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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
    Junior 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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