I’d like some help to understand how to derive the relationship of the intersection of a cone with a plane parallel to the cone axis. See below dimensions of the shapes involved.
What I want to define is shown in the next image.
The CAD is for illustration only as I'm doing these calculations on excel for multiple dimensions. So, I can define the height of the red-dotted shape (on the intersection), which is 3.122-1.1, but not sure how to get the length of that shape (I think is a hyperbola but not sure).
Any ideas? Let me know if you need more clarification.
If the base of the cone has diameter $D$ , and its height is $H$, then its semi-vertical angle is
$ \theta_c = \tan^{-1}\left( \dfrac{D}{2H}\right) $
The equation of the surface of the cone assuming the origin is at its apex is
$ x^2 + y^2 - z^2 \tan^2 \theta_c = 0 $
Cutting it with the plane $ x = x_0 $ (where $x_0 = 15.95$) results in the hyperbola
$ z^2 \tan^2 \theta_c - y^2 = x_0^2 $
Setting $ z = - H + \Delta z $ , where $\Delta z = 1.1 $, we get
$ (- H + \Delta z)^2 \tan^2 \theta_c - y^2 = x_0 ^ 2 $
Therefore the distance that you want is given by
$ \Delta y = 2 \sqrt{ ( - H + \Delta z)^2 \tan^2 \theta_c - x_0 ^ 2 } $
Which can be re-written as
$\Delta y = 2 \sqrt{ ( - H + \Delta z)^2 \left(\dfrac{D} { 2 H} \right)^2 - x_0 ^ 2 } $
Using $ H = 35.04 , D = 35.04 , x_0 = 15.95 , \Delta z = 1.1 $ results in
$\Delta y = 2 \sqrt{ ( - 35.04 + 1.1)^2 \left(\dfrac{35.04} {70.08} \right)^2 - 15.95 ^ 2 } = \boxed{11.59}$