Mixing
How to visualize a 2D cartesian point in homogeneous coordinates?
$begingroup$
Geometric primitives
Hi, I am trying to understand below text from a book on Computer Vision by Szeliski.
I am not able to visualize it combining cartesian and homogeneous coordinates. Below is my attempt in understanding visually. Kindly correct and confirm.
Further, if correct, what are A and B as shown in circle? Afaik,
$$
B = sqrt{(tilde x)^2 + (tilde y)^2 + (tilde w)^2} = 1?
$$
because, w=1 there?
linear-algebra linear-transformations homogeneous-spaces
asked Jan 25 at 18:00
Parthiban RajendranParthiban Rajendran
21027
$endgroup$
add a comment |
$begingroup$
Geometric primitives
Hi, I am trying to understand below text from a book on Computer Vision by Szeliski.
I am not able to visualize it combining cartesian and homogeneous coordinates. Below is my attempt in understanding visually. Kindly correct and confirm.
Further, if correct, what are A and B as shown in circle? Afaik,
$$
B = sqrt{(tilde x)^2 + (tilde y)^2 + (tilde w)^2} = 1?
$$
because, w=1 there?
linear-algebra linear-transformations homogeneous-spaces
asked Jan 25 at 18:00
Parthiban RajendranParthiban Rajendran
21027
$endgroup$
add a comment |
$begingroup$
Geometric primitives
Hi, I am trying to understand below text from a book on Computer Vision by Szeliski.
I am not able to visualize it combining cartesian and homogeneous coordinates. Below is my attempt in understanding visually. Kindly correct and confirm.
Further, if correct, what are A and B as shown in circle? Afaik,
$$
B = sqrt{(tilde x)^2 + (tilde y)^2 + (tilde w)^2} = 1?
$$
because, w=1 there?
linear-algebra linear-transformations homogeneous-spaces
asked Jan 25 at 18:00
Parthiban RajendranParthiban Rajendran
21027
$endgroup$
Geometric primitives
Hi, I am trying to understand below text from a book on Computer Vision by Szeliski.
I am not able to visualize it combining cartesian and homogeneous coordinates. Below is my attempt in understanding visually. Kindly correct and confirm.
Further, if correct, what are A and B as shown in circle? Afaik,
$$
B = sqrt{(tilde x)^2 + (tilde y)^2 + (tilde w)^2} = 1?
$$
because, w=1 there?
linear-algebra linear-transformations homogeneous-spaces
linear-algebra linear-transformations homogeneous-spaces
asked Jan 25 at 18:00
Parthiban RajendranParthiban Rajendran
21027
asked Jan 25 at 18:00
Parthiban RajendranParthiban Rajendran
21027
asked Jan 25 at 18:00
Parthiban RajendranParthiban Rajendran
21027
asked Jan 25 at 18:00
Parthiban RajendranParthiban Rajendran
21027
asked Jan 25 at 18:00
Parthiban RajendranParthiban Rajendran
21027
21027
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One way to look at it is that each point in the projective plane corresponds to a line through the origin of the three-dimensional Cartesian space,
and each such line corresponds to a point in the projective plane.
If you set $B = sqrt{tilde x^2 + tilde y^2 + tilde w^2} = 1$
then you will only reach points on the unit sphere around the origin of the three-dimensional Cartesian space. Every line through the origin intersects that sphere,
so you'll capture all the lines (and all the projective points) that way,
but except for $(0,0,1),$ none of the points on that sphere is actually on the plane $tilde w = 1.$
If you want to plot $(x,y)$ at $(tilde x, tilde y, 1)$ as in the figure,
then at each such point you have $A = tilde w = 1$
and $B = sqrt{tilde x^2 + tilde y^2 + tilde w^2} =
sqrt{x^2 + y^2 + 1}.$
I'm not sure what to do with those facts, however.
If you then take any point $(tilde x, tilde y, tilde w)$
on any of the lines through the origin in the three-dimensional space,
assuming you don't pick the origin $(0,0,0)$ itself, then there are just two cases to think about:
Case $tilde w = 0$: the line is in the $tilde x, tilde y$ plane and does not intersect the plane $tilde w$ at all.
