Zadatak 04 - Rotacione i translatorne površi

Zadatak 4

 
Nutzerbild von Jovana Ninkovic
Zadatak 4
von Jovana Ninkovic - Montag, 26. Mai 2014, 16:27
 

a)

X=sqrt((cos(u))^2 + (1)^2)*sin(v)

Y=u

Z=sqrt((cos(u))^2 + (1)^2)*cos(v)

0<=u<=2*pi

0<=v<=4*pi

X=cos(u)

Y=u

Z=1

0<=u<=2*pi

rotacija oko Y ose

b)

X=(4*pi - v)/(4*pi - 0) * cos(u) + (u - 0)/(2*pi - 0) * v*sin(v)

Y=(4*pi - v)/(4*pi - 0) * u + (u - 0)/(2*pi - 0) * 1

Z=(4*pi - v)/(4*pi - 0) * 1 + (u - 0)/(2*pi - 0) * v*cos(v)

cosinus

X=cos(s)

Y=u

Z=1

helix

X=v*sin(v)

Y=1

Z=v*cos(v)

0<=U<=2*pi

0<=V<=4*pi