA PRZEJŚCIOWA: |
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Amax2 = |
190,82 |
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3) WARUNEK ESTETYKI (WARUNEK KĄTA τ): |
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Amin3 = |
100,00 |
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To = |
89,68 |
m |
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Amax4 = |
300,00 |
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L= |
56,33 |
m |
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4) WARUNEK ESTETYKI (WIELKOŚĆ ODSUNIĘCIA ŁUKU H): |
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H = 0,5 m |
→ |
Amin5 = |
134,08 |
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L / 2 = |
28,17 |
m |
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H = 2,5 m |
→ |
Amax6 = |
200,39 |
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K = |
65,04 |
m |
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H = 0,2 m |
→ |
Amin7 = |
106,68 |
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5) WARUNEK KONSTRUKCYJNY (STOSOWANY NA ŁUKACH Z POSZERZENIEM): |
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40 / R = 0,11m |
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NIE UWZGLĘDNIAMY |
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6) WARUNEK KONSTRUKCYJNY (KOMFORTU JAZDY): |
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Amin8 = |
32,11 |
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134,08 |
≤ A ≤ |
190,82 |
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PRZY ZAŁOŻENIU PROPORCJI 1:2:1 |
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PRZY ZAŁOŻENIU PROPORCJI 1:1:1 |
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A = |
110,17 |
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A = |
134,93 |
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PRZYJMUJEMY PARAMETR KLOTOIDY RÓWNY A = |
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130 |
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1 |
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9 |
DŁUGOŚĆ KLOTOIDY: |
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L= |
56,33 |
m |
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L = A2 / R |
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KĄT |
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τ = |
0,094 |
rad |
γ = |
0,217 |
rad |
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DŁUGOŚĆ ŁUKU: |
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K = |
65,04 |
m |
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K = R · γ |
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OBLICZENIE PARAMETRÓW KLOTOIDY. |
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1) Odcięta X końca krzywej przejściowej: |
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56,28 |
m |
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2) Rzędna Y końca krzywej przejściowej: |
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1,76 |
m |
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3) Odcięta Xs środka krzywizny od początku krzywej przejściowej: |
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Xs = X - R · sin τ = |
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28,16 |
m |
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4) Odsunięcie H łuku koła od prostego kierunku trasy: |
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H = Y - R ( 1 - cos τ ) = |
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0,441 |
m |
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5) Dłudość stycznej T0 : |
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T0 = T + Xs = |
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89,68 |
m |
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T = R · tg α/2 = |
61,53 |
m |
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6) Odległość środka łuku Z0 od punktu przecięcia się stycznych: |
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6,69 |
m |
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2 |
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WARIANT 1 WIERZCHOŁEK 2 |
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W1 W2= |
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1+981,93 |
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1+981,93 |
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1981,93 |
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W2= AW1 +W1W2= |
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0+678,00 |
+ |
1+981,93 |
= |
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2+659,93 |
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2659,93 |
WARIANT 1 ŁUK 2 |
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PKP3= PKP2+W1W2-To-To2= |
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0+766,02 |
+ |
1+981,93 |
- |
0+089,68 |
- |
0+193,90 |
= |
2+464,37 |
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2464,37 |
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PŁ= PKP3 +L/2 = |
|
2+464,37 |
+ |
0+054,00 |
= |
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2+518,37 |
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|
2518,37 |
a) kąt zwrotu trasy: |
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α = |
50,02 |
[°] |
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KKP3= PŁ+L/2= |
|
2+518,37 |
+ |
0+054,00 |
= |
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2+572,37 |
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|
2572,37 |
b) prędkość projektowa: |
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vp = |
50 |
[km/h] |
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KKP4= KKP3+K2= |
|
2+572,37 |
+ |
0+149,56 |
= |
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2+726,27 |
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|
2726,27 |
c) promień łuku kołowego: |
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R = |
300 |
[m] |
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KŁ= KKP2+L/2= |
|
2+726,27 |
+ |
0+054,00 |
= |
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2+780,27 |
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|
2780,27 |
d) szerokość jezdni: |
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B = |
2,8 |
[m] |
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PKP4=KŁ+L/2= |
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2+780,27 |
+ |
0+054,00 |
= |
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2+834,27 |
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|
2834,27 |
e) pochylenie poprzeczne na prostej: |
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ip = |
2 |
[%] |
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f) pochylenie poprzeczne na łuku: |
|
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io = |
3 |
[%] |
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GDZIE: |
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1) WARUNEK DYNAMICZNY: |
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Amin1 = |
57,87 |
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2) WARUNEK GEOMETRYCZNY ZESPOŁU KRZYWA PRZEJŚCIOWA-ŁUK KOŁOWY KRZYWA PRZEJŚCIOWA: |
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Amax2 = |
280,31 |
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To = |
89,68 |
m |
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3) WARUNEK ESTETYKI (WARUNEK KĄTA τ): |
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Amin3 = |
100,00 |
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To2 = |
194 |
m |
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Amax4 = |
300,00 |
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L= |
108,00 |
m |
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4) WARUNEK ESTETYKI (WIELKOŚĆ ODSUNIĘCIA ŁUKU H): |
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H = 0,5 m |
→ |
Amin5 = |
134,08 |
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L / 2 = |
54,00 |
m |
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H = 2,5 m |
→ |
Amax6 = |
200,39 |
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K2 = |
153,90 |
m |
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H = 0,2 m |
→ |
Amin7 = |
106,68 |
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5) WARUNEK KONSTRUKCYJNY (STOSOWANY NA ŁUKACH Z POSZERZENIEM): |
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40 / R = |
0,13 |
NIE UWZGLĘDNIAMY |
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6) WARUNEK KONSTRUKCYJNY (KOMFORTU JAZDY): |
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Amin8 = |
32,11 |
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134,08 |
≤ A ≤ |
200,39 |
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PRZY ZAŁOŻENIU PROPORCJI 1:2:1 |
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PRZY ZAŁOŻENIU PROPORCJI 1:1:1 |
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A = |
161,83 |
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A = |
198,21 |
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PRZYJMUJEMY PARAMETR KLOTOIDY RÓWNY A = |
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180 |
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14 |
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3 |
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10 |
DŁUGOŚĆ KLOTOIDY: |
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L= |
108,00 |
m |
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L = A2 / R |
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KĄT |
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τ = |
0,180 |
rad |
γ = |
0,513 |
rad |
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DŁUGOŚĆ ŁUKU: |
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K = |
153,90 |
m |
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K = R · γ |
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OBLICZENIE PARAMETRÓW KLOTOIDY. |
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1) Odcięta X końca krzywej przejściowej: |
