# Lost in coordinate system

pnt p1 0 0 4
pnt p2 0 100 4
pnt p3 45 45 4
pnt p4 45 55 4
line l1 p1 p2
line l2 p3 p4
seta se1 l l1
seta se2 l l2
swep se1 ses1 tra 100 0 4
swep se2 ses2 tra 10 0 4

I expect these two surfaces to be on the same surface and they are not. I do not remember how to deal with. Please, guys and girls, help me.

orientation +y shows it clearly.

but strange enough, this way
pnt p1 0 0 0
pnt p2 0 100 0
pnt p3 45 45 0
pnt p4 45 55 0
line l1 p1 p2
line l2 p3 p4
seta se1 l l1
seta se2 l l2
swep se1 ses1 tra 100 0 0
swep se2 ses2 tra 10 0 0

all is as expected

The first sweep has an incline of 4/100, while the second 4/10.

Change the second line to `swep se2 ses2 tra 10 0 0.4` and the planes will be parallel.

It is really cool answer. I never see this information on incline in cgx manual. Would you tell me a place to read on incline thing.

It is from manual

“swep set set [tra dx dy dz div [a] ]”
and incline word is not even available in the cgx manual.

Incline is basically a synonym for slope or angle.

The translate (“tra”) subcommand of the sweep (“swep”) operation takes a vector dx dy dz as an argument. Since the displacement in the y direction is zero, only dx and dz matter in this case.

For two vectors to be parallel, the ratio between the components should be the same. For the one vector, dz/dx = 0.04, while for the other it’s 0.4.

In this case, dz/dx is the tangent of the angle between the displacement vector and the plane z=0.

Actually it is not what I expected, but this is good too. Thank you.