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How to find corners of square from it's center point?
The 2019 Stack Overflow Developer Survey Results Are InFind the position of a beacon using lat/lng/signal strength datapoint lying inside a squarePoint P(x,y) inside a square, differences of distances to corners knownDetermine Center Point based on 2 separate elipsesCalculating top-left point from center point and mouse pointer locationGiven a point on a sphere, how do I find the angles needed to 'point' at it's center?Find distance from point to line using trigonometryHow to compute the intercept of a square with a line given by two points?Find all points within the locus of a line segmentGiven the coordinates of a box, how can I find the coordinates of a box that is X% bigger than the original box
$begingroup$
It's has been a while since I used any basic trigonometry, so I'm here requesting some aiding. I have the center point of a square (as lng/lat pair which I believe I can use as x/y) and the side length, and I need to find all 4 corners coordinates from this origin.
How can I achieve this please?
geometry trigonometry
asked Oct 21 '15 at 2:55
dambrosdambros
11814
$endgroup$
add a comment |
$begingroup$
It's has been a while since I used any basic trigonometry, so I'm here requesting some aiding. I have the center point of a square (as lng/lat pair which I believe I can use as x/y) and the side length, and I need to find all 4 corners coordinates from this origin.
How can I achieve this please?
geometry trigonometry
asked Oct 21 '15 at 2:55
dambrosdambros
11814
$endgroup$
$begingroup$
Does the square require an orientation? The simplest answer is simply the four points (+/- 1/2 length +/- x, +/- 1/2 length +/- y)
$endgroup$
– fleablood
Oct 21 '15 at 3:11
$begingroup$
It doesn't require any orientation. It should be as simple as possible. Could you please elaborate you simplest answer? I am not that good with math...
$endgroup$
– dambros
Oct 21 '15 at 3:14
add a comment |
$begingroup$
It's has been a while since I used any basic trigonometry, so I'm here requesting some aiding. I have the center point of a square (as lng/lat pair which I believe I can use as x/y) and the side length, and I need to find all 4 corners coordinates from this origin.
How can I achieve this please?
geometry trigonometry
asked Oct 21 '15 at 2:55
dambrosdambros
11814
$endgroup$
It's has been a while since I used any basic trigonometry, so I'm here requesting some aiding. I have the center point of a square (as lng/lat pair which I believe I can use as x/y) and the side length, and I need to find all 4 corners coordinates from this origin.
How can I achieve this please?
geometry trigonometry
geometry trigonometry
asked Oct 21 '15 at 2:55
dambrosdambros
11814
asked Oct 21 '15 at 2:55
dambrosdambros
11814
edited Oct 21 '15 at 3:04
dambros
asked Oct 21 '15 at 2:55
dambrosdambros
11814
asked Oct 21 '15 at 2:55
dambrosdambros
11814
asked Oct 21 '15 at 2:55
dambrosdambros
11814
11814
$begingroup$
Does the square require an orientation? The simplest answer is simply the four points (+/- 1/2 length +/- x, +/- 1/2 length +/- y)
$endgroup$
– fleablood
Oct 21 '15 at 3:11
$begingroup$
It doesn't require any orientation. It should be as simple as possible. Could you please elaborate you simplest answer? I am not that good with math...
$endgroup$
– dambros
Oct 21 '15 at 3:14
add a comment |
$begingroup$
Does the square require an orientation? The simplest answer is simply the four points (+/- 1/2 length +/- x, +/- 1/2 length +/- y)
$endgroup$
– fleablood
Oct 21 '15 at 3:11
$begingroup$
It doesn't require any orientation. It should be as simple as possible. Could you please elaborate you simplest answer? I am not that good with math...
$endgroup$
– dambros
Oct 21 '15 at 3:14
$begingroup$
Does the square require an orientation? The simplest answer is simply the four points (+/- 1/2 length +/- x, +/- 1/2 length +/- y)
$endgroup$
– fleablood
Oct 21 '15 at 3:11
$begingroup$
Does the square require an orientation? The simplest answer is simply the four points (+/- 1/2 length +/- x, +/- 1/2 length +/- y)
$endgroup$
– fleablood
Oct 21 '15 at 3:11
$begingroup$
It doesn't require any orientation. It should be as simple as possible. Could you please elaborate you simplest answer? I am not that good with math...
$endgroup$
– dambros
Oct 21 '15 at 3:14
$begingroup$
It doesn't require any orientation. It should be as simple as possible. Could you please elaborate you simplest answer? I am not that good with math...
$endgroup$
– dambros
Oct 21 '15 at 3:14
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Assuming the length of the square is L, I hope the drawing below answers your question. I called the center point c, and each corner p1,p2,p3 and p4.
answered Oct 21 '15 at 3:26
NoChanceNoChance
3,76621321
$endgroup$
$begingroup$
Nice drawing...
