Coordinates of the centroid in a 3D rectangle
I have a 3D rectangle and I have to find the 3D coordinates of its centroid. I tried to take 4 vertices, one let's say the origin $o$ and the three adjacent vertices $a, b, c$.
I computed the distances between the pairs "origin-vertex" using this $sqrt{(o_x - a_x)^2+(o_y - a_y)^2+(o_z - a_z)^2}$ to compute the length, this $sqrt{(o_x - b_x)^2+(o_y - b_y)^2+(o_z - b_z)^2}$ to compute the width and this $sqrt{(o_x - c_x)^2+(o_y - c_y)^2+(o_z - c_z)^2}$ to compute the height.
(I'm doing this because my coordinates are not coplanar, otherwise, I could consider just the $x$ coordinate for the length, ...)
Then I added the coordinates of the origin, respectively $o_x$ to the $frac{length}{2}$, $o_y$ to the $frac{width}{2}$ and $o_z$ to the $frac{height}{2}$.
But what I'm getting is this
image
So what's the problem with my solution?
3d rectangles centroid
asked Nov 27 '18 at 9:25
User
105
|
show 1 more comment
I have a 3D rectangle and I have to find the 3D coordinates of its centroid. I tried to take 4 vertices, one let's say the origin $o$ and the three adjacent vertices $a, b, c$.
I computed the distances between the pairs "origin-vertex" using this $sqrt{(o_x - a_x)^2+(o_y - a_y)^2+(o_z - a_z)^2}$ to compute the length, this $sqrt{(o_x - b_x)^2+(o_y - b_y)^2+(o_z - b_z)^2}$ to compute the width and this $sqrt{(o_x - c_x)^2+(o_y - c_y)^2+(o_z - c_z)^2}$ to compute the height.
(I'm doing this because my coordinates are not coplanar, otherwise, I could consider just the $x$ coordinate for the length, ...)
Then I added the coordinates of the origin, respectively $o_x$ to the $frac{length}{2}$, $o_y$ to the $frac{width}{2}$ and $o_z$ to the $frac{height}{2}$.
But what I'm getting is this
image
So what's the problem with my solution?
3d rectangles centroid
asked Nov 27 '18 at 9:25
User
105
If the points aren’t coplanar, then you don’t have a rectangle.
– amd
Nov 27 '18 at 9:56
What I mean is that for example, the origin is $(37.5, 0.2, -0.3)$ and $a$ is $(39.4, 0.1, -0.1)$
– User
Nov 27 '18 at 10:05
1
If it really is a rectangle, you can just add up all of the vertex coordinates and divide by four or find the intersection of the diagonals, as one might in two dimensions.
– amd
Nov 27 '18 at 10:11
You say you have a rectangle, but your picture shows a box (a.k.a. a cuboid). Are you trying to find the centroid of one of the rectangular faces of the box, or the centroid of the whole box?
– Rahul
Nov 27 '18 at 11:46
User, you probably mean "not axis-aligned", when you wrote "not coplanar".
– Nominal Animal
Nov 28 '18 at 15:46
|
show 1 more comment
I have a 3D rectangle and I have to find the 3D coordinates of its centroid. I tried to take 4 vertices, one let's say the origin $o$ and the three adjacent vertices $a, b, c$.
I computed the distances between the pairs "origin-vertex" using this $sqrt{(o_x - a_x)^2+(o_y - a_y)^2+(o_z - a_z)^2}$ to compute the length, this $sqrt{(o_x - b_x)^2+(o_y - b_y)^2+(o_z - b_z)^2}$ to compute the width and this $sqrt{(o_x - c_x)^2+(o_y - c_y)^2+(o_z - c_z)^2}$ to compute the height.
(I'm doing this because my coordinates are not coplanar, otherwise, I could consider just the $x$ coordinate for the length, ...)
