Talk:Circular segment

No title

Write-up on Circular Segment well done. There is a simpler Trig approach

Removed the hand drawn diagram as it is not relevant to the derivation of the formula and it does not add anything to the article. 165.222.186.195 11:52, 13 September 2006 (UTC)

I don't see why the angle has to be less than 180 degrees

that is just a special case

a small additional problem

Is there a way of computing in an analytical way the length of radius R and possibly the angle Theta once the area of the segment (A) and the length and direction of the chord (c) are given?

It might be a very tiny problem

One sec^2...... pun intended

Isn't the area 1/2 r^2 theta, where theta is in radians. Don't think you need trig, unless i have misinterpreted this article —Preceding unsigned comment added by Addy-g-indahouse (talk • contribs) 11:00, 24 October 2007 (UTC)

That's the area of the circular sector. This is a circular segment. Look at the picture. --76.173.203.58 (talk) 06:53, 14 June 2009 (UTC)

The sentence below the chord length formula does not make sense. It should be removed. Bensij (talk) 09:21, 16 July 2008 (UTC)

Go ahead; nobody is stopping you. --76.173.203.58 (talk) 06:53, 14 June 2009 (UTC)

The area formula should be changed

The final formula for the area of a circular segment has a little problem: In the first case (where theta in radians) the sinus actually takes degrees as an argument! It should be changed to accept radians as an argument, why convert radians to degrees for the sinus? —Preceding unsigned comment added by Hiddensob (talk • contribs) 09:44, 7 May 2011 (UTC)

Changed it to the standard way of writing it. Also removed superfluous notations that theta is in radians. I think the problem is that early math students plug these into their calculators and don't understand why they don't work. What a dark mark on the general populous mathematics education... -72.45.211.194 (talk) 13:25, 20 May 2011 (UTC)

I don't think that an area formula for theta in degrees should be given at all. It's extraneous and confusing. If necessary, just remind people that theta is in radians. If someone doesn't know trig well enough to convert angles between degrees and radians, they are not going to understand the article anyway. Gsspradlin (talk) 00:53, 19 December 2013 (UTC)

Theta is in Radians everyone!

^ The comment that Theta is in radians is absolutely not superfluous! Not everyone needing to use this equation did Mathematics even beyond age 16 (which for some of us is a bloody long time ago)!! — Preceding unsigned comment added by 202.96.219.34 (talk) 07:15, 25 May 2011 (UTC)

incorrect formula

The 3rd formula "a = h(4c^2 + 3h) / (6c)" seems incorrect. I'm pretty sure pi should be involved somewhere. Consider the trivial case h = R and c = 2R when R = 1 (half a unit circle), area is then 19/12 which is not pi/2. — Preceding unsigned comment added by Cpt jeltz (talk • contribs) 07:36, 28 March 2012 (UTC)

I’m going to revive this ancient query because it may be worth reviving: first, I’ll rewrite the formula in its most recognizable form: $a\approx {\tfrac {2}{3}}c\cdot h+{\tfrac {h^{3}}{2c}}$ (your formula had a slight misprint). It’s an approximation formula, and a rather good one. The editors who removed it from the encyclopedia probably weren’t skilled mathematicians and 1) didn’t know that the chord length and segment height vs area are transcendentally related, so no formula is exact; 2) didn’t know the formula given was an approximation; 3) didn’t know the derivation of the formula to show its "goodness" of fit. Better approximations exist, far more complex and involving higher powers of the parameters, but this is the one commonly cited. As you note, the formula performs worst at the high end the range: Θ=Π. The area yielded is 19/12, which approximates Π/2 within 0.8%, that’s decent. But we usually have acute angles, so let’s have an angle of 0.5 radians: this time if c is 2, h is .1257, r is 4.042, the approximate area is .1680+, the actual area .1681-, an error of only 0.02%, pretty good. Again the relationship here is non-algebraic, so no formula can be exact. I think this can be restored to the encyclopedia, as usually, the measurements available are the easy ones to get: h and c, and that’s what we want to use.Sbalfour (talk) 15:49, 3 March 2021 (UTC)

I think I know how that formula must have been derived - it's from an approximation to arcsin, taking at least the first, third and fifth power terms of the Taylor series, discarding the little stuff and consolidating the rest. That'd make sense; if there's a better series over [0,pi] than the Taylor series, that might make more sense. The Taylor series doesn't actually converge very quickly. Sbalfour (talk) 21:07, 3 March 2021 (UTC)

By some experimentation, I came up with a plausible approximation of my own: $a\approx ch\cdot ({\tfrac {2}{3}}+{\tfrac {h}{12R}})$, from a 2-term substitution from the Taylor series for arcsin. R should be replaced with some expression in (c,h).Sbalfour (talk) 22:07, 3 March 2021 (UTC)

I'm going to lay out a derivation of sorts for the above approximation formula, since it's already been removed from the encyclopedia once, and I want to see mathematics to justify its re-inclusion.

