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I am trying to make a corner of a robot I am designing flush for aesthetic reasons as well as safety reasons but I'm not sure how to make the arch of the corner lay flush with the two lines that make up the edge.

Here is a photo of the side of the robot (top front of the starboard side). The green lines illustrate the perimeter, the gray arch shows the angle between the two green lines (which is 72.9 degrees), and the black arch is the arch that I would like to create. How can I find the correct radius for this black arch so it will connect the two sides without any angles? enter image description here

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Any radius can be used, to create an arc that will touch the two sides and make no angle. There must be some piece of information you're not telling us, is there? –  user22805 Oct 4 '12 at 2:48
    
Sorry, my geometry is not too sharp. What is in the picture is really all I know. If there is something else that I need, I can probably find it out... But I'm not sure what it is. –  Sponge Bob Oct 4 '12 at 2:51
    
OK, but what I'm saying is that you can use any sized curve you like, within reasonable limits, and make it work. –  user22805 Oct 4 '12 at 2:53

3 Answers 3

You're already using a drawing program like Autodesk Inventor so just use a tangent constraint and inventor will calculate the radius for you.

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One way to construct it is to choose the radius of the circle to start. If you call it $r$, draw lines parallel to each side of the angle and inboard by $r$. Where those lines intersect is the center of the circle.

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I do not know whether I am answering your question: there are many circular arcs that qualify.

Let $C$ be the centre of the desired circle. Let $A$ and $B$ be the two points of tangency, and let $O$ be where your two green lines meet at $72.9$ degrees. Let $2\theta$ be your angle, here $72.9$ degrees. So $\theta$ is $36.45$ degrees.

Let $r$ be the radius of the circle. Then $$r=(OA)\tan\theta.$$ This is because the radial line $CA$ meets the tangent line at right angles. Alternately, in terms of the distance $CO$, we have $$r=(CO)\sin\theta.$$

If you specify something else than $OA$ or $CO$ that determines $r$, I can add to this answer.

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