【正文】 T12=ST11= q= Ls/2=15mR’= R12+△R12=T12q/tga/2 = △R12= Ls2/(24R’)=R12= R’△R12= 驗算△R12= Ls2/(24R)=符合要求:即R11=緩和曲線常數(shù)及平曲線要素： β=Ls180/(2Rπ)= 176。JD13=S+ JD122T12+L12= m T13=ST12= q= Ls/2=R’= R13+△R13=T13q/tga/2 = △R13= Ls2/(24R’)=R13= R’△R13= 驗算△R13= Ls2/(24R)=符合要求:即R13=緩和曲線常數(shù)及平曲線要素： β=Ls180/(2Rπ)= 176。JD14=S+ JD132T13+L13= m T14=ST13= q= Ls/2=15mR’= R14+△R14=T14q/tga/2 = △R14= Ls2/(24R’)=R14= R’△R14= 驗算△R14= Ls2/(24R)=符合要求:即R14=緩和曲線常數(shù)及平曲線要素： β=Ls180/(2Rπ)= 176。JD15=S+ JD142T14+L14= m T15=ST14= q= Ls/2=R’= R15+△R15=T15q/tga/2 = △R15= Ls2/(24R’)=R15= R’△R15= 驗算△R15= Ls2/(24R)=符合要求:即R15=緩和曲線常數(shù)及平曲線要素： β=Ls180/(2Rπ)= 176。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。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。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。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。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。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。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。路線縱斷面圖上的設(shè)計標(biāo)高，即路基設(shè)計標(biāo)高，按照《規(guī)范》規(guī)定如下：三級公路采用路基邊緣標(biāo)高，在設(shè)置超高、加寬地段為設(shè)超高、加寬前該處邊緣標(biāo)高。（2）應(yīng)避免能看見近處和遠(yuǎn)處而看不見中間凹處的線形。（4）相鄰縱坡之代數(shù)差小時，應(yīng)盡量用用大的豎曲線半徑。（6）在積雪或冰凍地區(qū)，應(yīng)避免采用陡坡。（2）坡長限制①最大坡長的限制：《標(biāo)準(zhǔn)》對各級公路不同陡坡的最大坡長加以限制見下表。5．4平縱組合設(shè)計1．組合原則公路平、縱線形組合的基本原則是：（1）應(yīng)能在視覺上自然地誘導(dǎo)駕駛員的視線，并保持視覺的連續(xù)性；（2）平面與縱斷面線形的技術(shù)指標(biāo)應(yīng)大小均衡，不要懸殊太大，使線形在視覺上和心理上保持協(xié)調(diào)；（3）選擇組合得當(dāng)?shù)暮铣善露?，以利于路面排水和安全行車?．5設(shè)計標(biāo)高的計算豎曲線部分設(shè)計標(biāo)高的計算①計算豎曲線的基本要素：豎曲線長：L、切線長：T、外距：E②計算豎曲線的起、終點的樁號豎曲線的起點的樁號 = 變坡點的樁號－T豎曲線的終點的樁號 = 變坡點的樁號+T③計算豎曲線上任意點切線標(biāo)高及改正值 切線標(biāo)高 = 變坡點的標(biāo)高177。2．加寬值計算：本設(shè)計路線規(guī)定和要求，采用第1類假定汽車軸距加前懸為5，根據(jù)不同曲線半徑對應(yīng)的加寬值進行加寬，規(guī)定如下表：圓曲線半徑（m）2502002001501501001007070505030302525202015加寬值（m）0．40．60．81．01．21．41．82．22．56．2超高1．超高的方式：本設(shè)計采用繞內(nèi)邊緣線旋轉(zhuǎn)的超高方式。 