This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 |
This is wrong. The line parallel to one side of a triangle that divides it in half is sqrt(1/2) from the opposite corner, not 2/3, which is where the centroid is. - phma 21:06, 26 Jul 2004 (UTC)
Does anyone know what the formula for the centroid of a circle is? I tried to derive it, but it didn't work. I can't find it online, either. Can anyone tell me what it is?
-- Gscshoyru 17:42, 17 September 2005 (UTC)
Isn't it just the center?.... and if it isn't just use the formula in the article, with the circle centered at the origin. the y value is given by int(1/2*f(x)^2 from a to b) divided by the area. Xunflash 04:28, 30 October 2005 (UTC)
An anon at 129.110.8.39 (pc0839.utdallas.edu) added the reference:
Someone might want to check whether this is a worthwhile addition. Lunch 22:32, 1 September 2006 (UTC)
I didn't feel comfortable editing this page but felt I should point out that one image on this page does not appear for me.
The image Triangle_centroid_1.svg is on the page but appears not to contain anything. Clicking on the image yields http://en.wikipedia.org/wiki/Image:Triangle_centroid_1.svg which also appears not to contain anything. However, clicking on the image on that page yields a URL which does have a non-empty image http://upload.wikimedia.org/wikipedia/commons/8/85/Triangle_centroid_1.svg
-- 128.89.80.117 20:00, 29 October 2007 (UTC) Dan B.
This proof is not a proof. It is based on the statement that GBOC is a parallelogram, which is not proved.
Paolo.dL ( talk) 09:38, 19 February 2008 (UTC)
=== Proof that the centroid of a triangle divides each median in the ratio 2:1 === Let the medians AD, BE and CF of the triangle ABC intersect at G, the centroid of the triangle, and let the straight line AD be extended up to the point O such that : Then the triangles AGE and AOC are similar (common angle at A, AO is twice AG, AC is twice AE), and so OC is parallel to GE. But GE is BG extended, and so OC is parallel to BG. Similarly, OB is parallel to CG. The figure GBOC is therefore a parallelogram. Since the diagonals of a parallelogram bisect one another, the point of intersection D between the diagonals GO and BC is such that GD = DO, and : So, or This is true for every other median.
-- Jorge Stolfi ( talk) 03:49, 24 October 2008 (UTC)
From WordNet (r) 2.0 (August 2003) [wn]: centroid n : the center of mass of an object of uniform density
WordNet says, the center of mass is called centroid when the object has uniform density distribution. Is it ture?
Just a minor edit how to calculate centroid of a triangle using programming. I am sure someone will find it useful. SnegoviK 12:22, 24 February 2006 (UTC)
The first section distinguishes, clearly, between "centroid" (also called "barycenter" in geometry), and "barycenter" as used in physics. Then the section "Locating the centroid" describes a method which finds not the centroid (as defined at the start of the article), but the physics-style barycenter.
I realise that different people use the term "centroid" in different ways. But the article ought to be consistent. It should not define "centroid" in one way then use it in another. Maproom ( talk) 09:32, 9 July 2010 (UTC)
Casually flipping through, and I noticed that for this equation, there is no mention of what variable K is. Sum of masses? Size of left foot? Poor form, define your variables, people... —Preceding unsigned comment added by 67.40.180.206 ( talk) 23:49, 6 August 2010 (UTC)
In Geographic information systems, the term centroid may refer to other points than the geometric centre, primarily because of the desire that the centroid is inside the object. See for example [5]. Apus 08:24, 11 October 2006 (UTC)
I understand the argument that this is a different application of the term and I agree that perhaps a seperate page could be made for it, but is that really necessary? There are other examples in Wikipedia where a minor-use of a term is denoted on the same page where the majority-use of the term is explained. My understanding is that the test-range use of the term also stems from the maths term itself, so it has the same root. — Preceding unsigned comment added by Oceanbourne ( talk • contribs) 07:18, 20 April 2011 (UTC)
Does anyone know what the phrase
in the first sentence of the lede means? I've marked it with a clarification needed tag. Duoduoduo ( talk) 20:20, 25 March 2013 (UTC)
It doesn't mean "parts of equal area", which I thought at first, because e.g. for a triangle only three of the infinitude of lines through the centroid divide the area equally (see Bisection#Area bisectors and area-perimeter bisectors of a triangle). Duoduoduo ( talk) 22:23, 25 March 2013 (UTC)
The short section "Of a finite set of points" has been here since 22 November 2007. It was put here by editor Zeroin23a, who only edited Wikipedia from 22-25 November 2007. I don't think I've ever heard of the centroid of a set of points except here, and I suspect that it is Zeroin23a's original research. Any objection to my removing the section? Duoduoduo ( talk) 15:18, 26 March 2013 (UTC)
I was wondering if someone can find a reputable source for the problem of finding the smallest circle which inscribes a polygon in general? I have heuristically worked out a solution for triangles, but that would count as original research, so this cannot go into the article! Just to satiate anyone's curiosity the triangle problem can be solved as follows.
