Kolida KTS-442UT Series, Manual

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196
) 0 0000025 . 0 00078019 . 0 10666667 . 0 ( 64
   

6. 777
The example is a symmetrical transitional curve. N1=N2
，
E1=E2
⑷
calculate vector height
R

) cos 1 (

   
R E R
18 cos 1 ( 100 777 . 6
   
R
°20′06″)

1.700
In the symmetrical transitional curve
R R
2 1
  
⑸
calculate transitional point coordinate

sin R N N
m
 
=63.348-100sin18°20′06″=31.891
In the symmetrical transitional curve
N N
m m 2 1

⑹
Calculation of Tangent Distance
N R R D
m
LA LA ec
LA
R
1 1 2 1
) cot( ) ( cos )
2
tan(
     
 
LA
111°55′47″ ，
sin
1
cos

ec
,
tan
1
cot


D
1
100 * tan(111°55′47″/ 2) +1.7(1 / sin111°55′47″)
–1.7(1 / tan 111°55′47″) +31.891
=148.06015 + 1.8326 + 0.6844 +31.891
=182.468
D D
2 1

⑺
Calculation of coordinate of point KA1

1 1 1 1
cos
  
D N N
IP KA

1 1 1 1
sin
  
D E E
IP KA
Bearing from BP to IP1
 

1
74°03′16.6″

N
KA 1
1300 –182.468 * cos 74°03′16.6″=1249.872 m

E
KA 1
1750 –182.468 * sin 74°03′16.6″=1574.553 m
⑻
Calculation of Arc Length
) (
2 1
 
  
LA R L
=R(111°55′47″-2 * 18°20′06″)
=100( 75°15′35″
180


)