volume cone

[nick2013]

Hello all,

I trying to build layers on top of a cone which will have the same volume of the cone. The initial cone is equal to a (in the 2D figure), then the first layer b should have the same volume than a, and so on.



I understand that as the cone grow the layers will be thinner. But I don't know how to calculate them.

Can someone give me a hint on how to proceed?

Thanks a lot!

n.

 

[DrPhil]

Hello all,

I trying to build layers on top of a cone which will have the same volume of the cone. The initial cone is equal to a (in the 2D figure), then the first layer b should have the same volume than a, and so on.

View attachment 3103

I understand that as the cone grow the layers will be thinner. But I don't know how to calculate them.

Can someone give me a hint on how to proceed?

Thanks a lot!

n.

Suppose your starting cone a has a circular base of radius r and height h. Then the volume is

$\displaystyle \displaystyle V_a = \pi\ r^2\ h$

If you add a layer with the same volume, and the same ratio of radius to height, the new external Volume will be double if all linear dimensions are increased by a factor of the cube root of 2 (1.260):
The outer dimensions of layer b will be

$\displaystyle h \to 1.260\ h, \; \; \; \;r \to 1.260\ r$

For the next layer, the dimensions will be $\displaystyle \sqrt[3]{3}$ times the dimensions of a, etc.
The outer dimensions of layer c will be

$\displaystyle h \to 1.442\ h, \; \; \; \;r \to 1.442\ r$

 

[HallsofIvy]

This is a somewhat harder problem than you might think (well, harder than I thought when I started it!). Here is what I would do. Take the height of the first portion, labeled "a", to be "h" and the radius "r". Then it has volume $\displaystyle \pi r^2h$ and the line, indicating the side of the cone in your 2D picture, has slope -h/a. Take the height of the next section to b 'x'. It side is parallel to the first (if I am interpreting your picture correctly) and passes through (0, h+x). we let R be the base radius of that new cone, we have R/(h+x)= h/a so that $\displaystyle R= (h^2+ hx)/r$. That new cone has volume $\displaystyle \pi R^2(h+ x)= \pi\left(\frac{h^2}{r}(h+x)\right)(h+x)= \pi\frac{h^2}{r}(h+x)^2$ and we want that to be twice the volume of the first cone: $\displaystyle \pi\frac{h^2}{r}(h+x)^2= 2\pi r^2h$ so that, if I've done the algebra correctly, \(\displaystyle x= \sqr{\frac{2r^3}{h}}- h\). For the others, replace that "2" with "3", "4", etc.

 

[nick2013]

thank you very much

ps: I guess you meant the volume of the cone to be [FONT=MathJax_Math]V[FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]1/3 π[/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Math]h[/FONT][/FONT]

 

[DrPhil]

thank you very much

ps: I guess you meant the volume of the cone to be [FONT=MathJax_Math]V[FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]1/3 π[/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]h[/FONT][/FONT]

OOPS - I guess I was only thinking about ratios! Thanks for noticing.