数学图形之圆柱面

(1)圆柱面

vertices = dimension1:72 dimension2:72

u = from 0 to (2*PI) dimension1
v = from (-5) to (5) dimension2

r = 5

x = r*cos(u)
y = v
z = r*sin(u)

(2)任意角度圆柱面

vertices = dimension1:72 dimension2:72
u = from (-5) to (5) dimension2
v = from 0 to (2*PI) dimension1

a = rand2(0, 2*PI)
b = rand2(0, 2*PI)

c = sin(v)
d = cos(v)

e = sin(b)
f = cos(b)

g = sin(a)
h = cos(a)

x = f*h*d - f*g*c + e*u
y = g*d + h*c
z = -e*h*d + e*g*c + f*u

x = x*5
y = y*5
z = z*5

代码中会随机设置圆柱的旋转角度

(3)圆柱体

将圆柱面的两头封上,生成闭合的圆柱体

vertices = dimension1:72 dimension2:72

u = from 0 to (2*PI) dimension1
v = from (-5) to (5) dimension2

r = if(abs(v) < 4.9, 5, 0)

x = r*cos(u)
y = v
z = r*sin(u)

(4)由曲线生成的圆柱面

#http://www.mathcurve.com/surfaces/cylindrederevolution/cylindrederevolution.shtml
vertices = D1:100 D2:100
u = from 0 to (2*PI) D1
v = from 0 to (PI) D2
a = rand2(1, 10)
x = a*cos(u - v)
z = a*sin(u - v)
y = a*( u+v)

(5)鼓形

为了简单,对鼓的隆起使用了SIN函数.

vertices = dimension1:72 dimension2:72

u = from 0 to (2*PI) dimension1
v = from (-1) to (1) dimension2

r = 5
h = rand2(1, r)

t = h*cos(v*PI/2)
p = if(t > 0.01, r+t, 0) 

x = p*cos(u)
y = v*h
z = p*sin(u)

新的鼓形脚本,这个比上一个要圆润得多:

vertices = dimension1:72 dimension2:72

u = from 0 to (2*PI) dimension1
v = from (-PI/2) to (PI/2) dimension2

r = 5
h = rand2(1, r)

t = h*cos(v)
p = if(t > 0.01, r+t, 0) 

x = p*cos(u)
y = h*sin(v)
z = p*sin(u)

(6)齿轮

vertices = D1:65 D2: 4

u = from 0 to (2*PI) D1
v = from -1 to 1 D2

k = from 0 to 64 D1

m = mod(k, 2)

r = 10.0 + m*2

a = 0.9

b = in_range(v, -a, a)

x = r*sin(u)*b
z = r*cos(u)*b

y = if(b, v, v/3)*3

(7)刺柱

vertices = D1:129 D2:65

u = from 0 to (2*PI) D1
v = from (-20) to (20) D2

n = 4

a = from 0 to 128 D1
b = from 0 to 64 D2

t = (mod(a, n) + mod(b, n))/n*4

r = 10 + t

x = r*cos(u)
y = v
z = r*sin(u)

(8)蛏形

vertices = D1:100 D2:100
u = from 0 to (2*PI) D1
v = from -5 to 5 D2

r = 5

x = r*cos(u)
z = r*sin(u)
y = x*v

(9)指定方向的圆柱(极坐标原理)

vertices = dimension1:72 dimension2:72

u = from (0) to (5) dimension2
v = from 0 to (2*PI) dimension1

a = rand2(-2, 2)
b = rand2(-2, 2)
c = rand2(-2, 2)

ac = sqrt(a*a + c*c)
angleY = atan2(ac, b)
angleXZ = atan2(a,c)

m = cos(v)
n = sin(v)

e = sin(angleY)
f = cos(angleY)

g = sin(angleXZ)
h = cos(angleXZ)

x0 = m
y0 = u*f - n*e
z0 = u*e + n*f

x1 = x0*h + z0*g
y1 = y0
z1 = -x0*g + z0*h

x = x1 + a
y = y1 + b
z = z1 + c

(10)指定方向的圆柱(矩阵原理)

vertices = dimension1:72 dimension2:72

u = from (0) to (5) dimension2
v = from 0 to (2*PI) dimension1

a = rand2(-2, 2)
b = rand2(-2, 2)
c = rand2(-2, 2)

len = sqrt(a*a + b*b + c*c)
len = max(len, 0.00001)
px = a/len
py = b/len
pz = c/len

qx = 0
qy = pz
qz = -py
len = sqrt(qx*qx + qy*qy + qz*qz)
len = max(len, 0.00001)
qx = qx/len
qy = qy/len
qz = qz/len

wx = py*qz - pz*qy
wy = pz*qx - px*qz
wz = px*qy - py*qx

x0 = cos(v)
y0 = u
z0 = sin(v)

x = x0*wx + y0*px + z0*qx + a
y = x0*wy + y0*py + z0*qy + b
z = x0*wz + y0*pz + z0*qz + c

数学图形之圆柱面