Cone height ratios

[pavelcherepan]

Howdy all! I have yet another question, this time in the realm of mathematics. For a project I'm doing at work I need to create an optimal sampling method for a drill cone and so the basics of the problem are like this:

 

We have 3 cones nesting inside one another or a cone with 3 "layers". They all have the same base angle [latex]\alpha[/latex] but different radii and heights. If we use V as the volume of the bottom cone, then middle cone is 2V and the outer (largest) cone is 9/4V. Based on this I need to calculate percentages of the total height of the entire cone for the following:

 

h₃ - from the top of the outer cone to the top of middle cone

h₂ - from the top of middle cone to the top of bottom one

h₁ - from the top of inner cone to the ground

 

Probably it will be easier with a picture.

 

 

I've done calculations but for the ease of it I just used a section through the middle of the cone, so that I only have to deal with areas of triangles and in the end I got the following:

 

h₁ = 0.632*h, where h is the total height of the cone

h₂ = 0.262*h

h₃ = 0.106*h

 

Does this seem plausible? I could attach my calculations but those are pretty messy.

 

Thanks for the help!

 

 

 

 

[studiot]

 

Based on this I need to calculate percentages of the total height of the entire cone for the following:

 

Somewhat dificult to comment without knowing the %s involved.

 

[imatfaal]

You can do this with no further information

 

Volume of Cone = 1/3 pi r^2 h

 

so we know

[latex]

V= \frac{\pi r_1^2 h_1}{3}

[/latex]

 

[latex]

2V= \frac{\pi r_2^2 h_2}{3} - \frac{\pi r_1^2 h_1}{3}

[/latex]

 

[latex]

\frac{9}{4}V= \frac{\pi r_3^2 h_3}{3}-\frac{\pi r_2^2 h_2}{3}

[/latex]

 

As V is just a constant we can divide through by pi/3 to make it look nicer and rearrange

 

[latex]r_1^2h_1=W[/latex]

 

[latex]r_2^2h_2 = 2W+r_1^2h_1[/latex]

 

[latex]r_3^2h_3= \frac{9}{4} W+r_2^2h_2[/latex]

 

but we know that

 

[latex]h_i=r_i Tan (\alpha)[/latex]

 

We then have 3 equations in three unknowns (r₁ r₂ r₃) and W - which you can rearrange to give ratios of r₁:r₂ and r₁:r₃ etc. This allows you to get relative sizes of the r₁ r₂ r₃ and via the tangent function of the three heights

 

I do not get the same figures as you. But that would make sense as you cannot simplify to triangles as the triangle's area varies with r whereas the cone's volume varies with r^2. As the information you give us is to do with the volume then we must stick with that.

 

I can finish of the calcs if you are still confused

 

 

 

 

[latex] r_2^2h_2 = 2(r_1^2h_1)+r_1^2h_1 [/latex]

 

[latex] r_2^2h_2 = 3r_1^2h_1[/latex]

 

[latex] r_2^2 \cdot r_2 Tan(\alpha) = 3r_1^2\cdot r_1Tan(\alpha)[/latex]

 

[latex] r_2^2 \cdot r_2 = 3r_1^2\cdot r_1[/latex]

 

[latex] r_2^3 = 3r_1^3[/latex]

By the same method

[latex]r_3^3=\frac{21}{4}r_1^3[/latex]

[John Cuthber]

As Imatfall says, you can't just use the areas because the volumes are not proportional to the areas.

 

If you have two solid objects which are the same "shape" (i.e. all the angles are the same), then the volumes scale as the cube of any characteristic length- such as the height.

So, if the volumes of the whole cones are in the ratio A:B:C

And the enclosed volumes are

A

B - A

C - B

 

 

because they are the differences between the volumes of adjacent cones

the heights must be in the ratio

 

Cube root(A)

Cube root(B-A)

Cube root(C-B)

[assuming I got the arithmetic correct]

This also applies to other things, cuboids, spheres, Russian dolls, whatever.

