Interesting to note that, in most cases, when a given 2d Euclidian projection takes only finite space, the corresponding hyperbolic projection fills all available space. And vice versa, if a given hyperbolic projection only takes finite space, then the corresponding Euclidian projection fills all available space.

@ZenoRogue

3 жыл бұрын

Indeed, I also found this interesting (although there are counterexamples). (I would rather say "spherical projection" -- we are projecting a sphere to Euclidean space, and hyperbolic plane to Euclidean space; although "projection from ... to ..." would be probably the most clear)

@groszak1

3 жыл бұрын

Mercator projection is infinite vertically, band is infinite horizontally

@glitchy9613

2 жыл бұрын

did you mean to say spherical projection

Noticed that the distortions in spherical space are usually opposite what they are in hyperbolic space. Orthographic projection: distant objects get squished in spherical, stretched in hyperbolic Gnomonic projection: distant objects get squished in hyperbolic, stretched in spherical Stereographic projection: distant objects become smaller in hyperbolic, larger in spherical Mercator projection: objects get smaller further from the equator in hyperbolic, larger in spherical Mollweide projection: objects far from the chosen point along the equator form an ellipse in spherical, and a hyperbola in hyperbolic

I think my favorite is the orthogonal projection. I don't know why but it just seems more natural to me

@toimine8930

3 жыл бұрын

Orthogonal projection of the sphere is basically just a photo of the sphere from infinity metres away, so it looks the closest to what normal spheres look like (btw, if you were inside the hyperbolic space, a plane would look like Beltrami-Klein projection, no matter how close or far you were from it)

@palmberry5576

3 жыл бұрын

@@toimine8930 not really, it is what you would get if you had a flat camera, and then cast rays parallel to the normal of the camera’s lens.

@totally_not_a_bot

Жыл бұрын

​@@palmberry5576It's been way too long, but an infinitely distant focal point generates parallel focal lines, which is why lenses call near-orthographic focal lengths infinite.

@palmberry5576

Жыл бұрын

@@totally_not_a_bot yeah idk, I must have been tired when I wrote that or misunderstood then

@cheeseburgermonkey7104

Жыл бұрын

It's not too different from how we see things, its just that the camera's rays start at infinity

"Hmmm... what if... we made the big parts little and the little parts big?" -Nikolai Lobachevsky, staring at his wall map, 1829

@alexanderm5728

3 жыл бұрын

Equirectangular projection is actually my favourite. It's simple to code, lines of latitude and longitude are straight, it covers the entire Earth, and I think it looks like a nice compromise between the massively-area-distorting Mercator and the massively-shape-distorting Gall-Peters.

@TheSummoner

3 жыл бұрын

@@alexanderm5728 for hyperbolic space is awful tho

@palmberry5576

3 жыл бұрын

@@alexanderm5728 the only problem is that any height/ color maps you create will not have an equal distance between points

@groszak1

3 жыл бұрын

@@TheSummoner Why

The hyperbolical analog to a Goode homolosine projection: The spliced sinusoidal and Mollweide, available in a self-intersecting variation which is the hyperbolical analog to interrupted variation

I was confused by the video so I decided to read the description. Needless to say, that was a mistake XD

@neur303

3 жыл бұрын

😂 I'd love to understand it but I'm not sure I want to do the work ☺️

@alan2here

3 жыл бұрын

neur303 your computers screen is probably 2D and has no curvature, so when you try to show curved space it's a little tricky. One way it's trying to see all of the surface of ball at once, and the other it just doesn't fit. roguetemple.com/z/hyper/hyperrug.gif So you try rolling the orange peel out flat with a rolling pin and it looks good in places, and looks all bent still in others. But it's the best you can do anyway. These are all different ways of doing that. With there various pros and cons. Pro Tip: Put most of the distortion in the pacific as it's big and wet and same-y and nobody objects much to inaccuracies there. Don't do what the Mercator projection did and make Africa much smaller than it should be and the UK much larger.

I love how the first three make sense and then the rest are just *stretch*

This video is incredibly helpful. Thank you.

Awesome video! Finally someone using the description to describe what's going on in the video.

@ZenoRogue

3 жыл бұрын

Thanks! I do not like watching videos where people explain something, because I feel I could learn faster from a description. Great to see I am not the only one. Luckily KZread has a transcript feature, so this helps if I want to know what the video is about without watching it. The description is limited to 5000 characters though, so this does not work if you have lots to say.

@alan2here

3 жыл бұрын

I'm looking forward to a future reverse transcript feature. Where a machine synthesis of the content creators voice explains the information in the description at the relevant parts of the video. :)

@neur303

3 жыл бұрын

Yes thank you!

The hyperbolical analog to an Eckert III projection: The circle formula being √(1-x²) , it could be interpreted that similarly to the Minkowski hyperboloid the hyperbolical is √(1+x²). Plug into the formula to match the same way Eckert III does.