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107,7 |
m |
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2) Rzędna Y końca krzywej przejściowej: |
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6,47 |
m |
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3) Odcięta Xs środka krzywizny od początku krzywej przejściowej: |
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Xs = X - R · sin τ = |
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53,9 |
m |
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4) Odsunięcie H łuku koła od prostego kierunku trasy: |
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H = Y - R ( 1 - cos τ ) = |
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1,62 |
m |
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5) Dłudość stycznej T0 : |
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T0 = T + Xs = |
|
|
193,9 |
m |
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T = R · tg α/2 = |
139,96 |
m |
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6) Odległość środka łuku Z0 od punktu przecięcia się stycznych: |
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32,83 |
m |
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4 |
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WARIANT 2 WIERZCHOŁEK 1 |
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AW1= |
|
1+030,00 |
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1030,00 |
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PKP1= AW1-To= |
|
1+030,00 |
- |
0+161,78 |
= |
0+835,91 |
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|
835,91 |
WARIANT 2 ŁUK 1 |
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PŁ= PKP1 +L/2 = |
|
0+835,91 |
+ |
0+057,04 |
= |
0+889,91 |
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889,91 |
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KKP1= PŁ+L/2= |
|
0+889,91 |
+ |
0+057,04 |
= |
0+943,91 |
|
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|
943,91 |
a) kąt zwrotu trasy: |
|
|
α = |
50,08 |
[°] |
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KKP2= KKP1+K= |
|
0+943,91 |
+ |
0+149,56 |
= |
1+098,13 |
|
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|
|
1098,13 |
b) prędkość projektowa: |
|
|
vp = |
50 |
[km/h] |
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|
KŁ= KKP2+L/2= |
|
1+098,13 |
+ |
0+057,04 |
= |
1+152,13 |
|
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|
1152,13 |
c) promień łuku kołowego: |
|
|
R = |
300 |
[m] |
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|
PKP2=KŁ+L/2= |
|
1+152,13 |
+ |
0+057,04 |
= |
1+206,13 |
|
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|
|
1206,13 |
d) szerokość jezdni: |
|
|
B = |
2,75 |
[m] |
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|
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|
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e) pochylenie poprzeczne na prostej: |
|
|
ip = |
2 |
[%] |
|
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|
|
|
|
|
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
f) pochylenie poprzeczne na łuku: |
|
|
io = |
3 |
[%] |
|
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|
|
|
|
|
|
GDZIE: |
|
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1) WARUNEK DYNAMICZNY: |
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|
|
Amin1 = |
57,87 |
|
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To = |
194,09 |
m |
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|
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2) WARUNEK GEOMETRYCZNY ZESPOŁU KRZYWA PRZEJŚCIOWA-ŁUK KOŁOWY KRZYWA PRZEJŚCIOWA: |
|
|
|
|
|
|
|
Amax2 = |
280,47 |
|
|
|
|
L= |
108,00 |
m |
|
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|
|
|
|
|
|
|
|
3) WARUNEK ESTETYKI (WARUNEK KĄTA τ): |
|
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|
|
|
|
|
Amin3 = |
100,00 |
|
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|
|
L / 2 = |
54,00 |
m |
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|
|
Amax4 = |
300,00 |
|
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|
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K = |
154,22 |
m |
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|
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4) WARUNEK ESTETYKI (WIELKOŚĆ ODSUNIĘCIA ŁUKU H): |
|
|
|
|
|
H = 0,5 m |
→ |
Amin5 = |
134,08 |
|
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|
|
H = 2,5 m |
→ |
Amax6 = |
200,39 |
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|
H = 0,2 m |
→ |
Amin7 = |
106,68 |
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|
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5) WARUNEK KONSTRUKCYJNY (STOSOWANY NA ŁUKACH Z POSZERZENIEM): |
|
|
|
|
|
40 / R = |
0,13 |
NIE UWZGLĘDNIAMY |
|
|
|
|
|
|
|
|
|
|
|
|
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|
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6) WARUNEK KONSTRUKCYJNY (KOMFORTU JAZDY): |
|
|
|
|
|
|
|
Amin8 = |
32,11 |
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134,08 |
≤ A ≤ |
200,39 |
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PRZY ZAŁOŻENIU PROPORCJI 1:2:1 |
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PRZY ZAŁOŻENIU PROPORCJI 1:1:1 |
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A = |
161,93 |
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A = |
198,32 |
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PRZYJMUJEMY PARAMETR KLOTOIDY RÓWNY A = |
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180 |
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16 |
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5 |
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11 |
DŁUGOŚĆ KLOTOIDY: |
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L= |
108,00 |
m |
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L = A2 / R |
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KĄT |
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τ = |
0,180 |
rad |
γ = |
0,514 |
rad |
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DŁUGOŚĆ ŁUKU: |
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K = |
154,22 |
m |
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K = R · γ |
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OBLICZENIE PARAMETRÓW KLOTOIDY. |
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1) Odcięta X końca krzywej przejściowej: |
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107,65 |
m |
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2) Rzędna Y końca krzywej przejściowej: |
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6,47 |
m |
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3) Odcięta Xs środka krzywizny od początku krzywej przejściowej: |
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Xs = X - R · sin τ = |
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53,94 |
m |
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4) Odsunięcie H łuku koła od prostego kierunku trasy: |
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H = Y - R ( 1 - cos τ ) = |
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1,619 |
m |
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5) Dłudość stycznej T0 : |
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T0 = T + Xs = |
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194,09 |
m |
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T = R · tg α/2 = |
140,15 |
m |
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6) Odległość środka łuku Z0 od punktu przecięcia się stycznych: |
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32,91 |
m |
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6 |
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WARIANT 2 WIERZCHOŁEK 2 |
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W1 W2= |
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1+695,00 |
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1695,00 |
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W2= AW1 +W1W2= |
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1+030,00 |
+ |
1+695,00 |
= |
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2+530,91 |
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2530,91 |
WARIANT 2 ŁUK 2 |
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PKP3= PKP2+W1W2-To-To2= |
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1+206,13 |
+ |
1+695,00 |
- |
0+194,09 |
- |
0+214,64 |
= |
2+492,40 |
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2492,40 |
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PŁ= PKP3 +L/2 = |
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2+492,40 |
+ |
0+057,04 |
= |
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2+549,44 |
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2549,44 |
a) kąt zwrotu trasy: |
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α = |
55,45 |
[°] |
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KKP3= PŁ+L/2= |
|
2+549,44 |
+ |
0+057,04 |
= |
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2+606,48 |
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2606,48 |
b) prędkość projektowa: |
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vp = |
50 |
[km/h] |
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KKP4= KKP3+K2= |
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2+606,48 |
+ |
0+149,56 |
= |
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2+782,73 |
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2782,73 |
c) promień łuku kołowego: |
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R = |
300 |
[m] |
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KŁ= KKP2+L/2= |
|
2+782,73 |
+ |
0+057,04 |
= |
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2+839,77 |
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2839,77 |
d) szerokość jezdni: |
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B = |
2,75 |
[m] |
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PKP4=KŁ+L/2= |
