diameter calculation

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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.

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Dave,

In deference to Mr Dayman’s sensitivities [*], I have sent you a personal message.

MichaelG.
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[*] Sarcasm reared its ugly head in his post; so he’s obviously upset !

Edited By Michael Gilligan on 13/10/2019 08:50:41

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.

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.

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.

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 B-A 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

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

There's no hope for me!

I could never work to better than two places with a 10" Faber Log-Log 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

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 near-optimal 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.

That's excellent & elegant.

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.

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

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 off-the-shelf tailoring. Micro-economies aren't worth the effort in a home workshop, but pennies matter hugely when millions are made.

If maths is considered off-topic 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

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For the avoidance of doubt, Dave

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

Yes, you're right.

I am still struggling with the average of 1 and 2 being 2

Chris Gunn

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I think he rounded 1.5 up ... on the basis that 1 and 2 were expressed as integers.

MichaelG.

Thanks Michael, I knew I could rely on you.

CG