$a={\tfrac {R^{2}}{2}}\left(\theta -\sin \theta \right)$; the given formula for area

$a={\tfrac {1}{2}}R^{2}\left(2\arcsin {\tfrac {c}{2R}}-2({\tfrac {c}{2R}})({\tfrac {r-h}{r}})\right)$; subst. using $\theta =2\arcsin {\tfrac {\theta }{2}}$ and $\sin \theta =2sin{\tfrac {\theta }{2}}\cos {\tfrac {\theta }{2}}$

$a={\tfrac {1}{2}}R^{2}\left(2({\tfrac {c}{2R}}+\Delta )-2({\tfrac {c}{2R}})({\tfrac {r-h}{r}})\right)$; polynomial subst. using $\arcsin t=t+\Delta$

$a={\tfrac {1}{2}}R^{2}\left(2\Delta +{\tfrac {c\cdot h}{R^{2}}}\right)=R^{2}\Delta +{\tfrac {c\cdot h}{2}}$; distribute and combine inner terms

$a=R^{2}\left({\tfrac {1}{6}}({\tfrac {c}{2R}})^{3}+\Delta _{2}\right)+{\tfrac {c\cdot h}{2}}$; progressive polynomial expansion $\Delta ={\tfrac {1}{6}}({\tfrac {c}{2R}})^{3}+\Delta _{2}$

$a={\tfrac {c\cdot h}{6}}\left[{\tfrac {({\tfrac {c}{2}})^{2}}{2h\cdot R}}\right]+R^{2}\Delta _{2}+{\tfrac {c\cdot h}{2}}$; refactor to extract ${\tfrac {ch}{6}}$

$a={\tfrac {c\cdot h}{6}}\left[{\tfrac {2R\cdot h-h^{2}}{2h\cdot R}}\right]+R^{2}\Delta _{2}+{\tfrac {c\cdot h}{2}}$; subst.$({\tfrac {c}{2}})^{2}$from $R^{2}=({\tfrac {c}{2}})^{2}+(R-h)^{2}\Rightarrow ({\tfrac {c}{2}})^{2}=2R\cdot h-h^{2}$

$a={\tfrac {c\cdot h}{6}}\left[1-{\tfrac {h}{2R}}\right]+R^{2}\Delta _{2}+{\tfrac {c\cdot h}{2}}$; reduce fraction

$a=c\cdot h({\tfrac {2}{3}}-{\tfrac {h}{12R}})+R^{2}\Delta _{2}$; combine terms in $c\cdot h$

Some enquiry about the relationship between $c\cdot h({\tfrac {h}{12R}})$ and $R^{2}\Delta _{2}$ needs to be made. $\Delta _{2}$ is the residue of $\arcsin({\tfrac {c}{2R}})$ after taking the first two terms of the polynomial expansion. By sample trials, one may surmise that $R^{2}\Delta _{2}$ is nearly twice $c\cdot h({\tfrac {h}{12R}})$. So we may roughly combine the terms into:

$a\approx c\cdot h({\tfrac {2}{3}}+{\tfrac {h}{12R}})$

$a\approx c\cdot h({\tfrac {2}{3}})+{\tfrac {c\cdot h^{2}}{12\left[{\tfrac {h}{2}}+{\tfrac {c^{2}}{8h}}\right]}}$; want $a$ in terms of $c,h$, so subst. from $R={\tfrac {h}{2}}+{\tfrac {c^{2}}{8h}}$

What we want here is for the (usually) miniscule ${\tfrac {h}{2}}$ to be some fraction of ${\tfrac {c^{2}}{8h}}$, hence rewrite the denominator:

$a\approx c\cdot h({\tfrac {2}{3}})+{\tfrac {c\cdot h^{2}}{12\left[k\cdot {\tfrac {c^{2}}{8h}}\right]}}$; k>1 marginally

$a\approx c\cdot h({\tfrac {2}{3}})+{\tfrac {h^{3}}{{\tfrac {3}{2}}k\cdot c}}$; reduce and reorder fraction

What should $k$ be here? We’re approximating a function with polynomial behavior over most of its range except very near π. We don’t want to optimize best case or worst case, but something more like “best fit” case. That’s likely to occur a bit before the function devolves into non-polynomial behavior, so guess .75π. We can hypothecate k by equating the axiomatic area at $\theta =.75\pi$ to the approximation and solving for k:

$.9659=2\cdot .6682({\tfrac {2}{3}})+{\tfrac {.6682^{3}}{{\tfrac {3}{2}}k\cdot 2}}$; subst. the trial constants

$k={\tfrac {.2983}{3\cdot (.9659-.8909)}}\approx 1.326$; solve for k

But $k$ was multiplied by ${\tfrac {3}{2}}$ in the equation where it was a placeholder, hence, replace it with the hypothecated value:

$a\approx c\cdot h({\tfrac {2}{3}})+{\tfrac {h^{3}}{1.989\cdot c}}$

It's not a very pretty constant; it's actually a reciprocal function of π. The reference formula uses a value of 2 for the constant, because it is a "best fit" over most of the range, subtly implying that means 2.0 exactly, with no hint that that constant is conveniently truncated and is actually a transcendental number related to π. The constant 2 has a maximum relative error of .8% at Θ = π meaning that the second of only two significant digits cannot be fully relied upon. The constant needs to be a bit larger at the top end of the range, but while bumping the constant would make a better fit there (but worse elsewhere), it still won't be a good fit. The reason is that the area function is significantly non-polynomial near π in accordance with its arcsin component. We're essentially positing a rational polynomial function of "squaring the circle" to compute the area of the circle, which we know doesn't work very well. We need to shift from a paradigm of ${\tfrac {2c\cdot h}{3}}+\Delta$ to ${\tfrac {\pi R^{2}}{2}}-\Delta$ to compute or estimate the area of the circle because the latter delta is nearly quadratic near π while the former is non-polynomial.

