diameter calculation

[paul raynerParticipant @paulrayner36054]

Hi all

I'm wanting to drill 3 holes in a triangular fashion on my mill with the PCD function I've already set up the work piece with the quill in the centre, The centre of the holes are 65mm equally distant apart from each other. Question is how do I calculate the diameter of the circle needed? I could do it on paper with a compass etc but I want it more accurate than that. I've had a look at the machinery's manual but cannot seem to find anything.

thanks in advance for any help

regards

Paul

[paul raynerParticipant @paulrayner36054]

[Nicholas FarrParticipant @nicholasfarr14254]

Hi Paul, a little bit of Trigonometry will give you the answer, just substitute the 100 shown on the diagram below with 65.

Regards Nick.

Edited By Nicholas Farr on 10/10/2019 21:21:47

[vintage engineerParticipant @vintageengineer]

I will work it out in cad  for you. I make it 75.12mm diameter.

 

Edited By vintage engineer on 10/10/2019 21:37:15

[paul raynerParticipant @paulrayner36054]

Thanks Guys just the job, I will plot it out First job in the morning.

regards

Paul

[not done it yetParticipant @notdoneityet]

I make it only 75.06mm, not that it is likely to make any difference at all.

[DMBParticipant @dmb]

I prefer Nick's answer. Why 3 diff?

[Michael GilliganParticipant @michaelgilligan61133]

ndiy seems to be closer

MichaelG.

[Mike PooleParticipant @mikepoole82104]

75.0555349947 for rough work

Mike

[Nicholas FarrParticipant @nicholasfarr14254]

Hi Mike, have you been using my calculator?

Regards Nick.

[not done it yetParticipant @notdoneityet]

To be honest, as a mathematician one would only quote the result as 75mm – as one cannot provide an answer to more significant figures than the data supplied with any confidence.

That 65 value could be anywhere between 64.50 and 65.49 (extra significant figures shown to demonstrate the validity of the result after rounding to 3 sig. figs.). Had it been measured to greater precision, like 65.00mm, a more precise result could be provided with confidence.

In practice the result for 65mm could be anywhere between 74.5mm and 75.6mm. Hence 75mm, to 2 significant figures, is really the best one can offer.

Remember, the average of 1 and 2 is 2, but the average between 1.0 and 2.0 is 1.5. We often provide answers to greater precision than the real world measurements deserve!

Edited By not done it yet on 11/10/2019 08:02:52

[Mike PooleParticipant @mikepoole82104]

Obviously as a Myford user working to ten decimal places is normal practice

Mike

Edited By Mike Poole on 11/10/2019 08:08:17

[Nicholas FarrParticipant @nicholasfarr14254]

Hi NDIY, the OP asked the diameter for 65mm without stating a tolerance of any kind. The mathematical answer is exacting within the practicalities of calculation and from that the OP can fashion his own tolerance and I'm sure he should be able to get it very close to his needs.

Regards Nick

[paul raynerParticipant @paulrayner36054]

75.06 will do for my needs

thank you for all your help, It was a head scratcher for me!

regards

Paul

[Michael GilliganParticipant @michaelgilligan61133]

Posted by not done it yet on 11/10/2019 08:01:03:

[…]
We often provide answers to greater precision than the real world measurements deserve!

.

… I was tempted to show the full ‘resolution’ of the available calculator; but thought that might be excessive in these circles, so simply chose to endorse your answer.

That said; it does no harm to declare a theoretical target value, provided that you understand that reaching it may be impractical.

MichaelG.

.

Best I could get last night was:

(( 32.5 ÷ 0.866025403784 ) x 2 ) = 75.05553499468936

[SillyOldDufferModerator @sillyoldduffer]

Nick's excellent solution is for the simple case ie a circle defined by three points of an equilateral triangle. What's the solution when the circle is defined by 3 randomly placed points?

A mathematical solution must be possible: I drew the example above with QCAD, but most other CAD packages can draw circles specified by three points.

Please be gentle – maths doesn't come naturally to me…

Dave

[Michael GilliganParticipant @michaelgilligan61133]

I’ve got that tucked-away somewhere, Dave … it’s quite elegant.

I will post it later, if no-one beats me to it.

MichaelG.

.

 Saved myself some effort

Just search for ‘perpendicular bisector’

Edited By Michael Gilligan on 11/10/2019 09:24:10

[Anonymous]

Posted by SillyOldDuffer on 11/10/2019 09:06:46:

A mathematical solution must be possible

Simples!

