XmisterIS Posted January 2, 2015 Posted January 2, 2015 I have a simple problem (but on closer inspection it isn't so simple!).Consider two non-intersecting circles with locii C1 and C2 and with radii r1 and r2, respectively. Now, consider a third circle with locus C3 and radius r3. Find r3 as a function of the angle between the vectors C1C3 and C1C2 for all cases where the third circle touches the other two.Seems simple enough in concept ... Quote
Joeman Posted January 2, 2015 Posted January 2, 2015 i reckon it will form a triangle with r1+r3 as one side, r2+r3 as the other, and the distance between C1 and C2 as the third side. Quote
fq-craigus Posted January 2, 2015 Posted January 2, 2015 Erm....... Is this a recipe for a cake Quote
XmisterIS Posted January 2, 2015 Author Posted January 2, 2015 Lol!It's a continuum of triangles. The trouble is, I think it's probably partially differential because I think the solution is probably a symmetric hyperbolic equation, with one imaginary half. My differential calculus is a little rusty! Quote
Tiggie Posted January 2, 2015 Posted January 2, 2015 does adding counter steering to the equation help? Quote
Joeman Posted January 2, 2015 Posted January 2, 2015 whats the context of this anyway?? what crazy stuff are you upto? Quote
Joeman Posted January 2, 2015 Posted January 2, 2015 If im understanding the question correctly, it shouldn't be that bad... you know the length of all three sides of the triangle, so given the angle at C1 or C2 you can calculate the other two angles. Quote
XmisterIS Posted January 2, 2015 Author Posted January 2, 2015 If im understanding the question correctly, it shouldn't be that bad... you know the length of all three sides of the triangle, so given the angle at C1 or C2 you can calculate the other two angles. Not quite - there is a continuum of angles and if you think about it, you don't know the length of all three sides - because you don't know the radius of the third circle (which is what you are trying to find). Trying the cosine rule to form a set of simultaneous equations over which to integrate very quickly leads to nastiness! Quote
soll Posted January 2, 2015 Posted January 2, 2015 https://books.google.co.uk/books?id=W4acIu4qZvoC&pg=PA114&lpg=PA114&dq=Consider+two+non-intersecting+circles+with+locii&source=bl&ots=a8oxrjaca7&sig=HvjqYOTKnJEHJyDTYaM0P3me7yA&hl=en&sa=X&ei=FsGmVIDUCoGXOK_RgfAE&ved=0CC8Q6AEwAg#v=onepage&q=Consider%20two%20non-intersecting%20circles%20with%20locii&f=false Quote
Joeman Posted January 2, 2015 Posted January 2, 2015 If im understanding the question correctly, it shouldn't be that bad... you know the length of all three sides of the triangle, so given the angle at C1 or C2 you can calculate the other two angles. Not quite - there is a continuum of angles and if you think about it, you don't know the length of all three sides - because you don't know the radius of the third circle (which is what you are trying to find). Trying the cosine rule to form a set of simultaneous equations over which to integrate very quickly leads to nastiness! like this?http://www.mylnk.net/joeman/Circles.pngSo you know the angle at C1 and the length C1-C2.you also know r1 and r2. Quote
XmisterIS Posted January 2, 2015 Author Posted January 2, 2015 https://books.google.co.uk/books?id=W4acIu4qZvoC&pg=PA114&lpg=PA114&dq=Consider+two+non-intersecting+circles+with+locii&source=bl&ots=a8oxrjaca7&sig=HvjqYOTKnJEHJyDTYaM0P3me7yA&hl=en&sa=X&ei=FsGmVIDUCoGXOK_RgfAE&ved=0CC8Q6AEwAg#v=onepage&q=Consider%20two%20non-intersecting%20circles%20with%20locii&f=false Bingo! Thanks I thought it would be hyperbolic! Quote
JamBerryKing Posted January 2, 2015 Posted January 2, 2015 are you making a deathray or something? Quote
Tango Posted January 2, 2015 Posted January 2, 2015 When I used to work in a factory we were in the canteen one day queueing for lunch and one of the old girls was chatting to one of the old fellas in front of us. She was a lovely old girl, but definitely a sandwich or 2 short of a picnic and she was telling the old fella that her husband was an accountant. The old fella just shook his head slowly and said "it just don't add up".......and we all fell about laughing.......poor old girl just didn't see what we were laughing about..... Quote
XmisterIS Posted January 3, 2015 Author Posted January 3, 2015 are you making a deathray or something? Yup! Deathray! Quote
Joe85 Posted January 3, 2015 Posted January 3, 2015 When Joeman first answered "triangles" I thought there maybe a cleverly conceived joke in there that was beyond my comprehension. Now my brain aches. Quote
Grumpy Old Git Posted January 5, 2015 Posted January 5, 2015 https://books.google.co.uk/books?id=W4acIu4qZvoC&pg=PA114&lpg=PA114&dq=Consider+two+non-intersecting+circles+with+locii&source=bl&ots=a8oxrjaca7&sig=HvjqYOTKnJEHJyDTYaM0P3me7yA&hl=en&sa=X&ei=FsGmVIDUCoGXOK_RgfAE&ved=0CC8Q6AEwAg#v=onepage&q=Consider%20two%20non-intersecting%20circles%20with%20locii&f=false Wot e sed. Quote
XmisterIS Posted January 5, 2015 Author Posted January 5, 2015 https://books.google.co.uk/books?id=W4acIu4qZvoC&pg=PA114&lpg=PA114&dq=Consider+two+non-intersecting+circles+with+locii&source=bl&ots=a8oxrjaca7&sig=HvjqYOTKnJEHJyDTYaM0P3me7yA&hl=en&sa=X&ei=FsGmVIDUCoGXOK_RgfAE&ved=0CC8Q6AEwAg#v=onepage&q=Consider%20two%20non-intersecting%20circles%20with%20locii&f=false Wot e sed. It's not that complicated, it's just the rudiments of a hyperbolic differential, and they're like, pfft, whatever! Quote
Fozzie Posted January 5, 2015 Posted January 5, 2015 For those that are still like: A hyperbolic function is an analogue for circular and triangular functions.Go back to high school stuff where "cos t" and "sin t" with a radius in unit form creates a circle, a hyperbolic function finds the equilateral hyperbola (the right half)In the real world, a hyperbola can be seen with the triangular lighting effect from some down lighting. It's that shape.Using it with differentiation, which is just a way of taking real values and deciphering their rates of change. You can work out using the angle between the vectors where C3 will fall. I think... http://www.stasheverything.com/wp-content/uploads/2012/11/math-meme.jpeg Quote
XmisterIS Posted January 5, 2015 Author Posted January 5, 2015 That is the principle of it, yes. http://www.stasheverything.com/wp-content/uploads/2012/11/math-meme.jpeg When I was at Uni, we had a similar in-joke: "Hence find the mass of the Sun, using the approximation of a spherical horse in a vacuum". Quote
Guest Posted January 6, 2015 Posted January 6, 2015 I thought I was good at Maths until I read the original post !!! Quote
Grumpy Old Git Posted January 7, 2015 Posted January 7, 2015 Maths! - I was looking at using Trigonometry and 'J' notation to solve this! Quote
Grumpy Old Git Posted January 7, 2015 Posted January 7, 2015 Using pure Maths the answer is (and always will be) 42! Quote
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