# Calculating the trapezoid angle needed to fill the space to produce an n-gon cone

The situation is a bit convoluted. But essentially I am trying to create a bowl using a table saw. This table saw can do angled cuts (we call them mitered cuts) as well as tilt the table saw.

I am following this guide here which gives a table of the saw tilt and miter angle you would allegedly need to fit this shape.

There is only one problem. I think the math for saw tilt angle is a bit off. I realized this when trying to recreate this in cad.

Now he uploaded his own CAD file and I opened it. I am not really super familiar with that software but I am not convinced he tested every case like I can in mine.

For example with 3 sides and a 30 degree tilt he finds a miter angle of 40.89 and a saw blade tilt of 48.59 but I find that leaves a small gap and that the saw blade tilt needs to be more like 56.3 degrees.

I think the calculation for that which he defines as

a=polygonSideAngle (in this case 30 degrees)
splayx = tilt from 90 degrees
splayyc = atan(tan(splayx)/cos(a))
atan(tan(a)*cos(splayyc)


But doesnt really go into the derivation. I am fairly handy with geometry but I cant seem to sort it out myself due to the 3D nature of everything.

Because sides are tilted up that top face that must connect are on two different planes.

Here's how I am essentially seeing the problem. And it is possible I am modeling it wrong or my cad file is wrong. I dont know how to solve it because while I can derive in 2D that the angle between these sides should be (it should be the polygon angle) as soon as the top are tilted... I am not really sure what to do.

Its like I need to figure out the angle needed where when projected at the tilt angle will result in them being together.

I even brute force used my CAD model to estimate the values for side count of 3 to try and give myself an idea of what I might need to brute force this with. The grey line shows promise that this can be solved with the right trigonometry.

I am progressively realizing that linear algebra with vectors and projecting vectors onto a plane is probably a possible solution here. I can actually get my CAD software to calculate this angle this way but cannot drive geometry with that calculation. Even though a linear algebra solution would be great to see an algebraic definition that does not require any matrix/vector multiplications would be much more feasible to put into my CAD software.

The polygonal cone frustrum lateral surface is made up of $$n$$ wooden slabs of thickness $$h$$. Each of these identical slabs makes an inclination angle with the horizontal plane (this is called the Splay angle). This is shown in the OP images.

The bottom base of the frustum is an $$n$$-gon with circumradius $$r_B$$. The apothem of this regular polygon is given by

$$a_B = r_B \cos \left( \dfrac{\pi}{n} \right)$$

the splay angle determines the normal vector to the wooden slab. If the splay angle is denoted by $$\theta_S$$ then the outward-pointing unit normal vector to the outer surface plane is

$$\mathbf{n} = ( \sin(\theta_S) , 0 , -\cos(\theta_S) )$$

Now, assuming the first slab passes through the point $$(a_B, 0, 0)$$, then the equation of the outer surface is

$$\mathbf{n} \cdot (\mathbf{r} - (a_B, 0, 0) ) = 0$$

where $$\mathbf{r} = (x,y,z)$$.

And it follows that the inner surface of the same slab is given by

$$\mathbf{n} \cdot (\mathbf{r} - (a_B, 0, 0)) = - h$$

In addition, we have two joining planes surrounding this slabs which are vertical planes passing through the $$z$$-axis. Their equations are

$$\mathbf{n_1} \cdot \mathbf{r} = 0$$ and $$\mathbf{n_2} \cdot \mathbf{r} = 0$$

where

$$\mathbf{n_1} = \left( - \sin \left( \dfrac{\pi}{n} \right) , \cos \left( \dfrac{\pi}{n} \right) , 0 \right)$$

$$\mathbf{n_2} = \left( \sin \left( \dfrac{\pi}{n} \right) , \cos \left( \dfrac{\pi}{n} \right) , 0 \right)$$

with this information we can compute the tilt angle $$\theta_T$$.

The angle between the joining plane (whose unit normal vector is $$\mathbf{n_1}$$) and the outer surface plane (whose unit normal vector is $$\mathbf{n}$$) is given by

$$\cos \left(\dfrac{\pi}{2} - \theta_T \right) = - \mathbf{n} \cdot \mathbf{n_1}$$

Therefore,

$$\sin(\theta_T) = - ( \sin(\theta_S), 0, - \cos(\theta_S) ) \cdot \left( - \sin \left(\dfrac{\pi}{n} \right), \cos \left( \dfrac{\pi}{n} \right) , 0 \right) = \sin(\theta_S) \sin \left(\dfrac{\pi}{n} \right)$$

Next, we need to calculate the miter angle. This only depends on the splay angle.

If $$H$$ is the elevation of the a cross-section of the frustum above the bottom base, and $$a$$ is the apothem of this polygonal cross-section, then

$$\tan(\theta_S) = \dfrac{\Delta H}{\Delta a}$$

If $$s$$ is the side length of the cross-section polygon, then we know that

$$a = \dfrac{s}{2} \cot\left(\dfrac{\pi}{n}\right)$$

If $$\alpha$$ is the miter angle then

$$\tan(\alpha) = \dfrac{ \dfrac{\Delta s}{2} }{\dfrac{\Delta H }{ \sin(\theta_S)}}$$

Therefore,

$$\tan(\theta_S) = \dfrac{\Delta H}{\Delta s} \cdot 2 \tan \left(\dfrac{\pi}{n}\right) = \left( \dfrac{1}{\tan(\alpha)} \right) \left(\dfrac{\sin(\theta_S)}{2} \right) \left( 2 \tan\left( \dfrac{\pi}{n} \right) \right)$$

Therefore,

$$\tan(\alpha) = \cos(\theta_S) \tan \left(\dfrac{\pi}{n} \right)$$

I've built this frustum in a VBA (Visual Basic for Applications) code (for Microsoft Excel), and obtained the following view of the frustum.

I've verified that the two formulas derived above produce the correct values of the tilt and miter angles.

I've also built the polygonal cone frustum ($$n = 6$$) using Blender. The following three images show different views of the object.

• Thankyou so much. Especially the definitions of the planes involved are useful. I am however not completely following. Do you mean that the miter angle should be defined as atan(cos(splay)*tan(pi/n)) ? For example with 6 sides and a splay which is 30 degrees from horizontal I calculated with your formula a miter angle of 26.5 degrees which did not seem to work. Commented Aug 4 at 21:17
• Also just wanted to confirm that in this formula 𝑎=𝑠2cot(𝜋𝑛) what is that a? Its not the miter angle is it? Commented Aug 4 at 21:19
• Yes 26.5 is correct. I don't know what you mean with "did not seem to work"??? As for the second question, $a$ is the apothem (which is the length of the perpendicular drawn from the center of the polygon to a side. Commented Aug 4 at 22:36
• Sorry did not seem to work in my drawing of it. That is how I have been checking my work. Turns out that was an error on my part. Commented Aug 4 at 23:42
• But ok for a 6 sided vase tilted 45 deg I got a saw angle of 20.7 and a mitter of 22.2. That leaves a bit of a gap and I think the tilt should come out to about 22.2? Commented Aug 4 at 23:45