Equation of a rectangle's radius!

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A few days ago I got really excited when I found the equation for a retangles radius (distance from its center to its perimeter given an angle)

I have no idea if this is something very simple and stupid but my math teacher said it was impressive, I'm still in high school

It's like this:
(w/2)*sqrt(cos^2(theta)) + (h/2)*sqrt(sin^2(theta))

(Or my attempt at latex):
[tex]\frac{w}{2}\sqrt{cos^{2}(\theta)} + \frac{h}{2}\sqrt{sin^{2}(\theta)}[/tex]

gives the length of the line from the center of a rectangle (width w, height h) to its perimeter at angle of theta.

I graphed it also with x = theta (radians), for a rectangle of width 1 and height 1:
http://dl.dropbox.com/u/1828729/rr/graph.png [Broken] (it's a big image)

I needed to share this, excuse me if it's ridiculously simple.
 
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HallsofIvy

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That's NOT "ridiculously simple", it is, in fact, very clever.
 
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Basically what you have done is a polar coordinate construction of rectangle that's centered at origin It's a quite more complicated way to represent rectangles but nonetheless intresting.

There is an error though, you could try plugging in [tex]\pi/6[/tex] for the formula and you get the wrong answer. Your answer is close to

I found a paper in which I had worked on the same problem. I didn't generalize it for rectangles of different sizes but that would be easy enough to do so. The solution is a piecewise one since I didn't bother to try to combine these. You can view the attachment to see the graph of solution for square (with side length 2). As you can see how it involves reciprocal of sine and cosine instead of just sine and cosine.
 

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This equation does not work.

Consider an angle theta, such that [itex]0^\circ \le \theta \le 90^\circ[/tex]

The "radius" of a rectangle with this angle would lie in the upper right quadrant of the rectangle.

Let's assume [itex]\theta = 30^\circ[/tex]
This would result in one of three different situations:
1) the radius will intersect the upper right vertex of the rectangle
2) the radius will intersect the top of the rectangle
3) the radius will intersect the right side of the rectangle

Now, let's assume our radius is 2 units long.

With an angle of [itex]30^\circ[/tex], this means we have a 30-60-90 triangle with a height of 1 unit and a base of [itex]\sqrt{3}[/tex] units.

I've drawn this triangle inside three different rectangles (each representing the upper right quadrant of the "whole" rectangle) using random values for situations #2 & #3.
(See attachment)

The dimension of the "whole" rectangles are as such:
1) h = 2, w = 2[itex]\sqrt{3}[/tex]
2) h = 2, w = 4

3) h = 2.4, w = 2[itex]\sqrt{3}[/tex]


For situation #2, it's easy to see that this triangle will fit any upper right quadrant with a height of 1 unit and a width greater than or equal to [itex]\sqrt{3}[/tex]

So, in the case where the radius at a given angle intersects the top of the rectangle, you can increase the width of the rectangle without affecting the length of the radius. And, since w is increasing, the value of your equation is increasing, yet the length of the radius remains the same.
 

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Since no one addressed this problem, I'll give my solution...


In order to find the length of the line segment from the center of a rectangle to it's perimeter (or, what the OP calls the "radius" of a rectangle), given the angle of the line segment ([itex]\theta[/tex]) and the rectangle's dimensions (H and W), you must first determine whether the line segment intersects either the top or bottom (one of the widths), or one of the sides.


To simplify the problem, I find it easier to find a similar angle in the first quadrant. That is, an angle [itex]\beta[/tex] such that [itex]0^\circ \le \beta \le 90^\circ[/tex] which intersects the rectangle in the same way as the original angle [itex]\theta[/tex].


To do this, I use

[tex]\beta = \tan^{-1} \left| \tan(\theta) \right|[/tex]



Next, determine the angle of the diagonal from the rectangle's center to it's upper-right vertex:

[tex]\phi = \tan^{-1} \left( \frac{H}{L} \right)[/tex]



Finally, compare [itex]\beta[/tex] to [itex]\phi[/tex] and use the appropriate formula to determine the length of the "radius" R:

If [itex]\beta \ge \phi[/tex], then [itex]R = H/(2 \sin \beta)[/tex]

If [itex]\beta < \phi[/tex], then [itex]R = W/(2 \cos \beta)[/tex]
 

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