# Centroid

#### kris8888

[FONT=&quot]Hi everyone, [/FONT]

[FONT=&quot]I have a problem i can't figure out... [/FONT]

[FONT=&quot]I want to calculate the centroid of a tilted rectangle in a coordinate system (See image) based on the following informations [/FONT]

[FONT=&quot]I know the Y coordinate of corner A [/FONT]
[FONT=&quot]I know the slope of the line b to the X axis based on two known points[/FONT]
[FONT=&quot]I know lenght of a and b[/FONT]

[FONT=&quot]From a given point on line b I need to find the centroid?[/FONT]

[FONT=&quot]So given an example what would the centroid be[/FONT] [FONT=&quot]a = 10[/FONT]
[FONT=&quot]b = 15[/FONT]

[FONT=&quot]Y coordinate of A is 5[/FONT]

[FONT=&quot]The slope of line b to X can be calculated from two given points 0,10 and 10,15[/FONT]
[FONT=&quot]
From coordinate 0,10 how to[/FONT]
[FONT=&quot] calculate the centroid?[/FONT]

[FONT=&quot]All the best,[/FONT]

[FONT=&quot]Kris[/FONT]

#### Cervesa

line AD has equation $y = \dfrac{x}{2}+10 \implies \text{ point A } = (-10,5)$

let $\theta$ be the angle between line AD and the horizontal, and $\phi$ be the angle between line AD and diagonal AC. Label the centroid (intersection of the two diagonals) point E. (reference attached diagram)

$|AE| = \sqrt{5^2 + 7.5^2} = \dfrac{5\sqrt{7}}{2}$

$\tan{\theta} = \dfrac{1}{2}$, $\tan{\phi} = \dfrac{2}{3}$

$\tan(\theta+\phi) = \dfrac{7}{4}$

centroid x-value, $\bar{x} = -10 + \dfrac{5\sqrt{7}}{2}\cos(\theta+\phi) = -10 + \dfrac{5\sqrt{7}}{2} \cdot \dfrac{4}{\sqrt{65}}$

centroid y-value, $\bar{y} = 5 + \dfrac{5\sqrt{7}}{2}\sin(\theta+\phi) = 5 + \dfrac{5\sqrt{7}}{2} \cdot \dfrac{7}{\sqrt{65}}$

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#### kris8888

Thanks i completely see it now!!

#### kris8888

Hi Again,

Sorry for returning to an old topic. But I can't seem to get this to work properly I always end up quite a bit short and really need a bit of help. See attached image for concrete example

$$\displaystyle M = \dfrac{6.66 - 3.33}{5} = \dfrac{333}{500}$$

$$\displaystyle |AE| = \sqrt{(0.5 * 7.07)^2 + (0.5 * 18.03)^2} = \dfrac{\sqrt{3750658}}{200}$$

$$\displaystyle \tan(\theta) = \tan(M) = \tan(\dfrac{333}{500}) = 33.6636$$

$$\displaystyle \tan(\phi) = \tan(a/b) = \tan(7.07/18.03) = 21.4113$$

$$\displaystyle \tan(\theta + \phi) = \tan(33.6636 + 21.4113) = 55.0349$$

$$\displaystyle x coordinate A = \dfrac{0 - 3.33}{\dfrac{333}{500}} = -5$$

$$\displaystyle Centroid_x = x coordinate A + AE * \cos(\theta + \phi) = -5 + \dfrac{\sqrt{3750658}}{200} * \cos(55.0349) = 0.549285$$

$$\displaystyle Centroid_y = y coordinate A + AE * \sin(\theta + \phi) = 0 + \dfrac{\sqrt{3750658}}{200}* \sin(55.0349) = 7.93$$

This should have been (0 , 7.5) - what am I doing wrong  Last edited:

Double post

Last edited:

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