Then I added the coordinates of the origin, respectively $o_x$ to the $frac{length}{2}$, $o_y$ to the $frac{width}{2}$ and $o_z$ to the $frac{height}{2}$.
But what I'm getting is this
image
So what's the problem with my solution?
3d rectangles centroid
asked Nov 27 '18 at 9:25
User
105
I have a 3D rectangle and I have to find the 3D coordinates of its centroid. I tried to take 4 vertices, one let's say the origin $o$ and the three adjacent vertices $a, b, c$.
I computed the distances between the pairs "origin-vertex" using this $sqrt{(o_x - a_x)^2+(o_y - a_y)^2+(o_z - a_z)^2}$ to compute the length, this $sqrt{(o_x - b_x)^2+(o_y - b_y)^2+(o_z - b_z)^2}$ to compute the width and this $sqrt{(o_x - c_x)^2+(o_y - c_y)^2+(o_z - c_z)^2}$ to compute the height.
(I'm doing this because my coordinates are not coplanar, otherwise, I could consider just the $x$ coordinate for the length, ...)
Then I added the coordinates of the origin, respectively $o_x$ to the $frac{length}{2}$, $o_y$ to the $frac{width}{2}$ and $o_z$ to the $frac{height}{2}$.
But what I'm getting is this
image
So what's the problem with my solution?
3d rectangles centroid
3d rectangles centroid
asked Nov 27 '18 at 9:25
User
105
asked Nov 27 '18 at 9:25
User
105
edited Nov 27 '18 at 9:36
asked Nov 27 '18 at 9:25
User
105
asked Nov 27 '18 at 9:25
User
105
asked Nov 27 '18 at 9:25
User
105
105
If the points aren’t coplanar, then you don’t have a rectangle.
– amd
Nov 27 '18 at 9:56
What I mean is that for example, the origin is $(37.5, 0.2, -0.3)$ and $a$ is $(39.4, 0.1, -0.1)$
– User
Nov 27 '18 at 10:05
1
If it really is a rectangle, you can just add up all of the vertex coordinates and divide by four or find the intersection of the diagonals, as one might in two dimensions.
– amd
Nov 27 '18 at 10:11
You say you have a rectangle, but your picture shows a box (a.k.a. a cuboid). Are you trying to find the centroid of one of the rectangular faces of the box, or the centroid of the whole box?
– Rahul
Nov 27 '18 at 11:46
User, you probably mean "not axis-aligned", when you wrote "not coplanar".
– Nominal Animal
Nov 28 '18 at 15:46
|
show 1 more comment
If the points aren’t coplanar, then you don’t have a rectangle.
– amd
Nov 27 '18 at 9:56
What I mean is that for example, the origin is $(37.5, 0.2, -0.3)$ and $a$ is $(39.4, 0.1, -0.1)$
– User
Nov 27 '18 at 10:05
1
If it really is a rectangle, you can just add up all of the vertex coordinates and divide by four or find the intersection of the diagonals, as one might in two dimensions.
– amd
Nov 27 '18 at 10:11
You say you have a rectangle, but your picture shows a box (a.k.a. a cuboid). Are you trying to find the centroid of one of the rectangular faces of the box, or the centroid of the whole box?
– Rahul
Nov 27 '18 at 11:46
User, you probably mean "not axis-aligned", when you wrote "not coplanar".
– Nominal Animal
Nov 28 '18 at 15:46
If the points aren’t coplanar, then you don’t have a rectangle.
– amd
Nov 27 '18 at 9:56
If the points aren’t coplanar, then you don’t have a rectangle.
– amd
Nov 27 '18 at 9:56
What I mean is that for example, the origin is $(37.5, 0.2, -0.3)$ and $a$ is $(39.4, 0.1, -0.1)$
– User
Nov 27 '18 at 10:05
What I mean is that for example, the origin is $(37.5, 0.2, -0.3)$ and $a$ is $(39.4, 0.1, -0.1)$
– User
Nov 27 '18 at 10:05
1
1
If it really is a rectangle, you can just add up all of the vertex coordinates and divide by four or find the intersection of the diagonals, as one might in two dimensions.