The Equation of a circle has three unknowns. Three unique points give you three indpendent equations. Three equations and three unknowns allows a unique solution for the centre and radius, with a bit of algebraic manipulation.

Andrew

[Nicholas FarrParticipant @nicholasfarr14254]

Hi S.O.D., I've found this in one of my books, but it does give a couple of dimensions and an angle, but you've asking about random placed points and I suspected it would involve Algebra as Andrew has said. However I could never get to grips with Algebra, which seems that MichealG's search is full of.

Regards Nick.

[Kiwi BlokeParticipant @kiwibloke62605]

I'd do it using co-ordinate geometry. Not sure if this is what Andrew is suggesting. I'm a lousy mathematician, and I bet there's a far easier way (no, not CAD).

The perpendicular bisectors of two chords intersect at the circle's centre. You can easily find the co-ordinates of the mid-point of a chord and its slope. The equation of the bisector of the chord follows, straightforwardly. It's then a few small steps to solve the two equations of the two chord bisectors, to get the co-ordinates of the centre. The diameter is the distance from the centre to any of the three points. The equation for the circle follows.

Incidentally, the radius of the circumcircle of a triangle, R = a / 2 * sin A, where a is the length of the chord opposite the angle (vertex) A. Sometimes useful…

Oh, Nick just beat me to it…

Edited By Kiwi Bloke on 11/10/2019 10:07:27

[Michael GilliganParticipant @michaelgilligan61133]

Posted by Nicholas Farr on 11/10/2019 10:05:06:

[…]

Algebra, which seems that MichealG's search is full of.

.

Oh dear … maybe this will help: **LINK**

http://mathworld.wolfram.com/PerpendicularBisector.html

MichaelG.

.

Edit: or this : http://tgbasics.weebly.com/construct-a-circle-given-3-points.html

Edited By Michael Gilligan on 11/10/2019 10:48:42

[SillyOldDufferModerator @sillyoldduffer]

Posted by Nicholas Farr on 11/10/2019 10:05:06:

Hi S.O.D., I've found this in one of my books, but it does give a couple of dimensions and an angle, but you've asking about random placed points and I suspected it would involve Algebra as Andrew has said. However I could never get to grips with Algebra, which seems that MichealG's search is full of.

Regards Nick.

This simple example has left me in the dust! I've confirmed it works by drawing it with QCAD but why is the diameter of the circle given by dividing distance AC by sin( 41° ) ?

I don't understand the method. The calculation isn't solving the triangle ABC, or calculating the length of line AB.

I can do trigonometry on right angled triangles, but ABC isn't right angled.

Andrew says it's simples and I believe him. He said:

"The Equation of a circle has three unknowns. Three unique points give you three indpendent equations. Three equations and three unknowns allows a unique solution for the centre and radius, with a bit of algebraic manipulation." It's the 'bit of algebraic manipulation' that finishes me off. Michael's Wolfram and Weebly examples may help clear the fog.

Meanwhile I found this discussion on StackExchange. It goes with Quadratic equations and matrices.

It's like being back at school. I did a class called 'Maths for Scientists'. This was a euphemism. It should have been called "Remedial Maths for Silly Boys Who Should Have Chosen an Arts Subject".

Dave

Edited By SillyOldDuffer on 11/10/2019 11:27:23

[Kiwi BlokeParticipant @kiwibloke62605]

Dave (S.O.D), look up 'sine formula'. Not all trig needs right-angled triangles…

[Michael GilliganParticipant @michaelgilligan61133]

Dave,

I think it's useful to remember that this is an example in polar geometry … so the correct solution will only be obtained by polar 'construction'.

Any 'rectangular co-ordinates' solution will inevitably be an approximation: When I was using Autocad it worked to fourteen decimal places [it may have improved since then], but that's still an approximation.

MichaelG.

.

"Michael's Wolfram and Weebly examples may help clear the fog."

… I sincerely hope so.

Edited By Michael Gilligan on 11/10/2019 11:45:47

[Kiwi BlokeParticipant @kiwibloke62605]

Michael: 'I think it's useful to remember that this is an example in polar geometry'. You mean 'R – theta' geometry? Not really, although I agree a bit of construction is required to get to this, 'sine-formula'-like answer. Co-ordinate geometry is good, if only because it's easy to transfer that approach to machining co-ordinates.

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