Talk:Goat grazing problem

Analytic solution
Since the ostensible solution to the problem (other than the untenable integrals) is an iterative approximation anyway, I doubt I'd try to derive the lens area and grind out the numeric answer. I think I'd try to do some simpler possibly progressive geometric approximation, like approximating the arc with a series of line segments. How do you actually solve this problem expeditiously, such as on an engineering assignment? Sbalfour (talk) 00:30, 28 December 2020 (UTC)
 * Plausible and simple approximation added. Sbalfour (talk) 22:18, 29 December 2020 (UTC)

The problem is involuted
The original and best known problem of this kind involved a goat tied to a silo whose tether was the length of the circumference of the silo, and that problem appeared in the cited 1748 journal. The exterior grazing problem and solution as stated in the article is a drastically simplified version of the original. The interior grazing problem as stated in the article first appeared in the first issue of the renown ‘’American Mathematical Monthly’’ in 1894. We should describe and solve both problems as originally stated in the journals, which is not what we have. The entire article needs to be restructured and corrected to reflect the origin, solution and evolution of the problem historically. Sbalfour (talk) 05:03, 29 December 2020 (UTC)
 * Mostly done. Sbalfour (talk) 22:17, 29 December 2020 (UTC)

Practical solutions for interior grazing problem
I'm on the engineering side of mathematics. In my field, we need quick, accurate, understandable and verifiable solutions. I'm afraid none of the solutions given fulfills those tenets. The upper and lower limits in the approximation are not actually very tight. I have no way of verifying that the numeric solution given by the "lens area" method is any more accurate than the number given in the approximation section (I've found plenty of mathematical errors in Wikipedia). If I need a more accurate solution in the field than a number recorded in a spreadsheet, it's very difficult to reconstruct and solve the formulation.

The relationship between $$r$$, the radius of the cutting circle, and the grazing area is continuous, differentiable, monotonic and convex, at least in the relevant region. The area of a circular segment is cubic in the height of the segment. That's highly suggestive that a Taylor series polynomial of degree three would be a very accurate representation of the solution. That's easily obtainable by using one of the other formulations to plot 4 points of $$r$$ or ($$r$$ - $$R$$) versus the grazing area, then solving the resulting set of linear equations for the coefficients of the series.

An even better solution, is to find some simplification that would allow a reduction to a polynomial solution straight forwardly. That might not be the same as the Taylor series. That kind of solution is highly desirable for understandability and verifiability. In a meaningful sense a solution to this problem is still wanting. Sbalfour (talk) 20:09, 2 January 2021 (UTC)

Reduced case (no overlap)
There’s something obviously wrong, because no term in the equation for area represents the geometry of the silo - so we may assume the silo is reduced to a pole of negligible radius, and the grazing area is therefore a circle of radius r, which is $\pi$r2. But that’s not what the formula says. In any case, the proposed solution is not for the original problem, wherein the rope is longer than π or half the circumference of the silo (in fact equal to 2π assuming the silo has unit diameter). This section is not only erroneous, but irrelevant to the problem, and ought to be deleted. The erroneous material appears to have been added in good faith by user:Mnchnstnr at 15:20, 9 April 2016‎. Sbalfour (talk) 21:59, 2 January 2021 (UTC)
 * Done. Sbalfour (talk) 22:21, 2 January 2021 (UTC)

The interior grazing problem is linear
Astonishingly, grazing area as a function of tether length is linear within a few percent over at least 2/3 of the range of possible tether lengths (0 to the diameter of the enclosing circle), with only a slight deviation at the lower end. I actually graphed it out to see, and it's visually a straight line except very near the origin. Identification and quantification of some relevant geometrical property should therefore not only inform the nature of the problem, but make it usefully solvable as a trivial exercise.

Lets follow the implications a little bit: the slope of that nearly linear function is ~$$\Delta A / \Delta r$$. What we're really doing when we bump $$r$$ by $$\Delta r$$ is add in a slice at the mutual chord of the lenses. So, if the length of the chord (the diameter of the lenses) is $$c$$, then $$\Delta A = c\cdot \Delta r$$. I think this implies the rather obvious, that the relationship between the height of a circular segment and it's chord length is nearly linear except when the radius of the segment is very small.

