SEGMENT CALCULATION EQUATION WITHOUT ITERATION OR SOFTWARE

[MRSSPOCK]

Hi. I'm probably in the wrong area, (pardon the pun), but I don't know where else to post this. The question is, is there a formula that will let me quickly calculate the central angle required to construct a sector within a circle, where the segment resulting from that sector, comprises an area of a specific percentage of the full circle? That probably sounds a bit confused so I will try again. Imagine I have a circle of a known area, e.g. 100mm^2 Imagine this circle represents the bore of a pipe. The pipe has a guillotine type plate to constrict flow. Imagine I want the guillotine to rise a certain amount, thereby providing a flow area of 20mm^2. Is there a way I can relate this 20mm^2 area to the angle such a segment would create, if the ends of the chord were connected by lines to the circle centre? The description given is just as a means to better explain the question, rather than a real example, in case someone starts quoting information related to flow. That's not really the issue at all. It's only a mathematical exercise, that's all. Is there a way to do this other that iteration? Thanks

 

[LiteYear]

Sure, the geometry is straightforward once you realise that for a central angle <180°, the segment area is the sector area minus the isosceles triangle formed by the chord and the two radii. Then it's just a bit of equation rearrangement:

1. Area of circle: A = p*r*r
2. Area of sector: S = a*r*r/2 (a is central angle in radians)
3. Area of triangle: T = sin(a)*r*r/2
4. Area of segment: M = S-T = a*r*r/2 - sin(a)*r*r/2 = (r*r/2)*(a - sin(a))

To use your example, A = 100 and M = 100-20 = 80.

From (1), r=5.64 and from (4):

80 = (5.64*5.64/2)*(a - sin(a))
5.03 = a - sin(a)

Given the relative simplicity of this analysis so far, I suppose it's the non-invertibility of the final expression that is your actual concern. This is indeed a problem that is most readily solved by iterative approximation.

If you want a crude non-iterative estimate, I guess you could take the first two terms of the Taylor Series expansion of sin(a) = a-(a^3/3!). Since the a terms will cancel this will allow you to solve for a algebraically, however it will only be accurate near a=0 so you might need to adjust to get a better approximation.

 

[MRSSPOCK]

Thanks LiteYear.

Yes, I was hoping to derive an equation that would let me for example feed in the radius and a required percentage exposure, which would then return the required lift of the guillotine to reveal the required percentage.

e.g. I have a circle radius = 17.85mm

By how many millimetres would a guillotine need raised in order to expose 14.75% of the circle's area?



I always feel a bit disappointed when the only solution is via iteration.

It just feels to me that there should be a nice clean solution to such simple problems.

It almost feels like a shortfall in mathematics if you know what I mean.

I somehow believe there is a solution to such problems in another dimension, other than the one we are experiencing now, if that doesn't sound too David Ickey

Thanks for your input just the same.

 

[LiteYear]

Yeah, it does seem to be a gaping hole given maths sophistication in general. But such results are not uncommon - the only reason they seem that way is that the examples used in a standard maths education avoids them!

The reason the equation is not invertible in the first place is because it is not a one-to-one function - there are multiple inputs that give the same output. Therefore no inversion exists. Why you can't just restrict the domain to create a one-to-one function and then invert that... well that's a deep question! Unsatisfying yes, uncommon no. In fact, as soon as you involve pi or another transcendental number, it is optimistic to expect an exact solution!

The sector area problem itself is discussed in some detail here:

http://mathforum.org/library/drmath/view/61752.html

and in the linked solution page here (look at the very bottom):

http://mathforum.org/dr.math/faq/faq.circle.segment.html

The last reference in particular at least gives a fast, simple iterative solution based on Newton's Method.

 

[IRstuff]

Another option is an approximation approach. Plotting area of the segment as a function of the distance of the blade, h, below the center of the circle results in nearly a straight line from h = 0 to about h = 0.8*r. I've not figured out the general case, but using h = 0 and h = 0.675*r, results in an area estimate that's less than 1% in error. With some optimization, it looks like you could get an error of less than 0.8%.

TTFN
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