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UK diy (uk.d-i-y) For the discussion of all topics related to diy (do-it-yourself) in the UK. All levels of experience and proficency are welcome to join in to ask questions or offer solutions. |
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#1
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Triangular calculations
Given an equilateral triangle of side 185 mm, how do I calculate the
diameter or radius of the circle that intersects with the three corners of the triangle? My Zeus tables explain how to set up a drilling rig for a given diameter, but not the reverse. The formula would be helpful. A link to a web site of such useful information would be wonderful! TIA Richard |
#2
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Triangular calculations
Richard wrote:
Given an equilateral triangle of side 185 mm, how do I calculate the diameter or radius of the circle that intersects with the three corners of the triangle? My Zeus tables explain how to set up a drilling rig for a given diameter, but not the reverse. The formula would be helpful. A link to a web site of such useful information would be wonderful! TIA Richard More thought and less haste is an obvious answer (with apologies). http://www.ajdesigner.com/index_math.php Richard :-( |
#3
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Triangular calculations
In message , Richard
writes Given an equilateral triangle of side 185 mm, how do I calculate the diameter or radius of the circle that intersects with the three corners of the triangle? My Zeus tables explain how to set up a drilling rig for a given diameter, but not the reverse. The formula would be helpful. A link to a web site of such useful information would be wonderful! You mean like typing "diameter of a circle on an equilateral triangle" into google ? which yields various site such as this ... http://mathworld.wolfram.com/EquilateralTriangle.html Rocket science, eh ? -- geoff |
#4
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Triangular calculations
"geoff" wrote in message
... In message , Richard writes Given an equilateral triangle of side 185 mm, how do I calculate the diameter or radius of the circle that intersects with the three corners of the triangle? My Zeus tables explain how to set up a drilling rig for a given diameter, but not the reverse. The formula would be helpful. A link to a web site of such useful information would be wonderful! You mean like typing "diameter of a circle on an equilateral triangle" into google ? which yields various site such as this ... http://mathworld.wolfram.com/EquilateralTriangle.html Rocket science, eh ? plonk |
#5
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Triangular calculations
circumscribed radius of an equilateral triangle of side L = L * 0.5574
For 185mm side, this is 107mm |
#6
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Triangular calculations
In message , R D S
writes "geoff" wrote in message ... In message , Richard writes Given an equilateral triangle of side 185 mm, how do I calculate the diameter or radius of the circle that intersects with the three corners of the triangle? My Zeus tables explain how to set up a drilling rig for a given diameter, but not the reverse. The formula would be helpful. A link to a web site of such useful information would be wonderful! You mean like typing "diameter of a circle on an equilateral triangle" into google ? which yields various site such as this ... http://mathworld.wolfram.com/EquilateralTriangle.html Rocket science, eh ? plonk plonker ... -- geoff |
#7
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Triangular calculations
On 20 Jul, 14:47, Richard wrote:
Given an equilateral triangle of side 185 mm, how do I calculate the diameter or radius of the circle that intersects with the three corners of the triangle? My Zeus tables explain how to set up a drilling rig for a given diameter, but not the reverse. The formula would be helpful. A link to a web site of such useful information would be wonderful! TIA Richard If the equilateral triangle is converted into three equal sized triangles within the main one, using the centre of the circle as the common point where all three converge, you get three triangles with two 30deg angles and one of 120deg, and the lines from the side of the circle to the centre is the radius of the circle. Half one of these triangle, you get a right angled triangle with a 30deg and 60 deg angles. Using trig, the 185mm triangle edge now = 2 x the (A)djacent side of the triangle in relation to the 30deg angle. Formula is therefore (185/2) / cos(30) = 106.80. As a generic formula, r = L/2 / cos(30). http://www.google.co.uk/search?hl=en...grees%29&meta= I'm sure there is a simpler way, but that was an enjoyable challenge to me as someone who hadn't done any trig in about 20 years!! Matt |
