Wikipedia:Reference desk/Archives/Mathematics/2021 April 21

= April 21 =

Bisecting an angle in spherical coordinates
If I have two points in spherical coordinates $$(r_1,\theta_1,\phi_1)$$ and $$(r_2,\theta_2,\phi_2)$$, is there an elegant way to find the bisector $$(r_b,\theta_b,\phi_b)$$ (with respect to the origin) without first converting everything to cartesian coordinates?

The first thing I tried was simply to take the average of the radius and angles, but in the simple case of two coordinates with the same value of $$r$$ and $$\phi$$, differing only in $$\theta$$, the result also has the same $$\phi$$ (latitude), which isn't the same as a great-circle bisection.

I could do this by converting to cartesian, apply the angle bisector theorem, and then convert back to spherical, but that just seems kludgey. I can't figure out if there's a way to stay in spherical coordinates while doing it. My internet searches turned up nothing but perhaps I'm using the wrong search terms. ~Anachronist (talk) 05:38, 21 April 2021 (UTC)


 * Update: Basically this is the average of two vectors in spherical coordinates, where the tails of the vectors are at the origin. And it seems, from my searching, that vector addition isn't practical to do without first converting them to cartesian coordinates. ~Anachronist (talk) 06:04, 21 April 2021 (UTC)


 * I'm confused, do you want the angle bisector of the two vectors or do you want their midpoint? The midpoint is the average (V1+V2)/2. The angle bisector is different. You can talk about bisecting a line segment, which is the same as finding the midpoint, but you generally don't talk about bisecting two vectors unless you're talking about the angle bisector. --RDBury (talk) 06:50, 21 April 2021 (UTC)


 * I want the angle bisector of the two vectors in spherical coordinates, without converting to cartesian coordinates.
 * Restating the problem, hopefully with better precision: Given two vectors in spherical coordinates, $$v_1=(r_1,\theta_1,\phi_1)$$ and $$v_2=(r_2,\theta_2,\phi_2)$$ (both with tails at the origin), describing two points on the surface of a sphere on which there are terrain features (elevations defined by $$r_1$$ and $$r_2$$), what are the $$\theta_b$$ and $$\phi_b$$ values of vector $$v_b=(r_b,\theta_b,\phi_b)$$ that bisects the angle formed by $$v_1$$ and $$v_2$$? I already know that the length of $$v_b$$ is $$r_b=\frac{r_1+r_2}{2}$$. I suspect that $$\theta_b=\frac{\theta_1+\theta_2}{2}$$, but one cannot calculate the polar angle $$\phi_b$$ the same way. ~Anachronist (talk) 07:05, 21 April 2021 (UTC)
 * Take the two points to be $$(1,\pi/2,\pi)$$ and $$(1,\pi/4,\pi/2).$$ Then $$(\theta_1+\theta_2)/2 = 3\pi/8,$$ but my computations result in $$\theta_b=\pi/3.$$ --Lambiam 11:41, 21 April 2021 (UTC)


 * It appears to me that in the problem as stated, the magnitudes $$r_\ast$$ of the vectors are irrelevant; we may as well ask for the ray that bisects two given rays with the origin as their shared initial point. We can use points on the unit sphere as representatives of these rays, equivalent to setting all magnitudes equal to 1. I suspect that converting to Cartesian coordinates, taking the halfway point, projecting it out on the sphere and converting back is in fact the easiest route. --Lambiam 08:54, 21 April 2021 (UTC)


 * I was going to mention that but you beat me to it. Perhaps the OP is talking about the point D on the segment from v1 to v2 so that OD bisects angle v1Ov2. In other words calling the point D in the diagram given in the angle bisector theorem article the angle bisector, but I don't think I've ever heard the term used that way. An angle bisector, as I understand it, is a ray or a vector, which lies on the plane of a given angle, and splits the angle into two equal angles. If it's a vector then any vector in the same direction will also be an angle bisector, which is why magnitude doesn't matter in that case. But I guess it doesn't really matter; I'm not sure I can think of any interpretation of bisector which would be easy to compute in spherical coordinates. There exist formulas for the midpoint on a sphere for points given with latitude an longitude, because those have a relatively common practical application in navigation and cartography, but from what I've seen they aren't much of an improvement over converting to Cartesian coordinates; they still involve evaluating both trig and antitrig functions. The idea of averaging the coordinates will work as an approximation if the points are close together and not near the poles. --RDBury (talk) 09:29, 21 April 2021 (UTC)


