Radius of the ball
Number of splits

Distance from the center of the ball (to the right and to the left)
Length of the "chord" of the nth section

Consider the problem of dissecting a ball with parallel planes, with an arbitrary radius \(R\), into \(n\) parts equal to each other in volume.

We have already solved a similar dissection problem, but there we were talking about a flat circle.  Cutting a circle with straight lines into equal areas

Now, since we are talking about a three-dimensional figure, we will calculate how high it is necessary to cut the next piece from our "bun", so that the volumes of these "cuts" would be the same.

For any number of dissections of the ball, we always get a "hump" from the ball - a ball segment.

Its volume is calculated by the formula

\(V=h^2\pi(R-\cfrac{h}{3})\)

On the other hand, the volume of the ball is \(V=\cfrac{4}{3}R^3\pi\)

the nth part of this ball is \(V=\cfrac{4}{3n}R^3\pi\)

Equating, we get \(h^2\pi(R-\cfrac{h}{3})=\cfrac{4}{3n}R^3\pi\)

By converting, we get the following

\(\cfrac{4}{n}=3(\cfrac{h}{R})^2-(\cfrac{h}{R})^3\)

replace \(\cfrac{h}{R}=x\)

Got a beautiful cubic equation

\(\cfrac{4}{n}=3x^2-x^3\)

knowing how to solve an equation of the form \(x(x+a)^2+c=0\)

\(x_i=\cfrac{-4a}{3}cos(\cfrac{acos(-1-\cfrac{\cfrac{c}{2}}{\cfrac{-a^3}{3^3}})}{6}+\cfrac{k2\pi}{6})^2\)

after the transformations we will get

\(x_i=-4*cos^2(\cfrac{acos(-1+\cfrac{2}{n})+2k\pi}{6})+3\)

Or even simplifying

\(x_i=-2*cos(\cfrac{acos(-1+\cfrac{2i}{n})}{3}+\cfrac{4\pi}{3})+1\)

where \(i\) is the number of the part to be cut

The roots will be our solution, not forgetting to do the inverse transformation \(\cfrac{h}{R}=x\)

Let's check if there are five parts and the radius of the ball is equal to a single nominal value (1 m, or 1 km, or 1 parsec is not important)

You just take the result in the same dimensions

\\0.5742815\\0.86586220.5742815 span>
0.8658622

Length of the"chord" of the nth section

\\1.8097113\\1.98192541.809711 >3
1.9819254
Distance from the center of the ball (right and left)

Successful calculations!!

 

 

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