2010第三届优秀第一阶段题特一等

编号专用页2010年第三届“ScienceWord”杯数学中国数学建模网络赛题目 ： 全泊车，反之则不能安全泊车。在这个模型下我们得到了车位的长和宽，分别为 ( 填写参赛队号 ( 填写The”SmartCar”isaproblemabouthowtoparkacarsafelyintoacarparkingspace.Inordertosolvethisproblem,inthefirstce,wedefinethesafelyparkingaccordingtothecommondata.Underthiscircumstances,themodelonetakesminimumturningradiusasthecorecontentanditonlyconcernsthetwice-parkingsituation.Utilizingreversingmind,weconsiderthecardriveingfromtheparkingspacetotheinitialpositionortheotherwayaroundIntheconcretemodel,weexinthemodelconcreyandgetaoptimalresultwhichistheextremeparkingzoneandtheumsecurityparkingzone.Hence,thecarcanbeputinsafelywithinthesecurityzoneandviceversus.However,theconclusioncanonlybereachedontheconditionthatwecangetenoughevidenceandalsoit’llbealittledifferentfromthepracticalusage.So,intheendofthemodel,wemakeitmorehumaneandgivethedriver,carandweatheraneasierenvironmentsothatourmodelcanbemorepractical.Nevertheless,themodelstillhasitsconfinements,mainlybecauseweonlycareaboutthetwice-parkingsituation.Inthetwice-parkingsituation,onagreatextent,itconstraintsthemovementofthecarandtheminimumsecurityzone.Consideringthis,wemakeasecondmodel,mostlyforincreasingthehumanity.Inordertoexintheproblemthoroughly,westarttodiscussthethree-timeparking,underthiscondition,westillconsidertheminimumturningradiusasanimportantcoefficient.Sincewhenthefirst-parking,thecarhasalreadyutilizestheminimumturningradiusandwecanmakesureitsoptimalposition.Wecompleytakeadvantageofthe”tight-constraints”,theextremeconditiontomakethesecurityzoneassmallaspossible(seedetailsintheartical).Aftertakingthisintoaccount,obviously,wecanusemulti-parkingtoreachourgoal.Thealgorithmissimilartothethree-timeparking.Therefore,wecansolvebothproblemoneandproblemtwo.Fromwhathasbeendiscussedabove,twomodelsbothutilizes”tight-anassumption,toreachourgoalandgetthebestroute,hencesolvingthetwoproblems 11123456r=789θϕdγ23所示d，为汽车的最小转弯半径。4我们引入一般的转向时间的考虑，得出更加合理化的最小停车位的范围。得出一些数据的值：由图2所示，由几何关系可得：•r=?ROH r2cos2θ+(d1+?ROD r2cos2θ+d?ROG r2cos2θ+

tanϕ=d1+d1+d0ϕ=arctanrcosx2+y2=R2OH=r2cos2θ+(d1+d0)2=d02cot2θ+(d1+ Q[(d0cotθ+d0cotθx)cosα,(d0cotθ+d0cotθx)sin(x−xQ)2+(y−yQ)2=(d0cotθx−(x−(d0cotθ+d0cotθx)cos2α)+(y−(d0cotθ+d0cotθx)sin2α)=(d0cot2θx− 56QOBα.在后轴（AB）αPKP大圆 K

确定。

和 的交点记为(x,yx0xA

0[d0cotθx−d0(cotθcotθx)cosα−d]2+[y1−d0(cotθ+cotθx)sinα]2=[d0cotθx−?y1 (d0cotθx−d)2−[d0cotθx−d0(cotθ+cotθx)cosα−d]2+d0(cotθ+cotθx)sin2)x0xA时，xxAd0cotθdy2 d12+d02+2d1d0−d2+2d0dcotθ 能安全倒出，显然有：y1−y2≥0,即：y1≥y2.说明? 2w1=2ROD−2ROB+d= (rcosθ)2+d2−2rcosθ+2

l1=y2+d

? 下面，再考虑（即泊车两次）进入停车位的情况：次性前行停车的全过程。在前行停车时。车左侧扫过的区域的边界由左侧CG确? w2=2ROH−2ROB+d= r2cos2θ+(d1+d0)2−2rcosθ+CGC1G1NL6N N? l2 r2(1−cos2θ)−d2+2rdcosθ+d1+综合以上的两种方案，在此，我们作出断定，w1w2综合以上的两种方案，在此，我们作出断定，d1+d0>d2,这从本质上决定了ROD<ROH,从这点也可以给出比较好的预测，下面我们也将以实例举出。令目标函数z=l1−l2.21π/3≤θ≤2d≤3d≥43≥d0≥5r=0i61.5>d1>71.5>d2>lingomax(zl1l2.lyd

? d2+d2+2dd−d2+2ddcotθ+? 0 ? 极限停车宽度为：w1=2ROD−2ROB+d= (rcosθ)2+d22−2rcosθ+最小的停车区域的宽度：m(width)=l1最小的停车区域的长度：m(length)=l2S=w1+rcosθ−S0=s−d−7由以上的解知道一方案更好。现在以奥迪A6(2.0L125kw)为例，以下是两方案的,1.485m；最小转弯半径：r6.02m；前悬：d11.001m；后悬：d21.809m车辆前后轴距：d02.945m;车长：5.035m)(d1d03.946m>d21.089m).结合上面的341.7倍）最小的停车区域的宽度：m(width)=l1+0.2=最小的停车区域的长度：m(length)=l2+0.2=

d0+d1+

=8.1453=1.6177∈(1.5,m(width)=2.2570=1.2167m S=w1+rcosθ−d+0.1=S0=s−d−m(width)=m(length)m(width)都增大的情况下，当然车辆能够安2w1l1增加到w1+vt和l1+vt.这样完全可以保证车辆安全的停入泊车位。1全距离考虑在内，最后可以得到一次完成泊车时w0:根据模型一得1

w1=w1+00l1=l1+00851626364656789在模型一的基础上，我们可以记αMox=α,设αNox=ψ,OB=rcosθ.车置OM的中间，即：OA=BM.ABO为定点，MON。然处，O−→P的过程中车由模型一的初始位置经过两次运动，停段PN的中间。(ROCcosψ−αc,R(OC)sinψ−D(RODcosψ−αD,RODsinψ−G(ROHcosπ−αH+(ROA+ROB)cosψ,ROHsinπ−αH+(ROA+ROB)sin(ROGcosπ−αG+(ROA+ROB)cosψ,ROGsinπ−αG+(ROA+ROB)sin9运动到PN时，前端突出点是点H。可求得Hβ:[ROHcosβ,ROHsinβ],β∈[ψ+αH,α+ROGsinπαGROAROBsinψ=ROHsinβ,Hβy=ROHsinβ的交点，(x0y0)我们所要的目标函数：l=ROGsinπ−αG+(ROA+ROB)sinψ−ROCsinψ−0≤ψ<ROHcosπ−αH+(ROA+ROB)cosψ≥LINGOLmin.ψ所以车辆的各个部w=2RODcosψ−αD−2ROGcosπ−αG+2(ROA+ROB)cosψ+对于H的A6(2.0L125kwlminw6w后类似”