output velocity and angular velocity in a ModelTask's link export #31
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| math::Pose source2target( math::Pose(source2world - target2world) ); | ||
| math::Vector3 sourceInTarget_angular_vel(target2world.rot.RotateVectorReverse(sourceInWorld_angular_vel)); | ||
| math::Vector3 sourceInTarget_linear_vel (target2world.rot.RotateVectorReverse(sourceInWorld_linear_vel)); |
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It should not consider the velocities of the targetInWorld as well?
(I know that the target frame is usually a fixed point, but is this always the case?)
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I know that the target frame is usually a fixed point, but is this always the case?
I elected to generate the velocity of the source frame within the target frame considered-as a fixed frame at time t. Part for simplicity reasons, part because it's a possible convention, part because then one can use it with sourceFrame==targetFrame to get both velocities expressed in the body frame.
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Alright.
I just have doubt about the use of sourceFrame==targetFrame.
It means that we interpret the velocities as absolute (velocities of the sourceFrame related to a inertial frame, not the targetFrame) and represent them in targetFrame (for linear case) and sourceFrame (angular case).
The relative velocities of the sourceFrame related to the targetFrame should be zero if they are the same.
| math::Pose source2target( math::Pose(source2world - target2world) ); | ||
| math::Vector3 sourceInTarget_angular_vel(target2world.rot.RotateVectorReverse(sourceInWorld_angular_vel)); |
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My understanding of the angular velocity in RBS is rock-core/base-types#99:
So, If i got this right regarding the velocities:
The angular velocity is expressed in sourceFrame ( the angular velocity of the moving frame expressed on it own frame).
And the linear velocity is expressed in targetFrame (the linear velocity of the moving frame expressed on the fixed frame). Is that right?
It means that the it->second.source_link_ptr->GetRelativeAngularVel() would do the job.
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Yeah. I'll update to stick to that until we resolve rock-core/base-types#100
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Updated. Could you check @joaobrittoneto ? |
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Looks fine to me |
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