Something about the Mercator/band model just makes me happy

@ZenoRogue

3 жыл бұрын

It makes me happy that it makes you happy!

i think the sphere gnomonic projection ( 0:30 ) is the way they allow you to see in source games like hl2, portal, and gmod try going into gmod and taking out the camera: zoom all the way out and look down

@ZenoRogue

3 жыл бұрын

Indeed, the usual perspective projection is basically the gnomonic projection, and it looks quite bad for very wide FOV. Some games use Panini projection, which is better suited for very wide FOV: twitter.com/zenorogue/status/1314573353353609216?lang=en

@tristenarctician6910

3 жыл бұрын

@@ZenoRogue thanks

I love these kind of things. great work!!

Is there a nice description of a common correspondence that goes for each of these pairs? Like, a general recipe which, given a projection (from some class of nice projections?) from a sphere to euclidean space, produces a corresponding projection from hyperbolic space to euclidean space?

@ZenoRogue

3 жыл бұрын

For this video we are doing this mostly intuitively. Generally, if you have a formula which is true in spherical geometry, you can change sin and cos to sinh and cosh (when they are used for distances, NOT when they are used for angles) and change some signs (different sign of the curvature), and get a respective formula for hyperbolic geometry. For example, spherical Pythagorean theorem is cos(α)cos(β)=cos(γ) and the hyperbolic one is cosh(α)cosh(β)=cosh(γ). What I have written above is not precise, but with some practice it is quite easy to find the rules, and applying them to the formulas of given spherical projection yields a hyperbolic analog. (I like this simple writeup: math.stackexchange.com/questions/2768462/similarities-between-non-euclidean-geometries ) For most projections, one can also write the list of properties that it has (e.g., conformal, azimuthal) and if there is a unique projection which has these properties, there should be also a unique hyperbolic projection which has the same properties. This is still based on intuitions. If you want something formal, you could consider that spherical, Euclidean and hyperbolic are special cases of "geometry of curvature K" (respectively, K=+1, 0, - 1). Suppose you have the projection defined for a sphere of curvature 1, then you can also define it for a sphere of any different positive curvature (by scaling)*, and if the function is analytic, it should have a unique analytic continuation for negative curvatures -- this continuation should give the hyperbolic analog. I suppose this should work for all or most of the projections in this video, but I have not checked this. * let our projection for K=1 be given by f_1(x_s,y_s)=(x_e,y_e), where (x_s,y_s) are the coordinates of the point on the sphere, in stereographic projection, and (x_e,y_e) are the coordinates of the projected image. Then you can define this projection for a sphere of curvature 1/R^2 by f_{1/R^2}(x_s,y_s) = f_1(x_s*R, y_s*R)/R.

@drdca8263

3 жыл бұрын

@@ZenoRogue Thank you! This answers my question very nicely. Both thorough and accessible. I also like the writeup linked. Again, thank you!

Theres that one that cuts round the continents, projects them all locally, and arranges them nicely. An equivalent could be to cut round the hyper-rouge cells, project them as if it were flat so they're all the same size and shape, and then just move them apart far enough so that they don't overlap, there'd be lots of gaps between cells and more distant ones might need to rearrange somewhat arbitrarily/chaotically as you move. :)

@ZenoRogue

3 жыл бұрын

When you cut up a sphere and flatten it, you get an interrupted map; if you do the same with the hyperbolic plane, you get an overlapping map. Maybe you could move them away into completely different areas, but that would defeat the purpose.

@alan2here

3 жыл бұрын

It'd quickly get messy with distance from the centre, have to stagger them. I was thinking moving them far enough to get gaps to see the background colour (black) between the tiles as they wont but up against each other perfectly and then they wont need to overlap. I think most of the view would be of lets say distance 4 or so. It's the hyperbolic equivalent, I know it's not going to be very good.

@toimine8930

3 жыл бұрын

@@ZenoRogue how about, instead of keeping the centre uninterrupted and making the thing branch outwards, like you'd do with a sphere, keep a circle and make it branch inwards

@alan2here

3 жыл бұрын

toimine I think there's a couple of inverted projections in hyper-rouge for negative curvature, one comes to a central point and another comes to a central straight or curved line, like a horrifying smiling face.

@groszak1

3 жыл бұрын

What is hyper-rouge? Is rouge like another one of those terms used to describe games that are not quite roguelike?

ortographic projection appears like a more exagerated normal perspective of a sphere, at an first look it appears like the normal one but the countries get smashed when it comes closer to the border

@ZenoRogue

2 жыл бұрын

Orthographic projection is the normal projection (well, if you normally look at spheres from infinite distance).

