Math Question

I have to customize a bass for a friend and was wondering: what would the dimensions be for one of these if the circle has to go completely around a rectangle thats 9.5x5 centimeters?
In centimeters please.

The radius of the circle would have to be equal to the distance from the middle of the rectangle to one of its corners.

Let’s say we want to find the radius that intersects the upper right corner of the rectangle. You can draw a right triangle inside the rectangle, with the hypotenuse being the radius. The other two legs of the triangle would be half the short side of the rectangle (2.5cm) and half the long side (4.75cm).

Then it’s just Pythagoras: 2.5[sup]2[/sup] + 4.75[sup]2[/sup] = 28.8125
Square root of 28.8125 = 5.3677 (roughly).

So you’d need to draw a circle with a radius of 5.3677cm, or a diameter of 10.7354cm.

Soooooo… How many centimeters would the square need to be?

Well, the circle is just inscribed inside the square, right? So the square’s sides would have to be exactly the same as the circle’s diameter.

Oh, I didn’t think of that. So, would 9x9 be right?

Might something like this work better?

Check out this link.

The inscribed rectangle is exactly 9.5 x 5 units and fits perfectly within my string art. The outer rectangle is 38/3 x 20/3 units and each side is divided into 20 equal pieces. One cool thing is that the number 20 is arbitrary. It can be replaced with any multiple of four and the fit will still be perfect.

Last one, I promise.

Hexagons are pretty.

I’m sorry, the rectangle is 11.5x5.5 centimeters!

Edit: nevermind.

I realized that it’s not, oh well.

What does that even mean?

I’m having some trouble understanding exactly what you want, Wisthekiller. I thought you wanted to make a circle big enough to inscribe a 9.5x5 rectangle, but apparently that’s not it. Can you try explaining again (or better yet, drawing a picture.)

Just to further confuse matters, the figure described in the video isn’t even a circle; if the drawer is connecting equally spaced points along the edge of the square, it’s formed by four segments of parabola, and if the line segments crossing the square are of a fixed length he’s got himself parts of on astroid. (Search in each page for “envelope” to find the appropriate sections of the articles.)

ETA: there’s also a nice Java applet illustrating the parabola case near the bottom of this page.

I suck at drawing and apparently gave the wrong dimensions. The new dimensions are 11.5x5.5cm. I need to kmow how long the sides of the square to make the parabola/envelope/circle completely encircle the 11.5x5.5 rectangle. I hope this clears it up.

How tight of a fit do you need?

Sides the length of the diagonal of the rectangle (~12.75 cm) will have a little room left over (I think around 1 or 2 mm). Is this good enough?

Have you considered the other shapes I suggested for which there will be much less wasted space?

Assuming you want a circle around your rectangle, the radius of the circle is 6.37cm. You want a square that is 12.75cm a side.

The thing doesn’t actually form a circle, though. That would take some math to determine the proper locations along the outside of the boundary square.

Click here for a pic of what it would look like at 12.75. Is that good enough? Or are you making a watch?

By the way I think this looks a lot cooler. (pictured rectangle has the desired aspect ratio)

I think I got it, thanks guys!
…now to go make some money withthis information! MWAHAHAHAHA