So this line corresponds to one of the "points at infinity" that you can adjoin to the ordinary Cartesian plane in order to make it a projective plane.
Case $tilde w neq 0$: the line intersects the plane $tilde w$ at the point
$$(x,y,1) = left(frac{tilde x}{tilde w}, frac{tilde x}{tilde w}, frac{tilde w}{tilde w}right).$$
But something to remember is that if you're using a line through the origin to represent a projective point, any point on that line (other than the origin itself) is as good as any other point on that line.
answered Jan 25 at 19:53
David KDavid K
55.2k344120
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3087413%2fhow-to-visualize-a-2d-cartesian-point-in-homogeneous-coordinates%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One way to look at it is that each point in the projective plane corresponds to a line through the origin of the three-dimensional Cartesian space,
and each such line corresponds to a point in the projective plane.
If you set $B = sqrt{tilde x^2 + tilde y^2 + tilde w^2} = 1$
then you will only reach points on the unit sphere around the origin of the three-dimensional Cartesian space. Every line through the origin intersects that sphere,
so you'll capture all the lines (and all the projective points) that way,
but except for $(0,0,1),$ none of the points on that sphere is actually on the plane $tilde w = 1.$
If you want to plot $(x,y)$ at $(tilde x, tilde y, 1)$ as in the figure,
then at each such point you have $A = tilde w = 1$
and $B = sqrt{tilde x^2 + tilde y^2 + tilde w^2} =
sqrt{x^2 + y^2 + 1}.$
I'm not sure what to do with those facts, however.
If you then take any point $(tilde x, tilde y, tilde w)$
on any of the lines through the origin in the three-dimensional space,
assuming you don't pick the origin $(0,0,0)$ itself, then there are just two cases to think about:
Case $tilde w = 0$: the line is in the $tilde x, tilde y$ plane and does not intersect the plane $tilde w$ at all.
So this line corresponds to one of the "points at infinity" that you can adjoin to the ordinary Cartesian plane in order to make it a projective plane.
Case $tilde w neq 0$: the line intersects the plane $tilde w$ at the point
$$(x,y,1) = left(frac{tilde x}{tilde w}, frac{tilde x}{tilde w}, frac{tilde w}{tilde w}right).$$
But something to remember is that if you're using a line through the origin to represent a projective point, any point on that line (other than the origin itself) is as good as any other point on that line.
answered Jan 25 at 19:53
David KDavid K
55.2k344120
$endgroup$
add a comment |
$begingroup$
One way to look at it is that each point in the projective plane corresponds to a line through the origin of the three-dimensional Cartesian space,
and each such line corresponds to a point in the projective plane.
If you set $B = sqrt{tilde x^2 + tilde y^2 + tilde w^2} = 1$
then you will only reach points on the unit sphere around the origin of the three-dimensional Cartesian space. Every line through the origin intersects that sphere,
so you'll capture all the lines (and all the projective points) that way,
but except for $(0,0,1),$ none of the points on that sphere is actually on the plane $tilde w = 1.$
If you want to plot $(x,y)$ at $(tilde x, tilde y, 1)$ as in the figure,
then at each such point you have $A = tilde w = 1$
and $B = sqrt{tilde x^2 + tilde y^2 + tilde w^2} =
sqrt{x^2 + y^2 + 1}.$
I'm not sure what to do with those facts, however.
If you then take any point $(tilde x, tilde y, tilde w)$
on any of the lines through the origin in the three-dimensional space,
assuming you don't pick the origin $(0,0,0)$ itself, then there are just two cases to think about:
Case $tilde w = 0$: the line is in the $tilde x, tilde y$ plane and does not intersect the plane $tilde w$ at all.
So this line corresponds to one of the "points at infinity" that you can adjoin to the ordinary Cartesian plane in order to make it a projective plane.
