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Thread: Centroid

  1. #1
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    Centroid

    Hi everyone,

    I have a problem i can't figure out...

    I want to calculate the centroid of a tilted rectangle in a coordinate system (See image) based on the following informations

    I know the Y coordinate of corner A
    I know the slope of the line b to the X axis based on two known points
    I know lenght of a and b

    From a given point on line b I need to find the centroid?

    So given an example what would the centroid be

    Centroid-rectilt.png


    a = 10
    b = 15

    Y coordinate of A is 5

    The slope of line b to X can be calculated from two given points 0,10 and 10,15

    From coordinate 0,10 how to
    calculate the centroid?

    All the best,

    Kris
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  2. #2
    Member Cervesa's Avatar
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    Re: Centroid

    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, $\bar{y} = 5 + \dfrac{5\sqrt{7}}{2}\sin(\theta+\phi) = 5 + \dfrac{5\sqrt{7}}{2} \cdot \dfrac{7}{\sqrt{65}}$
    Attached Thumbnails Attached Thumbnails Centroid-rectangle.jpg  
    Thanks from kris8888
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  3. #3
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    Re: Centroid

    Thanks i completely see it now!!
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  4. #4
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    Re: Centroid

    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



    Centroid-centroid_2.jpg
    Last edited by kris8888; Sep 22nd 2019 at 02:17 AM.
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  5. #5
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    Re: Centroid

    Double post
    Last edited by kris8888; Sep 22nd 2019 at 02:22 AM.
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  6. #6
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    Re: Centroid

    Solved
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