平曲線要素計(jì)算

1、一、路線轉(zhuǎn)角、交點(diǎn)間距的計(jì)算（一）在地形圖上暈:出路線起終點(diǎn)及各路線交點(diǎn)的坐標(biāo):QP(23810,27180) JR(23996,26977)、JD(24684,26591) . JD(248480,25885)JD J 25350,25204)s ZD(26062,25783)(二) 計(jì)算公式及方法設(shè)起點(diǎn)坐標(biāo)為QD(Xq必)，第i個(gè)交點(diǎn)坐標(biāo)為刀2(X,，E)M = 1,2,3,4,則坐標(biāo)增 量 DX=Xj-Xi，DY =耳一阜I交點(diǎn)間距D = J(DX)2+(D汀象限角0 = arctanDY5%方位角A是由象限角推算的:象限D(zhuǎn)XDYAI+A = &n+A = 180 -0象限D(zhuǎn)XD

2、YAIIIA = 18O +&IV+A = 36O -01. QD與JD）之間:坐標(biāo)增量 DX = X|-Xo=23966-2381O=186>O"=片-； = 26977-27180 = -203<0交點(diǎn)間 距 D = J(DXy+(DY)= Jl 862+(-203)= 275.33加-203186象限角= arctanDYDX=arctan=47.502方位角 人=360 -0 = 360 -47.502' =312.4982. JDWa之間：坐標(biāo)增暈:DX=X. - X =24684 - 23966=688 > 0DY = Y,-Y, = 2

3、6591-26977 =-386 <0交點(diǎn)間距 D = J(DX)2+(DY$ = 688-+(-386/ = 788.89/DYDX象限角= arctan=arctan-386688= 29.294方位角 4 =360 -0 = 360 -29.294 =330.706, 轉(zhuǎn)角 a,=A, -人=330.706°-312.498 =18.208 3.刀2與JD3之間：坐標(biāo)增暈:DX=X-X? =24840 - 24684=156>0£>丫 = 丫3-匕=25885-26591 = -706<0交點(diǎn)間距 D = J(DX)2+(D 疔=156'

4、;+(-706)' = 723.03用-706156象限角0 = arctanDY=arctan= 77.54"方位角 £ =360 -0 = 360 -7754 =282460 轉(zhuǎn)角 a,=A, - A =282.46° -330.706 =-4&2464. JDjjJQ 之間：坐標(biāo)增量 DX = - X, =25350 - 24840=510 > 0DY = Y，-齊=25204-25885 = -681 <0交點(diǎn)間距 D = (DX/+(Dr)' = 510'+(-681/ =850.8川510象限角0 = arc

5、tanDYDX=arctan= 53.171方位角 4 =360 -& = 360 -53,171 =306.829 轉(zhuǎn)角 勺=4 一 4 =306.829-282.46 = 243695.丿2與ZD之間：坐標(biāo)增量 DX =X-Xj =26062 - 25350=712>0Dy = Y -片=25783 - 25204 = 579>0交點(diǎn)間距 D = J(DX)2+(Dy)2 = 7712'+579' = 917.706川DYDX象限角0 = arctan=arctan579712= 39.118方位角 4) = 0 = 39.118轉(zhuǎn)角 Q產(chǎn)人一人3 =

6、39118 -312.498 =92.289二、各平曲線要素的計(jì)算(-)JDi曲線要素計(jì)算取R = 800m,設(shè)計(jì)速度為60如/?JDi樁號(hào)為K0+275. 33,轉(zhuǎn)角a = 18.208r緩和曲線長度厶，則:1/3L,=0.036- = 0.036x_ = 9.72(.n)Ls > x3 = x3 = 50(m)3.6 3.6Ro(y)厶=上 800 = 8&89 800(m)99取整數(shù)，釆用緩和曲線長120m (公路工程技術(shù)標(biāo)準(zhǔn)規(guī)定:卩= 60jU時(shí)，最小緩和曲線長度為50m)120,120-2圓曲線內(nèi)移值A(chǔ)/?AR=H_ =二Z二 =0.75(m)24/? 2688x(/?

