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Old 2021-09-05, 20:06   #45
uau
 
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Quote:
Originally Posted by a1call View Post
If I understand correctly, you are describing a square base (area = 1 units square) with a free-formed dome (area = 2 units square). A sphere is the most optimum general minimized surface, but it has a circular cross-section.
I'm not requiring the added surface to match the boundary of the square, so for a small enough added surface area the solution would be a half sphere. However, on a unit square you could fit one with at most radius 0.5, which isn't big enough to use all the 2 available surface area.
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Old 2021-09-05, 20:25   #46
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no comment ;p
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Old 2021-09-05, 21:38   #47
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Originally Posted by firejuggler View Post
no comment ;p
Um....
Quote:
Originally Posted by Uncwilly View Post
https://en.wikipedia.org/wiki/Oxbow_lake
This may have an impact on how much land a 200 meter fence and a lake or river can encompass.
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Old 2021-09-05, 21:51   #48
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ah sorry. I should have known that someone already though of this.
Another example would be be mount st michel in france.
in that case it is the whole european-asian plate

Last fiddled with by firejuggler on 2021-09-05 at 21:53
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Old 2021-09-05, 21:54   #49
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Quote:
Originally Posted by firejuggler View Post
...
Another example would be be mount st michel in france.
Why use such a tiny island ? go for Afro-Eurasia. And with an island you don't need a fence, but for the sake of the problem you put it somewhere on the shore.
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Old 2021-09-05, 21:56   #50
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St michel island would be, in that case, the 'out' of the fence.
Also, the suez canal cut the Afro-Eurasia continent

Last fiddled with by firejuggler on 2021-09-05 at 21:59
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Old 2021-09-05, 23:24   #51
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Quote:
Originally Posted by Dr Sardonicus View Post
... there are fairly simple formulas for the quantities required, assuming the lake has radius 1.
And those formulae are?
Quote:
Originally Posted by Dr Sardonicus View Post
I'm a dunce at programming, but, with the formulas in hand, it took only a few minutes to write a Pari-GP script to print out the numbers.
And the code is?

And your answer is?
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Old 2021-09-06, 01:37   #52
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Quote:
Originally Posted by retina View Post
And those formulae are?
See post #32. You get an equation for half the lake sector angle like tan(x) * (\tau / 2 + 2*x) = 4, the solution to which probably isn't expressible with elementary functions so needs to be solved numerically.
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Old 2021-09-06, 14:03   #53
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Quote:
Originally Posted by retina View Post
And those formulae are?

And the code is?

And your answer is?
We assume the lake has radius 1, and the fence is a circular arc of length L which meets the lake at right angles. Let A be the center of the lake, B the center of the circle of which the fence is an arc, and P and Q the points where the fence meets the lake. Let \theta be the acute angle BAP (measured in radians). Then the angle ABP is \frac{\pi}{2}\;-\;\theta, and the radius BP has length \tan(\theta).

The angle of the arc of fence between BP and BQ is 2\pi\;-\;2(\frac{\pi}{2}\;-\;\theta)\;=\;\pi\;+\;2\theta so the length L of fence satisfies the equation

L\;=\;\tan(\theta)(\pi\;+\;2\theta)

The area enclosed by the sector of fence between the radii BP and BQ is \frac{1}{2}\tan^{2}(\theta)(\pi\;+\;2\theta)

The rest of the area enclosed by the fence and the edge of the lake is within the "kite-shaped" quadrilateral APBQA. If F is the intersection of PQ and AB, the quadrilateral is the union of the 4 triangles APF, AQF, BPF, and BQF. These can be rearranged into a rectangle of sides 1 and \tan(\theta), so the quadrilateral has area \tan(\theta). The circular sector of lake PAQ within the quadrilateral has area \frac{1}{2}1^{2}(2\theta)\;=\;\theta, so the area enclosed by the fence is

\frac{1}{2}\tan^{2}(\theta)(\pi\;+\;2\theta)\;+\;\tan(\theta)\;-\;\theta.

In the original version, the lake had radius 50 and the fence had length 200 = 4*50, so here we take L = 4, giving the equation \tan(\theta)(\pi\;+\;2\theta)\;=\;4. I don't know a closed-form solution, so I used Newton's method to obtain a numerical solution.
Code:
? {
t=Pi/4;
print("0 "t);
for(i=1,8,
f=tan(t)*(Pi+2*t)-4;
fp=Pi/(cos(t))^2 + 2*tan(t) + 2*t/(cos(t))^2;
t = t - f/fp;
print(i" "t)
);
print();
print("Radius of fence circle is "50*tan(t));
print();
print("Distance between center of lake and center of fence circle is "50/cos(t));
print();
print("Area enclosed by fence is "2500*((tan(t))^2/2*(Pi+2*t) + tan(t) - t))
}
0 0.78539816339744830961566084581987572105
1 0.72304343030496242371104078487746358835
2 0.71815938930962624688096481937315314626
3 0.71813340203617805545425687957399093198
4 0.71813340130850774616059434588664824093
5 0.71813340130850774559009248053397820187
6 0.71813340130850774559009248053397820152
7 0.71813340130850774559009248053397820152
8 0.71813340130850774559009248053397820152

Radius of fence circle is 43.688540881007952979946174346648823038

Distance between center of lake and center of fence circle is 66.397956326317025917614143737943306255

Area enclosed by fence is 4757.9476288799235830166949506623779518
?
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Old 2021-09-06, 14:15   #54
retina
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Quote:
Originally Posted by Dr Sardonicus View Post
Code:
Area enclosed by fence is 4757.9476288799235830166949506623779518
The units are missing, but nevermind. Mrs Fred will be very happy with her new 4757.9... m² rose garden next to the lake. I think Farmer Fred will have a very pleasant evening after erecting the fence.
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Old 2021-09-06, 17:31   #55
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Quote:
Originally Posted by retina View Post
The units are missing, but nevermind. Mrs Fred will be very happy with her new 4757.9... m² rose garden next to the lake. I think Farmer Fred will have a very pleasant evening after erecting the fence.
I think her name is Wilma. As for the pleasant evening, I wouldn’t count on that. He is probably cooking up a new scheme with Barney Rubble which will eventually get him into trouble.
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