ruby Ruby：获取旋转矩形的边界框的尺寸

Posted 2021-05-16

<img src="https://docs.google.com/drawings/d/1lV7Ektr4BnzyXdbiwo8MOAjQ339NpvDnjP1Qp5bQq1g/pub?w=488&amp;h=381">

require 'bigdecimal'

def bounding_box_dimensions(inner_rectangle_width, inner_rectangle_height, rotate_in_degrees)
  width  = BigDecimal.new inner_rectangle_width
  height = BigDecimal.new inner_rectangle_height
  angle  = BigDecimal.new((rotate_in_degrees * Math::PI) / 180, 5) # convert to radians

  # rectangle centre coords
  centre_x = width/2
  centre_y = height/2

  corners = [[0, 0], [0, height], [width, height], [width, 0]]

  corners.map! do |points|
    # translate rectangle centre to origin
    temp_x = points[0] - centre_x
    temp_y = points[1] - centre_y

    # do rotation
    rotated_x = (temp_x * Math::cos(angle)) - (temp_y * Math::sin(angle))
    rotated_y = (temp_x * Math::sin(angle)) + (temp_y * Math::cos(angle))

    # translate rectangle centre back to original place
    points[0] = rotated_x + centre_x
    points[1] = rotated_y + centre_y

    points
  end

  min_x, max_x = corners.minmax_by { |p| p[0] }.map { |p| p[0] }
  width        = max_x - min_x # distance in single dimension
  min_y, max_y = corners.minmax_by { |p| p[1] }.map { |p| p[1] }
  height       = max_y - min_y # distance in single dimension

  [
      BigDecimal.new(width).round(3, :default).to_s('F'),
      BigDecimal.new(height).round(3, :default).to_s('F')
  ]
end