超高橫坡度的確定本設(shè)計中設(shè)計車速為30km/h，假定為非積雪冰凍地區(qū)，即在一般情況下，超高橫坡度的可參照下表而確定：半徑（m）3502323015011080806015011060505030超高(%)2345678超高值的計算JDi1=2% i0=4% a= b/2= b= 查得： Bj= ib=8% L1=(i1/ib)ls= 樁號：KO+、KO+、KO+、hc= ai0（a+b）ib= hcˊ= ai0+(b/2 ) ib =hc〞=ai0（a+Bj）ib =JDi1=2% i0=4% a= b/2= b= x= ix=(x/ls)ib=%查得： Bj= ib=8% L1=(i1/ib)ls= Bjx=(x/ls) Bj=樁號：KO+、KO+、KO+、KO+hc=ai0（a+b）ib= hcˊ=ai0+(b/2 ) ib=hc〞=ai0（a+Bj）ib= 樁號：K0+hcx=(x/ls)hc= hcxˊ=ai0+ (b/2) i1=hcx〞=ai0（a+ Bjx）i1 =JDi1=2% i0=4% a= b/2= b= x= ix=(x/ls)ib=%查得： Bj= ib=8% L1=(i1/ib)ls= Bjx=(x/ls) Bj=樁號：KO+、KO+、KO+、KO+、KO+、KO+hc=ai0（a+b）ib= hcˊ=ai0+(b/2 ) ib=hc〞=ai0（a+Bj）ib= 樁號：K0+hcx=(x/ls)hc= hcxˊ=ai0+ (b/2) i1=hcx〞=ai0（a+ Bjx）(x/ls)i1 =JDi1=2% i0=4% a= b/2= b= x= ix=(x/ls)ib=%查得： Bj= ib=8% L1=(i1/ib)ls= Bjx=(x/ls) Bj=樁號：KO+、KO+、KO+、KO+、KO+hc=ai0（a+b）ib= hcˊ=ai0+(b/2 ) ib=hc〞=ai0（a+Bj）ib= 樁號：K0+hcx=(x/ls)hc= hcxˊ=ai0+ (b/2)（x/ls)ib=hcx〞=ai0（a+ Bjx）(x/ls)ib =JDi1=2% i0=4% a= b/2= b= x= ix=(x/ls)ib=%查得： Bj= ib=8% L1=(i1/ib)ls= Bjx=（x/ls）Bj=樁號：KO+、KO+、KO+、KO+、KO+hc=ai0（a+b）ib= hcˊ=ai0+(b/2 ) ib=hc〞=ai0（a+Bj）ib= 樁號：K0+hcx=(x/ls)hc= hcxˊ=ai0+ (b/2)（x/ls)ib=hcx〞=ai0（a+ Bjx）(x/ls)ib =JDi1=2% i0=4% a= b/2= b= x=10m ix=(x/ls)ib=%查得： Bj= ib=8% L1=(i1/ib)ls= Bjx=（x/ls）Bj=樁號：KO+、KO+、KO+、KO+、KO+、KO+、KO+hc=ai0（a+b）ib= hcˊ=ai0+(b/2 ) ib=hc〞=ai0（a+Bj）ib= 樁號：K0+hcx=(x/ls)hc= hcxˊ=ai0+ (b/2)（x/ls)ib=hcx〞=ai0（a+ Bjx）(x/ls)ib =JDi1=2% i0=4% a= b/2= b= x=10m ix=(x/ls)ib=%查得： Bj= ib=8% L1=(i1/ib)ls= Bjx=（x/ls）Bj=樁號：KO+、KO+、KO+、KO+hc=ai0（a+b）ib= hcˊ=ai0+(b/2 ) ib=hc〞=ai0（a+Bj）ib= 樁號：K0+hcx=(x/ls)hc= hcxˊ=ai0+ (b/2)（x/ls)ib=hcx〞=ai0（a+ Bjx）(x/ls)ib=JDi=2% i0=4% a= b/2= b= x=22m ix=(x/ls)ib=%查得： Bj= ib=8% L1=(i1/ib)ls= Bjx=（x/ls）Bj=樁號：KO+、KO+、KO+、KO+hc=ai0（a+b）ib= hcˊ=ai0+(b/2 ) ib=hc〞=ai0（a+Bj）ib= 樁號：K0+hcx=(x/ls)hc= hcxˊ=ai0+ (b/2)（x/ls)i1=hcx〞=ai0（a+ Bjx）(x/ls)i1 =JDi=2% i0=4% a= b/2= b= x= ix=(x/ls)ib=7%查得： Bj= ib=8% L1=(i1/ib)ls= Bjx=（x/ls）Bj=樁號：K1+、K1+、K1+、K1+、K1+hc=ai0（a+b）ib= hcˊ=ai0+(b/2 ) ib=hc〞=ai0（a+Bj）ib= 樁號：K1+hcx=(x/ls)hc= hcxˊ=ai0+ (b/2)（x/ls)i1=hcx〞=ai0（a+ Bjx）(x/ls)i1 =JDi1=2% i0=4% a= b/2= b= x= ix=(x/ls)ib=%查得： Bj= ib=8% L1=(i1/ib)ls= Bjx=(x/ls) Bj=樁號：K1+、K1+、K1+、K1+、K1+hc=ai0（a+b）ib= hcˊ=ai0+(b/2 ) ib=hc〞=ai0（a+Bj）ib= 樁號：K1+hcx=(x/ls)hc= hcxˊ=ai0+ (b/2) i1=hcx〞=ai0（a+ Bjx）i1 =JD1i1=2% i0=4% a= b/2= b= x= ix=(x/ls)ib=%查得： Bj= ib=8% L1=(i1/ib)ls= Bjx=(x/ls) Bj=樁號：K1+、K1+、K1+、K1+、K1+hc=ai0（a+b）ib= hcˊ=ai0+(b/2 ) ib=hc〞=ai0（a+Bj）ib= 樁號：K1+hcx=(x/ls)h