Panda3D 中不规则直柱体范围检测算法
以下是基于 Panda3D 的代码,可以精确计算出任意一个坐标点是否在不规则直柱体范围内:
from panda3d.core import *
from direct.showbase.ShowBase import ShowBase
class MyApp(ShowBase):
def __init__(self):
ShowBase.__init__(self)
# 8个点的坐标
points = [[67.85213470458984, -27.81583023071289, 0.0],
[44.97549819946289, -30.375598907470703, 0.0],
[44.97549819946289, -30.375598907470703, 52.0],
[71.8885726928711, -28.21024513244629, 52.0],
[68.07453155517578, -29.803424835205078, 0.0],
[45.19790267944336, -32.36319351196289, 0.0],
[45.19790267944336, -32.36319351196289, 52.0],
[72.11096954345703, -30.197839736938477, 52.0]]
# 创建不规则直柱体
geom = GeomVertexData('shape', GeomVertexFormat.getV3(), Geom.UHStatic)
vertex = GeomVertexWriter(geom, 'vertex')
for point in points:
vertex.addData3f(*point)
tris = [(0, 1, 4), (4, 1, 5), (1, 2, 5), (5, 2, 6),
(2, 3, 6), (6, 3, 7), (3, 0, 7), (7, 0, 4),
(0, 3, 1), (1, 3, 2), (4, 5, 7), (7, 5, 6)]
prim = GeomTriangles(Geom.UHStatic)
for tri in tris:
prim.addVertices(*tri)
prim.closePrimitive()
geom.addPrimitive(prim)
node = GeomNode('shape')
node.addGeom(geom)
self.shape = render.attachNewNode(node)
# 验证一个点是否在直柱体范围内
test_point = Point3(60.0, -30.0, 26.0)
if self.is_point_inside_shape(test_point):
print('Point is inside shape!')
else:
print('Point is outside shape!')
def is_point_inside_shape(self, point):
# 将点转换为局部坐标系中的坐标
local_point = self.shape.getRelativePoint(self.shape.getParent(), point)
# 创建一个射线,从该点向上射出
ray = CollisionRay(local_point, Vec3(0, 0, 1))
ray_node = CollisionNode('shape_ray')
ray_node.addSolid(ray)
ray_np = self.shape.attachNewNode(ray_node)
# 进行射线碰撞检测
queue = CollisionHandlerQueue()
base.cTrav.traverse(self.shape)
base.cTrav.addCollider(ray_np, queue)
# 如果碰撞队列中有碰撞信息,说明该点在直柱体范围内
if queue.getNumEntries() > 0:
return True
else:
return False
app = MyApp()
app.run()
代码中,我们通过GeomVertexData
和GeomTriangles
创建了一个不规则直柱体,并将其添加到了场景图中。然后我们实现了is_point_inside_shape
方法,该方法将一个点转换为局部坐标系中的坐标,并通过创建一个射线,并进行碰撞检测的方式,判断该点是否在直柱体范围内。最后,我们在MyApp
的构造函数中调用了is_point_inside_shape
方法,传入了一个测试点,以验证该方法的正确性。
原文地址: http://www.cveoy.top/t/topic/oK02 著作权归作者所有。请勿转载和采集!