$endgroup$
– fleablood
Oct 21 '15 at 3:30
$begingroup$
Thanks for the input, even I could follow it. One more question, is there a way I can correctly convert lng/lat to x/y?
$endgroup$
– dambros
Oct 21 '15 at 4:01
$begingroup$
@dambros, sorry I don't know about this.
$endgroup$
– NoChance
Oct 21 '15 at 4:04
$begingroup$
@fleablood, thx.
$endgroup$
– NoChance
Oct 21 '15 at 4:04
add a comment |
$begingroup$
Let L = length of side. Let (h,k) be the center point.
The simplest answer is $(h + frac 1 2 L, k + frac 1 2 L), (h + frac 1 2 L, k - frac 1 2 L),(h - frac 1 2 L, k - frac 1 2 L), (h - frac 1 2 L, k + frac 1 2 L)$
If the square must be rotated by the angle $theta$, The the solutions are $(h + frac 1 2 L sin theta, k + frac 1 2 L cos theta), (h - frac 1 2 L cos theta, k + frac 1 2 L sin theta),(h - frac 1 2 L sin theta, k - frac 1 2 L cos theta), (h + frac 1 2 L cos theta, k - frac 1 2 L sin theta)$
answered Oct 21 '15 at 3:29
fleabloodfleablood
73.9k22891
$endgroup$
add a comment |
$begingroup$
In case the square is rotated around the center by an angle theta, w.r.t the ground position (considered as when the sides are parallel to the x and y axis), the vertices are given as -
Let L = length of the side. Let (h,k) be the center point.
$(h - frac12L(sintheta + costheta), k - frac12L(sintheta - costheta)$,
$(h + frac12L(sintheta - costheta), k - frac12L(sintheta + costheta)$,
$(h + frac12L(sintheta + costheta), k + frac12L(sintheta - costheta)$,
$(h - frac12L(sintheta - costheta), k + frac12L(sintheta + costheta)$
Please beware, the answer provided by @fleablood seems to be wrong. Didn't work out for me. It fails in the very basic case of $theta=0$ and $theta=90$ or I'm misinterpreting the answer provided.
answered Mar 30 at 7:52
Kartik GuptaKartik Gupta
111
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assuming the length of the square is L, I hope the drawing below answers your question. I called the center point c, and each corner p1,p2,p3 and p4.
answered Oct 21 '15 at 3:26
NoChanceNoChance
3,76621321
$endgroup$
$begingroup$
Nice drawing...
$endgroup$
– fleablood
Oct 21 '15 at 3:30
$begingroup$
Thanks for the input, even I could follow it. One more question, is there a way I can correctly convert lng/lat to x/y?
$endgroup$
– dambros
Oct 21 '15 at 4:01
$begingroup$
@dambros, sorry I don't know about this.
$endgroup$
– NoChance
Oct 21 '15 at 4:04
$begingroup$
@fleablood, thx.
$endgroup$
– NoChance
Oct 21 '15 at 4:04
add a comment |
$begingroup$
Assuming the length of the square is L, I hope the drawing below answers your question. I called the center point c, and each corner p1,p2,p3 and p4.
answered Oct 21 '15 at 3:26
NoChanceNoChance
3,76621321
$endgroup$
$begingroup$
Nice drawing...
$endgroup$
– fleablood
Oct 21 '15 at 3:30
$begingroup$
Thanks for the input, even I could follow it. One more question, is there a way I can correctly convert lng/lat to x/y?
$endgroup$
– dambros
Oct 21 '15 at 4:01
$begingroup$
@dambros, sorry I don't know about this.
$endgroup$
– NoChance
Oct 21 '15 at 4:04
$begingroup$
@fleablood, thx.
$endgroup$
– NoChance
Oct 21 '15 at 4:04
add a comment |
$begingroup$
Assuming the length of the square is L, I hope the drawing below answers your question. I called the center point c, and each corner p1,p2,p3 and p4.
answered Oct 21 '15 at 3:26
NoChanceNoChance
3,76621321
$endgroup$
Assuming the length of the square is L, I hope the drawing below answers your question. I called the center point c, and each corner p1,p2,p3 and p4.
answered Oct 21 '15 at 3:26
NoChanceNoChance
3,76621321
answered Oct 21 '15 at 3:26
NoChanceNoChance
3,76621321
answered Oct 21 '15 at 3:26
NoChanceNoChance
3,76621321
answered Oct 21 '15 at 3:26
NoChanceNoChance
3,76621321
3,76621321
$begingroup$
Nice drawing...
$endgroup$
– fleablood
Oct 21 '15 at 3:30
$begingroup$
Thanks for the input, even I could follow it. One more question, is there a way I can correctly convert lng/lat to x/y?
