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Short version: How do i determine the profile of arc segment from two known points and curvature of the arc?

Long version: As an intro into modelling co-rotating twin-screw extruders i am trying to learn how the profiles of the screws are determined. For those of you who don't know, twin-screw extruders are machines which have two screws intermeshed in such a way that the action of a flight of one screw wipes the root of the other screw alternately. The below image shows the intermeshing of the screw (left) and the self-wiping action in a frontview perspective (right):

enter image description here enter image description here

For this self-wiping action, the screw needs a special geometry determined by a special set of rules.

In the analysis i am following The geometry of self-cleaning twin-screw extruders by C. Rauwendaal. The basic geometric properties of the barrel:

enter image description here enter image description here

where $\alpha_i$ is the intermesh angle, $I_h$ is the intermesh height, $C_l$ is the centerline distance and finally $D$ and $D_r$ are the screw and root diameter respectively. For a given screw diameter $D$, these are related through some geometric considerations: $$\frac{C_l}{D}=\cos\left(\frac{\alpha_i}{2}\right) \qquad \frac{I_h}{D} = \frac{1}{2}\sin\left(\frac{1}{2}\alpha_i\right) \qquad \frac{D_r}{D} = 2\frac{C_l}{D}-1$$ which basically implies that chosing an intermesh angle $\alpha_i$ fixes the centerline distance $C_l$, the intermesh height $I_h$ and the root diameter $D_r$.

Due to symmetry only a quarter of a screw needs to be examined. Such a quarter is further divisible into three sections:

  1. the tip -> $0<\alpha\leq\alpha_t/2$,
  2. the flank -> $\alpha_t/2<\alpha\leq\alpha_t/2+\alpha_f$
  3. the root -> $\alpha_t/2+\alpha_f<\alpha\leq\alpha_t/2+\alpha_f+\alpha_r/2$

where $\alpha$ is the angle from the horizontal, $\alpha_t$ is the tip angle, $\alpha_f$ is the flank angle, $\alpha_r$ is the root angle. As the two screws have the same cross-sectional shape, the root angle should equal the tip angle. Furthermore, the sum of the tip angle, root angle and flank angle should be: $$\frac{\alpha_r}{2}+\alpha_f+\frac{\alpha_t}{2}=\frac{\pi}{2}$$ Given the assumption $\alpha_r=\alpha_t$ as discussed above, this yields for the tip angle: $$\alpha_t=\frac{\pi}{2}-\alpha_f$$

For the case of co-rotating screws, it is given that the flank angle has to be equal to the intermesh angle, i.e. $\alpha_f = \alpha_i$. From this follows then the tip angle $\alpha_t$ and the root angle $\alpha_r$.

Having determined all these geometric properties, I would now like to be able to reconstruct the profile of the screw. Taking the center of the screw as the origin $O$, the root and tip of the screw are easily described as they have constant radii with respect to $O$. However, the flank area with respect to $O$ has a variable radius which ranges from $\frac{D_r}{2}$ at the root angle to $\frac{D}{2}$ at the tip angle. In other words:

$$r = \begin{cases} \frac{D}{2} & 0\leq\alpha<\frac{1}{2}\alpha_{t}\\ R\left(\alpha\right) & \frac{1}{2}\alpha_{t}\leq\alpha<\frac{1}{2}\alpha_{t}+\alpha_{f}\\ \frac{D_{r}}{2} & \frac{1}{2}\alpha_{t}+\alpha_{f}\leq\alpha<\frac{1}{2}\alpha_{t}+\alpha_{f}+\frac{1}{2}\alpha_{r} \end{cases}$$

However, in the article it is given that the shape of the flank is determined by an arc of a circle (with a different unknown origin $C$) with radius $C_l$.

What i would like to know is how I can determine the profile described by $R\left(\alpha\right)$ which is basically an arc segment between two points (i.e. $[r,\alpha]=[D/2, \alpha_t/2]$ and $[r,\alpha]=[D_r/2, \alpha_t/2+\alpha_f]$) with known curvature (radius $C_l$)?

Update: The answer of @ja27 gives a good and easy geometric method of constructing the arc in question:

enter image description here

However, as stated in the question (although maybe not clearly enough), I would like to be to describe the arc segment in terms of origin $O$ and angle $\alpha$ through function $R(\alpha)$, rather than origin $C$ and angle $\theta$.

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  • $\begingroup$ If you draw the correct triangle the problem becomes trivial. $\endgroup$ – ja72 Aug 3 '17 at 21:42
  • $\begingroup$ @ja72 I am looking forward to your answer then, it will be short and sweet! $\endgroup$ – nluigi Aug 3 '17 at 21:56
  • $\begingroup$ That depends. Is the arc with constant curvature, or variable? The title says a different thing from the contents. $\endgroup$ – ja72 Aug 3 '17 at 22:04
  • $\begingroup$ @ja72 The arc has a variable radius with respect to $O$ but constant radius with respect to $C$. $\endgroup$ – nluigi Aug 3 '17 at 22:08
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    $\begingroup$ @ja72 - i understand it is a bit difficult to visualize, i will try and draw a picture for you $\endgroup$ – nluigi Aug 4 '17 at 7:14
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Take the two end points A and B as a line segment and find the perpendicular bisector. The center of the arc is going to be on this line. Now find the point of the line which is a distance $r$ (the radius) from the end point.

pic1

  • You can find the distance $ m = \frac{1}{2} \| AB \| $ and then the distance $d = \sqrt{r^2-m^2}$ to find the point C.

  • Or you can find the angle $\theta = \sin^{-1} \frac{m}{r}$ and then set $d = r \cos \theta$.

Edit 1

I think the follow-up question asks to find the red distance $r$ and angle $\theta$ below from the blue $R$ and $\varphi$:

Polar1

These are found with a little trigonometry as

$$ \begin{aligned} r \sin \theta & = R \sin \varphi \\ r_O + r \cos\theta &= R \cos \varphi \end{aligned} \Rightarrow \begin{aligned} r &= \sqrt{ R^2+r_O^2-2 R r_O \cos \varphi} \\ \theta & = \tan^{-1} \left( \frac{R \sin \varphi}{R \cos\varphi -r_O} \right) \end{aligned} $$

where $r_O$ is the radius of the outer circle. The arc radius is $R=r_O+r_I$ it turns out (I think).

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  • $\begingroup$ Ok, this is a good first step and is indeed a bit trivial as it describes the arc wrt $C$ and $\theta$, how would it then be described wrt $O$ and $\alpha$ which would be my $R(\alpha)$? $\endgroup$ – nluigi Aug 4 '17 at 7:12
  • $\begingroup$ See edit in the answer for this. $\endgroup$ – ja72 Aug 4 '17 at 14:33

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