Michael Gilligan  13/10/2019 07:47:52 
14134 forum posts 615 photos  Posted by Jeff Dayman on 13/10/2019 00:34:16:
Agree Alan, it's getting ridiculous again, isn't it? Mind you, the egg cup I made this week was accurate to 22 millionths of an inch, and well worth the effort...... My goodness what a waste of time. The OP just wanted to know a shop grade calculation (and got one early on) . . Jeff, The question was answered, to the OP’s satisfaction, long ago. The discussion has evolved. Fact of Life ... That’s how it goes. Please give the whinging a rest. MichaelG. 
Michael Gilligan  13/10/2019 08:43:06 
14134 forum posts 615 photos  Posted by SillyOldDuffer on 12/10/2019 21:17:48:
. Oh dear […] . Dave, In deference to Mr Dayman’s sensitivities [*], I have sent you a personal message. MichaelG. [*] Sarcasm reared its ugly head in his post; so he’s obviously upset ! Edited By Michael Gilligan on 13/10/2019 08:50:41 
Nicholas Farr  13/10/2019 09:07:32 
1992 forum posts 950 photos  Hi, yes, the question was answered to the OP's satisfaction, If one finds that the evolved thread becomes of no more interest to them, then they are not forced to read it any further, but of course it still interests others. Regards Nick.

Hopper  13/10/2019 09:56:19 
3741 forum posts 76 photos  And we haven't even started to calculate how many angels can dance on the head of a pin of the diameter with three chords this size. I know the plain diameter formula has been posted on here before somewhere, but not encompassing the chords aspect. 
Gary Wooding  13/10/2019 10:38:21 
588 forum posts 141 photos  Just to dot all the 'I's, here is my 'proof' of Nick's solution. It depends on two properties of chords in circles, and the Sine rule of trigonometry. 1. All angles subtended on one side by a chord to any point on the circumference are equal, so in the diagram, angle C equals angle D. 2. If the chord is a diameter, the angle is 90°, and conversely, if the angle subtended by a chord is 90° then the chord is a diameter. 3. The Sine rule states that, for any triangle, Sin A/a = Sin B/b = Sin C/c In the diagram, A, B, and C, are the three random points, and the circle is their PCD. You want to calculate the length of its diameter. Choose any side of the triangle ABC (side AB in the diagram) and draw a perpendicular line from one end to to meet the circle. This is the dashed line BD. Because they both come from chord AB, angles C and D are equal, so SinC = SinD. But SinD = c/h, so h = c/SinD = c/SinC But angle ABD is a right angle, so h = the diameter of the circle. 
Nicholas Farr  13/10/2019 18:14:19 
1992 forum posts 950 photos  Hi Gary, a better way of proving the Sine rule works for the example 26 diagram is shown below. This is a half size one but the relationship between the angle and the length of the two sides shown are the same. This shows the integrity of the 41 degree angle and the lengths of the original sides and should help show those who don't quite understand how it works. The extended line BA becomes the diameter of the PCD for A, B and C and a new PCD drawn on the centre of this line will show the Right Angle Triangle with the red line at a right angle to line B and C and it will be noticed that the new PCD will pick up both ends of the red line and point B. Excuse the makeshift real world drawing, all the dimensions are close to those actually calculated. Regards Nick. Edited By Nicholas Farr on 13/10/2019 18:18:25 
Andrew Johnston  13/10/2019 18:43:24 
4893 forum posts 552 photos  Posted by Neil Wyatt on 12/10/2019 19:10:29:
Actually there are LOTS of algorithms that produce answers by paying off accuracy against speed. I doubt they use algorithms in the sense of reaching a definitive answer without a full search, although probably in the sense that the sequence is gauranteed to halt. I suspect the answers are more likely based on heuristics. If Google has an algorithm that solves the travelling salesman problem then I expect they're in line for the Fields Medal. Andrew 
Howard Lewis  13/10/2019 21:35:36 
2389 forum posts 2 photos  There's no hope for me! I could never work to better than two places with a 10" Faber LogLog Slide rule. And now am WELL out of practice. (Still have it somewhere! ) Hopper did not specify the units of the diameter, or the tolerance thereof, of the head of the pin on which we have to have our angels dancing! Howard 
Neil Wyatt  13/10/2019 21:58:34 
Moderator 16655 forum posts 687 photos 75 articles  Posted by Andrew Johnston on 13/10/2019 18:43:24:
Posted by Neil Wyatt on 12/10/2019 19:10:29:
Actually there are LOTS of algorithms that produce answers by paying off accuracy against speed. I doubt they use algorithms in the sense of reaching a definitive answer without a full search, although probably in the sense that the sequence is gauranteed to halt. I suspect the answers are more likely based on heuristics. If Google has an algorithm that solves the travelling salesman problem then I expect they're in line for the Fields Medal. Andrew Solving it accurately isn't the issue, it's just the time taken. There are algorithms that speed things up, for example by calculating a nearoptimal solution then testing alternatives based on it. The record for an accurate solution appears to be about 89,500 locations. en.wikipedia.org/wiki/Travelling_salesman_problem But there are LOTS of practical algorithms that trade absolute accuracy for huge increases in speed. 