7、)'24x800 2688x(800)'' '3總切線長幾先求 q = b_-A_ =型 _ _12- = 59.989(m)2240/?'2240x800-所以 7； = (R + AZ?) tan 2 + <7 = (800 + 0.75) tan 蘭竺 + 59.989 = 188.3 l(m)2 24曲線總長度厶n厶i 20)面R + 2厶 s而叫=374.225)5,五個(gè)基本樁號(hào)JDK0+274. 33-)Th188. 31ZHKO+087. 02+)厶s120. 00砒+)(厶廠 2L$ )KO+207. 02134. 22YH、KO+

8、341.24+) Ls120. 00HZKO+461.24-)187. 11QZ,KO+274. 13a1&208Eh =(R + A/?)secR = (800 + 0.75 sec800 = 10.97(m)2 2超距 D = 27'-=2x18&31-374.22 = 2.4(m)。所以，由 QZ 樁號(hào)算出 JDi 樁號(hào)為K0 + 274.13 + =K0 + 275.33,與原來的JDi樁號(hào)相同，說明計(jì)算無誤。2式中：M設(shè)緩和曲線后圓曲線內(nèi)移值，m;0緩和曲線終點(diǎn)緩和曲線角，rad O g緩和曲線起點(diǎn)到圓曲線原起點(diǎn)的距離，也稱為切線增值,R圓曲線半徑，m；緩和曲

9、線長，m ;a偏角，（。）T,一一設(shè)置緩和曲線的曲線切線長，L,設(shè)置緩和曲線的曲線長，m;E,-設(shè)置緩和曲線的外距，m;6一一設(shè)置緩和曲線的超距，m;0緩和曲線終點(diǎn)緩和曲線角，rado（二）JD2曲線要素計(jì)算JD?樁號(hào)為K1+061.82,轉(zhuǎn)角_4&246。，設(shè)計(jì)速度為60如/“此平曲線通過埋口，為了控制高程，決定以外距E值控制半徑的大小，在圖紙上用比例尺量取£沁，暫取厶“2叭代入公式：噸55 +茅丐=。求的 R = 3116m,取/? = 300mr緩和曲線長度厶則:Ls = 0.036 = 0.036 x = 25.92(m) R300V 60Ls n x3 = x3 =

10、 50(m)3.6 3.6R300R=-R =300 = 33.33 300(m)99取整數(shù)，采用緩和曲線長120m (公路工程技術(shù)標(biāo)準(zhǔn)規(guī)定:V = Skmlh時(shí)，最小緩和曲線長度為50m,并滿足上述假設(shè)）120'12()42圓曲線內(nèi)移值MSR =亙-H_ =-irr =2(m)24/?2688x(R)324x300 2688x(300/3總切線長7；先求q丄一厶=型_皿，=59.997他2240/?-2240x300'所以 7； = (R + AT?) tan - +(/ = (300 + 2) tan 蘭空 + 59.997 = 195.13(m) 24.曲線總長度厶力C

11、厶 120 crB = =0,22R 2x300L=a-2p R + 2L,=a R+L, = 372.6 l(m) f "180'180'5五個(gè)基本樁號(hào)JD,Kl+061. 82-)Th195. 13ZHKO+866. 69+) Ls120. 00HY,KO+986. 69+) (Lh 2L$)132. 61%Kl+119. 30+) Ls120. 00HZ.Kl+239. 30-)2 h186. 31QZ.Kl+052.99瓦=(/? + AR)sec蘭-R = (300+0.2)sec一300 = 30.90(m)2 2超距D = 27-1, =2x195,13

12、-372.61 = 17.65(m)o 所以，由 QZ?樁號(hào)算岀 JD?樁 號(hào)為K1 + 052.99 +空= KI + 061.82,與原來的JD?樁號(hào)相同，說明計(jì)算無誤。2式中：M設(shè)緩和曲線后圓曲線內(nèi)移值，m;0緩和曲線終點(diǎn)緩和曲線角，rad。9緩和曲線起點(diǎn)到圓曲線原起點(diǎn)的距離，也稱為切線增值,R-圓曲線半徑，m；L,緩和曲線長，m；a偏角，（。）T,一一設(shè)置緩和曲線的曲線切線長，m;L,設(shè)置緩和曲線的曲線長，m;E,-設(shè)置緩和曲線的外距，m;q一設(shè)置緩和曲線的超距，m0緩和曲線終點(diǎn)緩和曲線角，rado（三）jm曲線要素計(jì)算取/? = 400m,設(shè)計(jì)速度為6OW/,JD3 樁號(hào)為 K1+7

