Reverse coordinate mapping of object or a point
$begingroup$
Mapping Problem diagram/description
The screen size and coordinates (a,b,c,d) are known, and the Area of Interest (AOI) size and coordinates (x,y,w,z) are known.
If an object/coordinate/pixel, xx(1,1) is detected in AOI, how do I find or map the coordinates to xx(2,3)?
Note The AOI size/coordinates, and the location of the object/coordinate/pixel/point, even if changed, should result in the correct mapping of the detected object on the screen.
geometry
asked Dec 2 '18 at 11:59
Kshitij DhobleKshitij Dhoble
32
$endgroup$
add a comment |
$begingroup$
Mapping Problem diagram/description
The screen size and coordinates (a,b,c,d) are known, and the Area of Interest (AOI) size and coordinates (x,y,w,z) are known.
If an object/coordinate/pixel, xx(1,1) is detected in AOI, how do I find or map the coordinates to xx(2,3)?
Note The AOI size/coordinates, and the location of the object/coordinate/pixel/point, even if changed, should result in the correct mapping of the detected object on the screen.
geometry
asked Dec 2 '18 at 11:59
Kshitij DhobleKshitij Dhoble
32
$endgroup$
add a comment |
$begingroup$
Mapping Problem diagram/description
The screen size and coordinates (a,b,c,d) are known, and the Area of Interest (AOI) size and coordinates (x,y,w,z) are known.
If an object/coordinate/pixel, xx(1,1) is detected in AOI, how do I find or map the coordinates to xx(2,3)?
Note The AOI size/coordinates, and the location of the object/coordinate/pixel/point, even if changed, should result in the correct mapping of the detected object on the screen.
geometry
asked Dec 2 '18 at 11:59
Kshitij DhobleKshitij Dhoble
32
$endgroup$
Mapping Problem diagram/description
The screen size and coordinates (a,b,c,d) are known, and the Area of Interest (AOI) size and coordinates (x,y,w,z) are known.
If an object/coordinate/pixel, xx(1,1) is detected in AOI, how do I find or map the coordinates to xx(2,3)?
Note The AOI size/coordinates, and the location of the object/coordinate/pixel/point, even if changed, should result in the correct mapping of the detected object on the screen.
geometry
geometry
asked Dec 2 '18 at 11:59
Kshitij DhobleKshitij Dhoble
32
asked Dec 2 '18 at 11:59
Kshitij DhobleKshitij Dhoble
32
edited Dec 2 '18 at 12:24
Kshitij Dhoble
asked Dec 2 '18 at 11:59
Kshitij DhobleKshitij Dhoble
32
asked Dec 2 '18 at 11:59
Kshitij DhobleKshitij Dhoble
32
asked Dec 2 '18 at 11:59
Kshitij DhobleKshitij Dhoble
32
32
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If the units are the same in both axis systems, the transformation equations is just that of a translation
$$(x_s,y_s)=(x_a,y_a)+(1,2)$$ (where $(1,2)$ are the coordinates of the top-left corner of the AOI wrt the screen).
Reciprocally,
$$(x_a,y_a)=(x_s,y_s)-(1,2).$$
answered Dec 2 '18 at 14:49
Yves DaoustYves Daoust
124k671223
$endgroup$
add a comment |
$begingroup$
Combine a polar coordinate system with the line slope formula and the line distance formula.
Or use a land surveying system:
Any y-coordinate that is South of zero is negative.
Any x-coordinate that is West of zero is negative.
Point A is (y1, x1) .
Point B is (y2, x2) .
The direction of A to B is:
InvTan((x2 - x1)/(y2 - y1)) .
If (x2 - x1) is positive that is East else West.
If (y2 - y1) is positive that is North else South.
This procedure allows a quadrant direction to be determined between any two points.
The distance of A to B is:
SquareRoot of ((x2 - x1)^2 + (y2 - y1)^2) .
The combination of direction and distance is a vector but here as an inverse. The coordinates of a point is the location of the point.
Forwarding a point (or setting a point) from A to B is:
y2 = y1 + (Cos(Direction) * Distance) .
x2 = x1 + (Sin(Direction) * Distance) .
A North direction is a positive value added to y1 else negative.
An East direction is a positive value added to x1 else negative.
answered Dec 2 '18 at 13:43
S SpringS Spring
1343
$endgroup$
1
$begingroup$
There is absolutely no reason to use polar coordinates. This adds tons of confusion and complexity.
$endgroup$
– Yves Daoust
Dec 2 '18 at 14:45
add a comment |
$begingroup$
If the area-of-interest is a picture-in-picture feature then it is both scaled and translated.
Scaling multiplies all coordinates and both x and y coordinates by the same scalar. Scaling just makes the figure bigger or smaller but proportionally the same.
Translating adds the same x translation value to all x coordinates and adds the same y translation value to all y coordinates. Translating just moves the figure without changing its size.