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2+839,77 |
+ |
0+057,04 |
= |
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2+896,82 |
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2896,82 |
e) pochylenie poprzeczne na prostej: |
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ip = |
2 |
[%] |
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f) pochylenie poprzeczne na łuku: |
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io = |
3 |
[%] |
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GDZIE: |
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1) WARUNEK DYNAMICZNY: |
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Amin1 = |
57,87 |
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2) WARUNEK GEOMETRYCZNY ZESPOŁU KRZYWA PRZEJŚCIOWA-ŁUK KOŁOWY KRZYWA PRZEJŚCIOWA: |
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Amax2 = |
295,13 |
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To = |
194,09 |
m |
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3) WARUNEK ESTETYKI (WARUNEK KĄTA τ): |
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Amin3 = |
100,00 |
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To2 = |
214,64 |
m |
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Amax4 = |
300,00 |
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L= |
114,08 |
m |
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4) WARUNEK ESTETYKI (WIELKOŚĆ ODSUNIĘCIA ŁUKU H): |
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H = 0,5 m |
→ |
Amin5 = |
134,08 |
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L / 2 = |
57,04 |
m |
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H = 2,5 m |
→ |
Amax6 = |
200,39 |
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K2 = |
176,25 |
m |
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H = 0,2 m |
→ |
Amin7 = |
106,68 |
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5) WARUNEK KONSTRUKCYJNY (STOSOWANY NA ŁUKACH Z POSZERZENIEM): |
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40 / R = |
0,13 |
NIE UWZGLĘDNIAMY |
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6) WARUNEK KONSTRUKCYJNY (KOMFORTU JAZDY): |
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Amin8 = |
32,11 |
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134,08 |
≤ A ≤ |
200,39 |
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PRZY ZAŁOŻENIU PROPORCJI 1:2:1 |
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PRZY ZAŁOŻENIU PROPORCJI 1:1:1 |
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A = |
170,39 |
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A = |
208,69 |
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PRZYJMUJEMY PARAMETR KLOTOIDY RÓWNY A = |
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185 |
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7 |
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12 |
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DŁUGOŚĆ KLOTOIDY: |
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L= |
114,08 |
m |
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L = A2 / R |
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KĄT |
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τ = |
0,190 |
rad |
γ = |
0,588 |
rad |
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DŁUGOŚĆ ŁUKU: |
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K = |
176,25 |
m |
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K = R · γ |
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OBLICZENIE PARAMETRÓW KLOTOIDY. |
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1) Odcięta X końca krzywej przejściowej: |
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113,67 |
m |
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2) Rzędna Y końca krzywej przejściowej: |
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7,21 |
m |
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3) Odcięta Xs środka krzywizny od początku krzywej przejściowej: |
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Xs = X - R · sin τ = |
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56,97 |
m |
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4) Odsunięcie H łuku koła od prostego kierunku trasy: |
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H = Y - R ( 1 - cos τ ) = |
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1,807 |
m |
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5) Dłudość stycznej T0 : |
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T0 = T + Xs = |
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214,64 |
m |
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T = R · tg α/2 = |
157,67 |
m |
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6) Odległość środka łuku Z0 od punktu przecięcia się stycznych: |
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40,95 |
m |
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8 |
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WARIANT 1 WIERZCHOŁEK 3 |
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985 |
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W2 W3= |
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0+816,00 |
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816,00 |
WARIANT 2 ŁUK 3 |
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W3= W2 +W2W3= |
|
2+530,91 |
+ |
0+816,00 |
= |
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3+346,91 |
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3346,91 |
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PKP5= PKP4+W2W3-To2-To3= |
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2+896,82 |
+ |
3+346,91 |
- |
0+214,64 |
- |
#DIV/0! |
= |
#DIV/0! |
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|
#DIV/0! |
a) kąt zwrotu trasy: |
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α = |
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[°] |
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PŁ= PKP5+L/2 = |
|
#DIV/0! |
+ |
#DIV/0! |
= |
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#DIV/0! |
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|
#DIV/0! |
b) prędkość projektowa: |
|
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vp = |
|
[km/h] |
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KKP5= PŁ+L/2= |
|
#DIV/0! |
+ |
#DIV/0! |
= |
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#DIV/0! |
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|
#DIV/0! |
c) promień łuku kołowego: |
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R = |
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[m] |
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KKP6= KKP5+K3= |
|
#DIV/0! |
+ |
0+149,56 |
= |
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#DIV/0! |
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|
#DIV/0! |
d) szerokość jezdni: |
|
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B = |
|
[m] |
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KŁ= KKP6+L/2= |
|
#DIV/0! |
+ |
#DIV/0! |
= |
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#DIV/0! |
|
|
#DIV/0! |
e) pochylenie poprzeczne na prostej: |
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|
ip = |
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[%] |
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PKP6=KŁ+L/2= |
|
#DIV/0! |
+ |
#DIV/0! |
= |
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#DIV/0! |
|
|
#DIV/0! |
f) pochylenie poprzeczne na łuku: |
|
|
io = |
|
[%] |
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|
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B= PKP6 +W3B-To= |
|
#DIV/0! |
+ |
0+985,00 |
- |
0+194,09 |
= |
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#DIV/0! |
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#DIV/0! |
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1) WARUNEK DYNAMICZNY: |
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Amin1 = |
0,00 |
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GDZIE: |
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2) WARUNEK GEOMETRYCZNY ZESPOŁU KRZYWA PRZEJŚCIOWA-ŁUK KOŁOWY KRZYWA PRZEJŚCIOWA: |
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Amax2 = |
0,00 |
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3) WARUNEK ESTETYKI (WARUNEK KĄTA τ): |
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Amin3 = |
0,00 |
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To2 = |
214,64 |
m |
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Amax4 = |
0,00 |
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To3 = |
#DIV/0! |
m |
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4) WARUNEK ESTETYKI (WIELKOŚĆ ODSUNIĘCIA ŁUKU H): |
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H = 0,5 m |
→ |
Amin5 = |
0,00 |
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L= |
#DIV/0! |
m |
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H = 2,5 m |
→ |
Amax6 = |
0,00 |
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L / 2 = |
#DIV/0! |
m |
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H = 0,2 m |
→ |
Amin7 = |
0,00 |
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K3 = |
#DIV/0! |
m |
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5) WARUNEK KONSTRUKCYJNY (STOSOWANY NA ŁUKACH Z POSZERZENIEM): |
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40 / R = |
#DIV/0! |
NIE UWZGLĘDNIAMY |
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6) WARUNEK KONSTRUKCYJNY (KOMFORTU JAZDY): |
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Amin8 = |
0,00 |
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0,00 |
≤ A ≤ |
0,00 |
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PRZY ZAŁOŻENIU PROPORCJI 1:2:1 |
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PRZY ZAŁOŻENIU PROPORCJI 1:1:1 |
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A = |
0,00 |
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A = |