13、67. 19,轉(zhuǎn)角 a = 24.3691.緩和曲線長度S，則:Ls = 0.036 = 0.036 X = 19.44(m) R400V 60 厶 x3 = x3 = 50(m)3.6 3.6R400厶=-R =400 = 44-44-400( m)99公路工程技術(shù)標(biāo)準(zhǔn)規(guī)定：M = 60肋取整數(shù)，采用緩和曲線長70m (時(shí)，最小緩和曲線長度為50m）.2.圓曲線內(nèi)移值M70-7(/麗一 2688x(R廠 24x400 "2688x(400)33總切線長幾先求"卜為=牛顧嘉635）所以 7； = (R + A/?) tan - + 9 = (400 + 0.51) tan

14、竺型 + 34.991 = 121.47 l(m)24曲線總長度厶力L, 70卩=亠=0.08751R 2x400L=a-2B R + 2 厶=« /?+£, =240.123(m) f 180 s 180、5.五個(gè)基本樁號(hào)JD,K1+767. 19-)%121.471ZH,K1+645. 719+)厶70, 00砒K1+715. 719+）（厶一 2厶）100, 123Y%K1+815. 842+) G70. 00HZ,K1+885. 842-)知120. 061QZ,K1+765. 781a24 369Eh=(R + A/?)sec- - R = (400 + 0.51

15、) sec-400 = 974(m)2 2超距D = 27-1, =2x121.471-240.123 = 2.819(m)o 所以，由 QZ?樁號(hào)算出 JD? 樁號(hào)為K1 + 765.781 +芟匕= K1 + 767.19,與原來的JD?樁號(hào)相同，說明計(jì)算無2誤。式中:M-設(shè)緩和曲線后圓曲線內(nèi)移值，m;0緩和曲線終點(diǎn)緩和曲線角，rad Oq緩和曲線起點(diǎn)到圓曲線原起點(diǎn)的距離，也稱為切線增值,R-圓曲線半徑，m；厶一緩和曲線長，m；a偏角，（。）T,一一設(shè)置緩和曲線的曲線切線長，mS設(shè)置緩和曲線的曲線長，m;E,-設(shè)置緩和曲線的外距，m;%一一設(shè)置緩和曲線的超距，m;0緩和曲線終點(diǎn)緩和曲線角

16、rad O（四）JD4曲線要素計(jì)算設(shè)計(jì)速度為60如/JD2樁號(hào)為K2+615. 17b 轉(zhuǎn)角a =92.289在公路線形設(shè)計(jì)規(guī)范中明確規(guī)定：同向曲線之間的最短直線長度以不小于6V （m）為宜。為保證滿足此要求，此處平曲線半徑以切線長T值控制，r< D-6V-7； =850.8-360-121,471 = 369329(/;/), 取T = 369(zn), 暫時(shí)取厶=90("?)依據(jù)公式 % = (any/?- +L,I 2色-巾唔噸7代入相關(guān)數(shù)據(jù)取得R=31022(m),取R = 310m。緩和曲線長度厶則:W60Ls =0.036= 0.036x = 2592(m)R300

17、V 60 厶 > x3 = x3 = 50(m)3.6 3.6厶=蘭7? = 212310 = 34.44 - 310(m)99取整數(shù)，采用緩和曲線長90m (公路工程技術(shù)標(biāo)準(zhǔn)規(guī)定：M = 60肋強(qiáng)時(shí)，最小緩和曲線長度為50m）滿足上述假設(shè)。90“2圓曲線內(nèi)移值M曲亙= =1.089(.n)24R 2688x(7?)24x310 2688x(310/3總切線長7；先求"廠茅壽=罟一麗嘉皿96畑所以 7； =(R + A7?)tan# + q = (310+1.089)Uin 1 + 44.968 = 368.77(m)4曲線總長度厶J0 =如=9° =0.1452R 2x310L, =a 2fi)-R + 2Lv =a-R+厶=589,317(m) f ”180 s 180'5,五個(gè)基本樁號(hào)JDqK2+615. 171-)T»368.770ZH4K2+246. 401+) L$90, 00砒K2+336. 401+) (Lfi 2厶.)409. 317K2+745. 718+)90. 00HZ,K2+835.718-)»»294. 658QZ,K2+541.060Eh =(y? + A/?)sec-/? = (31O + l.O89)sec-3IO = 139.OO4(m)2 2超距 D