Helical cross-section as result of lathe/milling operation (cylinder intersection)

Question

I suspect a simple wooden toy "lead screw" was made by advancing a cylindrical rotary cutting tool ( Cylindrical End Mill Cutter) along the surface of the rotating wooden dowel (base cylinder), resulting in a helical cut (the axes of the cylinders are orthogonal (skew).

Videos of the manufacturing process close to what I suspect:

• https://youtu.be/5U9lJAgU1oE?t=31 (but: spherical cutter. radial, not tangent end mill)
• https://youtu.be/y5DOQWiexOQ?t=314 (cutter radial)
• https://youtu.be/pbaRRsG3BN4?t=9 (not radial, but "tilted")

I have tried to visualise/emulate the resulting geometry using multiple difference operations for cylinder primitives in (Open)JSCAD (see code at end of post) and adjusted the view manually:

What is the equivalent (elliptical?) shape that is the cross-section of the helical path?

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And: what is the contact surface/line/point of another, slightly smaller cylinder that is used as "lead screw nut" (having the same orientation as the cutting cylinder, i.e. orthogonal to the base cylinder) - a point contact on one of the helical edges?

Code for JSCAD

function main () {
    let main = cylinder({r: 3, h:10, center: true, fn: 64 });
    for (let i=0; i<36; i++)    {
        let cut = cylinder({r: 0.2, h:10, center: true});
        cut = translate([0,-3,0],cut);
        cut = rotate([0,90,i*3],cut);
        cut = translate([0,0,i*0.1],cut);
        main = difference(main, cut);
    }
    return main;
}

I think the underlying question may be about the surface created by a straight line moved along a spiral ( or helix):

(created with Blender: a mesh edge with Screw modifier)

Or the surface created by a helix that has been rotated (spin):

The cross-section of the "cutting" cylinder (End mill) is a circle of course, which is what an infinite number of cuts "converge" to (a cylinder with zero length).

Then the cross-section along the helix should be an ellipse (intersection of the hypothetical "cutting" cylinder (End mill cutter) and the plane orthogonal to the helix).

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It's not the same as moving a circle along the helix; to illustrate, I've reduced the cylinder's length:

My "straight line" theory does not apply either, I think these "lines" might be helices created by the intersection of the translated and rotated "cutting" cylinders.

So it seems this might be much more involved than I anticipated -- please don't spend too much time on this on my account. I was just curious to see whether the "cut" could be better created in 3D by "lofting" the equivalent cross-section along a helix.

Answer 1

If I am right, the cutting surface is described by a revolving vertical circle, the center of which describes an helix of vertical axis.

Parametrically:

$$\begin{cases}x=(R+r\cos u)\cos\ t,\\y=(R+r\cos u)\sin t,\\z=r\sin u+a^2t.\end{cases}$$

At time $0$, the plane normal to the helix is normal to the tangent vector $(0,R,a)$ and has the equation

$$Ry+az=0.$$

Hence the equation of the intersection of the cutting surface and the normal plane is given by the condition

$$R(R+r\cos u)\sin t+ar\sin u+a^2t=0.$$

This equation is transcendental in $t$, but can be solved for $u$.

$$(r\sin t)\cos u+(ar)\sin u=-(a^2t+R^2\sin t)$$

gives

$$r\sin u=\frac{-a(a^2t+R^2\sin t)\pm \sin t\sqrt{(r\sin t)^2+(ar)^2-(a^2t+R^2\sin t)^2}}{\sin^2t+a^2}$$

and $$r\cos u=\frac{-\sin t(a^2t+R^2\sin t)\mp a\sqrt{(r\sin t)^2+(ar)^2-(a^2t+R^2\sin t)^2}}{\sin^2t+a^2}.$$

Finally, the curve is given by the planar coordinates $\left(x,\dfrac{-ay+Rz}{\sqrt{a^2+R^2}}\right)$ obtained by rotating the coordinate frame. The final equation is terrible. It does not describe an ellipse.

$$\begin{cases}x=\left(R+\dfrac{-\sin t(a^2t+R^2\sin t)\mp a\sqrt{(r\sin t)^2+(ar)^2-(a^2t+R^2\sin t)^2}}{\sin^2t+a^2}\right)\cos t, \\y'=\\\dfrac1{\sqrt{a^2+R^2}}\left(-a\left(R+\dfrac{-a(a^2t+R^2\sin t)\pm\sin t\sqrt{(r\sin t)^2+(ar)^2-(a^2t+R^2\sin t)^2}}{\sin^2t+a^2}\right)+R\left(\dfrac{-a(a^2t+R^2\sin t)\pm \sin t\sqrt{(r\sin t)^2+(ar)^2-(a^2t+R^2\sin t)^2}}{\sin^2t+a^2}+a^2t\right)\right). \end{cases}$$

I leave the study of this curve to the future generations.

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Resolution of the trigonometric equation:

$$a\cos u+b\sin u=c\implies a^2(1-\sin^2u)=c^2-2bc\sin u+b^2\sin^2u$$

gives

$$\sin u=\frac{bc\pm a\sqrt{a^2+b^2-c^2}}{a^2+b^2}$$

and by symmetry

$$\cos u=\frac{ac\mp b\sqrt{a^2+b^2-c^2}}{a^2+b^2}.$$

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Disclaimer: done by hand, typos are not excluded.

Answer 2

Trying to understand the motions. At first I imagined that you were referring to a simple twisted tube of $(x,y,z)$ parametrization:

however no clue about where the straight straight line generators come from. So this is discarded. What appears to me is that a rotating milling cutter offset distance $b$ normally from cylinder axis and skewed angle $\alpha$ with respect to axis of the vertical wooden cylinder. It is mounted on a stationary tool post and mills out a one-sheeted Hyperboloid of revolution. If the tool post in addition moves with a helical pitch $ p= c\, \theta$ is the added torsion component around the vertical axis of cylinder. Required is a parametrization of the generated ruled surface.

Hyperboloid of one sheet is seen in parametrization when $c=0.$ When $z$ motion is imparted to the rotating milling cutter we are adding as pitch $p= 2 \pi c $ for each turn of the lathe.

$$r(u)= \sqrt{(u \cos \alpha)^2+b^2}$$

$$ (x,y,z)=(r(u) \cos \theta,r(u) \sin \theta,u \sin \alpha + c\, \theta). $$

EDIT1;

(In view of clarification)..The following surface is a helical channel groove made by a standard end-mill or router that can be CNC programmed:

Answer 3

With videos the question is more clear. We can consider the simplest router section to be a rectangular groove like a channel in a flat plate.

When wrapped around a cylinder isometrically we have a helical strip as channel bottom flanked by two developable helicoids on either side.

The cross section in general can be chosen arbitrary to suit any router shape that removes (wood) material by radial plunge and helical motion driven by lead screw.

The groove widths are $( w/\cos \alpha,w/\sin\alpha)$ along (circumferential,axial ) directions respectively.

If instead of a square router a ball end mill diameter $ 2w$ is chosen then we have the same radial dimension of groove but ellipses of axial dimension major axes $( 2w/\cos \alpha, 2w/\sin\alpha)$.

( to be continued)