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CMSC427/828E Spring 2000
Homework # 5
March 15, 2000
Assume the following notation: R means rotation counterclo ckwise with angle , Ti means translation to the lo cation (xi ; yi ), Ti+j means translation to the lo cation of (xi + xj ; yi + yj ), T(x;y ) means translation to the lo cation (x; y ), Sc means uniform scaling with factor c, and S(c 1 ;c 2 ) means that non-uniform scaling with c in the x direction, and c 2 in the y direction. This notation will b e used through out the homework.
φ (^) φ+θ−(π/2)
θ
x’
y’ x"
y"
Figure 1: Problem 1 Outline
x = x^0 + x^00 = a 1 cos + a 2 sin( + 2
= a 1 cos a 2 cos( + ) (1) y = y 0 y 00 = a 1 sin a 2 cos( +
2
= a 1 sin a 2 sin( + ) (2)
The previous metho d is straightforward, but as this is a graphics course, it is b etter to solve using transformations. It is a transformation of the origin to the p oint P.
P = R T(a 1 ;0) R T( a 2 ;0) O^
2 6 4
a 1 cos a 2 cos ( + ) a 1 sin a 2 sin( + ) 1
3 7 5
- (a)
R Sa =
2 6 4
cos sin 0 sin cos 0 0 0 1
3 7 5
2 6 4
a 0 0 0 a 0 0 0 1
3 7 5
2 6 4
a cos a sin 0 a sin a cos 0 0 0 1
3 7 5 (3)
Sa R =
2 (^64)
a 0 0 0 a 0 0 0 1
3 (^75)
2 (^64)
cos sin 0 sin cos 0 0 0 1
3 (^75)
2 (^64)
a cos a sin 0 a sin a cos 0 0 0 1
3 (^75) (4)
From equation 3 and equation 4, we reach R Sa = Sa R. Thus, uniform scaling, and rotation are commutative. (b) Now consider two rotations around and resp ectively,
R R =
2 6 4
cos sin 0 sin cos 0 0 0 1
3 7 5
2 6 4
cos sin 0 sin cos 0 0 0 1
3 7 5
2 (^64)
cos cos sin sin (sin sin cos cos ) 0 sin cos sin sin cos cos sin sin 0 0 0 1
3 (^75)
2 (^64)
cos ( + ) sin( + ) 0 sin( + ) cos( + ) 0 0 0 1
3 (^75)
= R( + ) (5)
Since addition is commutative =) R( + ) = R( + ).
R R = R R (6)
- In the left-handed system, p ositive rotations are cl ock w ise when lo oking from a p ositive axis toward the origin. This de nition of the p ositive rotations allows the same matrices of the right-handed system to b e used in the left-handed system without any mo di cations.
- Let's nd the matrix representation of a rotation followed by a translation:
T(x;y ) R =
2 6 4
cos sin x sin cos y 0 0 1
3 7 5 (9)
Now let's represent a translation followed by a rotation:
R T(x;y ) =
2 6 4
cos sin x cos y sin sin cos x sin + y cos 0 0 1
3 7 5 (10)
From equations 9 and 10, RT can b e represented by by a rotation followed by translation where
R T(x;y ) = T(x^0 ;y 0 ) R
2 6 4
1 0 x cos y sin 0 1 x sin + y cos 0 0 1
3 7 5
2 6 4
cos sin 0 sin cos 0 0 0 1
3 7 5
2 (^64)
1 0 x^0 0 1 y 0 0 0 1
3 (^75)
2 (^64)
cos sin 0 sin cos 0 0 0 1
3 (^75) (11)
where x^0 = x cos y sin , and y 0 = x sin + y cos . Also, recall from problem # 2 that a sequence of rotations can b e replaced by a single rotation, and also, a sequence of translations can b e replaced by a single translation. Note that in the following, T and R represent abstract translation, and abstract rotation, resp ectively. So assume that there is a general sequence S 1 ; S 2 ; : : : ; Sn , where each Si can b e either a R or a T. We want to prove that M = Sn Sn 1 : : : S 1 = T R. The pro of will b e done by induction. At t = 0, M = I , where I is the identity matrix. M = I = T R where the rotation is by angle 0, and the translation to the origin (0; 0). At t = 1, M = S 1 Mold = S 1 T R, if S 1 = T =) M = T T R = T R due to equation 7. Else, if S 1 = R =) M = RT R = T 0 RR = T R due to equation 11. Assume pro of is held for t = n, i.e.; M = Sn Sn 1 : : : S 1 = T R, then at t = n + 1, If Sn+1 = T =) M = T T R = T R, and similarly, if S 1 = R =) M = RT R = T 0 RR = T R due to equation 11. So, it holds for t = n + 1.
Thus, any sequence of rotations, and translations can b e represented by a rota- tion followed by a translation.
