6 ms·
Very well done, as always. As someone who is probably below the intended audience level my one complaint is that it wasn't clear to me why the fact that the squ
by hanoz 8y ago
Very well done, as always. As someone who is probably below the intended audience level my one complaint is that it wasn't clear to me why the fact that the square on the sphere has the same area as its projection on the cylinder implies that the total of all these areas is the same, i.e. why do we know the projections don't overlap, when each is cast from a light at a different height on the pole, as it were?
- appden 8y agoImagine the sphere’s surface is divided into one million little rectangles. Their total area is meant to approximate the surface area of the sphere. If each one of those were projected onto the cylinder, and we showed that each of those projected rectangles has the same area and that they don’t overlap, then the total area of all million of the projected rectangles would be there same as those on the sphere. Therefore, the cylinder has the same area, since it is comprised of all those projected rectangles.
- hanoz 8y agoThanks, so it's the overlapping issue I'm left struggling with (sorry, I elaborated on this in an edit after you'd already started replying, by the looks of it). It's not obvious to me they don't overlap, if the 'light' is at a different point for each one.
- coherentpony 8y agoWhen the sun shines through your blinds, do the shadows of each of the individual chinks overlap? How about throughout the day when the sun moves? :)
- hanoz 8y agoThey may do if I measure each shadow separately only when the sun is level with the slat which casts it - I don't know.
- istjohn 8y agoYou're intuition is correct that the rectangles would overlap if projected by a light like shadow puppets. Instead, imagine a series of lasers up and down the vertical axis positioned so that their beams are parallel to the plane that contains the sphere's equator. And imagine that the lasers spin 360 degrees around the vertical axis but their beams always stay parallel to the plane of the sphere's equator. That's the projection being described here.
- Twisol 8y agoRight. Effectively, we're projecting every horizontal slice of the sphere laterally to a fixed distance from the vertical axis. It's a stack of 1D projections, not a single 2D projection.
- Twisol 8y agoI realized after the fact that the more useful terms would be “orthogonal projection” and “perspective projection”. The novelty of the orthogonal projection at hand is that it’s projecting (flattening) in a cylindrical space. With an orthogonal projection, you can usually think of it as taking two planes and squashing whatever object your want to project between them. In the scenario here, the ambient space has been wrapped up, so one of these squashing planes has been wrapped into a cylinder, and the other has been wrapped into a line (the degenerate case). In either event, an orthogonal projection is indeed a collection of orthogonal projections of one dimension less. But that’s not really the whole picture.
- adrianratnapala 8y agoImagine if two non-overlapping patches on the sphere could overlap once they get projected. Now consider some single point in the overlapped region. It came from two distinct places on the original sphere, so this projection is not 1-to-1 but many-to-1. In principle there is nothing wrong with a many-to-1 projection. But in this particular case, I hope it's pretty intuitive that the projection from the sphere to the containing cylinder is indeed one-to-one except at the poles (where it is actually 1-to-many).