0,00 |
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PRZYJMUJEMY PARAMETR KLOTOIDY RÓWNY A = |
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185 |
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11 |
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18 |
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DŁUGOŚĆ KLOTOIDY: |
|
|
L= |
#DIV/0! |
m |
|
L = A2 / R |
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KĄT |
|
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τ = |
#DIV/0! |
rad |
γ = |
#DIV/0! |
rad |
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DŁUGOŚĆ ŁUKU: |
|
|
K = |
#DIV/0! |
m |
|
K = R · γ |
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OBLICZENIE PARAMETRÓW KLOTOIDY. |
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1) Odcięta X końca krzywej przejściowej: |
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#DIV/0! |
m |
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2) Rzędna Y końca krzywej przejściowej: |
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#DIV/0! |
m |
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3) Odcięta Xs środka krzywizny od początku krzywej przejściowej: |
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Xs = X - R · sin τ = |
|
|
#DIV/0! |
m |
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4) Odsunięcie H łuku koła od prostego kierunku trasy: |
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H = Y - R ( 1 - cos τ ) = |
|
|
#DIV/0! |
m |
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5) Dłudość stycznej T0 : |
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T0 = T + Xs = |
|
|
#DIV/0! |
m |
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T = R · tg α/2 = |
0,00 |
m |
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6) Odległość środka łuku Z0 od punktu przecięcia się stycznych: |
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#DIV/0! |
m |
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12 |
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CO DO DALSZEJ CZĘŚCI TO ZNACZY ODTĄD W DÓŁ NIE JESTEM PEWIEN CZY ARKUSZ POLICZY SAMODZIELNIE :) |
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2.3. TABELA ZAŁOMÓW. |
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2.3.1. WARIANT I. |
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50 |
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PIKIETAŻ |
ODLEGŁÓŚĆ |
POCHYLENIE [%] |
RÓŻNICA WYSOKOŚCI Δh [m] |
RZĘDNA ZAŁOMU |
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51,25 |
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WZNIESIENIE |
SPADEK |
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|
52,5 |
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|
|
0+000,00 |
- |
- |
- |
- |
57,50 |
|
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|
|
|
0 |
|
53,75 |
|
|
|
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|
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|
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|
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|
|
0+354,00 |
354,00 |
- |
0,34 |
1,2 |
56,30 |
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|
354 |
|
55 |
|
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0+600,00 |
246,00 |
1,50 |
- |
3,7 |
60,00 |
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600 |
|
56,25 |
|
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1+020,00 |
420,00 |
- |
0,48 |
2 |
58,00 |
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1020 |
|
57,5 |
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1+250,00 |
230,00 |
0,30 |
- |
0,700000000000003 |
58,70 |
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|
1250 |
|
58,75 |
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2+360,00 |
1110,00 |
0,15 |
- |
1,7 |
60,40 |
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2360 |
|
60 |
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2+925,00 |
565,00 |
- |
0,29 |
1,65 |
58,75 |
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2925 |
|
61,25 |
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3+102,74 |
177,74 |
- |
0,56 |
1 |
57,75 |
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3102,74 |
|
62,5 |
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63,75 |
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2.3.2. WARIANT II. |
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65 |
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66,25 |
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PIKIETAŻ |
ODLEGŁÓŚĆ |
POCHYLENIE [%] |
RÓŻNICA WYSOKOŚCI Δh [m] |
RZĘDNA ZAŁOMU |
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WZNIESIENIE |
SPADEK |
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0+000,00 |
- |
- |
- |
- |
57,50 |
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0 |
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0+440,00 |
440,00 |
- |
0,57 |
2,5 |
55,00 |
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440 |
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0+970,00 |
530,00 |
- |
0,38 |
2 |
53,00 |
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970 |
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1+430,00 |
460,00 |
1,26 |
- |
5,8 |
58,80 |
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|
1430 |
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1+680,00 |
250,00 |
- |
0,76 |
1,9 |
56,90 |
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|
1680 |
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1+910,00 |
230,00 |
0,13 |
- |
0,300000000000004 |
57,20 |
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1910 |
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2+700,00 |
790,00 |
0,51 |
- |
4 |
61,20 |
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2700 |
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3+338,38 |
638,38 |
- |
0,54 |
3,45 |
57,75 |
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3338,38 |
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19 |
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2.4. OBLICZENIA ŁUKÓW PIONOWYCH. |
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2.4.1. WARIANT I. |
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PIKIETA |
78 |
i1 |
i2 |
i |
R |
t |
f |
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0+354,00 |
8 |
0,34% |
1,50% |
1,84% |
3000 |
27,6 |
0,127 |
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0+600,00 |
7 |
1,50% |
0,48% |
1,98% |
3000 |
29,7 |
0,147 |
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1+020,00 |
8 |
0,48% |
0,30% |
0,78% |
3000 |
11,7 |
0,023 |
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2+360,00 |
7 |
0,15% |
0,29% |
0,44% |
3000 |
6,6 |
0,007 |
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2.4.2. WARIANT Ii. |
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PIKIETA |
78 |
i1 |
i2 |
i |
R |
t |
f |
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0+970,00 |
7 |
0,38% |
1,26% |
1,64% |
3000 |
24,6 |
0,101 |
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1+430,00 |
8 |
1,26% |
0,76% |
2,02% |
3000 |
30,3 |
0,153 |
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1+680,00 |
7 |
0,76% |
0,13% |
0,89% |
3000 |
13,35 |
0,030 |
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2+700,00 |
8 |
0,51% |
0,54% |
1,05% |
3000 |
15,75 |
0,041 |
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20 |
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2.5. DŁUGOŚĆ WIRTUALNA. |
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Lsp=Σl+Σiw/f x lw-Σis/f x ls |
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f = |
0,018 |
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2.5.1. WARIANT I. |
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OBLICZENIA W KIERUNKU ZGODNYM Z PIKIETAŻEM TRASY A-B |
|
OBLICZENIA W KIERUNKU PRZECIWNYM Z PIKIETAŻEM TRASY B-A |
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ODCINEK |
POCHYLENIE PODŁUŻNE [%] |
DŁUGOŚĆ ODCINKÓW WZNIESIEŃ ORAZ SPADKÓW NIESZKODLIWYCH [m] |
|
|
|
ODCINEK |
POCHYLENIE PODŁUŻNE [%] |
DŁUGOŚĆ ODCINKÓW WZNIESIEŃ ORAZ SPADKÓW NIESZKODLIWYCH [m] |
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ODCINEK 1 |
-0,34 |
354,00 |
- |
66,87 |
|
ODCINEK 1 |
0,34 |
354,00 |
66,87 |
- |
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ODCINEK 2 |
1,50 |
246,00 |
205,00 |
- |
|
ODCINEK 2 |
-1,50 |
246,00 |
- |
205,00 |
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ODCINEK 3 |
-0,48 |
420,00 |
- |
112,00 |
|
ODCINEK 3 |
0,48 |
420,00 |
112,00 |
- |
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ODCINEK 4 |
0,30 |
230,00 |
38,33 |
- |
|
ODCINEK 4 |
-0,30 |
230,00 |
- |
38,33 |
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ODCINEK 5 |
0,15 |
1110,00 |
92,50 |
- |
|
ODCINEK 5 |
-0,15 |
1110,00 |
- |
92,50 |
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ODCINEK 6 |
-0,29 |
565,00 |
- |
91,03 |
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ODCINEK 6 |
0,29 |
565,00 |
91,03 |
- |
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ODCINEK 7 |
-0,56 |
177,74 |
- |
55,30 |
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ODCINEK 7 |
0,56 |
177,74 |
55,30 |
- |
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SUMA = |
3102,74 |
335,83 |
325,19 |
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SUMA = |
3102,74 |
325,19 |
335,83 |
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Lsp= |
3113,38 |
m |
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Lsp= |
3092,10 |
m |
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DŁUGOŚĆ WIRTUALNA POTRZEBNA DO OCENY WARIANTU I WYNOSI: |
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L = |
3102,74 |
m |
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21 |
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Lsp=Σl+Σiw/f x lw-Σis/f x ls |
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f = |
0,018 |
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2.5.2. WARIANT II. |
