centre of area/centre of mass

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Reply 20

andymorl

thanks anyway making a lot more sense than it did previously

Reply 21

ghostwalker

Original post by andymorl

thanks

Yea I know the formula of the inner curve but it is a function of the radius i.e. x^2+y^2=r^2. I just thought it would be easier to integrate the polynomial in terms of x and y, I know my limits lol

Since the mathematics isn't your main focus, have a look at the Wolfram website - it can work out integrals for you.

Reply 22

andymorl

Thanks, good website. I know the basics of integration but not used it for some time. In relation to known the formula for the segment of the circle instead of the polynomial, how would you apply what you mentioned? I thought that when I estimated the polynomial equation that there would be a more simple way. :s-smilie:

Is a polynomial not the most simple way to integrate for such a problem?

Finding the centroid and area is only the start of the analysis. Any guidance or reference would be much appreciated.

Thank you again

Reply 23

ghostwalker

Original post by andymorl

You evaluate the results of your integration to the two end points of the section - subtract one from the other. Bit puzzled by your question, TBH.

Is a polynomial not the most simple way to integrate for such a problem?

They're certainly the most simple to handle in general.

Finding the centroid and area is only the start of the analysis. Any guidance or reference would be much appreciated.

You seem to have the centroid and area fairly well sorted. If you need more accuracy, you'll will need to brush up your integration skills. Beyond that, I don't know what you're trying to do (and probably couldn't help even if I did), nor what software you have available - you may find something on the internet, but I can't advise.

Reply 24

andymorl

That's just going over it again starting to make sense, not done polar coordinates for a good 4/5 years, but coming back. Just wondered any references for the problem state or a good programme that I can check my results against for the integration etc...

Reply 25

ghostwalker

Original post by andymorl

Wolfram can do any integration you're likely to come across; have a look at some of the examples in their website.

I don't know of any other significant software that's free. I use "graph" at http://www.padowan.dk/; it's fairly basic. Has been out a few years. There may be better stuff out there now, I've not checked for a long time.

If you don't mind paying some money you can go PRO with Wolfram, or look as things like Mathematica, Mathlab, etc., but there's a significant learning curve with those, I think, and probably overkill if this is your only usage.

The problem itself isn't any area I'm expert in, and you probably have a better chance of searching on the 'net for it, than I have. It's usually just a question of finding suitable keywords for your search.

Reply 26

andymorl

Thanks for the advice :yy:

Reply 27

andymorl

relating back to this problem, I can find the area between two circular curves using polar coordinates, which is fine for one part of the analysis. The second area I am investigating involves a non-continuous section of the arch with a circular inner radius of r^2=x^2+y^2 which is fine in polar, but the outer radius I can only really find from a polynomial trend line in excel, (which I could estimate myself but cannot justify) . Knowing the polar equation for the radius of the inner circle and finding the polynomial for the our to be equal to, y=-0.0009x^3 - 0.0034x^2 - 0.0829x + 11.931, ends up really messy when trying to convert to polar. This makes me wonder if it even possible, as I cannot find any reference which driving me insane.

Any guidance would be more than appreciated

p.s need to work to the 3rd degree for accuracy, or as I figure

(edited 8 years ago)

Reply 28

ghostwalker

Original post by andymorl

In what way is it non-continuous? Are the edges aligned with radii, or vertical, or something else.

Can't you use the piecewise linear approximation for the outer radius, as I previously suggested?

Are the inner and outer edges of these sections really curved? Or are they straight?

If you're going to use the polynomial approximation, then you may be better off working in cartesian co-ordinates.

What's "3rd degree for accuracy"? Is that 3 significant figures?

Reply 29

andymorl

In what way is it non-continuous? Are the edges aligned with radii, or vertical, or something else.

Like the pic I sent i.e. inside radius is circular outside increases from the springing to the crown section (material type remain constant). All edges aligned with radii

Can't you use the piecewise linear approximation for the outer radius, as I previously suggested?

Yeah it works but getting significant errors further in the analysis

Are the inner and outer edges of these sections really curved? Or are they straight?

really curved

If you're going to use the polynomial approximation, then you may be better off working in cartesian co-ordinates.

This is where the problem is, and I don't see how else it can be done without working polar

What's "3rd degree for accuracy"? Is that 3 significant figures?

3rd degree meaning to x^3 in the polynomial as this works best for regression as 2nd degree is not fitting (in terms of the excel trend line)

thanks

(edited 8 years ago)

Reply 30

andymorl

basically is it possible to find the area between to two curves expressed by

Outer radius

y = -0.0008x^3 - 0.0012x^2 - 0.0511x + 13.312

Inner Radius

y = -0.0009x^3 - 0.0034x^2 - 0.0829x + 11.931

between 0 radians at the vertical plane through the crown section to 1.031098 radians

Reply 31

ghostwalker

Original post by andymorl

If you can find where the 1.031098 radians line intersects each of the two polynomials - say s for the outer radius and t for the innner. Wolfram can help there.

Then, in cartesian, integral from 0 to t of "outer - inner" + integral from t to s of "outer - equation of radius line" would do it.

Reply 32

andymorl

Excellent! this is working perfectly and easy to programme into the spreadsheet :biggrin:

Thank you so much for your help and advice! really appreciated :smile:

Reply 33

ghostwalker

Original post by andymorl

Excellent! this is working perfectly and easy to programme into the spreadsheet :biggrin:

Thank you so much for your help and advice! really appreciated :smile:

Reply 34

andymorl

What method would you recommend to find hatched area and its centroid?

Integrating the polynomial of the curve and the linear horizontal function, are giving significant error as the segment of the circle considered becomes close to pi/2.

Is there a more simple and accurate method.... :s-smilie:

p.s. all dimensions given are known

(edited 8 years ago)

int.JPG24.5KB

Reply 35

ghostwalker

Original post by andymorl

p.s. all dimensions given are known

What about using the equation of the circle - should then be 100% accurate.
See Wolfram

and Wolf2

Reply 36

andymorl

I actually tried this equation and method but messed the integration up (never checked it on wolfram and assumed it was not possible).

Brilliant, yes the results are perfect :biggrin:

Thank you again for the advice :smile:

Math King

Hey ghostwalker, sorry to keep disturbing you, just unsure on one last thing...

Relating back to that last question (spandrel); I can find the area non problem, but when I am trying to locate the x(bar) centroid I am getting an answer appears to have significant error. I have attached my workings (sorry if the notations are all over) which I have been over many times and still get the same answer :s-smilie:

, is there anything obvious you can see I am doing wrong?

Thanks for your time again, really appreciated

p.s. I have also included a wolfram solution to the problem, which I am confident is the correct value of x-bar...