– amd
Nov 27 '18 at 10:11
If it really is a rectangle, you can just add up all of the vertex coordinates and divide by four or find the intersection of the diagonals, as one might in two dimensions.
– amd
Nov 27 '18 at 10:11
You say you have a rectangle, but your picture shows a box (a.k.a. a cuboid). Are you trying to find the centroid of one of the rectangular faces of the box, or the centroid of the whole box?
– Rahul
Nov 27 '18 at 11:46
You say you have a rectangle, but your picture shows a box (a.k.a. a cuboid). Are you trying to find the centroid of one of the rectangular faces of the box, or the centroid of the whole box?
– Rahul
Nov 27 '18 at 11:46
User, you probably mean "not axis-aligned", when you wrote "not coplanar".
– Nominal Animal
Nov 28 '18 at 15:46
User, you probably mean "not axis-aligned", when you wrote "not coplanar".
– Nominal Animal
Nov 28 '18 at 15:46
|
show 1 more comment
1 Answer
1
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Let's say you know one vertex of the cuboid, $vec{o}$, and three adjacent vertices $vec{a}$, $vec{b}$, and $vec{c}$:
$$bbox{vec{o} = left[begin{matrix} o_x \ o_y \ o_z end{matrix}right]} ,quad
bbox{vec{a} = left[begin{matrix} a_x \ a_y \ a_z end{matrix}right]} ,quad
bbox{vec{b} = left[begin{matrix} b_x \ b_y \ b_z end{matrix}right]} ,quad
bbox{vec{c} = left[begin{matrix} c_x \ c_y \ c_z end{matrix}right]}$$
All edges of the cuboid (where two faces meet) match one of the vectors
$$bbox{vec{e}_a = vec{a} - vec{o}} , quad
bbox{vec{e}_b = vec{b} - vec{o}} , quad
bbox{vec{e}_c = vec{c} - vec{o}}$$
and the eight vertices of the cuboid are at
$$begin{array}{rll}
vec{v}_{000} =& vec{o} & = vec{o} \
vec{v}_{001} =& vec{o} + vec{e}_a & = vec{a} \
vec{v}_{010} =& vec{o} + vec{e}_b & = vec{b} \
vec{v}_{011} =& vec{o} + vec{e}_a + vec{e}_b & = vec{a} + vec{b} - vec{o} \
vec{v}_{100} =& vec{o} + vec{e}_c & = vec{c} \
vec{v}_{101} =& vec{o} + vec{e}_a + vec{e}_c & = vec{a} + vec{c} - vec{o} \
vec{v}_{110} =& vec{o} + vec{e}_b + vec{e}_c & = vec{b} + vec{c} - vec{o} \
vec{v}_{111} =& vec{o} + vec{e}_a + vec{e}_b + vec{e}_c & = vec{a} + vec{b} + vec{c} - 2 vec{o} \
end{array}$$
The centroid for a cuboid (or a parallelepiped; the three edges do not need to be perpendicular to each other) is the sum of the vertex coordinates, divided by eight. If you sum the above, you'll find out that the centroid is at $vec{p}$,
$$bbox{vec{p} = frac{1}{2}left( vec{a} + vec{b} + vec{c} - vec{o} right) = left[begin{matrix}
frac{a_x + b_x + c_x - o_x}{2} \
frac{a_y + b_y + c_y - o_y}{2} \
frac{a_z + b_z + c_z - o_z}{2} \
end{matrix}right]}$$
answered Nov 28 '18 at 16:16
Nominal Animal
6,7702517
Thanks for the explanation and the solution!
