Please do more videos in Robotics, your explaining very good.
@MrLonglace4 жыл бұрын
this video really help me through my exam, thx a lot sir!
@liliw47813 жыл бұрын
Best video for Jacobian matrix!! Please keep making more video related robotics.
@Paul122M3 жыл бұрын
This intro is much better, the previous was in higher volume than the rest of the video.
@enhaozheng45393 жыл бұрын
Great videos! The presentation method is also great! What type of device do you use?
@jnine38073 жыл бұрын
You deserve more!
@veethikachourasiya30083 жыл бұрын
Thanks for the video, helped me alot
@omeroztoprak8693 жыл бұрын
awasomeee dudeeee
@vegan_cat Жыл бұрын
it was life saver thanks
@zannatulferdousbristy2353 жыл бұрын
good explanation
@AirAdventurer1944 жыл бұрын
(I'm not 100% on this; I'm opening this interpretation up for comments) I think the way you're doing Jacobian matrices, it's with a coordinate patch, mapping the generalized coordinates from the coordinate space into the configuration space (manifold) Q, and where the Jacobian (determinant of the Jacobian matrix) is zero, the vector field on the configuration space is zero, what is sometimes known as a critical point of a differential equation on the configuration space. There is another way of thinking about critical points of differential equations (vector fields) on configuration spaces, as a section of the tangent bundle of the configuration space, \xi:Q -> TQ. Where there is a critical point of the differential equation, one can take the Jacobian of \xi (*not* the Jacobian of the coordinate patch), and this becomes the "A matrix" of the state-space representation of a plant function about the critical point. *This* Jacobian matrix will have eigenvalues, some of which may have positive real part and some of which may have negative real part (let's assume none have zero real part, so the Hartman-Grobman Theorem applies), so we can put an open-loop controller and a unital feedback loop on the system near the critical point, making all the eigenvalues of the closed-loop system have negative real part ("stable eigenvalues"). We can then put the real part of all but two of the eigenvalues out between, say, -5 and -10 (called the "non-dominant eigenvalue"), and then, for the remaining two eigenvalues, put their real part, say, between 0 and -2 (called the "dominant eigenvalues"), and meet the design specifications of the control engineering problem with proper placement of the dominant eigenvalues. Let me know what you think about this interpretation of the underlying mathematics.
@Salid_Innovators3 жыл бұрын
Nice
@yinpozhu8043 Жыл бұрын
thank you!
@wweworld82983 жыл бұрын
if jacobian matrix is not square than how to calculate singularities.
@antonionakulabrigida2833 Жыл бұрын
If the jacobian matrix is not square than the singularities are the values of θ that reduce the "row" rank
@jonayethaque530 Жыл бұрын
You are too good to say
@spidymaster92213 жыл бұрын
thanks alot
@shukhratdad98914 жыл бұрын
Thanks for your video. It is easy understandable. If I have a quiz in future, can I connect by email? If yes, please give me yours