
#1
Dec1112, 11:44 AM

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I'm working to calculate the crosssectional area of a lathe turned feature machined with a radiused insert. My calculations have essentially led me to the equation for the area of a segment of a circle.
Area=r^2/2*(∏/180*Csin(C)) where r is the circle's radius and C is the central angle of the associated sector. In this particular case, I have a predetermined area and need to determine the central angle. How do I solve for C? In it's most basic form, I have: y=x+sin(x) Solve for x. Any help is appreciated. Thanks! 



#2
Dec1112, 03:14 PM

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Basically, you can't find a finite expression for the inverse here.
However, various appoximative techniques might be used. IF, for example, y is "sufficiently close to 0", a power series expansion about x=0 might zoom onto the solution fairly quickly. To show how this might be done: We have: [tex]y=x+x\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+++[/tex] when expanding the sine function in its power series (a finite segment of that highly accurate when x is close to zero) We now invert the power series, by assuming: [tex]x=a_{1}y+a_{2}y^{2}+a_{3}y^{3}+++[/tex] where the solution x(y) of the original equation boils down to determining the a's. Inserting the latter in the former, we get: [tex]y=2*(a_{1}y+a_{2}y^{2}+a_{3}y^{3}+a_{4}y^{4}+a_{5}y^{5}+++)\frac{1}{6}(a_{1}^{3}y^{3}+3a_{1}^{2}a_{2}y^{4}+3a_{1}^{2}a_{3}y^{5}+3a _{2}^{2}a_{1}y^{5})+\frac{1}{30}a_{1}^{5}y^{5}++[/tex] where terms of higher orders in y than 5 are dropped. Now, we simply compare coefficients in each power to "y", to determine the a's. We get: [tex]a_{1}=\frac{1}{2}, a_{2}=0, a_{3}=\frac{1}{48}, a_{4}=0, a_{5}=\frac{1}{640}[/tex] Thus, to fifth order accuracy, you have: [tex]x=\frac{1}{2}y\frac{1}{48}y^{3}+\frac{1}{640}y^{5}[/tex] 



#3
Dec1112, 03:50 PM

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Hmm, my coefficients were wrong.
We have: [tex]a_{3}=\frac{1}{96}[/tex] [tex]a_{5}=\frac{1}{4800}[/tex] or, I hope.. You'd better check for yourself. 



#4
Dec1912, 04:17 PM

P: 2

Segment of a circle calculation
arildno, I appreciate the response! Unfortunately, I can't assume y is close to zero in this situation. I agree with your guidance that a finite solution for x isn't possible. It seems that only ordered pairs are achievable (set a value for y, guess a value for x, and iterate until x converges on a solution). I generated a spreadsheet that does the iteration automatically. Thanks again for the help!
This situation is still a little tough to wrap my head around. It would seem that, with a finite area of a segment, I should be able to calculate the central angle. I've yet to intuitively explain why I can't calculate this feature when I have a real, finite, predefined segment area that I can measure. #continuouseducation 



#5
Dec1912, 05:02 PM

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#6
Dec2012, 06:41 AM

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However, a general formula, converging fast to any "y"value is not available. (But, to any given "y", you ought to be able to find a fast converging series) 



#7
Dec2012, 10:35 AM

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To generalize this, once you have a desired approximation X* for some "y" of yours (the closer X* is "y", the faster the convergence will be!), we may write:
[tex]x=X*+\epsilon[/tex] where epsilon is some "small function" of "y" and X* Inserting this into your equation yields: [tex]y=X*+\epsilon+\sin(X*+\epsilon)=X*+\sin(X*)(1\frac{\epsilon^{2}}{2}++)+\cos(X*)(\epsilon\frac{\epsilon^{3}}{3!}++)[/tex] Now, we arrange this to: [tex]\frac{yX*\sin(X*)}{1+\cos(X*)}=\epsilon+\frac{\sin(X*)}{1+\cos(X*)}(\frac{\epsilon^{2}}{2}++)+\frac{\cos(X*)}{1+\cos(X*)}(\frac{\epsilon^{2}}{3!}++)[/tex]  Remember now that the numerator of LHS is "close to zero", by assumption that X* is an approximate solution!. Breezing past, for now, the potential trouble of a "too small" denominator on LHS, we term LHS for "kappa", and expand "epsilon" in the power series. [tex]\epsilon=a_{1}\kappa+a_{2}\kappa^{2}+++[/tex] Thus, we get, trivially [itex]a_{1}=1[/itex], and to second order: [tex]0=a_{2}\kappa^{2}\frac{3\sin(X*)+\cos(X*)}{3!(1+\cos(X*))}\kappa^{2}[/tex] that is: [tex]a_{2}=\frac{3\sin(X*)+\cos(X*)}{3!(1+\cos(X*))}[/tex] and so on. Agreed so far? 



#8
Dec2012, 07:31 PM

P: 597

Is this the part where I ask if anyone's thought of using a different formula? The basic formula for the area of a segment is [itex]A_{segment} = A_{sector}  A_{triangle}[/itex].
If we know the length of the chord between the outer endpoints of the radii and the distance from the center to that chord, the area of the segment can be found by [itex]A_{segment} = \frac{x r^2}{2}  \frac{hc}{2} = \frac{x r^2  hc}{2}[/itex], where c is the length of the chord, h is the distance of the chord from the center, and x is in radians. In this case, [itex]x = \frac{2A_{segment}+hc}{r^2}[/itex]. If we can't find c and h, would it not be reasonable to say that [itex]\displaystyle y = x + sin(x) = x + \lim_{r→\infty}\sum_{n=0}^{r} \frac{(1)^n x^{n+1}}{(2n+1)!}[/itex]? Thus, our approximation for x gets increasingly better as r approaches infinity. Either way, remember that your answer for x will be in radians. 


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