|
6 | 6 | importwarnings |
7 | 7 |
|
8 | 8 | importnumpyasnp |
9 | | -importscipy.integrate |
10 | 9 |
|
11 | 10 | importmatplotlib.cbookascbook |
12 | 11 |
|
@@ -256,13 +255,77 @@ def point_at_t(self, t): |
256 | 255 | returntuple(self(t)) |
257 | 256 |
|
258 | 257 | defarc_area(self): |
259 | | -"""(Signed) area between curve an origin.""" |
260 | | -dB=self.differentiate(self) |
261 | | -defintegrand(t): |
262 | | -dr=dB(t).T |
263 | | -n=np.array([dr[1],-dr[0]]) |
264 | | -return (self(t).T @n)/2 |
265 | | -returnscipy.integrate.quad(integrand,0,1)[0] |
| 258 | +r"""(Signed) area swept out by ray from origin to curve. |
| 259 | +
|
| 260 | + Notes |
| 261 | + ----- |
| 262 | + Exact formula used for Bezier curves. |
| 263 | +
|
| 264 | + Now for an arbitrary bezier curve B(t), the area of the arc swept |
| 265 | + out by the ray from the origin to the curve, we simply need to compute |
| 266 | + :math:`\frac{1}{2}\int_0^1 B(t) \cdot n(t) dt`, where :math:`n(t) = |
| 267 | + u^{(1)}(t) \hat{x}_0 - u{(0)}(t) \hat{x}_1` is the normal vector |
| 268 | + oriented away from the origin and :math:`u^{(i)}(t) = \frac{d}{dt} |
| 269 | + B^{(i)}(t)` is the :math:`i`th component of the curve's tangent vector. |
| 270 | + (This formula can be found by applying the divergence theorem to |
| 271 | + :math:`F(x,y) = [x, y]/2`. |
| 272 | +
|
| 273 | + The control points of the curve are just its coefficients in a |
| 274 | + Bernstein expansion, so if we let :math:`P_i = [x_i, y_i]` be the |
| 275 | + :math:`i`th control point, then |
| 276 | +
|
| 277 | + .. math:: |
| 278 | +
|
| 279 | + \frac{1}{2}\int_0^1 B(t) \cdot n(t) dt \\ |
| 280 | + = \frac{1}{2}\int_0^1 B^{(0)}(t) \frac{d}{dt} B^{(1)}(t) |
| 281 | + - B^{(1)}(t) \frac{d}{dt} B^{(0)}(t) |
| 282 | + dt \\ |
| 283 | + = \frac{1}{2}\int_0^1 |
| 284 | + \left( \sum_{j=0}^n P_j^{(0)} b_{j,n} \right) |
| 285 | + \left( n \sum_{k=0}^{n-1} (P_{k+1}^{(1)} - |
| 286 | + P_{k}^{(1)}) b_{j,n} \right) |
| 287 | + - \left( \sum_{j=0}^n P_j^{(1)} b_{j,n} \right) |
| 288 | + \left( n \sum_{k=0}^{n-1} (P_{k+1}^{(0)} - |
| 289 | + P_{k}^{(0)}) b_{j,n} \right) |
| 290 | + dt |
| 291 | +
|
| 292 | + Where :math:`b_{\nu,n}(t)` is the :math:`\nu`'th Bernstein polynomial |
| 293 | + of degree :math:`n`, :math:`b_{\nu, n}(t) = {n \choose \nu} t^\nu {(1 - |
| 294 | + t)}^{n-\nu}`. |
| 295 | +
|
| 296 | + grouping :math:`t^l(1-t)^m` terms together for each :math:`l`, |
| 297 | + :math:`m`, we get that the integrand becomes |
| 298 | +
|
| 299 | + .. math:: |
| 300 | +
|
| 301 | + \frac{n}{2}\int_0^1 \sum_{j=0}^n \sum_{k=0}^{n-1} |
| 302 | + \frac{{n \choose j} {{n - 1} \choose k}} |
| 303 | + {{{2n - 1} \choose {j+k}}} |
| 304 | + [P_j^{(0)} (P_{k+1}^{(1)} - P_{k}^{(1)}) |
| 305 | + - P_j^{(1)} (P_{k+1}^{(0)} - P_{k}^{(0)})] |
| 306 | + b_{j+k,2n-1}(t) dt |
| 307 | +
|
| 308 | + Interchanging sum and integral, we notice that :math:`\int_0^1 b_{\nu, |
| 309 | + n}(t) dt = \frac{1}{n + 1}`, so we conclude that |
| 310 | +
|
| 311 | + .. math:: |
| 312 | +
|
| 313 | + \frac{1}{2}\int_0^1 B(t) \cdot n(t) dt |
| 314 | + = \frac{1}{4}\sum_{j=0}^n \sum_{k=0}^{n-1} |
| 315 | + \frac{{n \choose j} {{n - 1} \choose k}} |
| 316 | + {{{2n - 1} \choose {j+k}}} |
| 317 | + [P_j^{(0)} (P_{k+1}^{(1)} - P_{k}^{(1)}) |
| 318 | + - P_j^{(1)} (P_{k+1}^{(0)} - P_{k}^{(0)})] |
| 319 | + """ |
| 320 | +n=self.degree |
| 321 | +area=0 |
| 322 | +P=self.control_points |
| 323 | +dP=np.diff(P,axis=0) |
| 324 | +forjinrange(n+1): |
| 325 | +forkinrange(n): |
| 326 | +area+=_comb(n,j)*_comb(n-1,k)/_comb(2*n-1,j+k) \ |
| 327 | +* (P[j,0]*dP[k,1]-P[j,1]*dP[k,0]) |
| 328 | +return (1/4)*area |
266 | 329 |
|
267 | 330 | @classmethod |
268 | 331 | defdifferentiate(cls,B): |
|