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Math Help - Rhombus coordinates

  1. #1
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    Rhombus coordinates

    Can someone find a solution of this problem....


    I have 500mmX500mm rhombus ( Rhombus A ). This is divided into 2500 equal size rhombuses of size 10mmX10mm ( R1 - R2500) . Now what I need to find is if I place a point(x,y) in Rhombus A, I have to locate in which of the rhombus of R1-R2500 the point(x,y) is placed.

    Thanks in advance
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  2. #2
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    Hello prasadcad

    Welcome to Math Help Forum!
    Quote Originally Posted by prasadcad View Post
    Can someone find a solution of this problem....


    I have 500mmX500mm rhombus ( Rhombus A ). This is divided into 2500 equal size rhombuses of size 10mmX10mm ( R1 - R2500) . Now what I need to find is if I place a point(x,y) in Rhombus A, I have to locate in which of the rhombus of R1-R2500 the point(x,y) is placed.

    Thanks in advance
    There are a few details that you haven't included in your description, so in the attached diagram, I'm assuming that:

    • \theta is the acute angle between two adjacent sides of the rhombus


    • one side lies along the positive x-axis, with a vertex at the origin


    • the small rhombuses are numbered from the origin, in rows, so that R1 - R50 lie along the x-axis.

    Then let's suppose that the point (x, y) lies in the rhombus which is in the m^{th} column and the n^{th} row. Then, with the assumption I've made above, this is rhombus number 50(n-1)+m. It now remains to find the values of m and n.

    Next, you will need to make a decision about what to do if (x,y) lies on one or more of the edges of one of the small rhombuses. In my working below, I have in this case taken the rhombus to the right and/or above the point.

    In the diagram, then, you will see that the rhombus containing (x, y) is bounded by four lines, whose equations I have shown. The point (x,y) therefore satisfies the following inequalities:
    y \ge 10(n-1)\sin \theta

    y < 10 n\sin\theta

    y \le (x+10m+10)\tan\theta

    y > (x-10m)\tan\theta
    The values of m and n that satisfy these inequalities can be expressed using the floor function, where \lfloor z \rfloor, is the largest integer less than or equal to z. Then (if my working is correct - I have checked it with some simple data, but you need to check this with some data of your own) we get:
    n = \left\lfloor \frac{y}{10\sin\theta}\right\rfloor + 1

    m = \left\lfloor \frac{1}{10}\left(x - \frac{y}{\tan\theta}\right)\right\rfloor +1
    Grandad
    Attached Thumbnails Attached Thumbnails Rhombus coordinates-untitled.jpg  
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  3. #3
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    @Grandad
    Thanks for the reply.


    The details you have asked for are
    1. The acute angle is 52Degrees.
    2. X-Axis lies along the diagonal of the Rhombus A.


    with these values I am not getting appropiate results.
    What changes need to be done in deducing the equations for m and n?
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  4. #4
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    Hello prasadcad
    Quote Originally Posted by prasadcad View Post
    @Grandad
    Thanks for the reply.


    The details you have asked for are
    1. The acute angle is 52Degrees.
    2. X-Axis lies along the diagonal of the Rhombus A.


    with these values I am not getting appropiate results.
    What changes need to be done in deducing the equations for m and n?
    I have re-drawn the diagram so that the x-axis lies along the diagonal of the rhombus, and have written down the equations of the lines bounding the small rhombus containing the point (x, y). You will notice that I have now called the angle between the sides 2\theta. So in the case you mention, you will need \theta = 26^o.

    Using the same conventions that I assumed before, namely, that:

    • the small rhombuses are numbered from the bottom left-hand corner, with R1 - R 50 lying along the bottom edge;


    • if the point (x, y) lies exactly on one or more edges of a small rhombus, then we take the rhombus to the right and/or above it;

    then the rhombus containing (x, y) is number 50(n -1)+m, where:
    m = \left\lfloor\frac{x\tan\theta-y}{20\sin\theta}\right\rfloor+1
    and
    n = \left\lfloor\frac{x\tan\theta+y}{20\sin\theta}\rig  ht\rfloor+1
    Let me know if this works OK.

    Grandad
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    Hi Grandad,

    Thanks a lot for the solution. It absolutely works fine.

    But I need to understand how can we make it more generic. I had given some constant values like 500 mm X 500mm. If we consider the length of each side of Rhombus A is L1 and it is divided into P and small rhombus length is l1.

    In the question I had posted the values I had given were L1 = 500mm, P = 2500 and l1= 10.
    I would like to know how these values were used to deduce the equations? And how we can use it make it Generic equations?

    Please excuse for my little knowledge about geometry.
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  6. #6
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    Hello prasadcad
    Quote Originally Posted by prasadcad View Post
    Hi Grandad,

    Thanks a lot for the solution. It absolutely works fine.

    But I need to understand how can we make it more generic. I had given some constant values like 500 mm X 500mm. If we consider the length of each side of Rhombus A is L1 and it is divided into P and small rhombus length is l1.

    In the question I had posted the values I had given were L1 = 500mm, P = 2500 and l1= 10.
    I would like to know how these values were used to deduce the equations? And how we can use it make it Generic equations?

    Please excuse for my little knowledge about geometry.
    First, note that L_1, P and l_1 are not independent. The number of small rhombuses, P, will be the square of the number that will fit along one edge of the large rhombus. In other words:
    P = \left(\frac{L_1}{l_1}\right)^2
    and, of course, if P is to be an integer, then L_1 will have to be a multiple of l_1.

    The formula for the number of the small rhombus containing (x,y) that I gave you, 50(n-1) + m, will then become:
    \frac{L_1(n-1)}{l_1}+m
    and the formulae for m and n will become:
    m = \left\lfloor\frac{x\tan\theta-y}{2l_1\sin\theta}\right\rfloor+1
    and
    n = \left\lfloor\frac{x\tan\theta+y}{2l_1\sin\theta}\r  ight\rfloor+1
    Grandad
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