Neil Wyatt  13/10/2019 22:00:32 
Moderator 16655 forum posts 687 photos 75 articles  Posted by Nicholas Farr on 13/10/2019 18:14:19:
Hi Gary, a better way of proving the Sine rule works for the example 26 diagram is shown below. This is a half size one but the relationship between the angle and the length of the two sides shown are the same. This shows the integrity of the 41 degree angle and the lengths of the original sides and should help show those who don't quite understand how it works. The extended line BA becomes the diameter of the PCD for A, B and C and a new PCD drawn on the centre of this line will show the Right Angle Triangle with the red line at a right angle to line B and C and it will be noticed that the new PCD will pick up both ends of the red line and point B. Excuse the makeshift real world drawing, all the dimensions are close to those actually calculated. Regards Nick. Edited By Nicholas Farr on 13/10/2019 18:18:25 That's excellent & elegant. 
Nicholas Farr  13/10/2019 22:35:07 
1992 forum posts 950 photos  Hi Neil, thanks for your comment. The drawing is a bit rough and ready, but the 41 degree angle and the three points were set quite accurately and I think it shows what might seem to be an obscure triangle. Regards Nick.

duncan webster  13/10/2019 23:55:52 
2255 forum posts 32 photos  Posted by Gary Wooding on 13/10/2019 10:38:21:
Just to dot all the 'I's, here is my 'proof' of Nick's solution. It depends on two properties of chords in circles, and the Sine rule of trigonometry. 1. All angles subtended on one side by a chord to any point on the circumference are equal, so in the diagram, angle C equals angle D. 2. If the chord is a diameter, the angle is 90°, and conversely, if the angle subtended by a chord is 90° then the chord is a diameter. 3. The Sine rule states that, for any triangle, Sin A/a = Sin B/b = Sin C/c In the diagram, A, B, and C, are the three random points, and the circle is their PCD. You want to calculate the length of its diameter. Choose any side of the triangle ABC (side AB in the diagram) and draw a perpendicular line from one end to to meet the circle. This is the dashed line BD. Because they both come from chord AB, angles C and D are equal, so SinC = SinD. But SinD = c/h, so h = c/SinD = c/SinC But angle ABD is a right angle, so h = the diameter of the circle. That's really elegant, but just to be pedantic, don't you need to use the cosine rule to find angle C to start the whole thing off rather than sin rule 
SillyOldDuffer  14/10/2019 10:09:03 
4785 forum posts 1011 photos  Posted by Howard Lewis on 13/10/2019 21:35:36:
There's no hope for me! I could never work to better than two places with a 10" Faber LogLog Slide rule. And now am WELL out of practice. (Still have it somewhere! ) Hopper did not specify the units of the diameter, or the tolerance thereof, of the head of the pin on which we have to have our angels dancing! Howard Oh dear, I seem to have stirred up a hornets nest by putting my foot in it again! For the record, my comments about extreme accuracy of calculation relate to the properties of Polar and Cartesian coordinates, not to Paul's original question. By asking about a practical problem, Paul opened the door to some interesting mathematical solutions and related problems. I questioned the assertion that Polar Coordinates are intrinsically more accurate than Cartesian Coordinates by suggesting both systems are equivalent provided the calculations are done to a sufficiently large number of decimal places. (Not convinced this is correct and Michael has set me some homework!) I wasn't suggesting Paul or anyone else needs to do hard sums in their workshops! The maths might seem obscure and irrelevant, but it's vital to engineering. For example, a variation of Andrew's 'Travelling Salesman Problem' has a workshop application. It's used to find the most economic way of cutting shapes from sheet material, important when stuff is stamped from metal, and very common issue in offtheshelf tailoring. Microeconomies aren't worth the effort in a home workshop, but pennies matter hugely when millions are made. If maths is considered offtopic I apologise, but I've found value in all the posts, and don't mind at all that the conversation has opened out. Ta, Dave