[pavelcherepan]

Wow! Thanks a lot guys! I thought that using triangular sections and areas of those might not be the correct way but for some reason couldn't figure out a way to do it with volumes. Imatfaal, you're a legend! I should be fine to finish it from here.

Edited October 15, 2015 by pavelcherepan

[Sensei]

Volume of Cone = 1/3 pi r^2 h

 

so we know

[latex]V= \frac{\pi r_1^2 h_1}{3}[/latex]

Right.

 

[latex]2V= \frac{\pi r_2^2 h_2}{3} - \frac{\pi r_1^2 h_1}{3}[/latex]

Shouldn't be here

[latex]2V= \frac{\pi r_2^2 (h_1+h_2)}{3}[/latex]

??

 

[latex]\frac{9}{4}V= \frac{\pi r_3^2 h_3}{3}-\frac{\pi r_2^2 h_2}{3}[/latex]

Shouldn't be here

[latex]\frac{9}{4}V= \frac{\pi r_3^2 (h_1+h_2+h_3)}{3}[/latex]

??

 

According to this info h2 and h3 are partial heights:

 

"h3 - from the top of the outer cone to the top of middle cone

h2 - from the top of middle cone to the top of bottom one

h1 - from the top of inner cone to the ground"

Edited October 16, 2015 by Sensei

[pavelcherepan]

Sensei, you're right here, but using full heights of all three cones makes it easier to get to the final result. In my original calculations (which were obviously incorrect but only by a difference between square root and cubic root of the same number) I used partial heights and getting to the result was very tedious. It's much simpler with full heights as imatfaal has done and I can still get ratios I need.

[Sensei]

Sensei, you're right here, but using full heights of all three cones makes it easier to get to the final result. In my original calculations (which were obviously incorrect but only by a difference between square root and cubic root of the same number) I used partial heights and getting to the result was very tedious. It's much simpler with full heights as imatfaal has done and I can still get ratios I need.

If h2 and h3 are absolute, not relative (unlike your original post) then

 

[latex] 2V= \frac{\pi r_2^2 h_2}{3}[/latex]

and

[latex] \frac{9}{4}V= \frac{\pi r_3^2 h_3}{3} [/latex]

are true..

 

You never said that 2V is difference in volumes between the smallest cone and middle..

 

"If we use V as the volume of the bottom cone, then middle cone is 2V and the outer (largest) cone is 9/4V. "

 

so I assumed Vsmall=V, Vmiddle=2V, Vouter=9/4V

[pavelcherepan]

True here as well. My original conditions were that Vsmall = V, Vmiddle = 2V, Vouter = 9/4V, probably imatfaal misread it or i didn't explain too well, but what matters are not numbers themselves, but the method of solution. I've done calculations using the proper volumes, it's too easy once the method of arriving to the result has been shown so well.

[imatfaal]

We have 3 cones nesting inside one another or a cone with 3 "layers"

 

From the OP - this made me think it was a physical object with 3 objects nesting inside each other as did the picture; with object 1 being a standard cone with a circular base and object2 fitting nested over the top of object one to create a larger cone, and object three over both the others respectively. If they are just three cones then why use the word "nested" or "layers"

Right.

 

 

Shouldn't be here

[latex]2V= \frac{\pi r_2^2 (h_1+h_2)}{3}[/latex]

??

 

 

Shouldn't be here

[latex]\frac{9}{4}V= \frac{\pi r_3^2 (h_1+h_2+h_3)}{3}[/latex]

??

 

According to this info h2 and h3 are partial heights:

 

"h3 - from the top of the outer cone to the top of middle cone

h2 - from the top of middle cone to the top of bottom one

h1 - from the top of inner cone to the ground"

 

Your figures would only apply if the cones were solid - ie all three C1 (r1 h1 V) C2 (r2 h2 2V) and C3 (r3 h3 9v/4) are standard cones. The impression I got from the OP - and still do - is that C2 fits over C1, and C3 fits over C2.

 

But you are correct my r and h are all absolute - not partial; I converted to percentages later on in my handwritten stuff.