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OBLICZENIA W KIERUNKU ZGODNYM Z PIKIETAŻEM TRASY A-B |
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OBLICZENIA W KIERUNKU PRZECIWNYM Z PIKIETAŻEM TRASY B-A |
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ODCINEK |
POCHYLENIE PODŁUŻNE [%] |
DŁUGOŚĆ ODCINKÓW WZNIESIEŃ ORAZ SPADKÓW NIESZKODLIWYCH [m] |
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ODCINEK |
POCHYLENIE PODŁUŻNE [%] |
DŁUGOŚĆ ODCINKÓW WZNIESIEŃ ORAZ SPADKÓW NIESZKODLIWYCH [m] |
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ODCINEK 1 |
-0,57 |
440,00 |
- |
139,33 |
|
ODCINEK 1 |
0,57 |
440,00 |
139,33 |
- |
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ODCINEK 2 |
-0,38 |
530,00 |
- |
111,89 |
|
ODCINEK 2 |
0,38 |
530,00 |
111,89 |
- |
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ODCINEK 3 |
1,26 |
460,00 |
322,00 |
- |
|
ODCINEK 3 |
-1,26 |
460,00 |
- |
322,00 |
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ODCINEK 4 |
-0,76 |
250,00 |
- |
105,56 |
|
ODCINEK 4 |
0,76 |
250,00 |
105,56 |
- |
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ODCINEK 5 |
0,13 |
230,00 |
16,61 |
- |
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ODCINEK 5 |
-0,13 |
230,00 |
- |
16,61 |
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ODCINEK 6 |
0,51 |
790,00 |
223,83 |
- |
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ODCINEK 6 |
-0,51 |
790,00 |
- |
223,83 |
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ODCINEK 7 |
-0,54 |
638,38 |
- |
191,51 |
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ODCINEK 7 |
0,54 |
638,38 |
191,51 |
- |
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SUMA = |
3338,38 |
562,44 |
548,29 |
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SUMA = |
3338,38 |
548,29 |
562,44 |
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Lsp= |
3352,53 |
m |
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Lsp= |
3324,23 |
m |
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DŁUGOŚĆ WIRTUALNA POTRZEBNA DO OCENY WARIANTU I WYNOSI: |
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L = |
3338,38 |
m |
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22 |
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2. OBLICZENIA. |
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2.2. PUNKTY GŁÓWNE TRASY (PIKIETAŻ KILOMETRACJI) |
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2.1. OBLICZENIE ŁUKÓW POZIOMYCH. |
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WARIANT 1 WIERZCHOŁEK 1 |
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WARIANT 1 ŁUK 1 |
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AW1= |
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0+678,00 |
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678,00 |
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PKP1= AW1-To= |
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0+678,00 |
- |
0+038,49 |
= |
0+639,51 |
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|
639,51 |
a) kąt zwrotu trasy: |
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α = |
23,18 |
[°] |
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PŁ= PKP1 +L/2 = |
|
0+639,51 |
+ |
0+018,00 |
= |
0+657,51 |
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|
657,51 |
b) prędkość projektowa: |
|
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vp = |
50 |
[km/h] |
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KKP1= PŁ+L/2= |
|
0+657,51 |
+ |
0+018,00 |
= |
0+675,51 |
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|
675,51 |
c) promień łuku kołowego: |
|
|
R = |
100 |
[m] |
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KKP2= KKP1+K= |
|
0+675,51 |
+ |
0+004,46 |
= |
0+679,97 |
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|
679,97 |
d) szerokość jezdni: |
|
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B = |
5,5 |
[m] |
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KŁ= KKP2+L/2= |
|
0+679,97 |
+ |
0+018,00 |
= |
0+697,97 |
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|
697,97 |
e) pochylenie poprzeczne na prostej: |
|
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ip = |
2 |
[%] |
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PKP2=KŁ+L/2= |
|
0+697,97 |
+ |
0+018,00 |
= |
0+715,97 |
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|
715,97 |
f) pochylenie poprzeczne na łuku: |
|
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io = |
3 |
[%] |
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1) WARUNEK DYNAMICZNY: |
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Amin1 = |
57,87 |
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GDZIE: |
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2) WARUNEK GEOMETRYCZNY ZESPOŁU KRZYWA PRZEJŚCIOWA-ŁUK KOŁOWY KRZYWA PRZEJŚCIOWA: |
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Amax2 = |
63,61 |
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3) WARUNEK ESTETYKI (WARUNEK KĄTA τ): |
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Amin3 = |
33,33 |
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To = |
38,49 |
m |
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Amax4 = |
100,00 |
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L= |
36,00 |
m |
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4) WARUNEK ESTETYKI (WIELKOŚĆ ODSUNIĘCIA ŁUKU H): |
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H = 0,5 m |
→ |
Amin5 = |
58,82 |
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L / 2 = |
18,00 |
m |
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H = 2,5 m |
→ |
Amax6 = |
87,91 |
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K = |
4,46 |
m |
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H = 0,2 m |
→ |
Amin7 = |
46,80 |
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5) WARUNEK KONSTRUKCYJNY (STOSOWANY NA ŁUKACH Z POSZERZENIEM): |
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40 / R = 0,11m |
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NIE UWZGLĘDNIAMY |
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6) WARUNEK KONSTRUKCYJNY (KOMFORTU JAZDY): |
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Amin8 = |
26,22 |
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58,82 |
≤ A ≤ |
63,61 |
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PRZY ZAŁOŻENIU PROPORCJI 1:2:1 |
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PRZY ZAŁOŻENIU PROPORCJI 1:1:1 |
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A = |
36,72 |
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A = |
44,98 |
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PRZYJMUJEMY PARAMETR KLOTOIDY RÓWNY A = |
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60 |
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1 |
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9 |
DŁUGOŚĆ KLOTOIDY: |
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L= |
36,00 |
m |
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L = A2 / R |
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KĄT |
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τ = |
0,180 |
rad |
γ = |
0,045 |
rad |
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DŁUGOŚĆ ŁUKU: |
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K = |
4,46 |
m |
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K = R · γ |
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OBLICZENIE PARAMETRÓW KLOTOIDY. |
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1) Odcięta X końca krzywej przejściowej: |
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35,88 |
m |
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2) Rzędna Y końca krzywej przejściowej: |
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2,16 |
m |
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3) Odcięta Xs środka krzywizny od początku krzywej przejściowej: |
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Xs = X - R · sin τ = |
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17,98 |
m |
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4) Odsunięcie H łuku koła od prostego kierunku trasy: |
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H = Y - R ( 1 - cos τ ) = |
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0,540 |
m |
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5) Dłudość stycznej T0 : |
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T0 = T + Xs = |
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38,49 |
m |
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T = R · tg α/2 = |
20,51 |
m |
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6) Odległość środka łuku Z0 od punktu przecięcia się stycznych: |
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3257,37 |
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2,63 |
m |
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3234,70 |
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-22,67 |
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2 |
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WARIANT 1 WIERZCHOŁEK 2 |
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W1 W2= |
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1+981,93 |
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1+981,93 |
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1981,93 |
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W2= AW1 +W1W2= |
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0+678,00 |
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1+981,93 |
= |
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2+659,93 |
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2659,93 |
WARIANT 1 ŁUK 2 |
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PKP3= PKP2+W1W2-To-To2= |
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0+715,97 |
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1+981,93 |
- |
0+038,49 |
- |
0+078,54 |
= |
2+580,86 |