Case $tilde w neq 0$: the line intersects the plane $tilde w$ at the point
$$(x,y,1) = left(frac{tilde x}{tilde w}, frac{tilde x}{tilde w}, frac{tilde w}{tilde w}right).$$
But something to remember is that if you're using a line through the origin to represent a projective point, any point on that line (other than the origin itself) is as good as any other point on that line.
answered Jan 25 at 19:53
David KDavid K
55.2k344120
$endgroup$
add a comment |
$begingroup$
One way to look at it is that each point in the projective plane corresponds to a line through the origin of the three-dimensional Cartesian space,
and each such line corresponds to a point in the projective plane.
If you set $B = sqrt{tilde x^2 + tilde y^2 + tilde w^2} = 1$
then you will only reach points on the unit sphere around the origin of the three-dimensional Cartesian space. Every line through the origin intersects that sphere,
so you'll capture all the lines (and all the projective points) that way,
but except for $(0,0,1),$ none of the points on that sphere is actually on the plane $tilde w = 1.$
If you want to plot $(x,y)$ at $(tilde x, tilde y, 1)$ as in the figure,
then at each such point you have $A = tilde w = 1$
and $B = sqrt{tilde x^2 + tilde y^2 + tilde w^2} =
sqrt{x^2 + y^2 + 1}.$
I'm not sure what to do with those facts, however.
If you then take any point $(tilde x, tilde y, tilde w)$
on any of the lines through the origin in the three-dimensional space,
assuming you don't pick the origin $(0,0,0)$ itself, then there are just two cases to think about:
Case $tilde w = 0$: the line is in the $tilde x, tilde y$ plane and does not intersect the plane $tilde w$ at all.
So this line corresponds to one of the "points at infinity" that you can adjoin to the ordinary Cartesian plane in order to make it a projective plane.
Case $tilde w neq 0$: the line intersects the plane $tilde w$ at the point
$$(x,y,1) = left(frac{tilde x}{tilde w}, frac{tilde x}{tilde w}, frac{tilde w}{tilde w}right).$$
But something to remember is that if you're using a line through the origin to represent a projective point, any point on that line (other than the origin itself) is as good as any other point on that line.
answered Jan 25 at 19:53
David KDavid K
55.2k344120
$endgroup$
One way to look at it is that each point in the projective plane corresponds to a line through the origin of the three-dimensional Cartesian space,
and each such line corresponds to a point in the projective plane.
If you set $B = sqrt{tilde x^2 + tilde y^2 + tilde w^2} = 1$
then you will only reach points on the unit sphere around the origin of the three-dimensional Cartesian space. Every line through the origin intersects that sphere,
so you'll capture all the lines (and all the projective points) that way,
but except for $(0,0,1),$ none of the points on that sphere is actually on the plane $tilde w = 1.$
If you want to plot $(x,y)$ at $(tilde x, tilde y, 1)$ as in the figure,
then at each such point you have $A = tilde w = 1$
and $B = sqrt{tilde x^2 + tilde y^2 + tilde w^2} =
sqrt{x^2 + y^2 + 1}.$
I'm not sure what to do with those facts, however.
If you then take any point $(tilde x, tilde y, tilde w)$
on any of the lines through the origin in the three-dimensional space,
assuming you don't pick the origin $(0,0,0)$ itself, then there are just two cases to think about:
Case $tilde w = 0$: the line is in the $tilde x, tilde y$ plane and does not intersect the plane $tilde w$ at all.
So this line corresponds to one of the "points at infinity" that you can adjoin to the ordinary Cartesian plane in order to make it a projective plane.
Case $tilde w neq 0$: the line intersects the plane $tilde w$ at the point
$$(x,y,1) = left(frac{tilde x}{tilde w}, frac{tilde x}{tilde w}, frac{tilde w}{tilde w}right).$$
But something to remember is that if you're using a line through the origin to represent a projective point, any point on that line (other than the origin itself) is as good as any other point on that line.
answered Jan 25 at 19:53
David KDavid K
55.2k344120
answered Jan 25 at 19:53
David KDavid K
55.2k344120
answered Jan 25 at 19:53
David KDavid K
55.2k344120
answered Jan 25 at 19:53
David KDavid K
55.2k344120
55.2k344120
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3087413%2fhow-to-visualize-a-2d-cartesian-point-in-homogeneous-coordinates%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
