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 ANYANY 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:

x=r(1ϕ360)x=r\left(1-\dfrac{\phi}{360}\right)

θ=arccos(1ϕ360)\theta=\arccos \left(1-\dfrac{\phi}{360}\right)

h=ϕr2180(1ϕ720)h=\sqrt{\dfrac{\phi r^2}{180} \left(1-\dfrac{\phi}{720}\right)}

volume=π×ϕr2180(1ϕ720)×r2(1ϕ360)23volume=\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. xr=2πx2πr\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 2πr(1ϕ360)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(2πr(1ϕ360)2πr)r(1ϕ360)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

r(1ϕ2π)r\left(1-\dfrac{\phi}{2\pi}\right)



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

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

θ=arccos(r(1ϕ360)r)arccos(1ϕ360)\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(1ϕ2π)\arccos \left(1-\dfrac{\phi}{2\pi}\right).

IF ANYONE HAS A GEOMETRIC FORMULA FOR THIS ANGLE PLEASE COMMENT IT.



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

h=r2x2h=\sqrt{r^2-x^2}

h=r2r2(1ϕ360)2h=\sqrt{r^2-r^2\left(1-\dfrac{\phi}{360}\right)^2}

h=r2(r22r2ϕ360+r2ϕ2720×180)h=\sqrt{r^2-\left(r^2-\dfrac{2r^2 \phi}{360}+\dfrac{r^2\phi^2}{720\times 180}\right)}

h=r2ϕ180r2ϕ2720×180h=\sqrt{\dfrac{r^2 \phi}{180}-\dfrac{r^2\phi^2}{720\times 180}}

h=r2ϕ180(1ϕ720)h=\sqrt{\dfrac{r^2\phi }{180} \left(1-\dfrac{\phi}{720}\right)} or in radians r2ϕπ(1ϕ4π)\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=13hx2=π×ϕr2180(1ϕ720)×r2(1ϕ360)23volume=\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

13hx2=π×ϕr2π(1ϕ4π)×r2(1ϕ2π)23\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
4 years, 11 months ago

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Trevor Arashiro - 4 years, 11 months ago

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