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January 6

Daylight Savings Maximum Shift In Time

Dear Wikipedia volunteer,

I have already looked at many Wikipedia and Google searched articles on daylight savings. Here in New Zealand we shift our time by 1 hour twice a year for daylight savings. Is one hour standard world wide? I think I recall a state in Australia being 15 minutes shifted. I need to know the maximum shift by any country. Thank you :)

Gedium (talk) 02:48, 6 January 2018 (UTC)

First a linguistic point: there is only one S in "daylight saving" (the phrase "daylight-saving time" follows the same pattern as "cost-saving suggestion").

Next a standardization point: time zones and daylight-saving time practices depend entirely on local laws in each country or its subdivisions. There are no official world-wide standards, just common practices.

Now the answer: the tz database is a compilation of information about time zones and daylight-saving time practices throughout the world from a time several decades ago up to the present. Searching through its data tables, I find that a number of places have used a 2-hour shift temporarily in the past (most of them during World War II; the most recent case was Newfoundland in 1988), but they all stopped doing that. Today the only place shown with a 2-hour shift is Troll (research station) in Antarctica, which, presumably at the decision of the people who work there,switches between UTC and +2.

Similarly, the database lists a number of places that have used a shift of less than an hour temporarily in the past—specifically, 20, 30, and 40 minutes. But, again, there is only one place like that today: the Australian possession of Lord Howe Island, which shifts by 30 minutes. --76.69.117.217 (talk) 03:58, 6 January 2018 (UTC)

Perhaps it doesn't need to be said, but many places don't have DST at all, which I suppose would count as a 0-hour shift. Once on Car Talk there was a tongue in cheek proposal for what they called "double-dog DST"; it wasn't adopted anywhere though afaik. --RDBury (talk) 07:05, 6 January 2018 (UTC)

Some places, of course, have DST all year round. Even some of these may advance their clocks in the summer. The time in Vigo, Spain is identical to the time in Kirkenes, Norway, although the longitude difference is 40°. As noted here [1]

Except on the very longest days of summer, the sun never rises before 7 a.m. in the western region of Galicia; in the winter it doesn't rise until 9 -- 20 minutes later than in Copenhagen.

January 8

Distances on the Earth ellipsoid calculated by Googe Earth?

Hello you cunning people everywhere. I wonder how Google Earth can give instantly he distance between 2 points. I checked that they are not calculated on a sphere but on an ellipsoid. Ex : Google Earth meridian perimeter = 40 008 km (equal to the wikipedia one), equatorial perimeter = 40 066 km (wikipedia says 40 075).

I read here Ellipse#Circumference integrals formulas and here Elliptic_integral#Complete_elliptic_integral_of_the_second_kind that it's not easy to get a correct value of these distances. However, the distances are displayed INSTANTLY. You don't need to press the ENTER key to get the distance; you just move the mouse and the distance is there. How can it be done instantly! I thank you for your thingkings. Happy new year.--Jojodesbatignoles (talk) 12:39, 8 January 2018 (UTC)

A relevant article: Geographical distance#Ellipsoidal-surface formulae.

Hopefully someone else here will have a more specific answer about algorithms actually used, but if not, you may wish to ask over on the computing reference desk. -- ToE 14:17, 8 January 2018 (UTC)

(edit conflict × 2) ToE beat me to it, but also see Vincenty's formula. Computation of elliptic integrals can also be done to high precision very quickly via the arithmetic-geometric mean. –Deacon Vorbis (carbon • videos) 14:32, 8 January 2018 (UTC)

There are a few things to take into account here. First, you only need distances to within a few feet; it does no good to get much more precise than GPS. Second, the earth isn't an exact oblate spheroid so the ellipsoid distance is only an approximation anyway. Third, modern processors are very fast, so you can do a lot of computation in less time that it takes the internet to get the answer to you. Fourth, presumably most of the distances they need are precomputed; the length of a stretch of road isn't going to change (modulo a major earthquake) so you can easily look up the value rather than compute it each time. Anyway, is it documented that Google uses ellipsoidal distance? They could just be using spherical distance and applying a first order correction based on latitude. --RDBury (talk) 17:49, 8 January 2018 (UTC)

They have a 3D model of the Earth surface and then can calculate any distance just by numerically integrating along the path on this surface using an appropriate quadrature rule. Ruslik_Zero 20:15, 8 January 2018 (UTC)

Finding the minimum distance requires the calculus of variations though.--Jasper Deng (talk) 10:11, 10 January 2018 (UTC)

Jojo, how did you come up with 40 066 km? I just ran Google Earth Pro (the desktop application), created a path 0° 0° -> 0° 90° -> 0° 180° -> 0° -90° -> 0° 0° (or as close as I could get by clicking), and selected the Measurements tab which gave "Length: 40,075 kilometers". This corresponds to the 6 378 137.0 m Semi-major axis from WGS 84. (6378137.0 * 2π = 40 075 016.7 m) Your 40 066 km, besides being the subject of your question on Wikipédia:Oracle (the French equivalent of our Reference Desks), is given as the equatorial circumference in a number of Indian sources, such as this ISCE school atlas and this Civil Services Examination study book. 40 066 km corresponds to a 6376.7 km semi-major axis. I wonder which datum uses that. -- ToE 20:53, 8 January 2018 (UTC)

It could be a Google Earth bug, it draws some weird paths when you try to connect two points that are nearly 180° apart. For example I just tried to repeat your experiment in a different way, I connected on equator 0°1'E and 179°59'E and it routed me over the western hemisphere. Though, when I got it to work properly, I got 40075.02 km. 93.139.63.123 (talk) 05:12, 10 January 2018 (UTC)