First, determine if 2 or 3 points of the triangle will be on the circle's parameter. Two edges will be on the parameter when the second longest edge is shorter than half the longest edge. This can be clearly seen by selecting the midpoint of the longest edge, producing a circle with radius =1/2 the longest edge. This will inscribe the third vertex since the second longest edge is less than or equal to radius. Otherwise all three vertices will be on the circle's parameter. When that is the case you can analytically solve for some point (x,y) which produces the same distance from all three vertices, provided the triangle is not degenerate. Simply algebraically solve (y-y1)^2+(x-x1)^2=(y-y2)^2+(x-x2)^2=(y-y3)^2+(x-x3)^2 for x and y (although the solution is a bit ugly so I'm not going to post it here). Mouse7mouse9 20:31, 23 April 2013 (UTC)
I have moved the following coding here because it wasn't explained in the article. What language is it?
public static function getCentroid(vertices:Array):Point { var i:int, j:int; var A:Number = 0; var Cx:Number = 0; var Cy:Number = 0; for(i=0;i<vertices.length;i++){ j = (i+1)%vertices.length; A += (vertices[i].x * vertices[j].y) - (vertices[j].x * vertices[i].y); Cx += (vertices[i].x + vertices[j].x)*((vertices[i].x * vertices[j].y) - (vertices[j].x * vertices[i].y)); Cy += (vertices[i].y + vertices[j].y)*((vertices[i].x * vertices[j].y) - (vertices[j].x * vertices[i].y)); } A /= 2; Cx /= (6*A); Cy /= (6*A); return new Point(Cx,Cy); //it may be placed out of polygon }
Any comments about whether it should be included (with explanation and perhaps collapsed)? Dbfirs 17:37, 13 June 2013 (UTC)
But there is no mention of celestial mechanics or orbits in the Centroid article. Lori ( talk) 02:48, 15 October 2013 (UTC)
This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 |
This is wrong. The line parallel to one side of a triangle that divides it in half is sqrt(1/2) from the opposite corner, not 2/3, which is where the centroid is. - phma 21:06, 26 Jul 2004 (UTC)
Does anyone know what the formula for the centroid of a circle is? I tried to derive it, but it didn't work. I can't find it online, either. Can anyone tell me what it is?
-- Gscshoyru 17:42, 17 September 2005 (UTC)
Isn't it just the center?.... and if it isn't just use the formula in the article, with the circle centered at the origin. the y value is given by int(1/2*f(x)^2 from a to b) divided by the area. Xunflash 04:28, 30 October 2005 (UTC)
An anon at 129.110.8.39 (pc0839.utdallas.edu) added the reference:
Someone might want to check whether this is a worthwhile addition. Lunch 22:32, 1 September 2006 (UTC)
I didn't feel comfortable editing this page but felt I should point out that one image on this page does not appear for me.
The image Triangle_centroid_1.svg is on the page but appears not to contain anything. Clicking on the image yields http://en.wikipedia.org/wiki/Image:Triangle_centroid_1.svg which also appears not to contain anything. However, clicking on the image on that page yields a URL which does have a non-empty image http://upload.wikimedia.org/wikipedia/commons/8/85/Triangle_centroid_1.svg
-- 128.89.80.117 20:00, 29 October 2007 (UTC) Dan B.
This proof is not a proof. It is based on the statement that GBOC is a parallelogram, which is not proved.
Paolo.dL ( talk) 09:38, 19 February 2008 (UTC)
=== Proof that the centroid of a triangle divides each median in the ratio 2:1 === Let the medians AD, BE and CF of the triangle ABC intersect at G, the centroid of the triangle, and let the straight line AD be extended up to the point O such that : Then the triangles AGE and AOC are similar (common angle at A, AO is twice AG, AC is twice AE), and so OC is parallel to GE. But GE is BG extended, and so OC is parallel to BG. Similarly, OB is parallel to CG. The figure GBOC is therefore a parallelogram. Since the diagonals of a parallelogram bisect one another, the point of intersection D between the diagonals GO and BC is such that GD = DO, and : So, or This is true for every other median.
