Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.
Quote from "An Application of Geography to Mathematics: History of the Integral of the Secant", by V. Frederick Rickey and Philip M. Tuchinsky, Mathematics Magazine, v. 53, No. 3, May 1980, pp. 162–166:
“
Consider a cylinder tangent to the earth's equator and imagine the earth to "swal [swell] like a bladder." Then identify points on the earth with the points on the cylinder that they come into contact with. Finally, unroll the cylinder; it will be a Mercator map. This model has often been misinterpreted as the cylindrical projection (where a light source at the earth's center projects the unswollen sphere onto its tangent cylinder.
”
What exactly is the difference between these two things? Michael Hardy (talk) 01:54, 2 June 2011 (UTC)[reply]
It sounds like the distance of polar regions from the equator is the difference. In the cylindrical projection, the height of the north pole would be the (foreshortened) radius of the Earth, while with the Mercator projection, the height would be the (correct) 1/4th the circumference of the Earth. That should be about 1.57 times more. StuRat (talk) 02:23, 2 June 2011 (UTC)[reply]
Here's a diagram of one quarter of the globe and it's two projections:
Well, if that's what it means, it's wrong. The key special property of the Mercator projection is that it's a conformal mapping. In this case, what that means is, however much you have to stretch the horizontal scale at a given latitude, to make the parallel as long around as the cylinder is, that's how much you also have to stretch the vertical scale.
This means that small features at any latitude (except the pole itself) will appear undistorted, at least to first order. (The value of "small" depends on how much distortion you can tolerate, and on the latitude.) --Trovatore (talk) 02:45, 2 June 2011 (UTC)[reply]
Saying "The key special property of the Mercator projection is that it's a conformal mapping" seems to neglect the fact that many quite different map projections are conformal without having what should really be considered the key special property of the Mercator projection: that bearings on the map correspond to bearings on the earth. For example, the stereographic projection is conformal, but it doesn't preserve bearings at all. Michael Hardy (talk) 18:21, 4 June 2011 (UTC)[reply]
Oh, I think I see maybe what Rickey and Tuchinsky are getting at. The idea is, draw your globe on a rubber balloon. Put it inside an infinite rigid cylinder, radius equal to the original radius of the globe. Now start inflating it. You can see that the poles never do touch the side of the cylinder, no matter how much you blow up the balloon, but (assuming an infinitely inflatable balloon) any point other than the poles will eventually touch the cylinder and be mapped.
I suppose R and T are assuming that, for any given level of inflation, any point not yet touching the cylinder is equally stretched in all directions. I somewhat doubt that that is actually true for a physical balloon, because my intuition suggests that nearby points that are touching the cylinder are "pushing" the non-touching parts a little in the vertical direction, but I could be wrong. --Trovatore (talk) 02:57, 2 June 2011 (UTC)[reply]
"Trovatore", I don't see any explanation in your comments of the difference between the projection involving the balloon and the cylindrical projection. That the north and south poles are not part of the domain of the function is of course what I expect of both the cylindrical projection and the Mercator projection; I'm not sure why it's worth mentioning here. I think the latter is characterized by the conjunction of two things: (1) it's conformal (angle-preserving) and (2) curves of constant bearing appear as straight lines on the map; e.g. the curve that goes in a direction 71 degrees east of north, which in fact spirals infinitely many times around the north pole, although the arc length of the spiral is finite, is a straight line on the map. Michael Hardy (talk) 13:14, 2 June 2011 (UTC)[reply]
Actually, only the Mercator projection has a problem with showing the poles. They could be shown, as a line at the top and bottom of the map, under both the cylindrical projection and the middle projection I described above. StuRat (talk) 19:38, 2 June 2011 (UTC)[reply]
Well, but the cylindrical projection you described is not the same as the one Rickey and Tuchinsky described. They said the light source was at the center. What you've described is more like a light source along the central axis. Drill a hole between the poles and put a light stick through the middle, and you get your projection; put a light bulb in the actual center and you get the one R and T are talking about. --Trovatore (talk) 19:47, 2 June 2011 (UTC)[reply]
