如何在与直线的给定垂直距离处找到一个点?

你需要计算一个垂直于直线段的单位向量。避免计算斜率,因为这会导致被零误差除。

dx = x1-x2
dy = y1-y2
dist = sqrt(dx*dx + dy*dy)
dx /= dist
dy /= dist
x3 = x1 + (N/2)*dy
y3 = y1 - (N/2)*dx
x4 = x1 - (N/2)*dy
y4 = y1 + (N/2)*dx

你只需要计算正交向量并乘以n/2

vx = x2-x1
vy = y2-y1
len = sqrt( vx*vx + vy*vy )
ux = -vy/len
uy = vx/len

x3 = x1 + N/2 * ux
Y3 = y1 + N/2 * uy

x4 = x1 - N/2 * ux
Y4 = y1 - N/2 * uy

因为从2到1和1到3的向量是垂直的,所以它们的点积是0。

这会给你留下两个未知数:x从1到3(x13),y从1到3(y13)

用毕达哥拉斯定理得到另一个未知方程。

用代换法求解每一个未知数…

这需要平方和非平方,所以你会失去与方程相关的符号。

要确定标志,请考虑:

while x21 is negative, y13 will be positive
while x21 is positive, y13 will be negative
while y21 is positive, x13 will be positive
while y21 is negative, x13 will be negative

已知:点1:x1,y1

已知:点2:x2,y2

x21 = x1 - x2
y21 = y1 - y2

已知:距离1->3:n/2

方程A:勾股定理

x13^2 + y13^2 = |1->3|^2
x13^2 + y13^2 = (N/2)^2

已知:角度2-1-3:直角

向量2->1和1->3垂直

2->1点1->3为0

公式B:点积=0

x21*x13 + y21*y13 = 2->1 dot 1->3
x21*x13 + y21*y13 = 0

比率b/w x13和y13:

x21*x13 = -y21*y13
x13 = -(y21/x21)y13

x13 = -phi*y13

方程A:按比例求解y13

  plug x13 into a
phi^2*y13^2 + y13^2 = |1->3|^2

  factor out y13
y13^2 * (phi^2 + 1) = 

  plug in phi
y13^2 * (y21^2/x21^2 + 1) = 

  multiply both sides by x21^2
y13^2 * (y21^2 + x21^2) = |1->3|^2 * x21^2

  plug in Pythagorean theorem of 2->1
y13^2 * |2->1|^2 = |1->3|^2 * x21^2

  take square root of both sides
y13 * |2->1| = |1->3| * x21

  divide both sides by the length of 1->2
y13 = (|1->3|/|2->1|) *x21

  lets call the ratio of 1->3 to 2->1 lengths psi
y13 = psi * x21

  check the signs
    when x21 is negative, y13 will be positive
    when x21 is positive, y13 will be negative

y13 = -psi * x21

方程A:用比率求解x13

  plug y13 into a
x13^2 + x13^2/phi^2 = |1->3|^2

  factor out x13
x13^2 * (1 + 1/phi^2) = 

  plug in phi
x13^2 * (1 + x21^2/y21^2) = 

  multiply both sides by y21^2
x13^2 * (y21^2 + x21^2) = |1->3|^2 * y21^2

  plug in Pythagorean theorem of 2->1
x13^2 * |2->1|^2 = |1->3|^2 * y21^2

  take square root of both sides
x13 * |2->1| = |1->3| * y21

  divide both sides by the length of 2->1
x13 = (|1->3|/|2->1|) *y21

  lets call the ratio of |1->3| to |2->1| psi
x13 = psi * y21

  check the signs
    when y21 is negative, x13 will be negative
    when y21 is positive, x13 will be negative

x13 = psi * y21

凝结

x21 = x1 - x2
y21 = y1 - y2

|2->1| = sqrt( x21^2 + y^21^2 )
|1->3| = N/2

psi = |1->3|/|2->1|

y13 = -psi * x21
x13 =  psi * y21

我通常不会这样做,但我在工作中解决了它,并认为彻底解释它将帮助我巩固我的知识。

如果要避免SQRT,请执行以下操作:

in: line_length, cap_length, rotation, position of line centre

define points:
  tl (-line_length/2, cap_length)
  tr (line_length/2, cap_length)
  bl (-line_length/2, -cap_length)
  br (line_length/2, -cap_length)

rotate the four points by 'rotation'
offset four points by 'position'

drawline (midpoint tl,bl to midpoint tr,br)
drawline (tl to bl)
drawline (tr to br)