$endgroup$
– dambros
Oct 21 '15 at 4:01
$begingroup$
@dambros, sorry I don't know about this.
$endgroup$
– NoChance
Oct 21 '15 at 4:04
$begingroup$
@fleablood, thx.
$endgroup$
– NoChance
Oct 21 '15 at 4:04
add a comment |
$begingroup$
Nice drawing...
$endgroup$
– fleablood
Oct 21 '15 at 3:30
$begingroup$
Thanks for the input, even I could follow it. One more question, is there a way I can correctly convert lng/lat to x/y?
$endgroup$
– dambros
Oct 21 '15 at 4:01
$begingroup$
@dambros, sorry I don't know about this.
$endgroup$
– NoChance
Oct 21 '15 at 4:04
$begingroup$
@fleablood, thx.
$endgroup$
– NoChance
Oct 21 '15 at 4:04
$begingroup$
Nice drawing...
$endgroup$
– fleablood
Oct 21 '15 at 3:30
$begingroup$
Nice drawing...
$endgroup$
– fleablood
Oct 21 '15 at 3:30
$begingroup$
Thanks for the input, even I could follow it. One more question, is there a way I can correctly convert lng/lat to x/y?
$endgroup$
– dambros
Oct 21 '15 at 4:01
$begingroup$
Thanks for the input, even I could follow it. One more question, is there a way I can correctly convert lng/lat to x/y?
$endgroup$
– dambros
Oct 21 '15 at 4:01
$begingroup$
@dambros, sorry I don't know about this.
$endgroup$
– NoChance
Oct 21 '15 at 4:04
$begingroup$
@dambros, sorry I don't know about this.
$endgroup$
– NoChance
Oct 21 '15 at 4:04
$begingroup$
@fleablood, thx.
$endgroup$
– NoChance
Oct 21 '15 at 4:04
$begingroup$
@fleablood, thx.
$endgroup$
– NoChance
Oct 21 '15 at 4:04
add a comment |
$begingroup$
Let L = length of side. Let (h,k) be the center point.
The simplest answer is $(h + frac 1 2 L, k + frac 1 2 L), (h + frac 1 2 L, k - frac 1 2 L),(h - frac 1 2 L, k - frac 1 2 L), (h - frac 1 2 L, k + frac 1 2 L)$
If the square must be rotated by the angle $theta$, The the solutions are $(h + frac 1 2 L sin theta, k + frac 1 2 L cos theta), (h - frac 1 2 L cos theta, k + frac 1 2 L sin theta),(h - frac 1 2 L sin theta, k - frac 1 2 L cos theta), (h + frac 1 2 L cos theta, k - frac 1 2 L sin theta)$
answered Oct 21 '15 at 3:29
fleabloodfleablood
73.9k22891
$endgroup$
add a comment |
$begingroup$
Let L = length of side. Let (h,k) be the center point.
The simplest answer is $(h + frac 1 2 L, k + frac 1 2 L), (h + frac 1 2 L, k - frac 1 2 L),(h - frac 1 2 L, k - frac 1 2 L), (h - frac 1 2 L, k + frac 1 2 L)$
If the square must be rotated by the angle $theta$, The the solutions are $(h + frac 1 2 L sin theta, k + frac 1 2 L cos theta), (h - frac 1 2 L cos theta, k + frac 1 2 L sin theta),(h - frac 1 2 L sin theta, k - frac 1 2 L cos theta), (h + frac 1 2 L cos theta, k - frac 1 2 L sin theta)$
answered Oct 21 '15 at 3:29
fleabloodfleablood
73.9k22891
$endgroup$
add a comment |
$begingroup$
Let L = length of side. Let (h,k) be the center point.
The simplest answer is $(h + frac 1 2 L, k + frac 1 2 L), (h + frac 1 2 L, k - frac 1 2 L),(h - frac 1 2 L, k - frac 1 2 L), (h - frac 1 2 L, k + frac 1 2 L)$
If the square must be rotated by the angle $theta$, The the solutions are $(h + frac 1 2 L sin theta, k + frac 1 2 L cos theta), (h - frac 1 2 L cos theta, k + frac 1 2 L sin theta),(h - frac 1 2 L sin theta, k - frac 1 2 L cos theta), (h + frac 1 2 L cos theta, k - frac 1 2 L sin theta)$
answered Oct 21 '15 at 3:29
fleabloodfleablood
73.9k22891
$endgroup$
Let L = length of side. Let (h,k) be the center point.
