道路勘测设计章课后复习题与答案(5)
时间:2025-04-21
时间:2025-04-21
. .. .
.. .. .. .. 因);();(4.03.00.429-40.4290.429-30.42922===R T E ;且2
2ωR L T ==,代入后解得:m R )2000;1500(=
分下面三种情况计算:
(1)取半径m R 1750=
m R L 70%0.41750=⨯==ω;
m L T 352
==; m R T E 35.01750
23522
2=⨯== 设计高程计算
起点里程桩号=交点里程桩号—T 终点里程桩号=交点里程桩号+T =K1+520-35 = K1+520+35 = K1+485 = K1+555
里程桩号 切线高程 竖距R
x h 22
= 设计高程 起点 K1+485 429+35×2.5%=429.875 0202
==R
h 429.875+0=429.875 K1+500 429+20×2.5%=429.50 064.02152
==R
h 429.50+0.064 =429.564
K1+520 429+0×2.5%=429.00 35.02352
==R
h 429.00+0.35=429.35 K1+515 429+5×2.5%=429.125 257.02302
==R
h 429.125+0.257=429.382
终点 K1+555 429+35×1.5%=429.525 0202
==R
h 429.525+0=429.525 (2)取半径m R 1500=
m R L 60%0.41500=⨯==ω;