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2580,86 |
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PŁ= PKP3 +L/2 = |
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2+580,86 |
+ |
0+032,00 |
= |
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2+612,86 |
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2612,86 |
a) kąt zwrotu trasy: |
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α = |
50,02 |
[°] |
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KKP3= PŁ+L/2= |
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2+612,86 |
+ |
0+032,00 |
= |
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2+644,86 |
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2644,86 |
b) prędkość projektowa: |
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vp = |
50 |
[km/h] |
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KKP4= KKP3+K2= |
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2+644,86 |
+ |
0+149,56 |
= |
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2+668,17 |
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2668,17 |
c) promień łuku kołowego: |
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R = |
100 |
[m] |
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KŁ= KKP2+L/2= |
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2+668,17 |
+ |
0+032,00 |
= |
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2+700,17 |
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2700,17 |
d) szerokość jezdni: |
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B = |
5,5 |
[m] |
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PKP4=KŁ+L/2= |
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2+700,17 |
+ |
0+032,00 |
= |
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2+732,17 |
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2732,17 |
e) pochylenie poprzeczne na prostej: |
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ip = |
2 |
[%] |
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f) pochylenie poprzeczne na łuku: |
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io = |
3 |
[%] |
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GDZIE: |
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1) WARUNEK DYNAMICZNY: |
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Amin1 = |
57,87 |
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2) WARUNEK GEOMETRYCZNY ZESPOŁU KRZYWA PRZEJŚCIOWA-ŁUK KOŁOWY KRZYWA PRZEJŚCIOWA: |
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Amax2 = |
93,44 |
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To = |
38,49 |
m |
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3) WARUNEK ESTETYKI (WARUNEK KĄTA τ): |
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Amin3 = |
33,33 |
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To2 = |
79 |
m |
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Amax4 = |
100,00 |
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L= |
64,00 |
m |
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4) WARUNEK ESTETYKI (WIELKOŚĆ ODSUNIĘCIA ŁUKU H): |
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H = 0,5 m |
→ |
Amin5 = |
58,82 |
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L / 2 = |
32,00 |
m |
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H = 2,5 m |
→ |
Amax6 = |
87,91 |
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K2 = |
23,30 |
m |
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H = 0,2 m |
→ |
Amin7 = |
46,80 |
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5) WARUNEK KONSTRUKCYJNY (STOSOWANY NA ŁUKACH Z POSZERZENIEM): |
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40 / R = |
0,40 |
NIE UWZGLĘDNIAMY |
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6) WARUNEK KONSTRUKCYJNY (KOMFORTU JAZDY): |
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Amin8 = |
26,22 |
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58,82 |
≤ A ≤ |
87,91 |
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PRZY ZAŁOŻENIU PROPORCJI 1:2:1 |
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PRZY ZAŁOŻENIU PROPORCJI 1:1:1 |
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A = |
53,94 |
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A = |
66,07 |
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PRZYJMUJEMY PARAMETR KLOTOIDY RÓWNY A = |
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80 |
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14 |
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3 |
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10 |
DŁUGOŚĆ KLOTOIDY: |
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L= |
64,00 |
m |
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L = A2 / R |
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KĄT |
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τ = |
0,320 |
rad |
γ = |
0,233 |
rad |
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DŁUGOŚĆ ŁUKU: |
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K = |
23,30 |
m |
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K = R · γ |
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OBLICZENIE PARAMETRÓW KLOTOIDY. |
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1) Odcięta X końca krzywej przejściowej: |
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63,3 |
m |
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2) Rzędna Y końca krzywej przejściowej: |
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6,78 |
m |
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3) Odcięta Xs środka krzywizny od początku krzywej przejściowej: |
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Xs = X - R · sin τ = |
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31,9 |
m |
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4) Odsunięcie H łuku koła od prostego kierunku trasy: |
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H = Y - R ( 1 - cos τ ) = |
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1,70 |
m |
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5) Dłudość stycznej T0 : |
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T0 = T + Xs = |
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78,5 |
m |
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T = R · tg α/2 = |
46,65 |
m |
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6) Odległość środka łuku Z0 od punktu przecięcia się stycznych: |
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12,23 |
m |
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4 |
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WARIANT 2 WIERZCHOŁEK 1 |
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AW1= |
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1+031,89 |
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1031,89 |
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PKP1= AW1-To= |
|
1+031,89 |
- |
0+161,78 |
= |
0+953,28 |
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953,28 |
WARIANT 2 ŁUK 1 |
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PŁ= PKP1 +L/2 = |
|
0+953,28 |
+ |
0+032,00 |
= |
0+985,28 |
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985,28 |
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KKP1= PŁ+L/2= |
|
0+985,28 |
+ |
0+032,00 |
= |
1+017,28 |
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1017,28 |
a) kąt zwrotu trasy: |
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α = |
50,08 |
[°] |
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KKP2= KKP1+K= |
|
1+017,28 |
+ |
0+149,56 |
= |
1+040,69 |
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|
1040,69 |
b) prędkość projektowa: |
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vp = |
50 |
[km/h] |
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KŁ= KKP2+L/2= |
|
1+040,69 |
+ |
0+032,00 |
= |
1+072,69 |
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|
1072,69 |
c) promień łuku kołowego: |
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R = |
100 |
[m] |
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PKP2=KŁ+L/2= |
|
1+072,69 |
+ |
0+032,00 |
= |
1+104,69 |
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|
1104,69 |
d) szerokość jezdni: |
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B = |
5,5 |
[m] |
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e) pochylenie poprzeczne na prostej: |
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ip = |
2 |
[%] |
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f) pochylenie poprzeczne na łuku: |
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io = |
3 |
[%] |
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GDZIE: |
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1) WARUNEK DYNAMICZNY: |
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Amin1 = |
57,87 |
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To = |
78,61 |
m |
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2) WARUNEK GEOMETRYCZNY ZESPOŁU KRZYWA PRZEJŚCIOWA-ŁUK KOŁOWY KRZYWA PRZEJŚCIOWA: |
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Amax2 = |
93,49 |
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L= |
64,00 |
m |
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3) WARUNEK ESTETYKI (WARUNEK KĄTA τ): |
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Amin3 = |
33,33 |
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L / 2 = |
32,00 |
m |
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Amax4 = |
100,00 |
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K = |
23,41 |
m |
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4) WARUNEK ESTETYKI (WIELKOŚĆ ODSUNIĘCIA ŁUKU H): |
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H = 0,5 m |
→ |
Amin5 = |
58,82 |
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H = 2,5 m |
→ |
Amax6 = |
87,91 |
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H = 0,2 m |
→ |
Amin7 = |
46,80 |