 * The wiki source text in section, presenting the transformation from spherical to Cartesian coordinates, states twice, emphatically, " ". But it seems to me that $$\theta$$ and $$\phi$$ need to be swapped, both to obtain the converse of the spherical to Cartesian transformation given earlier in the section, and to conform to "the conventions", stated in the lead to be "the ISO convention" in which $$\theta$$ is the polar angle, the only angle that matters for the Cartesian $$z$$ value. Am I right? --Lambiam 09:40, 21 April 2021 (UTC)
 * I agree. The convention I'm familiar with says θ is the rotation around the z-axis (aka azimuthal angle), but I guess we're using φ for that in our articles on spherical, cylindrical and polar coordinates. (There's one section in the polar coordinates article that still uses θ though. And from what you're saying there are more scattered sections which aren't being consistent.) I'd suggest raising the issue at Wikipedia talk:WikiProject Mathematics so see if this was the result of a discussion there; admittedly I haven't checked in there in a while. Mathworld has a handy table for comparing the different "standards". (As the old saw goes, "The nice thing about standards is that you have so many to choose from.") I think I've seen more than my fair share of calculus texts and the notation I'm most familiar with is Apostle's. --RDBury (talk) 16:00, 21 April 2021 (UTC)
 * Yes, you're correct, the radius (magnitude) is irrelevant to the problem. It matters only in my application.
 * I think the mathematics convention, not the physics convention, should be used in Spherical coordinate system. As someone with a degree in physics myself, this Wikipedia article was the first time I was aware of a "physics convention". Where I went to school, I don't recall any inconsistency in spherical coordinate conventions between my physics and mathematics class. They should be swapped, in my opinion.
 * If there's no good way to find a vector that bisects an angle in spherical coordinates, so be it. I thought I'd ask. I suspected, after struggling with this, and finding nothing online except cartesian coordinate examples, that I may have been tilting at windmills. ~Anachronist (talk) 18:00, 21 April 2021 (UTC)
 * I've seen both conventions used. I've never noticed any systematic relationship between which convention is used and whether the subject is math or physics.  I think this is a case of caveat lector; in any particular writing, you need to check what the author means.  I have no strong preference as to which one to use in the article, but I do think it should point out that both conventions exist. --Trovatore (talk) 18:14, 22 April 2021 (UTC)

How To Check the Accuracy of a Watch With a Sextant and Stopwatch
I read in a book about how to use a sextant and a stopwatch to check the accuracy of a watch, say, "It's 15 seconds fast". That way one doesn't have to wait to check after making corrections every time. Or course, I can't find it. It had something to do with a degree every six minutes, I think. 2601:19C:4380:1A00:B0DB:E4F:F8CC:D0CB (talk) 13:58, 21 April 2021 (UTC)


 * I suppose it has something to do with the celestial sphere rotating by one degree every four minutes (relative to the reference frame of an Earth-bound observer). If "15 seconds fast" is with respect to standardized clock time (like the watch shows 12:35′11″ while a radio-controlled clock has 12:34′56″ EDT), I think this is not possible, since 00:00′00″ EDT is not related to any generally identifiable feature of the daily sky. So I presume this might be about a watch that gains 15 seconds per day. But for one day's difference, 15 seconds of time corresponds to the Sun moving by less than four arcseconds, so the sextant reading should be done fairly precisely. Given the limited precision of sextants, much less than a day between two sightings won't do. The observations can be done by two people, one holding the sextant and shooting the Sun, and another looking at the watch; a separate stopwatch is then not needed. One person will need the stopwatch. Take two sightings some 24 hours apart. The moment a sextant sighting is taken, start the stopwatch. Next, hold the stopwatch next to the watch and stop it at a convenient moment, say when the watch reaches a full minute. The sextant reading can be converted to a solar time of day, using that one degree corresponds to four clock minutes. Subtracting the elapsed time on the stopwatch from the watch time when the stopwatch was stopped gives the time of day according to the watch at the moment of sighting. We have now two times of day, that of the sextant and that of the watch. If the watch is accurate, the difference between these two times should be constant. By repeating the measurement the next day, a pronounced drift should become apparent. --Lambiam 15:29, 21 April 2021 (UTC)