#8
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Triangular calculations
wrote in message ... On 20 Jul, 14:47, Richard wrote: Given an equilateral triangle of side 185 mm, how do I calculate the diameter or radius of the circle that intersects with the three corners of the triangle? My Zeus tables explain how to set up a drilling rig for a given diameter, but not the reverse. The formula would be helpful. A link to a web site of such useful information would be wonderful! TIA Richard If the equilateral triangle is converted into three equal sized triangles within the main one, using the centre of the circle as the common point where all three converge, you get three triangles with two 30deg angles and one of 120deg, and the lines from the side of the circle to the centre is the radius of the circle. Half one of these triangle, you get a right angled triangle with a 30deg and 60 deg angles. Using trig, the 185mm triangle edge now = 2 x the (A)djacent side of the triangle in relation to the 30deg angle. Formula is therefore (185/2) / cos(30) = 106.80. As a generic formula, r = L/2 / cos(30). You beat me to it :-) I think it can be done (proofed) with no cos or sin involved IMHO. Adam |
#9
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Triangular calculations
In an earlier contribution to this discussion,
ARWadworth wrote: wrote in message ... On 20 Jul, 14:47, Richard wrote: Given an equilateral triangle of side 185 mm, how do I calculate the diameter or radius of the circle that intersects with the three corners of the triangle? My Zeus tables explain how to set up a drilling rig for a given diameter, but not the reverse. The formula would be helpful. A link to a web site of such useful information would be wonderful! TIA Richard If the equilateral triangle is converted into three equal sized triangles within the main one, using the centre of the circle as the common point where all three converge, you get three triangles with two 30deg angles and one of 120deg, and the lines from the side of the circle to the centre is the radius of the circle. Half one of these triangle, you get a right angled triangle with a 30deg and 60 deg angles. Using trig, the 185mm triangle edge now = 2 x the (A)djacent side of the triangle in relation to the 30deg angle. Formula is therefore (185/2) / cos(30) = 106.80. As a generic formula, r = L/2 / cos(30). You beat me to it :-) I think it can be done (proofed) with no cos or sin involved IMHO. Adam Yes it can. I did it using Pythagoras and interesting chords. It comes out as r=L/sqrt(3) - which is the same thing anyway because cos(30)=sqrt(3)/2 -- Cheers, Roger ______ Email address maintained for newsgroup use only, and not regularly monitored.. Messages sent to it may not be read for several weeks. PLEASE REPLY TO NEWSGROUP! |
#10
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Triangular calculations
On 21 Jul, 23:13, "Roger Mills" wrote:
In an earlier contribution to this discussion, ARWadworth *wrote: wrote in message .... On 20 Jul, 14:47, Richard wrote: Given an equilateral triangle of side 185 mm, how do I calculate the diameter or radius of the circle that intersects with the three corners of the triangle? My Zeus tables explain how to set up a drilling rig for a given diameter, but not the reverse. The formula would be helpful. A link to a web site of such useful information would be wonderful! TIA Richard If the equilateral triangle is converted into three equal sized triangles within the main one, using the centre of the circle as the common point where all three converge, you get three triangles with two 30deg angles and one of 120deg, and the lines from the side of the circle to the centre is the radius of the circle. Half one of these triangle, you get a right angled triangle with a 30deg and 60 deg angles. *Using trig, the 185mm triangle edge now = 2 x the (A)djacent side of the triangle in relation to the 30deg angle. Formula is therefore (185/2) / cos(30) = 106.80. *As a generic formula, r = L/2 / cos(30). You beat me to it :-) I think it can be done (proofed) with no cos or sin involved IMHO. Adam Yes it can. I did it using Pythagoras and interesting chords. It comes out as r=L/sqrt(3) - which is the same thing anyway because cos(30)=sqrt(3)/2 -- Cheers, Roger ______ Email address maintained for newsgroup use only, and not regularly monitored.. Messages sent to it may not be read for several weeks. PLEASE REPLY TO NEWSGROUP!- Hide quoted text - - Show quoted text - My 1988 GCSE Maths didn't stretch to the relationships of Pythagoras and trig. Why does sqrt(3)/2 = cos(30deg)? I can see that it does, I just can't see the Pythagoras on the way to getting there...... Is it because of the defined relationship in a triangle with 30 / 60 / 90 degree angles (making the sides = 106.8, 53.4, 92.5 in this case?) I should have done A level maths (perhaps I will one day) - its fascintating stuff! Matt |