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5) WARUNEK KONSTRUKCYJNY (STOSOWANY NA ŁUKACH Z POSZERZENIEM): |
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40 / R = |
0,40 |
NIE UWZGLĘDNIAMY |
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6) WARUNEK KONSTRUKCYJNY (KOMFORTU JAZDY): |
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Amin8 = |
26,22 |
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58,82 |
≤ A ≤ |
87,91 |
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PRZY ZAŁOŻENIU PROPORCJI 1:2:1 |
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PRZY ZAŁOŻENIU PROPORCJI 1:1:1 |
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A = |
53,98 |
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A = |
66,11 |
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PRZYJMUJEMY PARAMETR KLOTOIDY RÓWNY A = |
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80 |
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16 |
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5 |
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11 |
DŁUGOŚĆ KLOTOIDY: |
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L= |
64,00 |
m |
|
L = A2 / R |
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KĄT |
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τ = |
0,320 |
rad |
γ = |
0,234 |
rad |
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DŁUGOŚĆ ŁUKU: |
|
|
K = |
23,41 |
m |
|
K = R · γ |
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OBLICZENIE PARAMETRÓW KLOTOIDY. |
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1) Odcięta X końca krzywej przejściowej: |
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63,35 |
m |
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2) Rzędna Y końca krzywej przejściowej: |
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6,78 |
m |
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3) Odcięta Xs środka krzywizny od początku krzywej przejściowej: |
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Xs = X - R · sin τ = |
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31,89 |
m |
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4) Odsunięcie H łuku koła od prostego kierunku trasy: |
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H = Y - R ( 1 - cos τ ) = |
|
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1,705 |
m |
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5) Dłudość stycznej T0 : |
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T0 = T + Xs = |
|
|
78,61 |
m |
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|
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T = R · tg α/2 = |
46,72 |
m |
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6) Odległość środka łuku Z0 od punktu przecięcia się stycznych: |
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12,26 |
m |
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6 |
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|
WARIANT 2 WIERZCHOŁEK 2 |
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W1 W2= |
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1+680,85 |
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1680,85 |
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W2= AW1 +W1W2= |
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1+031,89 |
+ |
1+680,85 |
= |
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2+634,13 |
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2634,13 |
WARIANT 2 ŁUK 2 |
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PKP3= PKP2+W1W2-To-To2= |
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1+104,69 |
+ |
1+680,85 |
- |
0+078,61 |
- |
0+084,45 |
= |
2+622,48 |
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2622,48 |
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PŁ= PKP3 +L/2 = |
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2+622,48 |
+ |
0+032,00 |
= |
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2+654,48 |
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2654,48 |
a) kąt zwrotu trasy: |
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α = |
55,45 |
[°] |
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KKP3= PŁ+L/2= |
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2+654,48 |
+ |
0+032,00 |
= |
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2+686,48 |
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2686,48 |
b) prędkość projektowa: |
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vp = |
50 |
[km/h] |
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KKP4= KKP3+K2= |
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2+686,48 |
+ |
0+149,56 |
= |
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2+719,26 |
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|
2719,26 |
c) promień łuku kołowego: |
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R = |
100 |
[m] |
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KŁ= KKP2+L/2= |
|
2+719,26 |
+ |
0+032,00 |
= |
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2+751,26 |
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2751,26 |
d) szerokość jezdni: |
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B = |
5,5 |
[m] |
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PKP4=KŁ+L/2= |
|
2+751,26 |
+ |
0+032,00 |
= |
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2+783,26 |
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|
2783,26 |
e) pochylenie poprzeczne na prostej: |
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ip = |
2 |
[%] |
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f) pochylenie poprzeczne na łuku: |
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io = |
3 |
[%] |
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GDZIE: |
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1) WARUNEK DYNAMICZNY: |
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Amin1 = |
57,87 |
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2) WARUNEK GEOMETRYCZNY ZESPOŁU KRZYWA PRZEJŚCIOWA-ŁUK KOŁOWY KRZYWA PRZEJŚCIOWA: |
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Amax2 = |
98,38 |
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To = |
78,61 |
m |
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3) WARUNEK ESTETYKI (WARUNEK KĄTA τ): |
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Amin3 = |
33,33 |
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To2 = |
84,45 |
m |
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Amax4 = |
100,00 |
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L= |
64,00 |
m |
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4) WARUNEK ESTETYKI (WIELKOŚĆ ODSUNIĘCIA ŁUKU H): |
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H = 0,5 m |
→ |
Amin5 = |
58,82 |
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L / 2 = |
32,00 |
m |
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H = 2,5 m |
→ |
Amax6 = |
87,91 |
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K2 = |
32,78 |
m |
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H = 0,2 m |
→ |
Amin7 = |
46,80 |
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5) WARUNEK KONSTRUKCYJNY (STOSOWANY NA ŁUKACH Z POSZERZENIEM): |
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40 / R = |
0,40 |
NIE UWZGLĘDNIAMY |
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6) WARUNEK KONSTRUKCYJNY (KOMFORTU JAZDY): |
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Amin8 = |
26,22 |
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58,82 |
≤ A ≤ |
87,91 |
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PRZY ZAŁOŻENIU PROPORCJI 1:2:1 |
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PRZY ZAŁOŻENIU PROPORCJI 1:1:1 |
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A = |
56,80 |
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A = |
69,56 |
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PRZYJMUJEMY PARAMETR KLOTOIDY RÓWNY A = |
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80 |
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7 |
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12 |
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DŁUGOŚĆ KLOTOIDY: |
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L= |
64,00 |
m |
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L = A2 / R |
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KĄT |
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τ = |
0,320 |
rad |
γ = |
0,328 |
rad |
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DŁUGOŚĆ ŁUKU: |
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K = |
32,78 |
m |
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K = R · γ |
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OBLICZENIE PARAMETRÓW KLOTOIDY. |
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1) Odcięta X końca krzywej przejściowej: |
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63,35 |
m |
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2) Rzędna Y końca krzywej przejściowej: |
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6,78 |
m |
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3) Odcięta Xs środka krzywizny od początku krzywej przejściowej: |
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Xs = X - R · sin τ = |
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|
31,89 |
m |
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4) Odsunięcie H łuku koła od prostego kierunku trasy: |
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H = Y - R ( 1 - cos τ ) = |
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|
1,705 |
m |
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5) Dłudość stycznej T0 : |
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T0 = T + Xs = |
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|
84,45 |
m |
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T = R · tg α/2 = |
52,56 |
m |
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6) Odległość środka łuku Z0 od punktu przecięcia się stycznych: |
