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diameter calculation

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Michael Gilligan13/10/2019 07:47:52
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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...... devil

My goodness what a waste of time. The OP just wanted to know a shop grade calculation (and got one early on) .

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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.

Michael Gilligan13/10/2019 08:43:06
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Posted by SillyOldDuffer on 12/10/2019 21:17:48:
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Oh dear […]

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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

Nicholas Farr13/10/2019 09:07:32
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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.

Hopper13/10/2019 09:56:19
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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 Wooding13/10/2019 10:38:21
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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.

chords in a circle.jpg

Nicholas Farr13/10/2019 18:14:19
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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.

example26b003.jpg

Regards Nick.

Edited By Nicholas Farr on 13/10/2019 18:18:25

Andrew Johnston13/10/2019 18:43:24
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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. smile

Andrew

Howard Lewis13/10/2019 21:35:36
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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

Neil Wyatt13/10/2019 21:58:34
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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. smile

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 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.

Neil Wyatt13/10/2019 22:00:32
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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 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.

example26b003.jpg

Regards Nick.

Edited By Nicholas Farr on 13/10/2019 18:18:25

That's excellent & elegant.

Nicholas Farr13/10/2019 22:35:07
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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 webster13/10/2019 23:55:52
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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.

chords in a circle.jpg

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

SillyOldDuffer14/10/2019 10:09:03
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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 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

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

Michael Gilligan14/10/2019 15:02:46
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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!

[…]

.

.


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.

SillyOldDuffer14/10/2019 16:46:52
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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!

[…]

.

.


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...

blush

Ta,

Dave

Gary Wooding16/10/2019 12:18:37
588 forum posts
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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 Gunn17/10/2019 14:12:05
284 forum posts
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I am still struggling with the average of 1 and 2 being 2

Chris Gunn

Michael Gilligan17/10/2019 15:52:14
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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 Gunn17/10/2019 16:13:32
284 forum posts
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Thanks Michael, I knew I could rely on you.

CG

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