– User
Nov 29 '18 at 16:07
add a comment |
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1 Answer
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Let's say you know one vertex of the cuboid, $vec{o}$, and three adjacent vertices $vec{a}$, $vec{b}$, and $vec{c}$:
$$bbox{vec{o} = left[begin{matrix} o_x \ o_y \ o_z end{matrix}right]} ,quad
bbox{vec{a} = left[begin{matrix} a_x \ a_y \ a_z end{matrix}right]} ,quad
bbox{vec{b} = left[begin{matrix} b_x \ b_y \ b_z end{matrix}right]} ,quad
bbox{vec{c} = left[begin{matrix} c_x \ c_y \ c_z end{matrix}right]}$$
All edges of the cuboid (where two faces meet) match one of the vectors
$$bbox{vec{e}_a = vec{a} - vec{o}} , quad
bbox{vec{e}_b = vec{b} - vec{o}} , quad
bbox{vec{e}_c = vec{c} - vec{o}}$$
and the eight vertices of the cuboid are at
$$begin{array}{rll}
vec{v}_{000} =& vec{o} & = vec{o} \
vec{v}_{001} =& vec{o} + vec{e}_a & = vec{a} \
vec{v}_{010} =& vec{o} + vec{e}_b & = vec{b} \
vec{v}_{011} =& vec{o} + vec{e}_a + vec{e}_b & = vec{a} + vec{b} - vec{o} \
vec{v}_{100} =& vec{o} + vec{e}_c & = vec{c} \
vec{v}_{101} =& vec{o} + vec{e}_a + vec{e}_c & = vec{a} + vec{c} - vec{o} \
vec{v}_{110} =& vec{o} + vec{e}_b + vec{e}_c & = vec{b} + vec{c} - vec{o} \
vec{v}_{111} =& vec{o} + vec{e}_a + vec{e}_b + vec{e}_c & = vec{a} + vec{b} + vec{c} - 2 vec{o} \
end{array}$$
The centroid for a cuboid (or a parallelepiped; the three edges do not need to be perpendicular to each other) is the sum of the vertex coordinates, divided by eight. If you sum the above, you'll find out that the centroid is at $vec{p}$,
$$bbox{vec{p} = frac{1}{2}left( vec{a} + vec{b} + vec{c} - vec{o} right) = left[begin{matrix}
frac{a_x + b_x + c_x - o_x}{2} \
frac{a_y + b_y + c_y - o_y}{2} \
frac{a_z + b_z + c_z - o_z}{2} \
end{matrix}right]}$$
answered Nov 28 '18 at 16:16
Nominal Animal
6,7702517
Thanks for the explanation and the solution!
– User
Nov 29 '18 at 16:07
add a comment |
Let's say you know one vertex of the cuboid, $vec{o}$, and three adjacent vertices $vec{a}$, $vec{b}$, and $vec{c}$:
$$bbox{vec{o} = left[begin{matrix} o_x \ o_y \ o_z end{matrix}right]} ,quad
bbox{vec{a} = left[begin{matrix} a_x \ a_y \ a_z end{matrix}right]} ,quad
bbox{vec{b} = left[begin{matrix} b_x \ b_y \ b_z end{matrix}right]} ,quad
bbox{vec{c} = left[begin{matrix} c_x \ c_y \ c_z end{matrix}right]}$$
All edges of the cuboid (where two faces meet) match one of the vectors
$$bbox{vec{e}_a = vec{a} - vec{o}} , quad
bbox{vec{e}_b = vec{b} - vec{o}} , quad
bbox{vec{e}_c = vec{c} - vec{o}}$$
and the eight vertices of the cuboid are at
$$begin{array}{rll}
vec{v}_{000} =& vec{o} & = vec{o} \
vec{v}_{001} =& vec{o} + vec{e}_a & = vec{a} \
vec{v}_{010} =& vec{o} + vec{e}_b & = vec{b} \
vec{v}_{011} =& vec{o} + vec{e}_a + vec{e}_b & = vec{a} + vec{b} - vec{o} \
