ENB339 lecture 9: Image geometry and planar homography 

Ahmad Ghafouri you might like to take a look at my Robotic Vision MOOC, it runs again in October, see moocs.qut.edu.au

Thanks a lot. This is just a recording of a class room lecture. A much more polished version of this content can be found at robotacademy.net.au/masterclass/the-geometry-of-image-formation. You can also find individual short (5-10 minute) lessons on these topics and more at the Robot Academy.

Thank you a lot Prof. Peter.... I really need these courses for accomplishing my thesis. Thanks for your kindness and sharing your knowledge without any expectation.

Thank you. A more polished version can be found at A much more polished version of this content can be found at robotacademy.net.au/masterclass/the-geometry-of-image-formation/ and lessons on many related topics can be found at robotacademy.net.au by topic or by searching keywords

You're welcome. Check out robotacademy.net.au for more lectures on CV topics. The video you watched is an incomplete set of classroom lecture recordings.

Thank you so much for the response! I'm definitely heading towards the entire series! :)

Thanks for the kind feedback. I'm hoping to get a larger suite of material out this year, organized as lots of shorter segments. Will spread the word on this channel once that happens.

Will be looking forward to it; I have worked with your material in the three universities I've studied so far and it always proved very useful!

Joey, glad you liked it. A much more polished version of this content can be found at robotacademy.net.au/masterclass/the-geometry-of-image-formation/

Thanks. This is just a recording of a class room lecture. A much more polished version of this content can be found at robotacademy.net.au/masterclass/the-geometry-of-image-formation. You can also find individual short (5-10 minute) lessons on these topics and more at the Robot Academy.

Thanks. This is just a recording of a class room lecture. A much more polished version of this content can be found at @t

The Matlab function is part of the Machine Vision ToolBox: petercorke.com/wordpress/toolboxes/machine-vision-toolbox

You're welcome. Check out robotacademy.net.au for more lectures on CV topics. The video you watched is an incomplete set of classroom lecture recordings.

No. Image coordinates correspond to a line in 3D space, so you need additional information. For instance if the object is on the ground.

Exactly!

Only if the object is a plane. For a general 3D shape, sadly not.

Thanks for the help.

yes, I assume the pixels are square which they generally are

@@PeterCorke Thank you Peter. I was looking for this conversion for a while now. Got it from your lectures.

Hi Peter, The lecture is absolutely stunning. A lot of scattered information is found all over the internet but your lecture has it all. Just 2 doubts: 1. When you write the Camera matrix you write the inverse of the extrinsic parameter matrix. Is this a convention? 2. Isn't the matrix in 25:55 the inverse homography matrix?

Thanks, Peter. It clarifies my first question. But at 24:12 you mention that from plane 1p which is yellow we can go to 2p which is red using H. Where as at 25:55 you are going from the warped image which corresponds to 2p to 1p. Is the H different?

Nikhil, you are correct, I should use the inverse homography at 25:55. A better and correct version of this discussion can be found on the Robot Academy robotacademy.net.au/lesson/planar-homography at time 5:12. Thanks for pointing this out. The Robot Academy is the more polished version of the rough classroom lecture you were watching. Thanks again for your interest.

Thanks, Peter. I will go through the link. I work on Computational photography for which I have calibrated the camera too. Can we connect on LinkedIn so that we could discuss relevant topics if needed?

Sure thing.

You'll find it in a number of books: mine, Robotics, Vision & Control; Hartley & Zisserman; Intro to 3D, etc.

When invoke Homography function : H=homography(P1,P2) Why error message Undefined function or variable 'vgg_H_from_x_lin'. Error in homography (line 46) H = vgg_H_from_x_lin(p1, p2); is coming? Here P1 and P2 are 2by4 matrices

Generally 2 ways: either by knowing in advance all the camera parameters, or by calibration. Details in my book, chapter 11

svd decomposition ?

I'm not sure it can be zero for any realisable perspective camera.

@@PeterCorke yeah but why ?

It's not a matter of throwing away or adding. For every pixel in the output image we interpolate a value from the input image. So all output pixels are some mixture of pixels from the input image. This is covered in detail in the section on Image Warping, Sec 12.7.4 of the second edition

Hi, Peter, thank you for your reply. Given the age of this video I did not expect any reply, leave alone that quick one! It's impressive that you keep supporting your old works. Yes, I am aware about warping. Yet warping is not mentioned in this lecture, nevertheless the picture looks like warping was there. Do you mean that warping was not mentioned on order not to obscure the main point of the lecture which is Homography? Do you have similar lecture on Image Warping? I have an image of a plane rectangular surface distorted by perspective. I am interested in one point only. I know how to detect this point and get its coordinates within the distorted image. I need to calculate coordinates of this point in the un-destorted image. So I need to un-distort just one point. I guess that for this I do not need the whole power of Warping (including interpolation). But I need to know how Warp works to make my own warp to work with a singer point. Unless such one-point-warp function already exists

Yes, I was fortunate to visit long before the fire.