.net编程求出四个端点组成的两条直线的相交点
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.net编程求出四个端点组成的两条直线的相交点
'''求出两直线的交点
public function inters(byval p1 as point3d, byval p2 as point3d, byval p3 as point3d, byval p4 as point3d, optional byval type as boolean = true) as point3d
if math.ieeeremainder(math.abs(getangle(p1, p2) - getangle(p3, p4)), math.pi) = 0 then
return nothing
end if
dim a1, a2, b1, b2, c1, c2, c3, x1, x2, y1, y2, x, y, z as double
dim lp as point3d
x1 = p1.x
y1 = p1.y
x2 = p3.x
y2 = p3.y
a1 = p2.y - p1.y
b1 = p2.x - p1.x
a2 = p4.y - p3.y
b2 = p4.x - p3.x
c1 = a1 / b1
c2 = a2 / b2
c3 = (p2.z - p1.z) / b1
x = (y2 - c2 * x2 + c1 * x1 - y1) / (c1 - c2)
if b1 = 0 then
x = p1.x
c1 = 1.0e+20
c3 = c1
end if
if b2 = 0 then
x = p3.x
c2 = 1.0e+20
end if
y = c1 * (x - x1) + y1
z = c3 * (x - x1) - p1.z
if x - x1 = 0 then
y = c2 * (x - x2) + y2
end if
lp = new point3d(x, y, z)
dim d1, d2, d3, d4, od1, od2, od3, od4 as double
d1 = p1.distanceto(p2)
d3 = p3.distanceto(p4)
od1 = lp.distanceto(p1)
od2 = lp.distanceto(p2)
od3 = lp.distanceto(p3)
od4 = lp.distanceto(p4)
if type then
return lp
elseif od1 <= d1 and od2 <= d1 and od3 <= d3 and od4 <= d3 then
return lp
else
return nothing
end if
end function
望各位多指点。我没发现只好自已编了一个，但我怕出毛病。
[此贴子已经被作者于
lisp vba arx .net开发都会一点。可惜都不精。希望和大家共同学习进步。
'''求出两直线的交点
public function inters(byval p1 as point3d, byval p2 as point3d, byval p3 as point3d, byval p4 as point3d, optional byval type as boolean = true) as point3d
if math.ieeeremainder(math.abs(getangle(p1, p2) - getangle(p3, p4)), math.pi) = 0 then
return nothing
end if
dim a1, a2, b1, b2, c1, c2, c3, x1, x2, y1, y2, x, y, z as double
dim lp as point3d
x1 = p1.x
y1 = p1.y
x2 = p3.x
y2 = p3.y
a1 = p2.y - p1.y
b1 = p2.x - p1.x
a2 = p4.y - p3.y
b2 = p4.x - p3.x
c1 = a1 / b1
c2 = a2 / b2
c3 = (p2.z - p1.z) / b1
x = (y2 - c2 * x2 + c1 * x1 - y1) / (c1 - c2)
if b1 = 0 then
x = p1.x
c1 = 1.0e+20
c3 = c1
end if
if b2 = 0 then
x = p3.x
c2 = 1.0e+20
end if
y = c1 * (x - x1) + y1
z = c3 * (x - x1) - p1.z
if x - x1 = 0 then
y = c2 * (x - x2) + y2
end if
lp = new point3d(x, y, z)
dim d1, d2, d3, d4, od1, od2, od3, od4 as double
d1 = p1.distanceto(p2)
d3 = p3.distanceto(p4)
od1 = lp.distanceto(p1)
od2 = lp.distanceto(p2)
od3 = lp.distanceto(p3)
od4 = lp.distanceto(p4)
if type then
return lp
elseif od1 <= d1 and od2 <= d1 and od3 <= d3 and od4 <= d3 then
return lp
else
return nothing
end if
end function
自己编的，请大家评评。
lisp vba arx .net开发都会一点。可惜都不精。希望和大家共同学习进步。
望大侠指点。我对自己编的这个东东不放心。呵……如果程序有就太好了。
lisp vba arx .net开发都会一点。可惜都不精。希望和大家共同学习进步。