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14,90 |
m |
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8 |
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CO DO DALSZEJ CZĘŚCI TO ZNACZY ODTĄD W DÓŁ NIE JESTEM PEWIEN CZY ARKUSZ POLICZY SAMODZIELNIE :) |
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2.3. TABELA ZAŁOMÓW. |
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2.3.1. WARIANT I. |
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50 |
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PIKIETAŻ |
ODLEGŁÓŚĆ |
POCHYLENIE [%] |
RÓŻNICA WYSOKOŚCI Δh [m] |
RZĘDNA ZAŁOMU |
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51,25 |
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WZNIESIENIE |
SPADEK |
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52,5 |
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0+000,00 |
- |
- |
- |
- |
65,75 |
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0 |
|
53,75 |
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0+379,64 |
379,64 |
0,30 |
- |
1,12 |
66,87 |
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379,64 |
379,64 |
55 |
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0+973,69 |
594,05 |
|
0,41 |
2,42 |
64,45 |
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594,05 |
973,69 |
56,25 |
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1+979,29 |
1005,60 |
0,48 |
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4,8 |
69,25 |
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1005,6 |
1979,29 |
57,5 |
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2+646,19 |
666,90 |
|
0,82 |
5,5 |
63,75 |
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666,9 |
2646,19 |
58,75 |
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3+344,68 |
708,50 |
2,12 |
- |
15 |
78,75 |
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708,5 |
3354,69 |
60 |
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#VALUE! |
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2.3.2. WARIANT II. |
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#VALUE! |
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3344,68 |
#VALUE! |
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PIKIETAŻ |
ODLEGŁÓŚĆ |
POCHYLENIE [%] |
RÓŻNICA WYSOKOŚCI Δh [m] |
RZĘDNA ZAŁOMU |
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10,0100000000002 |
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WZNIESIENIE |
SPADEK |
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0+000,00 |
- |
- |
- |
- |
67,30 |
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0 |
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0+347,10 |
347,10 |
1,50 |
- |
5,2 |
72,50 |
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347,1 |
347,1 |
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1+422,80 |
1075,70 |
0,35 |
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3,75 |
76,25 |
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1075,7 |
1422,8 |
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1+887,50 |
464,70 |
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2,96 |
13,75 |
62,50 |
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464,7 |
1887,5 |
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2+440,40 |
552,90 |
0,45 |
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2,5 |
65,00 |
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552,9 |
2440,4 |
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3+257,37 |
794,25 |
1,73 |
- |
13,75 |
78,75 |
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794,25 |
3234,65 |
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2.4. OBLICZENIA ŁUKÓW PIONOWYCH. |
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2.4.1. WARIANT I. |
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PIKIETA |
78 |
i1 |
i2 |
i |
R |
t |
f |
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0+379,64 |
8 |
0,30% |
0,41% |
0,71% |
1500 |
5,325 |
0,009 |
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0+973,69 |
7 |
0,41% |
0,48% |
0,89% |
1000 |
4,45 |
0,010 |
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1+979,29 |
8 |
0,48% |
0,82% |
1,30% |
1500 |
9,75 |
0,032 |
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1,95 |
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2+440,40 |
7 |
0,82% |
2,12% |
2,94% |
1000 |
14,7 |
0,108 |
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2,94 |
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2.4.2. WARIANT Ii. |
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PIKIETA |
78 |
i1 |
i2 |
i |
R |
t |
f |
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0+347,10 |
8 |
1,50% |
0,35% |
1,15% |
1500 |
8,625 |
0,025 |
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1,725 |
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1+422,80 |
8 |
0,35% |
2,96% |
3,31% |
1500 |
24,825 |
0,205 |
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4,965 |
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1+887,50 |
7 |
2,96% |
0,45% |
3,41% |
1000 |
17,05 |
0,145 |
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3,41 |
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2+440,40 |
7 |
0,45% |
1,73% |
1,28% |
1000 |
6,4 |
0,020 |
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1,28 |
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2.5. DŁUGOŚĆ WIRTUALNA. |
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Lsp=Σl+Σiw/f x lw-Σis/f x ls |
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f = |
0,018 |
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2.5.1. WARIANT I. |
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OBLICZENIA W KIERUNKU ZGODNYM Z PIKIETAŻEM TRASY A-B |
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OBLICZENIA W KIERUNKU PRZECIWNYM Z PIKIETAŻEM TRASY B-A |
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ODCINEK |
POCHYLENIE PODŁUŻNE [%] |
DŁUGOŚĆ ODCINKÓW WZNIESIEŃ ORAZ SPADKÓW NIESZKODLIWYCH [m] |
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ODCINEK |
POCHYLENIE PODŁUŻNE [%] |
DŁUGOŚĆ ODCINKÓW WZNIESIEŃ ORAZ SPADKÓW NIESZKODLIWYCH [m] |
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ODCINEK 1 |
0,30 |
379,64 |
63,3 |
- |
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ODCINEK 1 |
-0,30 |
379,64 |
- |
63,2733333333333 |
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ODCINEK 2 |
-0,41 |
594,05 |
|
135,3 |
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ODCINEK 2 |
0,41 |
594,05 |
135,31 |
- |
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ODCINEK 3 |
0,48 |
1005,60 |
268,2 |
- |
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ODCINEK 3 |
-0,48 |
1005,60 |
- |
268,16 |
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ODCINEK 4 |
-0,82 |
666,90 |
- |
303,8 |
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ODCINEK 4 |
0,82 |
666,90 |
303,81 |
- |
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ODCINEK 5 |
2,12 |
708,50 |
834,46 |
- |
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ODCINEK 5 |
-2,12 |
708,50 |
- |
834,46 |
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SUMA = |
3354,69 |
1165,89 |
439,12 |
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SUMA = |
3354,69 |
439,12 |
1165,9 |
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Lsp= |
4081,46 |
m |
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Lsp= |
2627,92 |
m |
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DŁUGOŚĆ WIRTUALNA POTRZEBNA DO OCENY WARIANTU I WYNOSI: |
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L = |
3354,69 |
m |
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15 |
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Lsp=Σl+Σiw/f x lw-Σis/f x ls |
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f = |
0,018 |
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2.5.2. WARIANT II. |
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OBLICZENIA W KIERUNKU ZGODNYM Z PIKIETAŻEM TRASY A-B |
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OBLICZENIA W KIERUNKU PRZECIWNYM Z PIKIETAŻEM TRASY B-A |
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ODCINEK |
POCHYLENIE PODŁUŻNE [%] |
DŁUGOŚĆ ODCINKÓW WZNIESIEŃ ORAZ SPADKÓW NIESZKODLIWYCH [m] |
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ODCINEK |
POCHYLENIE PODŁUŻNE [%] |
DŁUGOŚĆ ODCINKÓW WZNIESIEŃ ORAZ SPADKÓW NIESZKODLIWYCH [m] |
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ODCINEK 1 |
1,50 |
347,10 |
289,25 |
- |
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ODCINEK 1 |
-1,50 |
440,00 |
- |
-289,25 |
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ODCINEK 2 |
0,35 |
1075,70 |
209,2 |
- |
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ODCINEK 2 |
-0,35 |
530,00 |
- |
209,16 |
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ODCINEK 3 |
-2,96 |
464,70 |
- |
764,2 |
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ODCINEK 3 |
2,96 |
460,00 |
764,17 |
- |
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ODCINEK 4 |
0,45 |
552,90 |
138,225 |
- |
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ODCINEK 4 |
-0,45 |
250,00 |
- |
138,23 |
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ODCINEK 5 |
1,73 |
794,25 |
763,36 |
- |
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ODCINEK 5 |
-1,73 |
230,00 |
- |
763,36 |
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SUMA = |
3234,65 |
1400,00 |
764,17 |
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SUMA = |
1910,00 |
764,17 |
821,50 |
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Lsp= |
3870,48 |
m |
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Lsp= |
1852,67 |
m |
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DŁUGOŚĆ WIRTUALNA POTRZEBNA DO OCENY WARIANTU I WYNOSI: |
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L = |
2861,58 |
m |
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16 |
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