Then the inverse operations are obvious.
answered Dec 2 '18 at 15:47
S SpringS Spring
1343
$endgroup$
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%2f3022554%2freverse-coordinate-mapping-of-object-or-a-point%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If the units are the same in both axis systems, the transformation equations is just that of a translation
$$(x_s,y_s)=(x_a,y_a)+(1,2)$$ (where $(1,2)$ are the coordinates of the top-left corner of the AOI wrt the screen).
Reciprocally,
$$(x_a,y_a)=(x_s,y_s)-(1,2).$$
answered Dec 2 '18 at 14:49
Yves DaoustYves Daoust
124k671223
$endgroup$
add a comment |
$begingroup$
If the units are the same in both axis systems, the transformation equations is just that of a translation
$$(x_s,y_s)=(x_a,y_a)+(1,2)$$ (where $(1,2)$ are the coordinates of the top-left corner of the AOI wrt the screen).
Reciprocally,
$$(x_a,y_a)=(x_s,y_s)-(1,2).$$
answered Dec 2 '18 at 14:49
Yves DaoustYves Daoust
124k671223
$endgroup$
add a comment |
$begingroup$
If the units are the same in both axis systems, the transformation equations is just that of a translation
$$(x_s,y_s)=(x_a,y_a)+(1,2)$$ (where $(1,2)$ are the coordinates of the top-left corner of the AOI wrt the screen).
Reciprocally,
$$(x_a,y_a)=(x_s,y_s)-(1,2).$$
answered Dec 2 '18 at 14:49
Yves DaoustYves Daoust
124k671223
$endgroup$
If the units are the same in both axis systems, the transformation equations is just that of a translation
$$(x_s,y_s)=(x_a,y_a)+(1,2)$$ (where $(1,2)$ are the coordinates of the top-left corner of the AOI wrt the screen).
Reciprocally,
$$(x_a,y_a)=(x_s,y_s)-(1,2).$$
answered Dec 2 '18 at 14:49
Yves DaoustYves Daoust
124k671223
answered Dec 2 '18 at 14:49
Yves DaoustYves Daoust
124k671223
answered Dec 2 '18 at 14:49
Yves DaoustYves Daoust
124k671223
answered Dec 2 '18 at 14:49
Yves DaoustYves Daoust
124k671223
124k671223
add a comment |
add a comment |
$begingroup$
Combine a polar coordinate system with the line slope formula and the line distance formula.
Or use a land surveying system:
Any y-coordinate that is South of zero is negative.
Any x-coordinate that is West of zero is negative.
Point A is (y1, x1) .
Point B is (y2, x2) .
The direction of A to B is:
InvTan((x2 - x1)/(y2 - y1)) .
If (x2 - x1) is positive that is East else West.
If (y2 - y1) is positive that is North else South.
This procedure allows a quadrant direction to be determined between any two points.
The distance of A to B is:
SquareRoot of ((x2 - x1)^2 + (y2 - y1)^2) .
The combination of direction and distance is a vector but here as an inverse. The coordinates of a point is the location of the point.
Forwarding a point (or setting a point) from A to B is:
y2 = y1 + (Cos(Direction) * Distance) .
x2 = x1 + (Sin(Direction) * Distance) .
A North direction is a positive value added to y1 else negative.
An East direction is a positive value added to x1 else negative.
answered Dec 2 '18 at 13:43
S SpringS Spring
1343
$endgroup$
1
$begingroup$
There is absolutely no reason to use polar coordinates. This adds tons of confusion and complexity.
$endgroup$
– Yves Daoust
Dec 2 '18 at 14:45
add a comment |
$begingroup$
Combine a polar coordinate system with the line slope formula and the line distance formula.
Or use a land surveying system:
Any y-coordinate that is South of zero is negative.
Any x-coordinate that is West of zero is negative.
Point A is (y1, x1) .
Point B is (y2, x2) .
The direction of A to B is:
InvTan((x2 - x1)/(y2 - y1)) .
If (x2 - x1) is positive that is East else West.
If (y2 - y1) is positive that is North else South.
This procedure allows a quadrant direction to be determined between any two points.
The distance of A to B is:
SquareRoot of ((x2 - x1)^2 + (y2 - y1)^2) .
The combination of direction and distance is a vector but here as an inverse. The coordinates of a point is the location of the point.
Forwarding a point (or setting a point) from A to B is:
y2 = y1 + (Cos(Direction) * Distance) .
x2 = x1 + (Sin(Direction) * Distance) .
A North direction is a positive value added to y1 else negative.
An East direction is a positive value added to x1 else negative.
answered Dec 2 '18 at 13:43
S SpringS Spring
1343
$endgroup$
1
$begingroup$
There is absolutely no reason to use polar coordinates. This adds tons of confusion and complexity.
$endgroup$
– Yves Daoust
Dec 2 '18 at 14:45
add a comment |
$begingroup$
Combine a polar coordinate system with the line slope formula and the line distance formula.
Or use a land surveying system:
Any y-coordinate that is South of zero is negative.
Any x-coordinate that is West of zero is negative.
Point A is (y1, x1) .
Point B is (y2, x2) .
The direction of A to B is:
InvTan((x2 - x1)/(y2 - y1)) .