vec{v}_{100} =& vec{o} + vec{e}_c & = vec{c} \
vec{v}_{101} =& vec{o} + vec{e}_a + vec{e}_c & = vec{a} + vec{c} - vec{o} \
vec{v}_{110} =& vec{o} + vec{e}_b + vec{e}_c & = vec{b} + vec{c} - vec{o} \
vec{v}_{111} =& vec{o} + vec{e}_a + vec{e}_b + vec{e}_c & = vec{a} + vec{b} + vec{c} - 2 vec{o} \
end{array}$$
The centroid for a cuboid (or a parallelepiped; the three edges do not need to be perpendicular to each other) is the sum of the vertex coordinates, divided by eight. If you sum the above, you'll find out that the centroid is at $vec{p}$,
$$bbox{vec{p} = frac{1}{2}left( vec{a} + vec{b} + vec{c} - vec{o} right) = left[begin{matrix}
frac{a_x + b_x + c_x - o_x}{2} \
frac{a_y + b_y + c_y - o_y}{2} \
frac{a_z + b_z + c_z - o_z}{2} \
end{matrix}right]}$$
answered Nov 28 '18 at 16:16
Nominal Animal
6,7702517
Thanks for the explanation and the solution!
– User
Nov 29 '18 at 16:07
add a comment |
Let's say you know one vertex of the cuboid, $vec{o}$, and three adjacent vertices $vec{a}$, $vec{b}$, and $vec{c}$:
$$bbox{vec{o} = left[begin{matrix} o_x \ o_y \ o_z end{matrix}right]} ,quad
bbox{vec{a} = left[begin{matrix} a_x \ a_y \ a_z end{matrix}right]} ,quad
bbox{vec{b} = left[begin{matrix} b_x \ b_y \ b_z end{matrix}right]} ,quad
bbox{vec{c} = left[begin{matrix} c_x \ c_y \ c_z end{matrix}right]}$$
All edges of the cuboid (where two faces meet) match one of the vectors
$$bbox{vec{e}_a = vec{a} - vec{o}} , quad
bbox{vec{e}_b = vec{b} - vec{o}} , quad
bbox{vec{e}_c = vec{c} - vec{o}}$$
and the eight vertices of the cuboid are at
$$begin{array}{rll}
vec{v}_{000} =& vec{o} & = vec{o} \
vec{v}_{001} =& vec{o} + vec{e}_a & = vec{a} \
vec{v}_{010} =& vec{o} + vec{e}_b & = vec{b} \
vec{v}_{011} =& vec{o} + vec{e}_a + vec{e}_b & = vec{a} + vec{b} - vec{o} \
vec{v}_{100} =& vec{o} + vec{e}_c & = vec{c} \
vec{v}_{101} =& vec{o} + vec{e}_a + vec{e}_c & = vec{a} + vec{c} - vec{o} \
vec{v}_{110} =& vec{o} + vec{e}_b + vec{e}_c & = vec{b} + vec{c} - vec{o} \
vec{v}_{111} =& vec{o} + vec{e}_a + vec{e}_b + vec{e}_c & = vec{a} + vec{b} + vec{c} - 2 vec{o} \
end{array}$$
The centroid for a cuboid (or a parallelepiped; the three edges do not need to be perpendicular to each other) is the sum of the vertex coordinates, divided by eight. If you sum the above, you'll find out that the centroid is at $vec{p}$,
$$bbox{vec{p} = frac{1}{2}left( vec{a} + vec{b} + vec{c} - vec{o} right) = left[begin{matrix}
frac{a_x + b_x + c_x - o_x}{2} \
frac{a_y + b_y + c_y - o_y}{2} \
frac{a_z + b_z + c_z - o_z}{2} \
end{matrix}right]}$$
answered Nov 28 '18 at 16:16
Nominal Animal
6,7702517
Let's say you know one vertex of the cuboid, $vec{o}$, and three adjacent vertices $vec{a}$, $vec{b}$, and $vec{c}$:
$$bbox{vec{o} = left[begin{matrix} o_x \ o_y \ o_z end{matrix}right]} ,quad