Finally, while the above exposition may justify the posited approximation function until a source can be found, it 1) does not prove that an asymptote exists for the area ratio; 2) nor prove that 2/3 exactly is that asymptote; 3) nor that the Δ of h³/2c is a constructive property of the geometry, i.e. that it should in some sense, be a component of a good or proper estimate.

Sbalfour (talk) 19:54, 5 March 2021 (UTC)

Messed up

This is all just made up. Do any of you know what you are doing? For example, there should not be any 180 degrees in the formulas and at the same time pi. You are mixing degrees and radians. Jfgrcar (talk) 16:37, 4 August 2013 (UTC)

Of course there should. I suppose you mean the arc length formula, $s={\frac {\alpha }{180}}\pi R$? Formulae that relate radius and arc length need the angle to be in radians. The article provides formulae with the angle $\alpha$ degrees as well as $\theta$ radians, presumably for the reader's convenience—and you find both 180 and π in the formula to convert degrees to radians. -- Perey (talk) 04:54, 13 September 2013 (UTC)

Angle from area and radius

What is the formula to get the angle if the area and the radius is known? — Preceding unsigned comment added by 115.70.29.185 (talk) 08:43, 21 December 2020 (UTC) Null string

It's highly unusual that one knows or measured the area directly, and also got the radius. Those two quantities are transcendentally related to the central angle, which is why I suspect this is a trick question. Here is the transcendental relationship: $a={\tfrac {1}{2}}r^{2}(\theta -\sin \theta )$ so, using a substitution from the Taylor series, $\sin \theta =\theta -{\tfrac {\theta ^{3}}{6}},\theta \approx {\sqrt[{3}]{\tfrac {12a}{r^{2}}}}$. Sbalfour (talk) 17:41, 2 March 2021 (UTC)

General order of presentation

There's only a few parameters of a circular segment: the chord length, segment height, arc length, and area. Then there's two incidental quantities, properties of the circle: radius and central angle. And that's it. Some are usually given or can be easily measured: chord length and height. The arc length may or may not be given, and can't easily be directly measured. The radius, central angle, and area are usually the unknowns, and must be computed. That gives us the structure of the presentation. First, givens; second, incidental quantities; third derived quantities (arc length and area).

I also think the presentation should be limited to one realization of each relationship; anyone with a modicum of algebra and trig can substitute arctan for arcsin, (r-h) for h, and etc. The article isn't a math textbook with every flavor an exam could ask about.

So I'm going to do a bit of reorg.

Sbalfour (talk) 17:02, 2 March 2021 (UTC)

I've substantially redrafted and simplified the presentation; just the basic formulae. There's a zillion perturbations of these, like you find in a Schaums mathematical reference - that's not the purpose of the encyclopedia. The article is about circular segments, not a mathematics cheat sheet. Sbalfour (talk) 17:01, 3 March 2021 (UTC)

A comment on completeness

Another editor queried, "what if I have the radius and area and want the angle?" and I answered him above. There are only six quantities associated with a circular segment: height, length, angle, radius, arclength, and area. Given any two, the others can/must be computed. That would be done by substituting and solving the requisite equations stated in the text for one of the other variables, in turn. As a proposition in linear algebra, there are only two independent variables, four dependent ones, and 6 equations relating them, some of which are redundant depending on what quantities are given (independent). In set theory, we have a set of 6 choose 2, which yields a set of (6*5)/(2*1) = 15 unique pairs. It is plausible for one to ask, given some pair, how to solve directly for the other four quantities. That's the kind of question that could occur on a geometry examination, but as an enlightening encyclopedia exposition, exhaustive and exhausting. At some level, it is more informative to assume some understanding of the relationships, than inundate the reader with information (so the article does not have 15 sets of four equations to solve for every possible pair of givens). Sbalfour (talk) 23:00, 5 March 2021 (UTC)

A region of two-dimensional space

Defining a circular segment as a region of two-dimensional space makes the "more formal" definition less specific than the informal one, by introducing the possibility that the space isn't flat. For example, the segment might be the shape of a soap bubble that's being blown through a D-shaped loop of wire. Wouldn't plane area be both more accurate, and more intuitive to readers?

If the term is also used in a non-euclidean context, we can simply change In geometry in the opening sentence to In plane geometry, after which there's no further need for phrases like two-dimensional space unless we want to explain what a plane is. Musiconeologist (talk) 17:29, 31 March 2024 (UTC)

I've now changed it to refer to a plane region, but kept In geometry for now. Musiconeologist (talk) 18:53, 31 March 2024 (UTC)