Doesn't sound like a bug. Per Geodesics on an ellipsoid, the shortest path between two points on the equator does not necessarily run along the equator. So had your 40 066 km come from measuring a path which skirted the equator? Funny if it was just a coincidence. I wonder where those Indian publishers came up with it. -- ToE 06:17, 10 January 2018 (UTC)

...for example. the shortest path from 0°N,0°E to 0°N,180°E goes through the pole. --CiaPan (talk) 10:37, 10 January 2018 (UTC)

January 9

Function for a catenary given unequal heights

All treatments of catenaries involving unequal heights (h1 and h2) seem to be only interested in arc length. What is the general function for any catenary connecting two unequal heights, y = f(x)? Arc length is a secondary concern. — Preceding unsigned comment added by 98.14.205.209 (talk) 00:55, 9 January 2018 (UTC)

The general formula for a catenary, allowing for scaling and translation, is

$y-k=a\cosh \left({\frac {x-h}{a}}\right).$

There are three parameters so requiring that the curve pass through two points does not allow you to solve for all three and get a specific equation. You need some additional piece of information, such as the arc length, to determine the curve. It might be possible to determine the catenary passing through three given points, but I don't know how complicated the expressions get. --RDBury (talk) 02:49, 9 January 2018 (UTC)

You have one degree of freedom in your problem, so, as RDBury said above, you need additional condition to solve for one free parameter.
Given a horizontal and vertical displacement you have a slope of a line segment, which shall become a chord of a catenary arc. You can have multiple chords with the same slope on a single catenary curve. Each of them can be scaled to your original problem thus giving different catenary arcs through the two given points. See images in Catenary # Mathematical description # Equation section.
So you can choose some additional constraint (say, the arc length, the maximum curvature, the height of minimum point etc.) to make a solution unique (which does not necessarily mean easy). --CiaPan (talk) 10:35, 10 January 2018 (UTC)

January 12

Life expectancy when you are a baby vs. when you are old

If you are just born into a society where your life expectancy is 80, then someone could say "...your life expectancy is 80..." right?

What if you're, say, 75? You've already survived childhood illness and car crashes in your '20s, right?

So, could someone say "...your life expectancy is now 85..."?*

• (All things being equal.)

Anna Frodesiak (talk) 01:40, 12 January 2018 (UTC)

The third paragraph of our article Life expectancy states, "Mathematically, life expectancy is the mean number of years of life remaining at a given age, assuming age-specific mortality rates remain at their most recently measured levels." According to that definition, your octogenarian has a life expectancy of 5 years, not 85.

Check out the last paragraph of Life expectancy#Life expectancy vs. life span, and that paragraph's last reference, Life Expectancy by Age, 1850–2011. A U.S. white male (the first table) born in 2000 had a Life Expectancy at Birth (LEB) of 74.8 years, while an 80-year-old in 2000 had a life expectancy of 7.6 years, though he had an LEB of only 56.34 years (back in 1920, when born). -- ToE 02:53, 12 January 2018 (UTC)

Thank you, ToE. Sorry about how I worded my question. Of course, I meant remaining life.

If that person you refer to had an LEB of only 56.34 years back in 1920, that was partially because they didn't anticipate a future improvement in healthcare etc., and partially because childhood disease then was more of a consideration, right?

My question would be about a baby born today and an 80-year-old today, assuming that nothing on Earth changes in terms of healthcare improvements, etc.

So, is the main consideration about the bullets the 80-year-old dodged in the past, and what ones are upcoming? And my guess about that (probably wrong) is that child mortality and wars and such are a lot lower now that before, so the 80-year-old is on track to drop dead within a couple of years if the expected lifespan is 80ish. Is that the way to reckon it? Cheers, Anna Frodesiak (talk) 03:10, 12 January 2018 (UTC)

Yes, comparing the octogenarian's future life expectancy of 7.6 years to his LEB of 56.34 years isn't really fair as by definition the values come from age-specific mortality rates at the time of prediction, not taking into account healthcare improvement. Still, back in 1920, an 80-year-old (white male; first table) has a further life expectancy of 5.47 years.

So as I understand the way you expressed your main consideration, then no, the 80-year-old today is expected to live longer than an 80-year-old back in 1920. But I think I know what you are getting at.

For a given year, look at the Life Expectancy at Birth. Then look at someone who is already at that age and see what their life expectancy is. And then compare those figures for different years. For instance, back in 1920 the LEB was 56.34, and a 56.34-year-old had a further life expectancy of between 22.22 and 15.25 years, which we could interpolate to 17.80 years, for an expected age at death of 74.14. But in 2000 when the LEB was 74.8, a 74.8-year-old had a further life expectancy of between 13.0 and 7.6, which interpolates to 10.4 years, for an expected age at death of 85.2.

So in 1920 an LEB=56.34-year-old could expect to live another 17.80 years, while in 2000 an LEB=74.8-year-old could only expect to live another 10.4 years.

Is that what you were getting at? -- ToE 03:59, 12 January 2018 (UTC)

Yes, totally. Thank you, ToE. I'm really glad I understand this now. I'd bet this doesn't occur to a lot of people who are 80 in a place where that is the expected lifespan. They probably think their time will be up any second. Thank you so much for helping me to understand this. You are very kind. (I'm thinking of England too because that is where I'm from originally.) :) Anna Frodesiak (talk) 08:13, 12 January 2018 (UTC)