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simplify logic of biggest diagonal

release/4.3a0
Fan Jiang 2021-10-13 13:46:33 -04:00
parent aa6b432911
commit 35061ca135
1 changed files with 7 additions and 6 deletions
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13
gtsam/geometry/SO3.cpp
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@ -262,20 +262,20 @@ Vector3 SO3::Logmap(const SO3& Q, ChartJacobian H) {
// when trace == -1, i.e., when theta = +-pi, +-3pi, +-5pi, etc.
// we do something special
if (tr + 1.0 < 1e-4) {
std::vector<double> diags = {R11, R22, R33};
size_t max_elem = std::distance(diags.begin(), std::max_element(diags.begin(), diags.end()));
if (max_elem == 2) {
if (R33 > R22 && R33 > R11) {
// R33 is the largest diagonal
const double sgn_w = (R21 - R12) < 0 ? -1.0 : 1.0;
const double r = sqrt(2.0 + 2.0 * R33);
const double scale = M_PI_2 / r - 0.5 / (r * r) * (R21 - R12);
omega = sgn_w * scale * Vector3(R31 + R13, R32 + R23, 2.0 + 2.0 * R33);
} else if (max_elem == 1) {
} else if (R22 > R11) {
// R22 is the largest diagonal
const double sgn_w = (R13 - R31) < 0 ? -1.0 : 1.0;
const double r = sqrt(2.0 + 2.0 * R22);
const double scale = M_PI_2 / r - 0.5 / (r * r) * (R13 - R31);
omega = sgn_w * scale * Vector3(R21 + R12, 2.0 + 2.0 * R22, R23 + R32);
} else {
// if(std::abs(R.r1_.x()+1.0) > 1e-5) This is implicit
// R33 is the largest diagonal
const double sgn_w = (R32 - R23) < 0 ? -1.0 : 1.0;
const double r = sqrt(2.0 + 2.0 * R11);
const double scale = M_PI_2 / r - 0.5 / (r * r) * (R32 - R23);
@ -283,8 +283,9 @@ Vector3 SO3::Logmap(const SO3& Q, ChartJacobian H) {
}
} else {
double magnitude;
const double tr_3 = tr - 3.0; // always negative
const double tr_3 = tr - 3.0; // could be non-negative if the matrix is off orthogonal
if (tr_3 < -1e-6) {
// this is the normal case -1 < trace < 3
double theta = acos((tr - 1.0) / 2.0);
magnitude = theta / (2.0 * sin(theta));
} else {