#11
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Triangular calculations
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#12
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Triangular calculations
dave wrote:
You say QED - but is it? Maybe being pedantic here but is this a proof as such? I mean what you have written is correct - but it's "just" a particular case that happens to give the result expected. As I say, not being pedantic but wonder if this really is a "proof"? I'd say that if anything is missing it's that Roger didn't /prove/ that the perpendicular from one corner of the equilateral triangle to the opposite side actually bisects the angle and the opposite side. It's obvious from the symmetry that it does, but can you prove it starting only from Euclid's axioms? I'm no mathematician but the definition of a cosine is, I believe, a series. Hmm... As usually taught you start with the simple right-angled triangle definitions - cos = adjacent/hypotenuse, etc. - and then extend that definition to allow angles outside the range 0 to 90 deg. Then the power series expansions can be derived and the trig functions generalised to allow complex number arguments. Hence if you can prove the bisection of the angle and side above and accept Pythagoras (which can be proved in several ways) you know that the 1 : root 3 : 2 triangle has angles of 30, 60 & 90 deg. and that cos 30 deg. = sqrt(3)/2, and so on. This was second form stuff when I was at school. It's a bit worrying to read things like My 1988 GCSE Maths didn't stretch to the relationships of Pythagoras and trig. Why does sqrt(3)/2 = cos(30deg)? (Discuss.) -- Andy |
#13
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Triangular calculations
On 23 Jul, 08:39, Andy Wade wrote:
dave wrote: You say QED - but is it? Maybe being pedantic here but is this a proof as such? I mean what you have written is correct - but it's "just" a particular case that happens to give the result expected. As I say, not being pedantic but wonder if this really is a "proof"? I'd say that if anything is missing it's that Roger didn't /prove/ that the perpendicular from one corner of the equilateral triangle to the opposite side actually bisects the angle and the opposite side. *It's obvious from the symmetry that it does, but can you prove it starting only from Euclid's axioms? I'm no mathematician but the definition of a cosine *is, I believe, a series. Hmm... *As usually taught you start with the simple right-angled triangle definitions - cos = adjacent/hypotenuse, etc. - and then extend that definition to allow angles outside the range 0 to 90 deg. *Then the power series expansions can be derived and the trig functions generalised to allow complex number arguments. *Hence if you can prove the bisection of the angle and side above and accept Pythagoras (which can be proved in several ways) you know that the 1 : root 3 : 2 triangle has angles of 30, 60 & 90 deg. and that cos 30 deg. = sqrt(3)/2, *and so on. This was second form stuff when I was at school. *It's a bit worrying to read things like My 1988 GCSE Maths didn't stretch to the relationships of Pythagoras and trig. *Why does sqrt(3)/2 = cos(30deg)? (Discuss.) -- Andy Perhaps only worrying in that its now 20 years ago and (despite being an accountant) I left maths far behind me so it may simply be my forgetfulness. Though I'm fairly sure that Pythagoras and trig were taught as sequential steps - first learn Pythagoras such that the dimensions of any right angled triangle can be worked out. Then learn trig in order that angles or lengths can be derived based on given information. So whilst we all knew SOHCAHTOA etc, and we all knew Asqr = Bsqr + Csqr, the two formula were not merged into the (entirely logical as I see it now) position above. That's not to say, though, that with a bit of thought and a simple explanation above I can't see how that works. Perhaps what I should have written was "my 1998 Maths GCSE didn't stretch to being explicit about formulae which merged Pyth and trig, and instead kept them separate (but highly related of course) to make a two step process rather than a single step" Matt |
