【文章內容簡介】 HY +(LLs) () HZ Ls 25YH (L2Ls)/2 (*25)/2QZ +D/2 JD15 K1 + 交點16樁號： JD16: K1 + Ls =25 α=82176。JD16=S+ JD152T15+L15= m T16=ST15= q= Ls/2=R’= R16+△R16=T16q/tga/2 = △R16= Ls2/(24R’)=R16= R’△R16= 驗算△R16= Ls2/(24R)=符合要求:即R16=緩和曲線常數(shù)及平曲線要素： β=Ls180/(2Rπ)= 176。 p=Ls2/(24R) Ls4/2688R3= q=(Ls/2) Ls3/(240R2) = T=(R+p)tg(α/2)+q= L=(α2β)πR/180+2Ls= m D=2TL= E=(R+p)sec(α/2)R= 主點樁樁號：JD16 K1 + T ZH +Ls 25HY +(LLs) () HZ Ls 25YH (L2Ls)/2 (*25)/2QZ +D/2 JD16 K1 + 交點17樁號： JD17: K2 + Ls =25 α=103176。JD17=S+ JD162T16+L16= m T17=ST16= q= Ls/2=R’= R17+△R17=T17q/tga/2 = △R17= Ls2/(24R’)=R17= R’△R17= 驗算△R17= Ls2/(24R)=符合要求:即R17=緩和曲線常數(shù)及平曲線要素： β=Ls180/(2Rπ)= 176。 p=Ls2/(24R) Ls4/2688R3= q=(Ls/2) Ls3/(240R2) = T=(R+p)tg(α/2)+q= L=(α2β)πR/180+2Ls= m D=2TL= E=(R+p)sec(α/2)R= 主點樁樁號：JD17 K2 + T ZH +Ls 25HY +(LLs) () HZ Ls 25YH (L2Ls)/2 (*25)/2QZ +D/2 JD17 K2 + 交點18樁號： JD18: K2 + Ls =25 α=58176。JD18=S+ JD172T17+L17=2 m T18=ST17= q= Ls/2=R’= R18+△R18=T18q/tga/2 = △R18= Ls2/(24R’)=R18= R’△R18= 驗算△R18= Ls2/(24R)=符合要求:即R18=緩和曲線常數(shù)及平曲線要素： β=Ls180/(2Rπ)= 176。 p=Ls2/(24R) Ls4/2688R3= q=(Ls/2) Ls3/(240R2) = T=(R+p)tg(α/2)+q= L=(α2β)πR/180+2Ls= m D=2TL= E=(R+p)sec(α/2)R= 主點樁樁號：JD18 K2 + T ZH +Ls 25HY +(LLs) () HZ Ls 25YH (L2Ls)/2 (*25)/2QZ +D/2 JD18 K2 + 交點19樁號： JD19: K2 + Ls =25 α=44176。JD19=S+ JD182T18+L18= m T19=ST18= q= Ls/2=R’= R19+△R19=T19q/tga/2 = △R19= Ls2/(24R’)=R19= R’△R19= 驗算△R19= Ls2/(24R)=符合要求:即R19=緩和曲線常數(shù)及平曲線要素： β=Ls180/(2Rπ)= 176。 p=Ls2/(24R) Ls4/2688R3= q=(Ls/2) Ls3/(240R2) = T=(R+p)tg(α/2)+q= L=(α2β)πR/180+2Ls= m D=2TL= E=(R+p)sec(α/2)R= 主點樁樁號：JD19 K2 + T ZH +Ls 25HY +(LLs) () HZ Ls 25YH (L2Ls)/2 (*25)/2QZ +D/2 JD19 K2 + 交點20樁號： JD20: K2 + Ls =30 α=124176。JD20=S+ JD192T19+L19= m T20=ST19= q= Ls/2=15mR’= R20+△R20=T20q/tga/2 = △R20= Ls2/(24R’)=R20= R’△R20= 驗算△R20= Ls2/(24R)=符合要求:即R20=緩和曲線常數(shù)及平曲線要素： β=Ls180/(2Rπ)= 176。 p=Ls2/(24R) Ls4/2688R3= q=(Ls/2) Ls3/(240R2) = T=(R+p)tg(α/2)+q= L=(α2β)πR/180+2Ls= m D=2TL= E=(R+p)sec(α/2)R= 主點樁樁號：JD20 K2 + T ZH +Ls 30HY +(LLs) () HZ Ls 30YH (L2Ls)/2 (*30)/2QZ +D/2 JD20 K2 + 交點21樁號： JD21: K2 + Ls =25 α=71176。JD21=S+ JD202T20+L20= m T21=ST20= q= Ls/2=R’= R21+△R21=T21q/tga/2 = △R21= Ls2/(24R’)=R21= R’△R21= 驗算△R21= Ls2/(24R)=符合要求:即R21=緩和曲線常數(shù)及平曲線要素： β=Ls180/(2Rπ)= 176。 p=Ls2/(24R) Ls4/2688R3= q=(Ls/2) Ls3/(240R2) = T=(R+p)tg(α/2)+q= L=(α2β)πR/180+2Ls= m D=2TL= E=(R+p)sec(α/2)R= 主點樁樁號：JD21 K2 + T ZH +Ls 25HY +(LLs) () HZ Ls 25YH (L2Ls)/2 (*25)/2QZ +D/2 JD21 K2 + 交點22樁號： JD22: K2 + Ls =25 α=44176。JD22=S+ JD212T21+L21= T22=ST21= q= Ls/2=R’= R22+△R22=T22q/tga/2 = △R22= Ls2/(24R’)=R22= R’△R22= 驗算△R22= Ls2/(24R)=符合要求:即R22=緩和曲線常數(shù)及平曲線要素： β=Ls180/(2Rπ)= 176。 p=Ls2/(24R) Ls4/2688R3= q=(Ls/2) Ls3/(240R2) = T=(R+p)tg(α/2)+q= L=(α2β)πR/180+2Ls= m D=2TL= E=(R+p)sec(α/2)R= 主點樁樁號：JD22 K2 + T ZH +Ls 25HY +(LLs) () HZ Ls 25YH (L2Ls)/2 (*25)/2QZ +D/2 JD22 K2 + 五、縱斷面設計縱斷面設計的主要任務就是根據(jù)汽車的動力特性、公路等級、地形、地物、水文地質，綜合考慮路基穩(wěn)定、排水以及工程經(jīng)濟性等，研究縱坡的大小、長短、豎曲線半徑以及與平面線形的組合關系，以便達到行車安全迅速、運輸經(jīng)濟合理及乘客感覺舒適的目的。路線縱斷面圖上的設計標高，即路基設計標高，按照《規(guī)范》規(guī)定如下：三級公路采用路基邊緣標高，在設置超高、加寬地段為設超高、加寬前該處邊緣標高。樁號Y(m)樁號Y(m)樁號Y(m)K0+K1+K1+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0+K1+K2+K0