Thank you all for these videos! I'm so excited that you all have put so much effort into making this information available with the book and such. Thanks.
@@physicsmadness no, he's not mistaken, because he mentioned that for a every point on the surface of a torus, there is one and only one ("unique") configuration for the robot arm with 2 joints. If you drew a disk on a flat plane and chose a point on that disk, you could have 2 configurations of the links of the robot arm (relative to each other) that will correspond to that point on the disk. Because of that, a point on the surface of a disk (if choosing that to [incorrectly] represent the C-space) you'll find that it's not associated with only a single, UNIQUE configuration of the links; you can check this by drawing both configurations, composed of different joint angles, but share the same endpoint of the outermost link. By overlaying the two distinct configurations, it should outline the shape of a parallelogram. Again, this shows that there is NOT a ONE-to-ONE correspondence between configurations and points on the configuration space (if that C-space were a disk instead of a torus).
Thanks for the video. I didn't understand how when specifying C, we have moved B away from the circumference of the point A. Isn't B supposed to be at the circumference of A?
Consider a joint between two rigid bodies. Each rigid body has mm degrees of freedom (m=3m=3 for a planar rigid body and m=6m=6 for a spatial rigid body) in the absence of any constraints. The joint has ff degrees of freedom (e.g., f=1f=1 for a revolute joint or f=3f=3 for a spherical joint). How many constraints does the joint place on the motion of one rigid body relative to the other? Write your answer as a mathematical expression in terms of m and f. ans?
In the video, it is said that there is only one number needed to specify location of point C. I could not understand it. How can we specify location of point C with one value ?
See the location of point C is represented by 3 numbers but it's position has to be in the circle that A and B formed which means it has to be at a particular distance from A and B ,and whatever the position might be in that circle, it IS going to be represented by 3 coordinates. It's not the coordinates of point C that is decreasing, its the degrees freedom of it's location. Edit: I'm replying 7 months later, you probably would have mastered robotics until now. xD
Simply take any object (e.g. pen, shoe) & fix its x & y co-ordinates (i.e. keep it constant), then you'll observe that you only need one more number to specify the location of the object.
Thank you all for these videos! I'm so excited that you all have put so much effort into making this information available with the book and such. Thanks.
this is some hardest way to explane DoF
I found it helpful
Can't understand 1:46. The CSpace should be a disk, right? How come it became a torus?
u r right....i guess the first joint is orthogonal to the second...he has mistaken it...it should actually be an annular disc...
It is the result of S1xS1. 1 Joint rotation is the angle on the torus from above, and the second is the angle within the torus itself.
@@physicsmadness no, he's not mistaken, because he mentioned that for a every point on the surface of a torus, there is one and only one ("unique") configuration for the robot arm with 2 joints.
If you drew a disk on a flat plane and chose a point on that disk, you could have 2 configurations of the links of the robot arm (relative to each other) that will correspond to that point on the disk.
Because of that, a point on the surface of a disk (if choosing that to [incorrectly] represent the C-space) you'll find that it's not associated with only a single, UNIQUE configuration of the links;
you can check this by drawing both configurations, composed of different joint angles, but share the same endpoint of the outermost link.
By overlaying the two distinct configurations, it should outline the shape of a parallelogram.
Again, this shows that there is NOT a ONE-to-ONE correspondence between configurations and points on the configuration space (if that C-space were a disk instead of a torus).
Thanks for the video. I didn't understand how when specifying C, we have moved B away from the circumference of the point A. Isn't B supposed to be at the circumference of A?
Thought this was going to be beginner friendly
Consider a joint between two rigid bodies. Each rigid body has mm degrees of freedom (m=3m=3 for a planar rigid body and m=6m=6 for a spatial rigid body) in the absence of any constraints. The joint has ff degrees of freedom (e.g., f=1f=1 for a revolute joint or f=3f=3 for a spherical joint). How many constraints does the joint place on the motion of one rigid body relative to the other? Write your answer as a mathematical expression in terms of m and f. ans?
here the Answer is m-f.
as per the equation given over here, f+c=m . and we want the value of c in terms of m and f.
In the video, it is said that there is only one number needed to specify location of point C. I could not understand it. How can we specify location of point C with one value ?
See the location of point C is represented by 3 numbers but it's position has to be in the circle that A and B formed which means it has to be at a particular distance from A and B ,and whatever the position might be in that circle, it IS going to be represented by 3 coordinates. It's not the coordinates of point C that is decreasing, its the degrees freedom of it's location.
Edit: I'm replying 7 months later, you probably would have mastered robotics until now. xD
a circle is parametrically represented with just theta...any curve is intrinsically one dimensional..
@@sparshmecwan2962 but still your answer is helping me with same question
But professor mentions point c wrt a as an outer parameter of the B sphere
Simply take any object (e.g. pen, shoe) & fix its x & y co-ordinates (i.e. keep it constant), then you'll observe that you only need one more number to specify the location of the object.
thank you mister . its helpful
hey nice video ,can you do a video for 5 DOF bipedal robot .thanks
if have one send me the link.
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