Yes, but they don't seem to be using the term properly, IMHO. Isn't a projection assumed to be normal to the target surface, unless otherwise specified ? StuRat (talk) 20:31, 2 June 2011 (UTC)[reply]
That's "projection" in the sense of a vector space (more precisely, an inner product space). Cartography uses the term a little differently. Any way of mapping the globe onto a flat surface is called a "projection". --Trovatore (talk) 22:05, 2 June 2011 (UTC)[reply]
Right, the cylindrical projection they're talking about, with the light source at the center, is not conformal. A point of latitude θ gets projected by the light source to a point of height tan(θ). That means the vertical stretching is proportional to d/dθ tan(θ) equals sec2(θ), whereas the horizontal stretching is only sec(θ). --Trovatore (talk) 19:15, 2 June 2011 (UTC)[reply]
That establishes easily that the cylindrical projection is not conformal. But how does "swelling like a bladder" result in anything other than the cylindrical projection? Michael Hardy (talk) 20:40, 2 June 2011 (UTC)[reply]
Well, I'm shakier on that part of it. R and T's method works if you assume that (i) once the rubber touches the walls of the cylinder, it sticks and doesn't move and (ii) for the part of the rubber that hasn't yet touched the cylinder walls, at any point, the tension is the same in all directions. I can easily throw (i) into my intuitive picture (I'm already allowing infinitely stretchable rubber) but I'm less clear on (ii). (However I certainly don't see any reason the method would give you R and T's "cylindrical projection"; why would you think it would?) --Trovatore (talk) 21:17, 2 June 2011 (UTC)[reply]
It hadn't occurred to me that the cylinder was a solid thing that would stop the spherical balloon from expanding. I pictured the sphere continuing to expand, so that progressively less of it is inside the cylinder and most of it is outside, but retaining its spherical shape. Each point on the sphere would pass through the cylinder once, and the point on the cylinder where it passes through would be the points it's mapped to. If R & T had something else in mind, they were certainly cryptic about it. Michael Hardy (talk) 22:00, 2 June 2011 (UTC)[reply]
Wouldn't that mean that, since the width at the poles is infinitely stretched, that the height at the poles is also infinitely stretched ? Do they handle this by not going all the way to the poles ? StuRat (talk) 02:53, 2 June 2011 (UTC)[reply]
Yes, to make a Mercator map finite in the vertical direction, you do have to cut out a ring around each pole. --Trovatore (talk) 02:58, 2 June 2011 (UTC)[reply]
Mercator distortionRight. I'd like to see a Mercator map with very small neighborhoods of the poles removed. I think it would be a good illustration of the shortcomings of that projection. Anyone know where we can see something like this? (i.e. even less polar neighborhoods removed than this example from mercator projection). SemanticMantis (talk) 16:00, 2 June 2011 (UTC)[reply]
Well, all flat projections have issues. It's not possible to make a flat projection that preserves all distances, say, because from those you can recover the metric tensor and thereby the curvature tensor. The Mercator projection has some excellent properties if you understand how to read it. It really does show correctly how things are shaped, as long as those "things" are not too large based on their latitude.
In the Seventies or so there were people trying to make a political point out of this, claiming that the Mercator projection was designed to inflate the importance of "northern" peoples. I think that was basically crankery. I read Arno Peters's little handbook on the subject, and it was plain that he just didn't understand the mathematics. --Trovatore (talk) 19:21, 2 June 2011 (UTC)[reply]
There aren't many features to map very near to the poles, so such a map would be kind of boring... just endless Arctic sea at the top and endless Antarctic plateau at the bottom, both at an ever huger scale. 86.160.216.165 (talk) 23:26, 2 June 2011 (UTC)[reply]
Antarctica has lots of interesting features, including mountains, glaciers, underground (or under-ice ?) lakes, colonies of penguins, research stations, routes taken by explorers (and their associated remains), magnetic south pole, etc. StuRat (talk) 23:06, 3 June 2011 (UTC)[reply]
The standard Mercator projection is quite an inappropriate way of mapping Antarctica. Michael Hardy (talk) 00:06, 4 June 2011 (UTC)[reply]
Wikepedia calculates by formula the force T exerted by a swinging pendulum.
what is the unit of measurement for this force T ?
given a length of 25 inches to the bob , has the vector T force of this magnitude ever been put to any useful purposes?- eg battery recharging/electrical generation?
