资源描述：
《重庆大学 机自实验班 计算机辅助设计与制造习题答案》由会员上传分享，免费在线阅读，更多相关内容在行业资料-天天文库。
1、PRINCIPLESOFCAD/CAM–TUTORIAL11.AlinesegmenthastwoendpointsAandB.IfthislinesegmentistoberepresentedbyacubicBeziercurvemodelwith4controlpointsP0,P1,P2,andP3,provethatP0,P1,P2,andP3areco-linear.(2002paper)Solution:TheparametricequationsforthelinesegmentABis:r(u)=A+u(B-A)with0u1Base
2、donthecharacteristicsofcubicBeziercurve,wehaveP0=r(0)=A;P3=r(1)=B3(P1-P0)=B-A3(P3–P2)=B-ASolvingtheequations,wehaveP1=A+(B-A)(1/3)=r(1/3)P2=A+(B-A)(2/3)=r(2/3)Therefore,P0,P1,P2,andP3areco-linear.2.ConstructaquadraticBeziercurvemodel(equation)withthreecontrolpoints,V0,V1,andV2.S
3、olution:2Thegeneralformofaquadraticcurvemodelisr(u)=a+bu+cuwith0u1ForaquadraticBeziercurvewithcontrolpointsV0,V1,andV2,wehavethefollowing:r(0)=V0;r(1)=V2;r’(0)=2(V1–V0);r’(1)=2(V2–V1)Wethereforehave:a=V0;a+b+c=V2;b=2(V1–V0)Solve,wehavea=V0;b=2(V1–V0);c=V0–2V1+V22Therefore,r(u)=V
4、0+2(V1–V0)u+(V0–2V1+V2)uwith0t1121V0Orr(u)=(1–u)2V+2u(1–u)V+V=2012uu1220V1100V213.ConstructacompositecurveinterpolatingthreepointsP0,P1,andP2,usingcubicFergusoncurvemodel.2Thecompositecurveshouldsatisfyparametric-Ccondition.Estimatetheendtangentsusing(a)Quadraticpolynomialendcon
5、ditionP1(b)Free-endconditionr0(u)r1(u)Solution:P0P2ThreepointsneedtwocubicFergusoncurvesr0(u)andr1(u)01r0(u)=UCSr1(u)=UCS01S=[P0P1t0t1]S=[P1P2t1t2]2SincethecompositecurvesatisfiesparametricC–condition,wehavet1=(3P2–3P0–t0–t2)/4.Toobtaint0andt2(a)Usingquadraticpolynomialendcondit
6、ionbasedonthefirst3points(fort0)andthelast3points(fort2).Inthiscase,thetwopolynomialsarethesame,i.e.,2r(u)=a+bu+cuwith0u1Therefore,wehaver(0)=a=P0(1)r(1)=a+b+c=P2(2)AtP1,thevalueofuisestimatedusingu1=
7、P1–P0
8、/(
9、P1–P0
10、+
11、P2–P1
12、)2Therefore,r(u1)=a+bu1+cu1(3)Solveequations(1),(2),and
13、(3)fora,b,andc.Finally,t0=r’(0)=b;t2=r’(1)=b+2c(b)Usingthefree-endcondition,i.e.,r0’’(0)=0andr1’’(1)=0Wehave2t0+t1=3(P1–P0)(1)2t2+t1=3(P2–P1)(2)2Atthesametime,theparametricC–conditiongivesust1=(3P2–3P0–t0–t2)/4(3)Solveequations(1),(2),and(3)toobtaint0,t1,andt2.t0=(6P1–P2–5P0)/4；
14、t1=(P2–P0)/2；t2=(5P2–3P1+P0)/424.Constructtwoqu
