The angle of engagement of a milling tool is not often thought about directly, but usually indirectly when considering the behavior of a tool when entering a corner (Chatter!! or Breakage!).
Some sites show how to calculate the Angle of Engagement (We Will TOO!), but mainly you get generic graphical representations that look like this:
I personally find it frustrating when I’m interested in a deep dive in machining theory or would like to decide how deep to go. When faced with the above, theres no where to go because you’re in the shallow end of the information pool. So all you can do is machine like Rock Chipper.
We’re gonna try to be more useful than that.
Skip Ahead to the Answer:
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When a tool manufacturer provides you with cutting data, the angle of engagement is fully defined with the manufacturer provided parameters and can be a useful tool to understand tool load.
Modern “advanced” CAM roughing cycles are more in-tune with tool engagement and work to improve corner machining by preventing tool load spikes by doing things like maintaining a constant engagement angle, reducing step over and feed rate reductions.
Traditional tool paths calculations are based on a constant step over provided by the the programer which leads to load spikes as the internal radii being machined approaches the size of the tool. (CNC operators usually have to baby the machining and maybe wear earplugs)
The following diagram hopefully helps visualize the angle of engagement in several scenarios: .75″ Diameter, 6% step over
How do we compute the angle of engagement if we want to??!?
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We’ll start with defining our primary variables.
\[\] \begin{aligned} \Theta_{ae} = \text{Angle of Engagement} \end{aligned} \[R_T = \text{Radius of Tool}\] $$SO = \text{Step Over}$$
Next we’ll breakdown our cutting operation into triangles. The goal is to produce Right Triangles (has a 90° Corner).
We have created a few more pieces so we will define any secondary variables to aid us in calculation and start creating our important relationships.
Secondary Variables:
\[R_C = \text{Left over portion of Line}\]
Our main interest is \(\Theta_{ae}\), so let us define it. Using our good ol’ friend and mnemonic SOH-CAH-TOA, \[\cos(\Theta_{ae}) = \frac{R_C}{R_T}\]
\(R_C\) is not currently known, but we can find it.
The Red line and the Blue line are the same length, \(R_T\).
So going from tool center to the tangent point of the cutter (Red Line), that means that the red line, \(R_T\), is made up of \(R_C\) and \(SO\). So, \[R_T = R_C + SO\]
Rearranging, we get \[R_C = R_T – SO\] We can, now, substitute the expression for \(R_C\) into the equation for \(\Theta_{ae}\). \[\cos(\Theta_{ae}) = \frac{R_T – SO}{R_T}\]
We can simplify the equation a bit by separating the common denominator. \[\cos(\Theta_{ae}) = \frac{R_T}{R_T} – \frac{SO}{R_T}\] which simplifys to \[\cos(\Theta_{ae}) = 1 – \frac{SO}{R_T}\]
To solve for \(\Theta_{ae}\) we take the inverse cosine/\(\arccos\)/\(\cos^{-1}\) of both sides, \[\cos^{-1}(\cos(\Theta_{ae})) = \cos^{-1}(1 – \frac{SO}{R_T})\]
Giving us our final equation for \(\Theta_{ae}\), \[\boxed{\Theta_{ae} = \cos^{-1}(1 – \frac{SO}{R_T})}\]In case you skipped here: \[\Theta_{ae} = \text{Angle of Engagement}\] \[R_T = \text{Tool Radius}\] \[SO = \text{Step Over}\]
If you prefer to keep everything as a percentage of step over. We’ll make an additional relationship: \[SO = \frac{SO_{\%}}{100} * 2R_T\]
YouTube Tutorial:
Link to PDF of work from the video tutorial:
My next post will likely be an online calculator to compute the Angle of Engagement.
What is it good for? Where do we go from here? Directly, I haven't found the angle of engagement to be that useful by itself, but it is necessary intermediate if you want to calculate the force on the tool or develop other cutting relationships.
This whole site started because a coworker asked me for a calculator to help him determine the step over necessary to maintain a constant angle of engagement for a specified corner radius. I couldn't find it online, so I derived it myself and I felt the need to put it "Out There". The math is pretty hairy and non-linear. I will show the mathematics and probably make an calculator in a future post.