Wikipedia:Reference desk/Archives/Mathematics/2007 November 22

= November 22 =

Another form of tesseracts?
When you look at an unidimensional object exactly from in front of it, it appears to be zero-dimensional (like a thread which, seen from exactly ahead, seems a point). When you look at a bidimensional object from in front, it seems unidimensional (like when you only see the longitudinal edge of a bidimensional sheet of paper). When you look at a tridimensional object from in front, it looks like a bidimensional object (like a cube which, seen from in front, seems to be a square). Therefore, one could conclude that objects seem to have one dimension less than they really have if seen from exactly in front.

My question is: do you think this would apply to quadridimensional objects as well? This would imply that 4-dimensional figures, seen from exactly in front, would look like cubes - in other words, the shadow cast by a four-dimensional object would be cubic. Would this also imply that 4-dimensional objects, instead of looking like tesseracts, would be rather more similar to two pyramids stuck together at the base with their vortexes pointing in different directions?

I hope I have explained my question well, if there is a part which I haven't made clear enough, just ask and I'll try to explain it better. Thanks. -- Danilot (talk) 08:11, 22 November 2007 (UTC)


 * Vision depends on obtaining a 2D image from a 3D object. The image is formed by projection of points of the 3D image (usually on the surface, unless it is transparent or translucent) on a thin, and therefore approximating an ideal 2D, "screen" (which can be a retina, or a photographic plate). By the process of projection, you thus lose one dimension, along the line of sight.
 * Now if the 3D object is very flat, and thus approximating a flat 2D object, like a straight sheet of paper, it can be maneuvered in such a way that the direction in which it has a very small extent is perpendicular to the line of sight. Then you also get a small extent in the image.
 * In a 4D world, 4D hyperobjects might be likewise projected on a 3D hyperscreen. Since we live in a 3D world, we cannot experience this, but only imagine it. An attempt to visualize this, though, is given in the form of animated images in the image gallery of our article Tesseract. If the projection sometimes looks like a cube, it is because the original is a hypercube, so whenever one set of axes is in the direction of the line of sight, the others are more or less perpendicular to the line of sight. For most other 4D shapes, like the pentachoron, you never get a cube-shaped 3D projection. --Lambiam 09:30, 22 November 2007 (UTC)

Projecting a 2D object in the x-y plane upon the z-w plane in a 4D x-y-z-w space gives the point (0,0), because the equation of the x-y-plane is z=w=0. In this case you loose two dimensions along the plane of sight. Bo Jacoby (talk) 19:05, 24 November 2007 (UTC).

Simple sums
7+8=?

5+7=?

1+8=?

8-5=?

4-2=?

1-5=?

7-1=?

8-7=?

9-4=?

9-1=?

5+1=? —Preceding unsigned comment added by 86.137.243.55 (talk) 19:14, 22 November 2007 (UTC)


 * If you are really unable to do these yourself from memory, you can always ask Google to do them for you. See an example: 7+8.  moink (talk) 20:25, 22 November 2007 (UTC)


 * I would prefer not to use a calculator, that would be cheating. —Preceding unsigned comment added by 86.137.243.55 (talk) 20:51, 22 November 2007 (UTC)
 * And asking other people to do it for you isn't? asyndeton (talk) 20:56, 22 November 2007 (UTC)
 * Bwahahahaha, this is priceless. Using a calculator is cheating, but posting a bunch of addition and subtraction problems on the reference desk is A-OK. —Keenan Pepper 03:02, 24 November 2007 (UTC)
 * See addition, subtraction and negative and non-negative numbers. In particular, you should acquaint yourself with the Addition table. -- Meni Rosenfeld (talk) 11:43, 23 November 2007 (UTC)

snooker
A 'snooker' table (measuring 8 metres by 4m) with 4 'pockets' (measuring 0.5m and placed at diagonal slants in all 4 corners) contains 10 balls (each with a diameter of 0.25m) placed at the following coords:

2m,1m...(white ball) ...and red balls... 1m,5m... 2m,5m... 3m,5m 1m,6m... 2m,6m... 3m,6m 1m,7m... 2m,7m... 3m,7m

The white ball is then shot at a particular angle from 0 to 360 degrees (0 being north, and going clockwise). Just to make it clear, a ball is 'potted' if at least half of the ball is in area of the 'pocket'

Assuming the balls travel indefinitely (i.e. no loss of energy via friction, air resistance or collisions), answer the following:

a: What exact angle/s should you choose to ensure that all the balls are potted the quickest? b: What is the minimum amount of contacts the balls can make with each other before they are all knocked in? c: Same as b, except that each ball - just before it is knocked in - must not have hit the white ball on its previous contact (must be a red instead of course). d: What proportion of angles will leave the white ball the last on the table to be potted? —Preceding unsigned comment added by 86.137.243.55 (talk) 19:15, 22 November 2007 (UTC)


 * We don't answer homework questions here, although if you let us know what specific aspect of the problem you're having trouble with, we can help guide you through it. If I were you, I would begin with a diagram, plotting the positions of the pockets and the initial position of each ball. moink (talk) 20:32, 22 November 2007 (UTC)


 * What are the laws of physics in this virtual Brobdingnagian snooker universum? When two balls collide, should the effect of rotational momentum be taken into account? I'm afraid that this is beyond normal analysis; perhaps simulation could give a result, but to set that up (unless someone has already programmed all the building blocks) is a matter of weeks rather than days. --Lambiam 21:45, 22 November 2007 (UTC)


 * Well, consider how much study has gone into something like the three body problem, and note that this is much more complicated. The amount of collisions and discontinuities to resolve to analytically determine the best angle is rather large. I'm not sure that it would even be possible for a computer to resolve analytically (the amount of cases to check might be of a very large order), though a simulation (which would permit a guess and try approach) is certainly possible. - Rainwarrior (talk) 07:15, 23 November 2007 (UTC)