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It only takes a minute to sign up. Connect and share knowledge within a single location that is structured and easy to search. Use-cases are for example in game engines and simulations to compute the position of objects relative to one another, or for example in robotics my use-case to compute the position of an object relative to the gripper of a robot arm.

Weirdly enough, I haven't yet found a good Python library to perform such conversions. Particularly one that depends on the scientific Python stack numpyscipymatplotlibSign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more.

Utilities for Coordinate Transformations via homogeneous coordinates Ask Question. Asked 4 days ago. Active 3 days ago. Viewed 20 times. What I would like reviewed is general code quality and clarity of the documentation. The code: kinematics. This is a 6-dimensional vector consisting of the origin's position and the frame's orientation xyz Euler Angles : [x, y, z, alpha, beta, gamma].

The adjusted trial balance shows account balances after adjustments For performance reasons, it is better to sequentially apply multiple transformations to a vector or set of vectors than to first multiply a sequence of transformations and then apply them to a vector afterwards. For performance reasons, it is better to sequentially apply multiple transformations to a vector than to first multiply a sequence of transformations and then apply them to a vector afterwards.

Improve this question. Reinderien FirefoxMetzger FirefoxMetzger 1, 4 4 silver badges 8 8 bronze badges. Add a comment. Active Oldest Votes. It's "homogeneous", not "homogenious". Otherwise this is pretty sane. Improve this answer. Reinderien Reinderien Sign up or log in Sign up using Google. Sign up using Facebook. Sign up do i love him accurate quiz Email and Password.

Post as a guest Name. Email Required, but never shown. The Overflow Blog. Podcast Getting Dev and Ops to actually work together. Podcast A director of engineering explains scaling from dozens of…. Featured on Meta. Stack Overflow for Teams is now free for up to 50 users, forever.In Euclidean space geometrytwo parallel lines on the same plane cannot intersect, or cannot meet each other forever. It is a common sense that everyone is familiar with.

However, it is not true any more in projective space, for example, the train railroad on the side picture becomes narrower while it moves far away from eyes. Finally, the two parallel rails meet at the horizon, which is a point at infinity. The Cartesian coordinates of a 2D point can be expressed as x, y. What if this point goes far away to infinity? The parallel lines should meet at infinity in projective space, but cannot do in Euclidean space. Mathematicians have discoverd a way to solve this issue.

To make 2D Homogeneous coordinates, we simply add an additional variable, winto existing coordinates. Therefore, a point in Cartesian coordinates, X, Y becomes x, y, w in Homogeneous coordinates.

As mentioned before, in order to convert from Homogeneous coordinates x, y, w to Cartesian coordinates, we simply divide x and y by w ; Converting Homogeneous to Cartesian, we can find an important fact. Therefore, these points are "homogeneous" because they represent the same point in Euclidean space or Cartesian space. In other words, Homogeneous coordinates are scale invariant. Therefore, two parallel lines meet at x, y, 0which is the point at infinity. Homogeneous coordinates are very useful and fundamental concept in computer graphics, such as projecting a 3D scene onto a 2D plane.

Railroad gets narrower and meets at horizon.Join Stack Overflow to learn, share knowledge, and build your career. Connect and share knowledge within a single location that is structured and easy to search. I am trying to find the transformation matrix H so that i can multiply the x,y pixel coordinates and get the x,y real world coordinates.

**Rotation Matrix - Interactive 3D Graphics**

Here is my code:. Homographies are 3x3 matrices and points are just pairs, 2x1so there's no way to map these together. Instead, homogeneous coordinates are used, giving 3x1 vectors to multiply.

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However, homogeneous points can be scaled while representing the same point; that is, in homogeneous coordinates, kx, ky, k is the same point as x, y, 1. From the Wikipedia page on homogeneous coordinates :. Given a point x, y on the Euclidean plane, for any non-zero real number Zthe triple xZ, yZ, Z is called a set of homogeneous coordinates for the point.

By this definition, multiplying the three homogeneous coordinates by a common, non-zero factor gives a new set of homogeneous coordinates for the same point.

In particular, x, y, 1 is such a system of homogeneous coordinates for the point x, y. For example, the Cartesian point 1, 2 can be represented in homogeneous coordinates as 1, 2, 1 or 2, 4, 2. The original Cartesian coordinates are recovered by dividing the first two positions by the third.

## Programmer’s guide to homogeneous coordinates

Thus unlike Cartesian coordinates, a single point can be represented by infinitely many homogeneous coordinates. So we need a way to map these homogeneous coordinates, which can be represented an infinite number of ways, down to Cartesian coordinates, which can only be represented one way. Luckily this is very easy, just scale the homogeneous coordinates so the last number in the triple is 1. Homographies multiply homogeneous coordinates and return homogeneous coordinates.

