zdx411 发表于 2005-10-5 22:07:00

输入两点和弧长，画弧，谁能帮我，谢谢。

输入两点和弧长，画弧，谁能帮我，弧的方向列所谓，画两个也可以。VBA，lisp,都可以。谢谢。

ahlzl 发表于 2005-10-5 23:59:00

<P>我以前写的。还有LISP，请到LISP版：<A href="dispbbs.asp?boardid=3&amp;id=42791" target="_blank" >dispbbs.asp?boardid=3&amp;id=42791</A></P>
<P><FONT face=宋体 size=2>Sub ddARC()<BR>Dim ddARC As AcadArc<BR>Dim S, L, R, a0, a1, fx, flx, c, angs, ange As Double<BR>Dim pa, pb, cen As Variant<BR>Const PI = 3.1415926535<BR><BR>pa = ThisDrawing.Utility.GetPoint(, "请输入圆弧起点：")<BR>pb = ThisDrawing.Utility.GetPoint(pa, "请输入圆弧终点：")<BR>S = ThisDrawing.Utility.GetDistance(pa, "请输入圆弧弧长：")<BR><BR>L = dis(pa, pb)<BR>b = ThisDrawing.Utility.AngleFromXAxis(pa, pb)<BR><BR>If S &lt;= L Then<BR>MsgBox "您要画的圆弧并不存在，请再执行一次程序！"<BR>End<BR>End If<BR><BR>a0 = 2<BR>a1 = a0<BR>Do<BR>a0 = a1<BR>fx = Sin(a0 / 2) / a0 - L / (2 * S)<BR>flx = (Cos(a0 / 2) * a0 * 0.5 - Sin(a0 / 2)) / (a0 * a0)<BR>a1 = a0 - fx / flx<BR>Loop While Abs(a1 - a0) &gt; 0.0000000001<BR><BR>R = S / a1<BR>c = b - a1 * 0.5 + 90 * PI / 180<BR>cen = ThisDrawing.Utility.PolarPoint(pa, c, R)<BR>angs = c + PI<BR>ange = angs + a1<BR><BR>Set ddARC = ThisDrawing.ModelSpace.AddArc(cen, R, angs, ange)<BR>End Sub<BR><BR>Public Function dis(pa, pb As Variant) As Double<BR>dis = ((pa(0) - pb(0)) ^ 2 + (pa(1) - pb(1)) ^ 2 + (pa(2) - pb(2)) ^ 2) ^ 0.5<BR>End Function<BR></FONT></P>

ahlzl 发表于 2005-10-9 18:08:00

<P>发完贴子，就不来了，这样的人以后谁愿意帮你！</P>

galayan 发表于 2005-10-10 15:44:00

<P>正需要这样的程序呢</P>
<P>感谢！</P>

3xxx 发表于 2013-9-18 18:28:43

很受用啊。谢谢。

睡醒的蜗牛 发表于 2013-10-8 13:59:26

支持xxxxxx

清风明月名字 发表于 2013-10-8 22:40:40

谢谢楼主的分享！很有意思的代码，收藏备用。
页: [1]
查看完整版本: 输入两点和弧长，画弧，谁能帮我，谢谢。