The simplest answer is $(h + frac 1 2 L, k + frac 1 2 L), (h + frac 1 2 L, k - frac 1 2 L),(h - frac 1 2 L, k - frac 1 2 L), (h - frac 1 2 L, k + frac 1 2 L)$
If the square must be rotated by the angle $theta$, The the solutions are $(h + frac 1 2 L sin theta, k + frac 1 2 L cos theta), (h - frac 1 2 L cos theta, k + frac 1 2 L sin theta),(h - frac 1 2 L sin theta, k - frac 1 2 L cos theta), (h + frac 1 2 L cos theta, k - frac 1 2 L sin theta)$
answered Oct 21 '15 at 3:29
fleabloodfleablood
73.9k22891
answered Oct 21 '15 at 3:29
fleabloodfleablood
73.9k22891
answered Oct 21 '15 at 3:29
fleabloodfleablood
73.9k22891
answered Oct 21 '15 at 3:29
fleabloodfleablood
73.9k22891
73.9k22891
add a comment |
add a comment |
$begingroup$
In case the square is rotated around the center by an angle theta, w.r.t the ground position (considered as when the sides are parallel to the x and y axis), the vertices are given as -
Let L = length of the side. Let (h,k) be the center point.
$(h - frac12L(sintheta + costheta), k - frac12L(sintheta - costheta)$,
$(h + frac12L(sintheta - costheta), k - frac12L(sintheta + costheta)$,
$(h + frac12L(sintheta + costheta), k + frac12L(sintheta - costheta)$,
$(h - frac12L(sintheta - costheta), k + frac12L(sintheta + costheta)$
Please beware, the answer provided by @fleablood seems to be wrong. Didn't work out for me. It fails in the very basic case of $theta=0$ and $theta=90$ or I'm misinterpreting the answer provided.
answered Mar 30 at 7:52
Kartik GuptaKartik Gupta
111
$endgroup$
add a comment |
$begingroup$
In case the square is rotated around the center by an angle theta, w.r.t the ground position (considered as when the sides are parallel to the x and y axis), the vertices are given as -
Let L = length of the side. Let (h,k) be the center point.
$(h - frac12L(sintheta + costheta), k - frac12L(sintheta - costheta)$,
$(h + frac12L(sintheta - costheta), k - frac12L(sintheta + costheta)$,
$(h + frac12L(sintheta + costheta), k + frac12L(sintheta - costheta)$,
$(h - frac12L(sintheta - costheta), k + frac12L(sintheta + costheta)$
Please beware, the answer provided by @fleablood seems to be wrong. Didn't work out for me. It fails in the very basic case of $theta=0$ and $theta=90$ or I'm misinterpreting the answer provided.
answered Mar 30 at 7:52
Kartik GuptaKartik Gupta
111
$endgroup$
add a comment |
$begingroup$
In case the square is rotated around the center by an angle theta, w.r.t the ground position (considered as when the sides are parallel to the x and y axis), the vertices are given as -
Let L = length of the side. Let (h,k) be the center point.
$(h - frac12L(sintheta + costheta), k - frac12L(sintheta - costheta)$,
$(h + frac12L(sintheta - costheta), k - frac12L(sintheta + costheta)$,
$(h + frac12L(sintheta + costheta), k + frac12L(sintheta - costheta)$,
$(h - frac12L(sintheta - costheta), k + frac12L(sintheta + costheta)$
Please beware, the answer provided by @fleablood seems to be wrong. Didn't work out for me. It fails in the very basic case of $theta=0$ and $theta=90$ or I'm misinterpreting the answer provided.
answered Mar 30 at 7:52
Kartik GuptaKartik Gupta
111
$endgroup$
In case the square is rotated around the center by an angle theta, w.r.t the ground position (considered as when the sides are parallel to the x and y axis), the vertices are given as -
Let L = length of the side. Let (h,k) be the center point.
$(h - frac12L(sintheta + costheta), k - frac12L(sintheta - costheta)$,
$(h + frac12L(sintheta - costheta), k - frac12L(sintheta + costheta)$,
$(h + frac12L(sintheta + costheta), k + frac12L(sintheta - costheta)$,
$(h - frac12L(sintheta - costheta), k + frac12L(sintheta + costheta)$
Please beware, the answer provided by @fleablood seems to be wrong. Didn't work out for me. It fails in the very basic case of $theta=0$ and $theta=90$ or I'm misinterpreting the answer provided.
answered Mar 30 at 7:52
Kartik GuptaKartik Gupta
111
answered Mar 30 at 7:52
Kartik GuptaKartik Gupta
111
answered Mar 30 at 7:52
Kartik GuptaKartik Gupta
111
answered Mar 30 at 7:52
Kartik GuptaKartik Gupta
111
111
add a comment |
add a comment |
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$begingroup$
Does the square require an orientation? The simplest answer is simply the four points (+/- 1/2 length +/- x, +/- 1/2 length +/- y)
$endgroup$
– fleablood
Oct 21 '15 at 3:11
$begingroup$
It doesn't require any orientation. It should be as simple as possible. Could you please elaborate you simplest answer? I am not that good with math...
$endgroup$
– dambros
Oct 21 '15 at 3:14