Wikipedia:Reference desk/Archives/Mathematics/2017 February 15

= February 15 =

Suicide rates among existential nihilists
This question has been moved to the Humanities desk.

Ellipse arc from focus angle calculation
Can you calculate the area of a fan in an ellipse if you know its angle and shape? Arc length and shape? Thanks.

— Preceding unsigned comment added by Someone with a Question (talk • contribs) 04:01, 15 February 2017 (UTC)


 * Do you mean you want to approximate the shape of the fan blades on a ventilation fan as an ellipse, in order to calculate the area ? I doubt if the shape is very close to an ellipse. An easier method might be to use the fan guard as a grid, if it's rectangular, and count the number of squares (or rectangles) with the fan blade behind them.  Be sure to include portions of a grid unit, too, like 1/10th.  Then determine the grid unit size and multiply by that. StuRat (talk) 04:07, 15 February 2017 (UTC)

No I want to use it in orbit calculations.


 * I'm not sure what an ellipse fan is, in that context. StuRat (talk) 04:46, 15 February 2017 (UTC)

I just realized used the wrong word. My question has changed anyway: Can you determine the arc length if you know the angle from the focus (not the center) & if possible the other way around (if you know the arc length)


 * Well, you would need to know the shape of the ellipse in some way, like it's equation or major and minor axis lengths and directions. If you know that, it seems doable, to me. StuRat (talk) 05:27, 15 February 2017 (UTC)

Thanks. But how? Is there a specific formula or function. — Preceding unsigned comment added by Someone with a Question (talk • contribs) 05:33, 15 February 2017 (UTC)

In polar form, the equation for the distance as a function of the angle is of the form:

$$r = \frac{A}{1 + B\cos(\theta)}$$

An infinitesimal arc length element $$ds$$ can be calculated using Pythagoras' theorem, we have

$$ds^2 = r^2 d\theta^2 + dr^2$$

Therefore, you need to calculate the integral:

$$s = \int_{\theta_1}^{\theta_2}\sqrt{\left(\frac{dr}{d\theta}\right)^2 + r^2}d\theta$$

Count Iblis (talk) 06:01, 15 February 2017 (UTC)

Hi. Thanks for the answer. However, the second step is a little confusing. Could you please explain a little more? (especially the second step)

Our question is similar to this http://math.stackexchange.com/questions/433094/how-to-determine-the-arc-length-of-ellipse. but we need the angle from the focus. (If thats possible)


 * If you failed to understand Count Iblis' calculation of the arc length, you may want to give a read to this tutorial about that kind of calculus first. Tigraan Click here to contact me 14:13, 15 February 2017 (UTC)

Do note that it's not going to be possible with only elementary functions. You need elliptic integrals for this.--Jasper Deng (talk) 18:42, 16 February 2017 (UTC)

What do you get
If you arrange n rows of circles to make the nth triangular number and connect the circles with lines, you'll have the triangular tiling.

But what if you arrange n layers of spheres to make the nth tetrahedral number. (See the image at the top of the article for the n=5 example of what I mean.) What 3-dimensional tiling will you get if you connect each sphere with a line?? Because the tetrahedron doesn't tesselate 3D space, a tetrahedral tiling isn't a valid answer. Georgia guy (talk) 22:59, 15 February 2017 (UTC)
 * First, to get a tiling like you're talking about you need to fill in the faces, not just connect the edges. The arrangement of spheres is called the face-centered cubic packing and is covered in the articles Close-packing of equal spheres (math viewpoint) and Cubic crystal system (crystallography viewpoint), though we don't have an article for it on it's own which seems a bit of a lapse. Between the two articles you should be able glean that the tiling in question uses alternating tetrahedra and octahedra. --RDBury (talk) 01:14, 16 February 2017 (UTC)
 * Choosing planes to fill in the space between the lines connecting the sphere centres has some choice involved, leading to different shapes, which may not all be the same. If one takes the approach of extending the tangent planes at the kissing points between the spheres until they intersect, I expect one ends up quite naturally at a regular tesselating shape, which I'm guessing is the rhombic dodecahedron.  Note that this approach would lead to hexagonal tiling rather than triangular tiling in the two-dimensional case.  —Quondum 19:08, 17 February 2017 (UTC)