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1、计算机图形学基础华东理工大学计算机系·谢晓玲习题7.4/P2277.4将图7-40中物体ABCDEFGH进行如下变换的变换矩阵，写出复合变换后图形各顶点的规范化齐次坐标，并画出复合变换后的图形。①平移，使点C与点p(1,-1,0)重合。②绕z轴旋转60°。解：①T1=T(0,-2,0)=1000010000100-201ABCDEFGH112-1P(1,-1,0)xyz图7-40②T2=Rz(60)=1/2√3/200-√3/21/20000100001∴T=T1·T2=T(0,-2,0)Rz(60)=1/2√3/200-√3/21/2000010√3-101ABC
2、DEFGH112-1P(1,-1,0)xyzA0001√3-101B10011/2+√3√3/2-101C11011/2+√3/2√3/2-1/201D0101·T=√3/2-1/201E0021√3-121F10211/2+√3√3/2-121G11211/2+√3/2√3/2-1/221H0121√3/2-1/221ABCDEFGH112-1P(1,-1,0)xyzD’C’四舍五入得：A’2-101B’2001C’1001D’1-101E’2-121F’2021G’1021H’1-121ABCDEFGH112-1P(1,-1,0)xyz图7-40A’G112-1
3、xyz-22B’E’F’G’H’习题7.4/P2277.4将图7-40中物体ABCDEFGH进行如下变换的变换矩阵，写出复合变换后图形各顶点的规范化齐次坐标，并画出复合变换后的图形。①平移，使点C与点p(1,-1,0)重合。②相对于点p(1,-1,0)绕z轴旋转60°。解：①T1=T(0,-2,0)②T2=T(-1,1,0)Rz(60)T(1,-1,0)∴T=T1·T2=T(0,-2,0)T(-1,1,0)Rz(60)T(1,-1,0)ABCDEFGH112-1P(1,-1,0)xyz图7-40T=T(0,-2,0)T(-1,1,0)Rz(60)T(1,-1,0)=
4、100010001/2√3/200100001000100-√3/21/200010000100010001000100-201-110100011-101=1/2√3/200-√3/21/2000010(1+√3)/2-(3+√3)/201ABCDEFGH112-1P(1,-1,0)xyzT=T(0,-2,0)T(-1,1,0)Rz(60)T(1,-1,0)=1/2√3/200-√3/21/2000010(1+√3)/2-(3+√3)/201A0001(1+√3)/2-(3+√3)/201B10011+√3/2-3/201C11011-101D0101·T=1/2
5、-1-√3/201E0021(1+√3)/2-(3+√3)/221F10211+√3/2-3/221G11211-121H01211/2-1-√3/221ABCDEFGH112-1P(1,-1,0)xyz四舍五入得：A’1-201B’2-101C’1-101D’0-201E’1-221F’2-121G’1-121H’1-221ABCDEFGH112-1P(1,-1,0)xyz图7-40A’G112-1xyz-22B’C’D’E’F’G’H’习题7.5/P2287.5将图7-41中四面体ABCD进行如下变换的变换矩阵，写出复合变换后图形各顶点的规范化齐次坐标，并画出复
6、合变换后的图形。①关于P点整体放大2倍。②关于y轴进行对称变换。解：①T1=T(-2,2,-2)S(2)T(2,-2,2)②T2=TFy()∴T=T(-2,2,-2)S(2)T(2,-2,2)TFy()A(2,0,0)B(2,1,0)C(0,1,0)D(1,1,1)P(2,-2,2)xyz图7-41100020001000-1000T=0100·0200·0100·010000100020001000-10-22-2100012-2210001-2000=020000-202221A2001-2221B2101·T=-2421C01012421D11110401∴得
7、A’(-2,2,2)、B’(-2,4,2)、C’(2,4,2)、D’(0,4,0)A(2,0,0)B(2,1,0)C(0,1,0)D(1,1,1)P(2,-2,2)xyzA(-2,2,2)B(-2,4,2)C(2,4,2)D(0,4,0)P(2,-2,2)xyz7.6假定直线AB的两个端点为A(0,0,0)和B(2,2,2)，试写出绕AB旋转30°的三维复合变换矩阵。第一步，使得AB与z轴重合①TRx：绕x轴旋转α(45°)TRx(45°)=100001/√21/√200-1/√21/√200001B2221·TRx(45)=202√21B’习题7.6/P228