Michael Gilligan  14/10/2019 15:02:46 
14134 forum posts 615 photos  Posted by SillyOldDuffer on 14/10/2019 10:09:03:
[…] For the record, my comments about extreme accuracy of calculation relate to the properties of Polar and Cartesian coordinates, not to Paul's original question. By asking about a practical problem, Paul opened the door to some interesting mathematical solutions and related problems. I questioned the assertion that Polar Coordinates are intrinsically more accurate than Cartesian Coordinates by suggesting both systems are equivalent provided the calculations are done to a sufficiently large number of decimal places. (Not convinced this is correct and Michael has set me some homework!) I wasn't suggesting Paul or anyone else needs to do hard sums in their workshops! […] . .
What I intended to assert was not that “Polar Coordinates are intrinsically more accurate than Cartesian Coordinates” But that ”for situations where Polar Coordinates properly specify the geometry, their use is intrinsically more accurate than a conversion to Cartesian Coordinates” ... It’s a subtle difference, but important. MichaelG. 
SillyOldDuffer  14/10/2019 16:46:52 
4785 forum posts 1011 photos  Posted by Michael Gilligan on 14/10/2019 15:02:46:
Posted by SillyOldDuffer on 14/10/2019 10:09:03:
[…] For the record, my comments about extreme accuracy of calculation relate to the properties of Polar and Cartesian coordinates, not to Paul's original question. By asking about a practical problem, Paul opened the door to some interesting mathematical solutions and related problems. I questioned the assertion that Polar Coordinates are intrinsically more accurate than Cartesian Coordinates by suggesting both systems are equivalent provided the calculations are done to a sufficiently large number of decimal places. (Not convinced this is correct and Michael has set me some homework!) I wasn't suggesting Paul or anyone else needs to do hard sums in their workshops! […] . .
What I intended to assert was not that “Polar Coordinates are intrinsically more accurate than Cartesian Coordinates” But that ”for situations where Polar Coordinates properly specify the geometry, their use is intrinsically more accurate than a conversion to Cartesian Coordinates” ... It’s a subtle difference, but important. MichaelG. Arrgh! '... It’s a subtle difference, but important.' Certainly is. It's so easy to misread other people's posts! I am kicking myself... Ta, Dave 
Gary Wooding  16/10/2019 12:18:37 
588 forum posts 141 photos  Posted by duncan webster on 13/10/2019 23:55:52:
........... but just to be pedantic, don't you need to use the cosine rule to find angle C to start the whole thing off rather than sin rule
Yes, you're right. 
Chris Gunn  17/10/2019 14:12:05 
284 forum posts 16 photos  I am still struggling with the average of 1 and 2 being 2 Chris Gunn 
Michael Gilligan  17/10/2019 15:52:14 
14134 forum posts 615 photos  Posted by Chris Gunn on 17/10/2019 14:12:05:
I am still struggling with the average of 1 and 2 being 2 Chris Gunn . I think he rounded 1.5 up ... on the basis that 1 and 2 were expressed as integers. MichaelG.

Chris Gunn  17/10/2019 16:13:32 
284 forum posts 16 photos  Thanks Michael, I knew I could rely on you. CG 
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