JD Mather Posted July 8, 2009 Posted July 8, 2009 Why the hassle? Why let wrong solutions stand if it is indeed wrong? Is it right or is it wrong. If it is right then I'm missing something basic and the question is still open for me. The problem is not solved for me. Educate me. Show me what I'm missing. Quote
eldon Posted July 8, 2009 Posted July 8, 2009 If you were to go through Cad Tutor and delete all answers that did not answer the question and wrong answers, you would make the number of posts far fewer. Life is too short for that, and you would discourage folk from having a try at solving other's problems, which is what this forum is about. Quote
SEANT Posted July 8, 2009 Posted July 8, 2009 Rob-GB’s method works if the Chord Perpendicular Bisector is extended to intersect with the original circles horizontal centerline. See attached. Circle solution_Extended.dwg Quote
PeteUK Posted July 8, 2009 Author Posted July 8, 2009 Jeez, calm down people! JDM's answer seemed correct to me, but I didn't fully understand how he'd got there, so meant for him to clarify (ie he answered my question 99%). I wasnt slagging his answer off! Eldon and Seant - thanks I kind of understood your method Eldon, but the example by Seant let me fully understand what you explained. Thanks to you all - but stop bickering! Quote
JD Mather Posted July 8, 2009 Posted July 8, 2009 Rob-GB’s method works if the Chord Perpendicular Bisector is extended to intersect with the original circles horizontal centerline. See attached. Thanks for the clarification. Quote
Rob-GB Posted July 8, 2009 Posted July 8, 2009 Just looked back in! May I just clarify my original answer and in doing so maybe afford a bit of help to future readers of this thread. Firstly my answer was based on the belief that Pete needed to find the centre point of a second cicle that passed through the four red circled points; in that case my answer was correct. However, as it appears that Pete required the original circle to "flow" through to the lower point on each side separately, J D Mather is correct in his solution, may well be his experience as an educator that afforded this insight. The solution to either lies in the same basic geometry. For the second circle the centre point will lay on a vertical line through the original centre point; while for the "flow" to be maintained the centre point will lay on a horizontal line through the original centre point. As has been shown the intersection of either is found by extending a line at right angles to a line between the two points to be connected by an arc. Pete is happy with the solution and good debate has been had, well done all. P.S. have used the word "flow" but in my field we use "easement" or "easing" but thats carpenter & joiners for you.:wink: ATB Rob. Quote
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