#15
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Triangular calculations
John Rumm coughed up some electrons that declared:
I did GCE O level maths in '83 or '84, and don't recall trig getting much beyond basic right angle triangle stuff, with a possible requirement to be able to use the sine or cosine rules. I was not sufficiently a fan of maths (at the time) to take it at A level. The AO-level maths did cover a bit more of this sort of thing, also in 83 for me. A level maths covered a lot more and definately was the level physics at uni assumed. I do remember having quite a nasty shock when doing maths at university to find there was a whole world about trig identities and equivalents that I knew absolutely nothing about! (Much of the difficulty stemming from the fact that there were several maths groups one could be in, and I happened to get the one where 95% of the students who had done A level maths, and the very obtuse Welshman teaching, thought he could race through it all as a "quick bit of revision"). After a few weeks of that I decided to switch groups and found that the alternative was a running at a much more sensible pace (the first group had almost finished course in the first four weeks!) Wave mechanics and associated maths blew my mind - never did get the hand of that stuff. Never used any of it since, being a sysadmin and programmer. I fused a braincell trying to do 3d-trig on my roof last month. Fortuanately the basics were still there but I had to work lots of things out the hard way, starting from the beginning. Cheers Tim |
#16
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Triangular calculations
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#17
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Triangular calculations
"John Rumm" wrote in message
et... wrote: On 23 Jul, 08:39, Andy Wade wrote: Hence if you can prove the bisection of the angle and side above and accept Pythagoras (which can be proved in several ways) I think that is one of the few mathematical proofs I ever mastered! ;-) This was second form stuff when I was at school. It's a bit worrying to read things like My 1988 GCSE Maths didn't stretch to the relationships of Pythagoras and trig. Why does sqrt(3)/2 = cos(30deg)? (Discuss.) I did GCE O level maths in '83 or '84, and don't recall trig getting much beyond basic right angle triangle stuff, with a possible requirement to be able to use the sine or cosine rules. I was not sufficiently a fan of maths (at the time) to take it at A level. I was lucky in that I enjoyed doing my GCE O level and did go on to do an A level. I bumped into my A level teacher 20 years after doing my A level and took the time to tell her she was the best teacher I ever had (she was no longer teaching and was interviewing my wife for a job when I bumped into her). I have just dragged some of my old schoolbooks out of the loft to help a friends lad out who is starting A level next term. She used to do maths things that were not on the course, just for fun. Learning that any recurring decimal can be written as a fraction is my favourite eg .123123123 recurring or any other numbers (I have used 3 digits here but you can use more eg .1234512345) will always be a fraction if (time to use letters for numbers and remember that these are recurring decimals) ..abcabcabc will be the fraction abc/999 ..abcdabcdabcd will will be the fraction abcd/9999. Adam |
#18
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Triangular calculations
In message , ARWadworth
writes I did GCE O level maths in '83 or '84, and don't recall trig getting much beyond basic right angle triangle stuff, with a possible requirement to be able to use the sine or cosine rules. I was not sufficiently a fan of maths (at the time) to take it at A level. I was lucky in that I enjoyed doing my GCE O level and did go on to do an A level. I bumped into my A level teacher 20 years after doing my A level and took the time to tell her she was the best teacher I ever had (she was no longer teaching and was interviewing my wife for a job when I bumped into her). I have just dragged some of my old schoolbooks out of the loft to help a friends lad out who is starting A level next term. by far my best teacher at "O" level was "Nut" Hayes, who was B.Sc. Physics (Failed), stirred his tea with his comb etc ... but was prolly the best teacher I came across Respect ... as one would say now -- geoff |
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