PeteUK Posted July 7, 2009 Posted July 7, 2009 Hi I've got a circle with the radius of 42.15mm but then it changes half way round to meet two points 40mm out from the centreline. Basically I need to draw a second circle to meet the points (circled red) so that the shape stretches out from being a circle to these two points. I have the position of the four points that I want the circle to meet, but cant remember how to figure out the centrepoint of the second circle or the radius of the second circle. Hopefully the image below will help understand what I mean. Quote
Zorg Posted July 7, 2009 Posted July 7, 2009 Hi I've got a circle with the radius of 42.15mm but then it changes half way round to meet two points 40mm out from the centreline. Basically I need to draw a second circle to meet the points (circled red) so that the shape stretches out from being a circle to these two points. I have the position of the four points that I want the circle to meet, but cant remember how to figure out the centrepoint of the second circle or the radius of the second circle. Hopefully the image below will help understand what I mean. maybe you need the ellipse tool? Quote
JD Mather Posted July 7, 2009 Posted July 7, 2009 Using a parametric software like Autodesk Inventor reveals that there are an infinite number of solutions - you need more information. However, if you continued with the same radius there is a unique solution. (see attached, looks like I put in the wrong radius but that is not relevant to the problem statement) Quote
eldon Posted July 7, 2009 Posted July 7, 2009 I've got a circle with the radius of 42.15mm I have the position of the four points that I want the circle to meet There is no vertical dimension to position the lower points. If you want one circle to encompass the four points, then you could use the 3P option to draw a circle. Quote
JD Mather Posted July 7, 2009 Posted July 7, 2009 I have the position of the four points that I want the circle to meet... So, what is the distance from the horizontal centerline to the horizontal 40mm lines? With that information a unique solution can be found. If you had zipped and attached your file here your question would already be answered. The (reference) dimension is dependant on the vertical dimension. see attached. Quote
PeteUK Posted July 7, 2009 Author Posted July 7, 2009 Apologies, I should have included that. The vertical dimension between the original circle centreline and the '80mm' line is 20.2mm Quote
JD Mather Posted July 7, 2009 Posted July 7, 2009 Apologies, I should have included that. The vertical dimension between the original circle centreline and the '80mm' line is 20.2mm General geometric solution for the two arcs tangent. Trim the access. If you are not looking for the tangent solution I think more information is needed. I get R95.968023256 mm Quote
eldon Posted July 7, 2009 Posted July 7, 2009 Here we go. A circle through three points (choose any three out of your four known points) gives a radius of 42.54 Quote
eldon Posted July 8, 2009 Posted July 8, 2009 To put the solution a bit more formally, the centre of a circle lies on the perpendicular bisector of a chord. Quote
PeteUK Posted July 8, 2009 Author Posted July 8, 2009 General geometric solution for the two arcs tangent. Trim the access. If you are not looking for the tangent solution I think more information is needed. I get R95.968023256 mm This is the correct answer (thanks!) but how did you figure it out? I don't follow your description above. Quote
eldon Posted July 8, 2009 Posted July 8, 2009 With an answer like that, it is obvious that I did not understand the problem. Could you please post a screen shot, so that I can see how it worked out. Thank you. Quote
PeteUK Posted July 8, 2009 Author Posted July 8, 2009 Apologies if I didn't describe the problem well enough. The circle above the horizontal centreline is correct, but once it crosses this centreline, it should then change radius to meet the second point. However the detail between the end of the 'upper' circle and the new arc should be smooth (i.e. no valley created). The answer JD Mather is 99% correct but the answer has been rounded up so its not 100%. Hopefully he can explain how he got the answer so I can run through it myself. I've attached a dwg. file which will hopefully let people understand the problem. Problem.dwg Quote
eldon Posted July 8, 2009 Posted July 8, 2009 Now I see the problem, and it didn't take long to find a graphical solution. You have to use the property in AutoCAD that if you first draw a line, then immediately draw an arc, if you give a null response to the start point of the arc, it is drawn tangentially to the line. So, offset the diameter upwards. Then draw a line from the offset line to the diameter line. Start drawing an arc, giving a null response to the start point (just press enter) and click on the required point below. I think you are being a bit harsh on JD Mather being only 99% correct, because the solution is not an exact integer, and depends on the amount of decimal places. I got 95.968023255814, but for more decimal places you would have to evaluate the trigonomical expression. (20.2² + 2.15²) / (2 * 2.15) Quote
Rob-GB Posted July 8, 2009 Posted July 8, 2009 To put the solution a bit more formally, the centre of a circle lies on the perpendicular bisector of a chord. Which is what my earlier post shows. Basic geometry. Quote
eldon Posted July 8, 2009 Posted July 8, 2009 I know. But isn't it nice when great minds think alike, and someone else puts the same point in a different way. Makes it easier for folk to follow the thought pattern Quote
JD Mather Posted July 8, 2009 Posted July 8, 2009 Which is what my earlier post shows. Basic geometry. What your "solution" shows is not the solution to the problem. It is wrong. Basic geometry. Quote
JD Mather Posted July 8, 2009 Posted July 8, 2009 I know. But isn't it nice when great minds think alike, and someone else puts the same point in a different way. How is it alike? My mind must not work like this. I don't see and tangency in Rob's "solution". Quote
eldon Posted July 8, 2009 Posted July 8, 2009 How is it alike? My mind must not work like this. I don't see and tangency in Rob's "solution". Why the hassle? The problem is solved and nobody gets extra points. Quote
JD Mather Posted July 8, 2009 Posted July 8, 2009 The answer JD Mather is 99% correct but the answer has been rounded up so its not 100%. Hopefully he can explain how he got the answer so I can run through it myself. The graphical solution I presented is 100% correct (or at least as far as the CAD software is concerned). You should be able to figure it out from the graphical solution with no calculations (or further explanation) needed. Quote
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