Sbalfour (talk) 07:21, 3 January 2021 (UTC)

Exterior grazing problem solution
There are several items remaining TBD in ths section:
 * justification for substitution tan(x) = x
 * Done. Sbalfour (talk) 17:27, 2 February 2021 (UTC)


 * derivation of formula for area of wedge
 * Done. Sbalfour (talk) 17:27, 2 February 2021 (UTC)


 * derivation of tan transcendental equation for x
 * Done. Sbalfour (talk) 16:32, 7 February 2021 (UTC)


 * derivation of integral expression for dθ (use mnop)
 * Done. Sbalfour (talk) 16:32, 7 February 2021 (UTC)


 * combine second and third steps into upper and lower bounds of single integral
 * Done. Sbalfour (talk) 17:27, 2 February 2021 (UTC)


 * separation of r and R for clarification
 * Done. Sbalfour (talk) 17:27, 2 February 2021 (UTC)


 * analytic solution f(r,R) for any R greater than pi*r
 * Done. Sbalfour (talk) 17:27, 2 February 2021 (UTC)


 * replace K(phi) with linear or other approximation in terms of R/r in final summation
 * Mostly done. Sbalfour (talk) 16:33, 4 February 2021 (UTC)

A bit ago I discovered that if $$x = \frac{R}{r}-\frac{\pi}{2}$$, then $$\phi = \frac{1}{x-\frac{1}{x}} = \frac{x}{x^2-1}$$. That expression is the first two terms in the continued fraction expansion of $$-tan(x) = \cfrac{x}{1-\cfrac{x^2}{3-\cfrac{x^2}{5-\cfrac{x^2}{...}}}}$$ The estimate for $$\phi$$ is quite plausible over nearly the entire range of $$\phi$$, except within .03 radians of $$\frac{R}{r}=\pi$$.
 * May need a high level copyedit for composition and technical diction now that everything that was originally intended to be in, is there; this is supposed to be an article about the goat problem, not a how-to exposition. Details of the solution like approximation of a transcendental quantity, mechanical algebraic transformations, and numerical computation, should/could be moved into a sidebar, footnote, condensed, or removed from the article.  The solution should outline the major steps: decomposition of the problem into computable areas that can be reassembled into the needed ones; selection of polar coordinate system for representation; approximation of a transcendental quantity; integration and analytical geometry to find the areas; finally, algebraic and numerical solution
 * this article is an orphan? No other main article pages link to this one, hmmm... Sbalfour (talk) 17:13, 7 February 2021 (UTC)

Testing $$\tan(t)=t+\tfrac{t^3}{3}$$ solves to x'=4.4933409+ giving $$\phi$$ of .21905-, off by 7 ten-thousandths versus 17 for the linear estimate and 32 for the rational function. (correct=.21898) But not nearly close enough to give a better rounded result to 5 places.

Sbalfour (talk) 08:12, 31 January 2021 (UTC)

Sbalfour (talk) 16:33, 4 February 2021 (UTC)

Maths aint much fun
The solution could be stated more simply, convincing readers that maths they learnt back when is useful for problems like this. The external goat grazing problem is a simple O level use of polar coordinates, involving a very simple parameterisation. Remarkably, the integral comes out, and adding wedges and subtracting the circular area, using a radius of 1 for that, it is straightforward to derive the formula for any R>=pi (R=rope length), being 0.5*pi*R^2 + (R^3 - (R-t)^3)/3 + (R - pi) where t is the solution of tan(pi-t) = (R-t) For R = 2*pi, the original 1748 problem, we have pi - t = 1.351816804 giving Area = 117.5958136 This needs multiplication by r^2 where r=actual circle radius, giving A = 76255.66 sq yds.

Presumably someone would have to submit this as a paper for Wikipedia to allow it. HTH A1jrj (talk) 10:46, 27 June 2021 (UTC)

Interior grazing approx solution arithmetic wrong
"the harmonic mean rather than the arithmetic mean is preferable, i.e. 1.189..." But the harmonic mean of √2 and 1 is 4-2√2 ≈ 1.1716 Shcha (talk) 12:43, 22 June 2023 (UTC)


 * I just came came up to the exact same error while reading the article, the correct harmonic mean should be
 * $$\frac{2}{\frac{1}{1} + \frac{1}{\sqrt{2}}}$$ = 1.172 Enoxos (talk) 18:17, 22 July 2023 (UTC)
 * The section had no citations, and the idea that it was "useful" seems exceedingly dubious, so I've just removed it. --JBL (talk) 20:05, 23 July 2023 (UTC)
 * The rationale of the approximation is justifiable and correct, albeit a bit not clear. Perhaps a better diagram will enlighten the reader about the choice of the upper and lower limits of the approximation. As such, the approximation should not be removed as it serves a practical approximation to an exceedingly challenging and rigorous problem. When and if the approximation posts again, I will construct a better diagram. Enoxos (talk) 00:17, 28 July 2023 (UTC)