If (x2 - x1) is positive that is East else West.
If (y2 - y1) is positive that is North else South.
This procedure allows a quadrant direction to be determined between any two points.
The distance of A to B is:
SquareRoot of ((x2 - x1)^2 + (y2 - y1)^2) .
The combination of direction and distance is a vector but here as an inverse. The coordinates of a point is the location of the point.
Forwarding a point (or setting a point) from A to B is:
y2 = y1 + (Cos(Direction) * Distance) .
x2 = x1 + (Sin(Direction) * Distance) .
A North direction is a positive value added to y1 else negative.
An East direction is a positive value added to x1 else negative.
answered Dec 2 '18 at 13:43
S SpringS Spring
1343
$endgroup$
Combine a polar coordinate system with the line slope formula and the line distance formula.
Or use a land surveying system:
Any y-coordinate that is South of zero is negative.
Any x-coordinate that is West of zero is negative.
Point A is (y1, x1) .
Point B is (y2, x2) .
The direction of A to B is:
InvTan((x2 - x1)/(y2 - y1)) .
If (x2 - x1) is positive that is East else West.
If (y2 - y1) is positive that is North else South.
This procedure allows a quadrant direction to be determined between any two points.
The distance of A to B is:
SquareRoot of ((x2 - x1)^2 + (y2 - y1)^2) .
The combination of direction and distance is a vector but here as an inverse. The coordinates of a point is the location of the point.
Forwarding a point (or setting a point) from A to B is:
y2 = y1 + (Cos(Direction) * Distance) .
x2 = x1 + (Sin(Direction) * Distance) .
A North direction is a positive value added to y1 else negative.
An East direction is a positive value added to x1 else negative.
answered Dec 2 '18 at 13:43
S SpringS Spring
1343
edited Dec 2 '18 at 13:49
answered Dec 2 '18 at 13:43
S SpringS Spring
1343
answered Dec 2 '18 at 13:43
S SpringS Spring
1343
answered Dec 2 '18 at 13:43
S SpringS Spring
1343
1343
1
$begingroup$
There is absolutely no reason to use polar coordinates. This adds tons of confusion and complexity.
$endgroup$
– Yves Daoust
Dec 2 '18 at 14:45
add a comment |
1
$begingroup$
There is absolutely no reason to use polar coordinates. This adds tons of confusion and complexity.
$endgroup$
– Yves Daoust
Dec 2 '18 at 14:45
1
1
$begingroup$
There is absolutely no reason to use polar coordinates. This adds tons of confusion and complexity.
$endgroup$
– Yves Daoust
Dec 2 '18 at 14:45
$begingroup$
There is absolutely no reason to use polar coordinates. This adds tons of confusion and complexity.
$endgroup$
– Yves Daoust
Dec 2 '18 at 14:45
add a comment |
$begingroup$
If the area-of-interest is a picture-in-picture feature then it is both scaled and translated.
Scaling multiplies all coordinates and both x and y coordinates by the same scalar. Scaling just makes the figure bigger or smaller but proportionally the same.
Translating adds the same x translation value to all x coordinates and adds the same y translation value to all y coordinates. Translating just moves the figure without changing its size.
Then the inverse operations are obvious.
answered Dec 2 '18 at 15:47
S SpringS Spring
1343
$endgroup$
add a comment |
$begingroup$
If the area-of-interest is a picture-in-picture feature then it is both scaled and translated.
Scaling multiplies all coordinates and both x and y coordinates by the same scalar. Scaling just makes the figure bigger or smaller but proportionally the same.
Translating adds the same x translation value to all x coordinates and adds the same y translation value to all y coordinates. Translating just moves the figure without changing its size.
Then the inverse operations are obvious.
answered Dec 2 '18 at 15:47
S SpringS Spring
1343
$endgroup$
add a comment |
$begingroup$
If the area-of-interest is a picture-in-picture feature then it is both scaled and translated.
Scaling multiplies all coordinates and both x and y coordinates by the same scalar. Scaling just makes the figure bigger or smaller but proportionally the same.
Translating adds the same x translation value to all x coordinates and adds the same y translation value to all y coordinates. Translating just moves the figure without changing its size.
Then the inverse operations are obvious.
answered Dec 2 '18 at 15:47
S SpringS Spring
1343
$endgroup$
If the area-of-interest is a picture-in-picture feature then it is both scaled and translated.
Scaling multiplies all coordinates and both x and y coordinates by the same scalar. Scaling just makes the figure bigger or smaller but proportionally the same.
Translating adds the same x translation value to all x coordinates and adds the same y translation value to all y coordinates. Translating just moves the figure without changing its size.
Then the inverse operations are obvious.
answered Dec 2 '18 at 15:47
S SpringS Spring
1343
answered Dec 2 '18 at 15:47
S SpringS Spring
1343
answered Dec 2 '18 at 15:47
S SpringS Spring
1343
answered Dec 2 '18 at 15:47
S SpringS Spring
1343
1343
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%2f3022554%2freverse-coordinate-mapping-of-object-or-a-point%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