The spherical projections of cylindrical equal-area and Mollweide are stretched, in order to compensate for the squished distortion in the higher parallels. The hyperbolical analog of the vertical stretching would be horizontal stretching instead, wouldn't it? In case of cylindrical equal-area there could be a standard parallel parameter, scaling by sin(x) horizontally and csc(x) (which is (sin(x))⁻¹) vertically. There is also a cylindrical equidistant analog of standard parallel with no vertical scaling (because the scale is correct vertically), and a Mercator analog which scales the projection vertically same as horizontally. The hyperbolical analogs would use hyperbolical functions instead of the trigonometrical ones.

I think I thought up (rediscovered) Azimuthal Equidistant projection of the sphere as a child but didn't know what to call it or understand that it would be distorted. :) Now I really understand. It might be nice with the centre on the north pole, perhaps be like the UN map.

@Nerdule

3 жыл бұрын

The azimuthal projection centered on the North Pole, in fact, literally is the map used in the UN flag!

@alan2here

3 жыл бұрын

Cheers Eldritch! :)

Was Werner also showboating in making a heart shaped one?

Mercator is the only non circle one that didn't get weird when it rotated, and that is why I like it the best

Really nice work! I have two questions: 1. Did you use some generic procedure for converting between the map projections of the sphere to the map projections of the hyperbolic plane, kind of like how you can use the Noether procedure to arrive at the conserved quantity associated with a given continuous symmetry in physics? If so, what procedure did you use? 2. What software did you use to animate this?

@ZenoRogue

3 жыл бұрын

Thanks! 1. I have answered this question in another comment ("Is there a nice description of a common correspondence that goes for each of these pairs?" by drdca) 2. the HyperRogue engine, aka RogueViz.

@kristoferkrus

3 жыл бұрын

@@ZenoRogue Thanks!

My favorite is Lambert's azimuthal equal-area.

0:45 to 1:30 (hyperbolic) - When you stare off in the distance and the weed starts to hit hard 🙂

You are now a hyperbolic paraboloid moderator, what tiling systems on a hyperbolic paraboloid will be soon supported in HyperRogue?

What ways to project hyperbolical surface to a HyperBolic ParaBoloid and how many projections to HBPB are you aware of

honestly the unwrapped spherical projections look weirder than the unwrapped hyperbolic projections

What is a hyperbolical analog to a Wiechel projection?

Nice

I have a question about the world of oyur game. If it's a flat infinite plane, then how does the day/night cycle work?

@ZenoRogue

3 жыл бұрын

It does not work. The world is a hyperbolic plane (more precisely, the floor level is an equidistant surface), not a flat plane. Every sun or star is visible in a very small part of the world. There is eternal day in some places and eternal night in some places. See here: kzread.info/dash/bejne/nphl1KV7gLbblag.html

@benthomason3307

3 жыл бұрын

@@ZenoRogue so it's some suns and stars attatched to another plane parallel to the floor? does that mean your world is hyperbolic horizontally but euclidean vertically?

@ZenoRogue

3 жыл бұрын

@@benthomason3307 They are all just in a fixed spot, no need to attach them to anything. The world is hyperbolic in all directions, otherwise the perspective would work differently.

@benthomason3307

3 жыл бұрын

@@ZenoRogue ah. Have you decided on what exactly keeps them hovering in place instead of crashing to the ground?

2:33 What is a green on left? Bug?

@ZenoRogue

Жыл бұрын

You mean green above and below the projection? The projection is not defined there. Not sure why it displays green there in some frames but it does not have any meaning, you can call it a bug.

@Luigicat11

Жыл бұрын

@@ZenoRogue I think that's a cell or two being stretched over the top and bottom, but you'd probably know better than I would.

band model gang rise up

flat earthers? no, i'm a GNOMONIC earther

The hyperbolical analog to a Robinson projection: Input user's infinite array of latitude scales

The description is really.....discriptive

I would really like to see what an azimuthal equal-circumferences projection of the hyperbolic plane looks like.

@ZenoRogue

3 жыл бұрын

You mean, an azimuthal projection where a circle around the center with circumference C is mapped to a circle around the center also with circumference C? This is the orthographic projection / Gans model, which is included in the video.

@TheSummoner

3 жыл бұрын

@@ZenoRogue - I didn’t know that, only that geodesics were hyperbolas in it. Thank you again for answering!

Reply to this comment to discuss which hyperbolic and spherical projection is the best and which is cringe. My opinion: Hyperbolic Orthographic is best and Hyperbolic Gans and anything that looks like it is cringe If you defend hyperbolic (co)sinusoidal or both Collignon projections you're also cwige

alright, now project hyperbolic space to spherical

@MattMcIrvin

Жыл бұрын

There is a useful projection that maps hyperbolic space conformally to a hemisphere. It's a basis for some of these. The Beltrami-Klein disk is an orthographic projection of that, and the Poincaré disk is a stereographic projection of that. There's also a half-plane model that is a different stereographic projection of the hemisphere.

The hyperbolical analog to an Equal Earth projection: Input user's polynomial of latitude spacing, adjust longitudes for equal area