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Maybe it is a stupid question but I can't seem to wrap my head around it. If I have a cone where the first circular face has a radius R1 and the second circular face has radius R2 and the height between those surfaces is H, how can I calculate the corresponding radius of the cone to every height between h = 0 and h = H?

Thanks in advance!

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Since the slope of a cone is a straight line, there is a linear relationship between the difference in radius and the difference in height:

$$(r-R_1) = constant \times (h-H_1)$$

Here $H_1$ is the $h$-value at the $R_1$ point, let's assume it corresponds to $h=0$. This implies:

$$H_1=0$$

So for $h=0$ we get $r-R_1=0$, so obviously $r=R_1$ in that case, just as expected.

And for $h=H$ we should get $r=R_2$. So we have $R_2-R_1 = constant \times H$. We can now solve $constant = \dfrac{R_2-R_1}{H}$, and substitute that in the first formula which gives the formula for $r$: $$r=R_1 + \dfrac{R_2-R_1}{H} h$$

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If $R_2$ is the small radius and $R_1$ is the big radius, you can find any radius in between by this formula: $$r=R_2+\frac{hH}{R_1-R_2}$$ This is possible because the cone angle is the same and its tangent is $\frac{H}{R_1-R_2}$

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  • $\begingroup$ This doesn't seem right. Plugging in $h=H/2$ should give $r=(R_1+R_2)/2$, but it doesn't. $\endgroup$ – Cuspy Code Feb 22 '18 at 18:13
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Let $r$ and $R$ be the radii of small and large bases and $H$ the height of the truncated cone. The radius $\bar{r}$ of the circle at the height $h$ from the large base is: $$\bar{r}=r+\frac{(H-h)(R-r)}{H}.$$

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