sword wrote:
先以方程做出曲线,旋转生成圆锥面,
通过偏距生成一条曲线,距离为管子的半径。
管子以扫描生成,不能以螺旋线扫描生成(因为那样要指定螺距)
扫描线通过一个螺旋面同圆锥面求交线得到。
关键是螺旋面的生成,
螺旋面以var sec swp生成,在这之前要生成基准图形,
指定升程与螺旋角的关系,在截面中加入relation
得到交线后,就好办了。
boby水手,是不是这样做的?
不过基准图形我没看懂,能说明一下吗?

真难,做不出来,扫描时relation 无法加上


总算有点象了。 






elei_027 wrote:
版主,不要忘了加分。
想不到学习也能加分.
![]()



by 孤峰醉酒


上班很忙,`看了boby的提示,匆匆忙忙用了45分鐘作出來了,版主快給我加分呀!!!

-by boby

andy wrote:
版主為甚麼沒有加分![]()
-by boby
-by boby
boby wrote:
1.一抛物圆锥面
曲线方程:
/* For cartesian coordinate system, enter parametric equation
/* in terms of t (which will vary from 0 to 1) for x, y and z
/* For example: for a circle in x-y plane, centered at origin
/* and radius = 4, the parametric equations will be:
/* x = 4 * cos ( t * 360 )
/* y = 4 * sin ( t * 360 )
/* z = 0
/*-------------------------------------------------------------------
x = 1 + (4 * t)
y = 2 + (3 * t) + (5 * t ^2)
z = 0
-by boby
给一分算了
-by boby
by 孤峰醉酒
by 孤峰醉酒

!斑竹,还能给加分不?收获不小!


-by boby


-by bobyboby wrote:
3.管路的升程(x)和转的圈数(y)按下图









-by boby
,
-by bobyoo00pp wrote:
呵呵????
严肃批评:这是你做的?抄来的?(抄lrdyye的)。
先扣一分,如果不服可PM给我![]()
by boby







,
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