1. ## 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.

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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.

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:
$n = \left\lfloor \frac{y}{10\sin\theta}\right\rfloor + 1$

$m = \left\lfloor \frac{1}{10}\left(x - \frac{y}{\tan\theta}\right)\right\rfloor +1$

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?

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.

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?

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?

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:
$m = \left\lfloor\frac{x\tan\theta-y}{2l_1\sin\theta}\right\rfloor+1$
$n = \left\lfloor\frac{x\tan\theta+y}{2l_1\sin\theta}\r ight\rfloor+1$