Inverse Kinematics Problem
hello,
Would someone please help explain to me how inverse kinematics works? At the last summer camp there was a robot arm which had a function that you could increment x,y and z functions and i am trying to do that with my crustcrawler 5DOF arm. Could someone give me some tips on how to learn this stuff? (I am taking Calculus AP now and i think that Jacobian matrices are a little over my head at this point). Thanks!-Signol01
Would someone please help explain to me how inverse kinematics works? At the last summer camp there was a robot arm which had a function that you could increment x,y and z functions and i am trying to do that with my crustcrawler 5DOF arm. Could someone give me some tips on how to learn this stuff? (I am taking Calculus AP now and i think that Jacobian matrices are a little over my head at this point). Thanks!-Signol01
Comments
··· Here's some information... but's in spanish, if u don't understand·just ask
··· Direct Kinematic, Inverse Kinematic
Best regards
Goten
www.learnaboutrobots.com
From the above website, here is what I understand:
Forward Kinematics:
The forward kinematics problem is stated as follows: Given the angles at each of the robots joints, where is the robot's hand (Xhand, Yhand,
I know how to do vector addition easy enough. This would give me an endpoint. Then when i try to do inverse kinematics (take the endpoint and work backwards to find the 3 joint angles) I have one equation with three unknowns which i have no clue how to solve.
For instance i took a theoretical arm with 3 segments of 4, 3 and 1 units. The three joints would be set to the following angles, 50,130,130. This would then yeild for the x component of the point the equation 4cos50+3+cos50 =x. Then for the y component it would be 4sin50+0-sin50 (50 was gotten through the corresponding angles in two parallel lines and the fact that it mirrors the first joint). The end point was then found to be (6.21,2.3). Then when you work backwards I am stuck in the equation 4cosA+3cosB+cosC= 6.21 and 4sinA+3sinB+sinC=2.3. I am at a loss to where to go from here.-signol01
Reread the site looks alot easier, have a few questions (dunno if you know just thought i would throw these out there) when it has the theta hand why is that parameter important in the end position of the hand? Can it be done without this parameter or is it essential? Also how would this parameter be determined? Also does the example go over a three jointed hand? If so would i have to modify it in any way if the third segment actually had a length? It looks as though it only uses a two segmented arm and i don't think that i could apply the law of cosines to three segments (i don't know its been a while since i went over that law in trig). Thanks!-Signol
100 posts!!!!