(disregard and force required to start the pendulum movement)
this enquiry is not homework but of general interest
(edit conflict)You're talking about the formulas in Conical pendulum? The units of measurement depend on what units you want to use for all the other variables - if you're working in SI units, then m would be in kilograms, v in metres per second, r in metres, and g in meters per second squared, meaning that T would be in kilograms metres per second squared, or in other words Newtons. But that force just comes from two things - the downwards gravitational force of the Earth on the bob, and the tension from the string. And that force doesn't do any work, since it always acts perpendicular to the direction of motion (see also Work (physics)#Zero work). If you try to extract any energy from the pendulum, you'll just result in the pendulum slowing down, meaning you need to add more energy in to get it spinning at its original speed again. You probably want to read up on Work (physics) to see what you need to have extractable energy from a moving system. Confusing Manifestation(Say hi!) 04:29, 2 June 2011 (UTC)[reply]
Indeed, even pendulum clocks (such as grandfather clocks, most tower clocks, and shelftop pendulum clocks) use the pendulum only to regulate the period of the motion of a wheel, with the power being supplied from somewhere else, usually from a heavy weight lowered slowly. – b_jonas 22:02, 4 June 2011 (UTC)[reply]
With regards to a measure theory problem I was discussing with a lecturer recently, it was claimed in the space of seconds that
.
Why this is is not obvious to me. I have of course worked with limit/sum inequalities before (the standard one I think is bounding log(x) as the integral of 1/x above and below by sums of 1/n) - however, I can't seem to picture why this one is true. I am sure it is entirely obvious and I am simply out of practice - I have not done anything like this in a few years! - but could anyone quickly explain intuitively to me why the inequality holds? The variable y is meant to take any value, and while the integrand didn't specify what it was with respect to, it looks to be 'z' fairly obviously, and I guess we can lose the square roots, I just added them for the sake of completeness. I have tried to draw a picture of the integral to obtain the inequality but I couldn't quite piece the argument together. Could anyone lend a hand? I'm sure it will seem obvious to me as soon as someone has pointed out why it was true so don't worry too much about the background theory, I just need to see why it is true (hopefully in a way that I could apply to other integrals similar in nature if needs be).
Estimate the integral by its upper and lower Riemann sums wrt the partition consisting of integer points. Sławomir Biały (talk) 19:14, 2 June 2011 (UTC)[reply]
Ah smashing, of course. Thanks! :) Typeships17 (talk) 19:21, 2 June 2011 (UTC)[reply]
Your question makes no sense. It may help if you give an example of what you mean by "consecutive integer of x" having a sum y. 86.160.216.165 (talk) 21:50, 2 June 2011 (UTC)[reply]
Seems pretty straightforward to me: there is a set of 13 consecutive integers that sum to 0, and the OP wants to know what the largest is.
Oh yeah, sorry, I failed to understand what "consecutive integer of 13" referred to! 86.160.216.165 (talk) 01:56, 3 June 2011 (UTC)[reply]
I don't think the phrase "consecutive integer of 13" occurs here. Rather, I'd parse it like this:
largest consecutive integer of {13 whose sum is 0}
Thus, the sum of 13 consecutive integers is 0. What is the largest of them? Michael Hardy (talk) 02:46, 3 June 2011 (UTC)[reply]
-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 is a sequence of 13 consecutive integers whose sum is 0. The largest is 6. Widener (talk) 07:13, 3 June 2011 (UTC)[reply]
On a language note, the reason this question is hard to parse is that it doesn't make sense to talk of a consecutive integer. So, instead of how it reads, it should read "What is the largest consecutive integer of 13 consecutive ones whose sum is 0". --86.8.139.65 (talk) 18:59, 5 June 2011 (UTC)[reply]
