Thread: Conversion of sphere to square

I was messing around with geometry a bit today and came across a problem that I do not understand.

I created a circle with a radius of 2. The circumference of such a circle is 12.56, and the area is 12.56

Now...I tried to 'fit a square' into the circle, but not in a typical way. I decided to bend the square to fit perfectly into the circle with no space left over...effectively creating a 4 'sided' circle. Now...each 'side' of the circle/square is 3.14, or pi, since 12.56/4 = pi. Now...if you try to find the area of this circle-square using the formula for the area of a square, you get 3.14^2 = 9.86

This value is different from the 12.56 of the traditional area of the circle formula. The problem I have with this is that no matter how you re-arrange the sides of a square, if the perimeter is the same, the area will be the same. So...why is it that when you re-arrange how the perimeter is 'counted' from a circle to a square, the area changes?

How do you "fit" the square? If the sides are no longer straight lines, its no longer a square in 2D space.

All you have proven is that a square of the same circumference as the circle occupies less area. The shape of a circle minimizes the circumference for a given area.

I'm not trying to demonstrate some revolutionary idea, lol....I'm just curious why what I said is true.

Basically the shape I am describing is a 4-way 'pinched' circle. The square idea was just to help imagine what I am talking about, but no, it would not look like a square.

I just do not understand how two objects with the same perimeter could possess different areas. How could 'pinching' a circle on 4 corners, and generating an angle at them (forming a square) reduce the area? It seems off to me, but the math says its true.

You mean sort of like these?

2.jpg

Hmmm...yea, those do look like what I am thinking of.

I essentially do not understand why flattening a circle on 4 sides (turning it into a square) manages to decrease the inner area. Unless, my calculations were wrong and that is not the case? I did quadruple check the numbers though....

" I just do not understand how two objects with the same perimeter could possess different areas."

One could be the two objects are geometrically different one is a triangle and the other is a sphere.

Another is the object orientation based on dimensional space.

In the process of "bending" the sides of the square to fit the circle, you stretched the surrounded area - the area inside the original square.

The transformation of the internal areas - square to circle of equivalent perimeter or back - is non-linear. Area is not preserved.

But why is it not linear? It seems perfectly natural for it to be linear to me, however in reality it is not. Is there some law of geometry specifying what goes on in this process, or some sort of detailed proof of it?

The classical isoperimetric problem dates back to antiquity. The problem can be stated as follows: Among all closed curves in the plane of fixed perimeter, which curve (if any) maximizes the area of its enclosed region? This question can be shown to be equivalent to the following problem: Among all closed curves in the plane enclosing a fixed area, which curve (if any) minimizes the perimeter?

Isoperimetric inequality - Wikipedia, the free encyclopedia

But why is it not linear? It seems perfectly natural for it to be linear to me,

Try this intuition: The area near some stretches of the sides of the square is being stretched much more than the area near others. You aren't moving the perimeter near the corners nearly as much, for example. Depending on how you set the square to start, you are gaining area rapidly along stretches in the middle of the square's sides, not gaining much area right near the corners or wherever the sides of the square start close to the eventual circle, right? So the gain in area is not linearly coupled to the length of the perimeter - some equal length stretches of square perimeter are going to cover much more area, as they are bent to fit, than others.

A square with a perimeter of, oh, lets say 12 (side lengths will be 3) will have an area An equilateral triangle(regular polygon) with perimeter 12 will have side lengths of 4, and as such an area of which is clearly less than that of the square. A Hexagon of perimeter 12 will have an area of which is clearly larger than the square, and nearly 3/2's the size of the triangle. A circle of circumference 12 will have an area of which is clearly larger than all the other Euclidean polygons (regular or irregular).

The geometry of each shape is different, and as such, yields a different area. Think of a long thin rectangle, and think of the huge perimeter it will have relative to the tiny area it will surely have. Its simply that the area and perimeter aren't dependent upon each other, simply correlated to each other (for regular polygons where each individual polygon is dependent, but a given perimeter will have a different area across different polygons)

Make sense?

If you want a simple way to visualize why different shapes with the same perimeter can have different areas, just take a loop of string. Lay it out flat on the table with no crossings and those shapes will always have the same perimeter. Now, it should be clear that very thin shapes will have less area than more round ones.

drawing a cirkel with a calliper you don,t do 1/4 pi *A^2. That would be a square with sides A pinned down where the diagonals cross and use the square as a calliper with four pencilpoints making a rotation 1/4 pi radials.

It,s 2 Pi *R for circumference, (or 1 rev * R ).

It,s basically reinventing a wheel or paintroller and make a sketch for the wheel. Make a paintroller from this as more complete designsketch add a side view as how technical drawings are made.

A rectangle D*W with W the width for the roller and D the diameter.

Then roll paint on the roller, put it to the rectangle W*D to start from there: one rev (or 2 pi rad), the constructed rectangle :

A = W *2pi * D. For different rollers W can be adapted or D but not make a roller with a different PI*D . Painting with such a roller the dynamic angle can be adapted to the wall surface but not with staring at it.

Take a rectangular sheet of paper. Draw a squiggly line from one corner to an adjacent corner. If you cut along that line, you increase the perimeter, because the squiggly line is longer than the straight edge. You also decrease the area by removing some of the paper.

The viete equation 2/pi =rth 2/2 +.........(http://en.wikipedia.org/wiki/Viète's_formula)

can be read with 2 as 2*1. Then it can be read with 1 for mathematical unit for area of a rectangle with straight angles ór reading 2 as 2* "the sum of two sides" for the same rectangle or others (circumference).
If I,m not mistaken the equation fits both. Offcourse Pi as 1/2 pi radials then or ninety degree angle.

But instead of converting (2/pi)/2 to 1/straight angle also converting the equation to 4/2pi with 2 pi for radials exchangable for 1 rev, 1/2 pi for quarter rev ...

Thanks for the replies! This is starting to make sense now....this notion was never taught to me in school or ever mentioned in my life, but I feel as if it is very important to know for some reason.

I am again learning more outside of class than I ever did in class : )

It,s basically an effect of perspective.