-- Jorge Stolfi ( talk) 03:49, 24 October 2008 (UTC)
From WordNet (r) 2.0 (August 2003) [wn]: centroid n : the center of mass of an object of uniform density
WordNet says, the center of mass is called centroid when the object has uniform density distribution. Is it ture?
Just a minor edit how to calculate centroid of a triangle using programming. I am sure someone will find it useful. SnegoviK 12:22, 24 February 2006 (UTC)
The first section distinguishes, clearly, between "centroid" (also called "barycenter" in geometry), and "barycenter" as used in physics. Then the section "Locating the centroid" describes a method which finds not the centroid (as defined at the start of the article), but the physics-style barycenter.
I realise that different people use the term "centroid" in different ways. But the article ought to be consistent. It should not define "centroid" in one way then use it in another. Maproom ( talk) 09:32, 9 July 2010 (UTC)
Casually flipping through, and I noticed that for this equation, there is no mention of what variable K is. Sum of masses? Size of left foot? Poor form, define your variables, people... —Preceding unsigned comment added by 67.40.180.206 ( talk) 23:49, 6 August 2010 (UTC)
In Geographic information systems, the term centroid may refer to other points than the geometric centre, primarily because of the desire that the centroid is inside the object. See for example [5]. Apus 08:24, 11 October 2006 (UTC)
I understand the argument that this is a different application of the term and I agree that perhaps a seperate page could be made for it, but is that really necessary? There are other examples in Wikipedia where a minor-use of a term is denoted on the same page where the majority-use of the term is explained. My understanding is that the test-range use of the term also stems from the maths term itself, so it has the same root. — Preceding unsigned comment added by Oceanbourne ( talk • contribs) 07:18, 20 April 2011 (UTC)
Does anyone know what the phrase
in the first sentence of the lede means? I've marked it with a clarification needed tag. Duoduoduo ( talk) 20:20, 25 March 2013 (UTC)
It doesn't mean "parts of equal area", which I thought at first, because e.g. for a triangle only three of the infinitude of lines through the centroid divide the area equally (see Bisection#Area bisectors and area-perimeter bisectors of a triangle). Duoduoduo ( talk) 22:23, 25 March 2013 (UTC)
The short section "Of a finite set of points" has been here since 22 November 2007. It was put here by editor Zeroin23a, who only edited Wikipedia from 22-25 November 2007. I don't think I've ever heard of the centroid of a set of points except here, and I suspect that it is Zeroin23a's original research. Any objection to my removing the section? Duoduoduo ( talk) 15:18, 26 March 2013 (UTC)
I was wondering if someone can find a reputable source for the problem of finding the smallest circle which inscribes a polygon in general? I have heuristically worked out a solution for triangles, but that would count as original research, so this cannot go into the article! Just to satiate anyone's curiosity the triangle problem can be solved as follows.
First, determine if 2 or 3 points of the triangle will be on the circle's parameter. Two edges will be on the parameter when the second longest edge is shorter than half the longest edge. This can be clearly seen by selecting the midpoint of the longest edge, producing a circle with radius =1/2 the longest edge. This will inscribe the third vertex since the second longest edge is less than or equal to radius. Otherwise all three vertices will be on the circle's parameter. When that is the case you can analytically solve for some point (x,y) which produces the same distance from all three vertices, provided the triangle is not degenerate. Simply algebraically solve (y-y1)^2+(x-x1)^2=(y-y2)^2+(x-x2)^2=(y-y3)^2+(x-x3)^2 for x and y (although the solution is a bit ugly so I'm not going to post it here). Mouse7mouse9 20:31, 23 April 2013 (UTC)
I have moved the following coding here because it wasn't explained in the article. What language is it?
public static function getCentroid(vertices:Array):Point { var i:int, j:int; var A:Number = 0; var Cx:Number = 0; var Cy:Number = 0; for(i=0;i<vertices.length;i++){ j = (i+1)%vertices.length; A += (vertices[i].x * vertices[j].y) - (vertices[j].x * vertices[i].y); Cx += (vertices[i].x + vertices[j].x)*((vertices[i].x * vertices[j].y) - (vertices[j].x * vertices[i].y)); Cy += (vertices[i].y + vertices[j].y)*((vertices[i].x * vertices[j].y) - (vertices[j].x * vertices[i].y)); } A /= 2; Cx /= (6*A); Cy /= (6*A); return new Point(Cx,Cy); //it may be placed out of polygon }
Any comments about whether it should be included (with explanation and perhaps collapsed)? Dbfirs 17:37, 13 June 2013 (UTC)
But there is no mention of celestial mechanics or orbits in the Centroid article. Lori ( talk) 02:48, 15 October 2013 (UTC)