Lines Matching refs:r_0

33 Let two circles be $C_0, r_0$ and $C_1, r_1$ where $C$ is the center and $r$ is the radius. For any
36 $r_t = (1-t) \cdot r_0 + t \cdot r_1 > 0$ (note that radius $r_t$ has to be *positive*). If
42 2. $r_0 = r_1$ so the gradient is a single strip with bandwidth $2 r_0 = 2 r_1$.
46 They are easy to handle so we won't cover them here. From now on, we assume $C_0 \neq C_1$ and $r_0
49 As $r_0 \neq r_1$, we can find a focal point $C_f = (1-f) \cdot C_0 + f \cdot C_1$ where its
50 corresponding linearly interpolated radius $r_f = (1-f) \cdot r_0 + f \cdot r_1 = 0$.
51 Solving the latter equation gets us $f = r_0 / (r_0 - r_1)$.
54 swap $C_0, r_0$ with $C_1, r_1$, compute swapped gradient $t_s$ as if $r_1 \neq 0$, and finally set
68 always the bigger one (note that $f \neq 1$, otherwise we'll swap $C_0, r_0$ with $C_1, r_1$).
91 3. we still need to handle the swapped case (we swap $C_0, r_0$ with $C_1, r_1$ if $r_1 = 0$);
117 1. Let $C'_0, r'_0, C'_1, r'_1 = C_0, r_0, C_1, r_1$ if there is no swapping and $C'_0,
118     r'_0, C'_1, r'_1 = C_1, r_1, C_0, r_0$ if there is swapping.
128   2. let $\hat x_t =  -\sqrt{\hat x^2 - \hat y^2} - \hat x / r_1$ if we've swapped $r_0, r_1$,
155 (1/a) \cdot \left(-b \pm \sqrt{b^2 - 4ac}\right)$$ It precomputes $a = 1 - (r_1 - r_0)^2, 1/a, r1 -
157 $c = x^2 + y^2 - r_0^2$ costs 3 multiplications and 2 additions. And the final $t$ costs 5 more
160 if it saves the $0.5 \cdot (1/a), 4a, r_0^2$ and $(r_1 - r_0) r_0$ multiplications, there are still