Great lecture. I do this for living. I really did not need to learn. But there is something in your teaching that kept me engaged. Superbly explained and very well executed lecture. Respect!
Simply awesome.. I searched throughout the internet for transformation matrices but nothing is as simple and lucid as your video.. Great work.. n thank a lot..
Views are just standard notations ascribed to objects by engineers for uniformity. Any can be your front or side but you might prefer to choose the side with most detail as your front.
I have question about rotation matrix Say I have a point P under coordinate 1, and I want to know P's location under coordinate 0, then I actually have to rotate coordinate 0 to and let it line up with coordinate 1 to find out the angle, am I right? And when I rotate, I should always rotate it counter clockwise, am I right? Am a little bit confused here, hope you can clarify that for me. Thanks.
There are some conventions including a rule called the "right hand rule" that come in the later lectures which should help with confusion about "which direction" to rotate reference frames and so on. For this simple example, I'd suggest thinking about basic geometry and trigonometry. For the way the axes are drawn, the x-coordinate of a point in reference frame 1 will contribute a total of cos(theta) * x to the x-coordinate in reference frame 0. A y-coordinate of a point in reference frame 1 will contribute a total of -sin(theta) * x to the x-coordinate in reference frame 0. That's where the top line of the rotation matrix comes from: [cos(theta) -sin(theta)]. You use this equation to work out the x-coordinate of point p in reference frame 0 based on its x and y coordinates in reference frame 1: px in reference frame 0 = cos(theta) * px in reference frame 1 - sin(theta) * py in reference frame 1.
Great lecture. I do this for living. I really did not need to learn. But there is something in your teaching that kept me engaged. Superbly explained and very well executed lecture. Respect!
Thank you so much making perfect sense of the horrible jumble that we had in school!
Simply awesome.. I searched throughout the internet for transformation matrices but nothing is as simple and lucid as your video.. Great work.. n thank a lot..
Very clear and straightforward lecture. Thank you very much! I also borrowed this for a robotics class. Dr. Milford is an excellent lecturer.
Thx man.
That darn angle, finally a video with the arrow pointing out the way of the angle!
thanks for your videos. it helped a lot me in my robotic class
Really great tutorial
Really helpful !! Thanks!
I just wanna ask for the slides, are they available for downloading?
what makes a value negative ? and also when we are looking at this 2D graphs are we looking at a top view or side view?
Views are just standard notations ascribed to objects by engineers for uniformity. Any can be your front or side but you might prefer to choose the side with most detail as your front.
I have question about rotation matrix
Say I have a point P under coordinate 1, and I want to know P's location under coordinate 0, then I actually have to rotate coordinate 0 to and let it line up with coordinate 1 to find out the angle, am I right?
And when I rotate, I should always rotate it counter clockwise, am I right?
Am a little bit confused here, hope you can clarify that for me. Thanks.
There are some conventions including a rule called the "right hand rule" that come in the later lectures which should help with confusion about "which direction" to rotate reference frames and so on.
For this simple example, I'd suggest thinking about basic geometry and trigonometry. For the way the axes are drawn, the x-coordinate of a point in reference frame 1 will contribute a total of cos(theta) * x to the x-coordinate in reference frame 0. A y-coordinate of a point in reference frame 1 will contribute a total of -sin(theta) * x to the x-coordinate in reference frame 0. That's where the top line of the rotation matrix comes from: [cos(theta) -sin(theta)].
You use this equation to work out the x-coordinate of point p in reference frame 0 based on its x and y coordinates in reference frame 1:
px in reference frame 0 = cos(theta) * px in reference frame 1 - sin(theta) * py in reference frame 1.
***** Thanks, I'll search right hand rule and take a look. Appreciated :)
***** thank you i find the answer before asking the question
thanks so useful how can I download the PDF
Thank you sir
would you like sen to me the slides of these lecture?
They are not have knack for teaching grammar :)