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# centre of area/centre of mass

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Why bother with a post grad course - waste of time? 17-10-2016
1. thanks anyway making a lot more sense than it did previously
2. (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.
3. 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.

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
4. (Original post by 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?
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.
5. 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...
6. (Original post by 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...
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.
8. 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
9. (Original post by 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
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?
10. 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
11. basically is it possible to find the area between to two curves expressed by

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

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
12. (Original post by andymorl)
basically is it possible to find the area between to two curves expressed by

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

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
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.
13. Excellent! this is working perfectly and easy to programme into the spreadsheet

14. (Original post by andymorl)
Excellent! this is working perfectly and easy to programme into the spreadsheet

15. 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....

p.s. all dimensions given are known
Attached Images

16. (Original post by 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....

p.s. all dimensions given are known
What about using the equation of the circle - should then be 100% accurate.
See Wolfram

and Wolf2
17. 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

Thank you again for the advice
18. Math King
19. 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 , 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...
Attached Images

20. (Original post by andymorl)
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 , 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...
Sorry for butting in, but should your denominator not be ? i.e: you're missing the factor of a half?

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