46 COMPUTER AIDED ENGINEERING DESIGN
RRxy= TQ10 00
0–1 0 0
00–10
00 01, =0.707 0 – 0.707 0
01 0 0
0.707 0 0.707 0
00 0 1, =100 0
010 0
001–2
000 11 0⎡⎣⎢
⎢
⎢⎤⎦⎥
⎥
⎥⎡⎣⎢
⎢
⎢⎤⎦⎥
⎥
⎥⎡⎣⎢
⎢
⎢⎤⎦⎥
⎥
⎥→Rα ααα
() = ααcos –sin 0 0
sin cos 0 0
0010
0001⎡⎣⎢
⎢
⎢⎤⎦⎥
⎥
⎥Finally,
′
′
′⎡⎣⎢
⎢⎤⎦⎥
⎥⎡⎣⎢
⎢
⎢⎤⎦⎥
⎥
⎥⎡⎣⎢
⎢
⎢⎤⎦⎥
⎥
⎥⎡⎣⎢
⎢
⎢⎤⎦⎥
⎥
⎥P
P
PT
1
2
3–1 –1=100 0
010 0
001–2
000 110 00
0–1 0 0
0 0 –1 0
00 010.707 0 – 0.707 0
01 0 0
0.707 0 0.707 0
00 0 1cos –sin
sin cos– (^1) αα 00
αα 000
0010
0001
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
0.707 0 – 0.707 0
01 0 0
0.707 0 0.707 0
00 0 1
10 00
0–1 0 0
0 0 –1 0
00 01
100 0
010 0
001–2
000 1
0021
1011
0011
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
T
(i) When α = β = 54.73°, rotation of P 3 around Q 1 Q 3 by α will bring the plane S 1 on the top of
S 2 as shown in Figure 2.19 (b). Hence
′
′
′
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
P
P
P
1
2
3
0021
1011
0011
.7887 –.5773 –.2113 0
.5773 .5773 .5773 0
–.2113 –.5773 .7887 0
.4226 1.1547 .4226 1
00 21
10 11
.2113 .5773 1.2113 1
(ii) For α = – (180° – β), the two planes S 1 * and S 2 are hinged about Q 1 Q 3 such that they are in
the same plane as shown in Figure 2.19 (c)
′
′
′
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
P
P
P
1
2
3
0021
1011
0011
.2113 .5773 –.7887 0
–.5773 –.5773 –.5773 0
–.7887 .5773 .2113 0
1.5773 –1.1547 1.5773 1
00 21
10 11
.7887 –.5773 1.7887 1
(iii) For α = – (90° – β),S 1 * and S 2 are hinged about Q 1 Q 3 and are perpendicular to each other
(Figure 2.19d)
′
′
′
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
P
P
P
1
2
3
0021
1011
0011
.9082 .4083 –.0918 0
–.4083 .8165 –.4083 0
–.0918 .4083 .9082 0
.1835 –. .1835 1
00 21
10 11
.0918 –.4083 1.0918 1
8165