So in order to map them back to Cartesian world, you just need to divide by the last coordinate to scale them and then rip the first two numbers out. Use the built-in OpenCV function convertPointsFromHomogeneous to convert your points from homogeneous 3-vectors to Cartesian 2-vectors. Stack Overflow for Teams — Collaborate and share knowledge with a private group.

Create a free Team What is Teams? Learn more. Opencv homography to find global xy coordinates from pixel xy coordinates Ask Question. Asked 3 years, 9 months ago. Active 2 years, 8 months ago. Viewed 6k times. Here is my code: import cv2 import numpy as np from numpy. Am i doing something wrong somewhere? Improve this question. Vivek Annem Vivek Annem 63 1 1 silver badge 3 3 bronze badges. What are these supposed to represent? Anyways, where you compute the "real world" x,y coordinates, you will get homogenous points which are equivalent when they're scaledin other words, they may be scaled and would still be considered the same point.

But you need x,y points, not scaled ones, so you need to divide by the scale. The three-vector is all scaled by the same amount, so you can use the last entry for the scaling factor. I am sorry. I didnt quite get the scaling part.

I think I understand what you meant. Let me check and see if I am getting the right values. Try out the code real quick and see if it solves your problemif it does I'll add a more elaborate answer. But you need the homogenous points to end with a 1 to be equivalent to x,y coordinates, so you need to divide by s from the homogenous coordinates.Why would you care about some homogeneous coordinates, whatever they are?

Well, if you work with geometry: 3D-graphics, image processing, physical simulation, the answer is obvious. Knowing the mathematics behind your framework of choice lets you write more efficient code.

I once had to speed up one. NET application that bends pictures and simply re-implementing some primitive operations the right way gave me times performance boost.

This is the one rare example of mathematical magic when a small complication benefits in enormous simplification. One little obscurity pays off in terms of unification and homogenization.

I think, learning this particular piece of geometry is a valuable experience in its own right. And you know how it works. More experience, higher level, better loot. So if you do work with 3D graphics, you might notice that it is quite common to write 3D points as a tuple of 4 numbers.

And that is not the worst answer ever, because setting it to 1 indeed makes everything work like in good old Cartesian coordinates. But the fourth coordinate is much more interesting than that. Cartesian coordinates are just the first 3 numbers of homogeneous coordinates divided by the fourth.

So if it is 1then homogeneous coordinates is basically the same thing as Cartesian. But the smaller it gets, the further the point in Cartesian coordinates travels from the null.

What if the fourth coordinate is 0? Intuition tells, that it should be further from the 0 than every other point. Every other point in Euclidean space that is. Homogeneous coordinates indeed denote points not only in Euclidean or, more general, affine space, but in projective space that includes and expands affine one.

From the pragmatic point of view, this lets us compose 3D-scene in a manner that every object that can be reached would fit in affine space with the coordinates x, y, z, 1and all the objects that can never be reached will belong to projective extension x, y, z, 0. In this regard point on that extension in a way set a general direction and not a specific point in Euclidean space.

A ray that starts at null and has no length, has no ending, only the direction. Sometimes we can, but this might not always work as expected.

### Homogeneous Coordinates

The thing is, floating point numbers usually used to store geometry are not that large as they seem. Yes, they can technically store numbers from 0.

Numbers with different exponent loose precision on every operation, and numbers with the same exponent only have 23 meaningful binary digits. This roughly denotes a range from 0 to If you want to have millimeter-sized details in every part of your scene, this means your scene should not be larger than 8 kilometers.

Good enough for a third-person shooter, but not for a space simulator. Using projective space gives you more options. In fact, we are only starting to get into the benefits.

There are two kinds of projection in Euclidean space: central and parallel. In projective space they are the same. You see, in affine space you can set a center for a central projection very-very far away from the scene you want to render. This will make disproportion very small. Just set the central projection to x, y, z, 0 and this will automatically turn it into parallel.

I remember on my first year in college we were studying quadric surfaces and one of the exercises allegedly made up to help us learn their classification was to make an album.The x axis for the camera frame runs to the right from the origin along the top of the field of view. The y axis for the camera frame runs downward from the origin along the left side of the field of view.

This is cool, but what I really want to know are the coordinates of the quarter relative to the base frame of my two degree of freedom robotic arm. If I know the coordinates of the object relative to the base frame of my two degree of freedom robotic arm, I can then use inverse kinematics to command the robot to move the end effector to that location.

A tool that can help us do that is known as the homogeneous transformation matrix. We need to find the rotation matrix portion i. We also need to find the displacement vector portion the rightmost column. We first need to look at how we can rotate the base frame to match up with the camera frame of the robotic arm.

