# Solving for radius of a combined shape of a cone and a cylinder where the cone is base is concentric with the cylinder?

I have a solid that is a combined shape of a cylinder and a concentric cone

(a round sharpened pencil would be a good example)

Know values are:

Total Volume = 46,000

Height to Base Ratio = 2/1 (Height = cone height + cylinder height) (Base = Diameter)

Angle of cone slope = 30 degrees (between base and slope of cone)

How do you solve for the Radius?

2 let say the ratio is from the orignal length

Know value:

Total Volume = V = 46,000

Original Height to Base Ratio = t = 2/1 (Height = cone height + cylinder height + distance shortened) (Base = Diameter)

Angle of cone slope = θ = 30 degrees (between base and slope of cone)

distance shortened from original Height x = 3

(Let h be the height of the cylinder. Then h+rtanθ+x=2(2r)

How do you solve for the Radius?

• Do you know the equations for the volume of a cylinder and cone? Also, can you clarify what is the height to base ratio? Is that the total height including both structures, or just the cone? Is the "base" in that case the radius? Jul 24, 2013 at 18:53
• Height is the Total Height cone plus cylinder. Jul 24, 2013 at 20:48
• Base is the Diameter Jul 24, 2013 at 20:48
• what is that after this line "2 let say the ratio is from the orignal length". Is that your attempt? A different question?
– MonK
Mar 23, 2015 at 13:24

I assume that by height to base ratio you mean the ratio of total height (cylinder plus cone) to the diameter of the base.

Let $\theta$ be the angle between the base of the cone and the sloping walls. Then the height of the cone is $r\tan\theta$. In our case $\tan\theta=\frac{1}{\sqrt{3}}$. Kind of a stubby cone. We keep on writing $\tan\theta$ instead of $\frac{1}{\sqrt{3}}$, it makes typing easier.

Let $h$ be the height of the cylinder. Then $h+r\tan\theta=2(2r)$, and therefore $h=r(4-\tan\theta)$.

The combined volume is $\pi r^2h+\frac{1}{3}\pi r^3\tan\theta$. Substituting for $h$, and simplifying a little, we find that the volume is $$\pi\left(4-\frac{2}{3}\tan\theta \right)r^3.$$ Now we can set this equal to $46000$ and solve for $r$. I get something close to $15.94$, but do check!

• You are welcome. Unfortunately I do not understand the last three comments. Any question will need to be written out at greater length, perhaps as an addendum to the original question. Jul 24, 2013 at 22:13
• lets say the ratio is from the orignal length of the pencil and you have the distance the pencil was shortened lets say x=3 Let h be the height of the cylinder. Then h+rtanθ+x=2(2r) Jul 24, 2013 at 22:48
• Yes, if the ratio is from some sort of original length, where $x$ is known, then you would do the same thing with relation as you have written it. So we end up with volume the one I wrote down, minus $\pi xr^2$. That gives us an unpleasant cubic equaition for $r$. In principle we could use the Cardano formula, in practice I would solve the cubic equation using some numerical method (like bisection, or Newton Method). Jul 24, 2013 at 23:54

Cone height = $\sqrt 3 R$.

Cylinder height = Total height - Cone heght

$$= (4 -\sqrt3 R )$$

Total Volume

$$V_{total} = V_{cone}+ V_{cyl } = \pi R^2 [ (4- \sqrt 3) R ] + \frac\pi3 R^2 \cdot \sqrt3 R = 46000.$$

from which only unknown $R^3, R$ can be calculated.