线性路径厚度
我可以使用一些帮助来获得线性路径实体的厚度。创建线性路径模型(在本例中为金属闪光部件)按预期工作,但它没有厚度,这也是我喜欢的。到目前为止,我还没有找到任何关于如何执行此操作的有用提示。
如果我需要走另一条路,请指教。
代码部分:
var b1 = Math.Cos(Utility.DegToRad(45)) * 30;
var h1 = Math.Tan(Utility.DegToRad(45)) * 30;
var l1 = new LinearPath(Plane.YZ,
new Point2D(0, 0),
new Point2D(60, 0),
new Point2D(60, -120),
new Point2D(60 + b1, -120 - h1)
);
var s1 = l1.ExtrudeAsSolid(new Vector3D(500, 0, 0), 1);
s1.Color = System.Drawing.Color.AliceBlue;
s1.ColorMethod = colorMethodType.byEntity;
return s1;
I can use some help to get thickness to a linear path solid. Creating the linear path model (in this case a metal flashing part) works as intended, but it has no thickness, which I like to have as well. I haven't so far found any useful hints on how to do this.
If I need to go another way, please advise.
Code part:
var b1 = Math.Cos(Utility.DegToRad(45)) * 30;
var h1 = Math.Tan(Utility.DegToRad(45)) * 30;
var l1 = new LinearPath(Plane.YZ,
new Point2D(0, 0),
new Point2D(60, 0),
new Point2D(60, -120),
new Point2D(60 + b1, -120 - h1)
);
var s1 = l1.ExtrudeAsSolid(new Vector3D(500, 0, 0), 1);
s1.Color = System.Drawing.Color.AliceBlue;
s1.ColorMethod = colorMethodType.byEntity;
return s1;
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我会尝试这样做:
1 单位
的挤压向量和零
的轮廓偏差(LinearPath
实体不受偏差影响)。I would try this instead:
An extrusion vector of
1 unit
and a contour deviation ofzero
(LinearPath
entity is not affected by deviation).