 * Back in the day, when people had to navigate by the stars to avoid crashing into something and sinking the ship they were on, celestial navigation was a subject unto itself and aspiring ship's officers would spend months in a classroom learning its arcane secrets; you can still find the textbooks on places like archive.org. One of biggest advances came in the mid 1700's with Harrison's Marine chronometer; you need to know the time accurately to determine longitude, latitude being much easier to discover. (See Longitude (book) for details.) Conversely, given accurate knowledge of you latitude and longitude you can determine the time, or at least GMT since you'd need to consult a map to find out which time zone you're in. I don't know know the details, but I assume other information was needed as well, accurate star maps etc. Given sufficient time and motivation I assume a mathematician could work out a lot of the science from first principles, but it's really kind of a specialized field of knowledge which mostly falls outside what is considered mathematics. --RDBury (talk) 16:37, 21 April 2021 (UTC)
 * Although GMT was defined in the 17th century, time zones didn't exist until the second half of the 1800s. --JBL (talk) 18:48, 21 April 2021 (UTC)

What is the name of the center of population type that cares about finding the point X, that will minimize the distance between X and the person most distant from X?
What is the name of the center of population type that cares about finding the point X, that will minimize the distance between X and the person most distant from X? Wikipedia have information about Geometric median, Median center, Mean center, but none of those seems to be what I am thinking about.

Edit:If there is a tie, then you try to minimize the distance between the point X and the second most distant person from X, and then if still a tie the third most distant person from X and this goes on.....

One example: If we are talking about russia and a person most distant from a point Y at russia is 4000km from it and a person most distant from a point Z at russia is 6000km from it, point Y would better fit as being the type of center of population I am wanting to find. 2804:7F2:59B:D224:D8D3:25BF:F2C1:C3CB (talk) 17:32, 21 April 2021 (UTC)


 * You want the centroid, or center of mass. ~Anachronist (talk) 18:10, 21 April 2021 (UTC)
 * No, certainly not. --JBL (talk) 18:49, 21 April 2021 (UTC)
 * That minimizes the total and average; the OP wants to minimize the extreme. —Tamfang (talk) 00:42, 24 April 2021 (UTC)
 * Also no: e.g., for three points, the Fermat point of the triangle minimizes the sum of the distances, not the centroid. --JBL (talk) 02:36, 24 April 2021 (UTC)
 * Is the center of the smallest enclosing circle what you are looking for? —Kusma (𐍄·𐌺) 19:09, 21 April 2021 (UTC)
 * Under the usual Euclidean metric, there is a unique point minimizing the maximal distance to points of a given nonempty bounded set, but in other metrics (e.g. the taxicab metric) several points may share this property. (For example, for the set {{nowrap|{(1, 1), (3, 3)$)$}}, each of (1, 3), (2, 2) and (3, 1) is central.) They then all have the same extreme distance, though. The term Chebyshev centre is used for this, but note that this term is also used with a different meaning. There is a relation with morphological erosion. Given a bounded nonempty subset S of a topological space, if d is the supremal distance in S from its Chebyshev centre, then erosion by a disk of radius d wipes out S. If S contains its Chebyshev centre, then no smaller disk wipes it out entirely. --Lambiam 20:27, 21 April 2021 (UTC)
 * Talking about ties, at this case you minimize the distance between X and the second most distant person from X, and then the third.... My question didnt had this rule, because I forgot to add it, I edited the question text to add it.2804:7F2:59B:D224:D8D3:25BF:F2C1:C3CB (talk) 22:33, 21 April 2021 (UTC)
 * One can think of this as the decision problem of selecting the best spot for, say, a fire station in the face of uncertainty where a fire will break out, aiming to minimize the worst-case response time (assumed to increase monotonically with distance). This decision problem can be modelled as a game against nature in which one applies a minimax criterion (see ). I have not seen tie-breaking considerations in this context. With the tie-breaking rule, the optimality criterion is still minimax if the value domain does not simply consists of the non-negative numbers, but is a set of reversely sorted tuples of non-negative numbers, where the order used for comparison is the lexicographic order on Cartesian products. In the case of the Euclidean distance metric, ties cannot occur, since the intersection of two distinct disks of some given radius is contained in a disk with a smaller radius. Working with continuous real numbers in a real-word context where the numbers are obtained by measurement, the likelihood of a tie is infinitesimally small. The tie-breaking rule does not resolve all ties; for the example using the taxicab metric, all three possible centres have the same distances to all points. --Lambiam 07:23, 22 April 2021 (UTC)