Which way is z 0 pointing? Using the right hand rule, take your four fingers and point them towards the x 0 axis. Your palm faces towards y 0. Your thumb points towards z 0which is upwards out of the page or upwards out of the dry erase board. Which way is z c pointing? Using the right hand rule, take your four fingers and point them towards the x c axis.

Your palm faces towards y c. Your thumb points towards z cwhich is downwards into the page or downward into the dry erase board. Now, stick your thumb in the direction of x 0. Your thumb points is the axis of rotation, while your four fingers indicate the direction of positive rotation.

The standard rotation matrix for rotation around the x-axis is…. Now that we have the rotation matrix portion of the homogeneous transformation matrix, we need to calculate the displacement vector from the origin of frame 0 to the origin of frame c.

Grab your ruler and measure the distance from the origin of frame 0 to the origin of frame c along the x 0 axis. Grab your ruler and measure the distance from the origin of frame 0 to the origin of frame c along the y 0 axis.

Both reference frames are in the same z plane, so there is no displacement along z 0 from frame 0 to frame c. We can now convert a point in the camera reference frame to a point in the base frame of the robotic arm. I remeasured the width of the field of view in centimeters and calculated it as Then place a quarter on the dry erase board. You should see that the x and y coordinates of the quarter relative to the base frame are printed to the screen.

The text on the screen should now display the coordinates of the quarter in the base frame coordinates. The first time I ran this code, my x coordinate was accurate, but my y coordinate was not. In the end, my displacement value was Thus, here was my final equation for converting camera coordinates to robotic base frame coordinates:.

Up until now, we have assumed that the camera lens is always parallel to the underlying surface like this. You can imagine that in the real world, this situation is quite common. Consider this humanoid robot for example. This robot might have some sort of camera inside his helmet.

Any surface that he looks at that is in front of his body will likely be from some angle. Also, in real-world use cases, a physical grid is usually not present. Instead, the structured light technique is used to project a known grid pattern on a surface in order to help the robot determine the coordinates of an object.Blender Stack Exchange is a question and answer site for people who use Blender to create 3D graphics, animations, or games.

It only takes a minute to sign up. Connect and share knowledge within a single location that is structured and easy to search. I get a translation of the mesh by 1 in object-Z. So somewherevert. I'm hoping for a basic perspective transform, in which the w of every coordinate is set to its zand then, in normalization, the whole coordinate is divided through by its w before using its x,y,z.

But, OK, nothing happens. Is there any way of getting at the implicit 'w' of the homogeneous vector, or do I have to divide it by hand? From Perspective projection of transformation matrix wiki. We can express this in homogeneous coordinates as:.

However others will fail with zero divide error. One property of homogeneous coordinates is that they allow you to have points at infinity infinite length vectorswhich is not possible with 3D coordinates.

What use does this have? Well, directional lights can be though of as point lights that are infinitely far away. When a point light is infinitely far away the rays of light become parallel, and all of the light travels in a single direction.

This is basically the definition of a directional light. Playing around with meshes using above Not sure how useful the answer is.

In first example in question above, worth noting m. Converting 3d vector to 4d sets w to 1. Which I assume is done for us since it doesn't spit the dummy multiplying a 4x4 with a 3x1. Sign up to join this community. The best answers are voted up and rise to the top.

Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. Can you work with homogeneous coordinates in the Python API?Join Stack Overflow to learn, share knowledge, and build your career. Connect and share knowledge within a single location that is structured and easy to search. One approach making use of broadcasted elementwise divisions.

We can use a[:,-1,None] or a[:,-1][:,None] or a[:,-1]. With a[:,[-1]]we are keeping the number of dims intact, letting us perform the broadcasting divisions. Another with np. Stack Overflow for Teams — Collaborate and share knowledge with a private group.

Create a free Team What is Teams? Learn more. Usefull way of reverting Homogeneous coordinates back to 2d? Ask Question. Asked 3 years, 10 months ago. Active 1 year, 3 months ago. Viewed 4k times.

Is there some numpy sugar for reverting Homogeneous coordinates back to 2d coordinates. So this: [[4,8,2], 6,3,2]] becomes this: [[2,4], [3,1. Jan Klaas Jan Klaas 6 6 silver badges 16 16 bronze badges. What are Homogeneous coordinates? Is there a non-sugary way to implement it? I can put it different. I want to divide the first two elements of the array by the last element. And the output also should be 2 elements and 3 elements as the example : — Jan Klaas May 11 '17 at Did either of the posted solutions work for you?

Add a comment. Active Oldest Votes. Divakar Divakar k 15 15 gold badges silver badges bronze badges. Edited it to include an example for multiple coordinates.

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