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Generalized formulas: turning the circle into the cone

Recently, I saw a problem on Facebook asking for the height of a cone created by removing a certain degree of a circle. So I decided to go above and beyond and derive a generalized form. IF YOU HAVE \(ANY\) QUESTIONS, PLEASE ASK BECAUSE READINGS NOTHING WITHOUT UNDERSTANDING.

In this note, I will discuss a generalized formula for the height of a cone, angle of repose, radius of a cone's base, and the volume of a cone given the angle cut from the circle.

here is a list of the formulas:


\(\theta=\arccos \left(1-\dfrac{\phi}{360}\right)\)

\(h=\sqrt{\dfrac{\phi r^2}{180} \left(1-\dfrac{\phi}{720}\right)}\)

\(volume=\dfrac{\pi \times \sqrt{\dfrac{\phi r^2}{180} \left(1-\dfrac{\phi}{720}\right)} \times r^2\left(1-\dfrac{\phi}{360}\right)^2}{3}\)

For those who aren't familiar with how to make a cone from a circle, click here to see how to make a cone from a circle Just start watching from where the video starts (it should start at3:45)

Refer to the above picture for reference to these terms. First, we begin by finding the radius of the cone. Because all circles are similar, the ratio of the cone base's circumference to the circle's circumference will be equivalent to the ratio of the cone's radius to the circles radius. \(\dfrac{x}{r}=\dfrac{2\pi x}{2\pi r}\).

The cone base circumference=the circumference of the Pac-Man like circle (whats the mathematical term for this anyone?). Now to solve for the circumference of the cone's base in terms of r. The portion removed will be equivalent to \(2\pi r(1-\dfrac{\phi}{360})\). Thus multiplying the ratio of the radii (or circumference since they're equivalent) by the radius of the Pac-Man, we get

\(x=r\left(\dfrac{2\pi r(1-\dfrac{\phi}{360})}{2\pi r}\right)\Rightarrow r\left(1-\dfrac{\phi}{360}\right)\)

Or in radians


If that was confusing, don't worry, everything after this is pretty straight forward.

Next, we will solve for \(\theta\) aka, angle of repose. Because \(\cos \theta =\dfrac{x}{r}\). We can solve for \(\theta\) by using \(\arccos \left(\dfrac{x}{r}\right)=\theta\). Thus we get

\(\theta=\arccos \left(\dfrac{r\left(1-\dfrac{\phi}{360}\right)}{r}\right)\Rightarrow \arccos \left(1-\dfrac{\phi}{360}\right)\).

Or in radians

\(\arccos \left(1-\dfrac{\phi}{2\pi}\right)\).


Next, to solve for the height, we will simply use the Pythagorean theorem.



\(h=\sqrt{r^2-\left(r^2-\dfrac{2r^2 \phi}{360}+\dfrac{r^2\phi^2}{720\times 180}\right)}\)

\(h=\sqrt{\dfrac{r^2 \phi}{180}-\dfrac{r^2\phi^2}{720\times 180}}\)

\(h=\sqrt{\dfrac{r^2\phi }{180} \left(1-\dfrac{\phi}{720}\right)}\) or in radians \(\sqrt{\dfrac{r^2\phi }{\pi} \left(1-\dfrac{\phi}{4\pi}\right)}



Finally, solving for the volume is trivial (this isn't THAT useful because it's basically the normal process)

\(volume=\frac{1}{3}hx^2=\dfrac{\pi \times \sqrt{\dfrac{\phi r^2}{180} \left(1-\dfrac{\phi}{720}\right)} \times r^2\left(1-\dfrac{\phi}{360}\right)^2}{3}\)

Or in radians

\(\frac{1}{3}hx^2=\dfrac{\pi \times \sqrt{\dfrac{\phi r^2}{\pi} \left(1-\dfrac{\phi}{4\pi}\right)} \times r^2\left(1-\dfrac{\phi}{2\pi}\right)^2}{3}\)

Note by Trevor Arashiro
3 years, 3 months ago

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If you have any questions, please add them to this comment by hitting "reply". All questions are worth asking so feel free to ask anything.

Trevor Arashiro - 3 years, 3 months ago

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