bbox{vec{a} = left[begin{matrix} a_x \ a_y \ a_z end{matrix}right]} ,quad
bbox{vec{b} = left[begin{matrix} b_x \ b_y \ b_z end{matrix}right]} ,quad
bbox{vec{c} = left[begin{matrix} c_x \ c_y \ c_z end{matrix}right]}$$
All edges of the cuboid (where two faces meet) match one of the vectors
$$bbox{vec{e}_a = vec{a} - vec{o}} , quad
bbox{vec{e}_b = vec{b} - vec{o}} , quad
bbox{vec{e}_c = vec{c} - vec{o}}$$
and the eight vertices of the cuboid are at
$$begin{array}{rll}
vec{v}_{000} =& vec{o} & = vec{o} \
vec{v}_{001} =& vec{o} + vec{e}_a & = vec{a} \
vec{v}_{010} =& vec{o} + vec{e}_b & = vec{b} \
vec{v}_{011} =& vec{o} + vec{e}_a + vec{e}_b & = vec{a} + vec{b} - vec{o} \
vec{v}_{100} =& vec{o} + vec{e}_c & = vec{c} \
vec{v}_{101} =& vec{o} + vec{e}_a + vec{e}_c & = vec{a} + vec{c} - vec{o} \
vec{v}_{110} =& vec{o} + vec{e}_b + vec{e}_c & = vec{b} + vec{c} - vec{o} \
vec{v}_{111} =& vec{o} + vec{e}_a + vec{e}_b + vec{e}_c & = vec{a} + vec{b} + vec{c} - 2 vec{o} \
end{array}$$
The centroid for a cuboid (or a parallelepiped; the three edges do not need to be perpendicular to each other) is the sum of the vertex coordinates, divided by eight. If you sum the above, you'll find out that the centroid is at $vec{p}$,
$$bbox{vec{p} = frac{1}{2}left( vec{a} + vec{b} + vec{c} - vec{o} right) = left[begin{matrix}
frac{a_x + b_x + c_x - o_x}{2} \
frac{a_y + b_y + c_y - o_y}{2} \
frac{a_z + b_z + c_z - o_z}{2} \
end{matrix}right]}$$
answered Nov 28 '18 at 16:16
Nominal Animal
6,7702517
answered Nov 28 '18 at 16:16
Nominal Animal
6,7702517
answered Nov 28 '18 at 16:16
Nominal Animal
6,7702517
answered Nov 28 '18 at 16:16
Nominal Animal
6,7702517
6,7702517
Thanks for the explanation and the solution!
– User
Nov 29 '18 at 16:07
add a comment |
Thanks for the explanation and the solution!
– User
Nov 29 '18 at 16:07
Thanks for the explanation and the solution!
– User
Nov 29 '18 at 16:07
Thanks for the explanation and the solution!
– User
Nov 29 '18 at 16:07
add a comment |
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If the points aren’t coplanar, then you don’t have a rectangle.
– amd
Nov 27 '18 at 9:56
What I mean is that for example, the origin is $(37.5, 0.2, -0.3)$ and $a$ is $(39.4, 0.1, -0.1)$
– User
Nov 27 '18 at 10:05
1
If it really is a rectangle, you can just add up all of the vertex coordinates and divide by four or find the intersection of the diagonals, as one might in two dimensions.
– amd
Nov 27 '18 at 10:11
You say you have a rectangle, but your picture shows a box (a.k.a. a cuboid). Are you trying to find the centroid of one of the rectangular faces of the box, or the centroid of the whole box?
– Rahul
Nov 27 '18 at 11:46
User, you probably mean "not axis-aligned", when you wrote "not coplanar".
– Nominal Animal
Nov 28 '18 at 15:46