One way to classify the toratopes was in terms of unshrinkable curves and surfaces. Here, the middle-cuts are useful. It seems that the global rule is:
IF THE CUT WITH CERTAIN COORDINATE HYPERPLANE RESULTS IN TWO DISTINCT SURFACES OF LOWER DIMENSIONS, THOSE SURFACES CAN'T BE SHRUNK.
[b]S[/b]_n+1 =
[cos a_n 0 ] [S_n]
[ 0 sin a_n ] [ 1 ]
PWrong wrote:One way to classify the toratopes was in terms of unshrinkable curves and surfaces. Here, the middle-cuts are useful. It seems that the global rule is:
IF THE CUT WITH CERTAIN COORDINATE HYPERPLANE RESULTS IN TWO DISTINCT SURFACES OF LOWER DIMENSIONS, THOSE SURFACES CAN'T BE SHRUNK.
What about the torinder? You can wrap a string around the tube, but if you pull it all the way to the edge, it will come off and shrink.
The new coordinate system must also be orthogonal; in other words, the dot product of a position vector and the partial derivative of that vector with respect to any of the angular coordinates must equal zero. We see this for spherical coordinates, for example.
In 6D, there are 3 possibilities for (5+1) split, 2 for (4+2) split and 1 for (3+3) split, 6 in total.
In 7D, there are 6+3+2 = 11, in 8 there are 11+6+3+(2*3/2) = 23, in 9D 23+11+6+3*2 = 46, and so on.
PWrong wrote:The difference is the values the angles range between. If you use the conventional coordinate system, the first angle (azimuthal, east-west) runs from 0 to 2 pi, and all the other angles (polar angles, like north-south and garp-marp) from 0 to pi. The north pole is at 0, and the south pole is at pi.
Under your system, the polar angles go from -pi/2 to pi/2, with the equator at 0.
I'm not sure what's wrong with this, but when I was trying to integrate something once with the wrong coordinates it didn't work properly. Anyway, this stuff has already been defined, there's no point changing it now. http://planetmath.org/encyclopedia/HypersphericalCoordinates.html
Let's get back to the tiger. The equations for the tiger are the sum of two circles and a glome. But this means the tiger probably has two sets of equations, like the glome. How can we find the other set?
On the subject of cutting objects, what happens if we play with the radii and angles, as well as the cartesian coordinates? For instance, say we take a radius from the tiger and let it go to zero.
x = A * cos a + C * cos a * cos c
y = A * sin a + C * sin a * cos c
z = B * cos b + C * cos b * sin c
w = B * sin b + C * sin b * sin c
If A=0 or B=0, we have the equations for a circle plus a glome. This is the circle*sphere torus.
If we let C=0, we get a duocylinder.
If we hold the angle c fixed at any value, we also get a duocylinder, with
r1= (A + C cos c), and r2 = (B + C sin c)
If we let a = 0, we get a torus, with its centre on (A, 0, 0)
This suggests that if we hold either a or b fixed, we'll always get a rotated torus.
The only unexpected shape here is the circle*sphere torus. I didn't think that would be related to the tiger much.
Plus, if we let all the coordinates just run from 0 to 2*pi, we get the same results, albeit with multiple covering. (which can be dealt with through some simple identities).
The page you refer to shows merely a convention, and it even mentions that mathematicians and physicians don't agree on it.
The thing is that my approach requires to treat both sin and cos the same, as the hypersphere equation used in tiger equations, for example, require splitting both cos and sin parts.
PWrong wrote:I'm not sure what you mean. Wouldn't you get the same thing either way?
I'm more worried about which angles are azimuthal now. There's a 6D sphere, ((2+2)+2) that can be made in two ways. You can take a circle and replace one component with a (2+2) glome. Or you can take a sphere, and replace each component with a circle. We could label this as (2+2+2). Either way you get the same set of equations, but the first set has five azimuthal angles, and the other has one polar angle.
The good news is there's only one correct answer. The volume of ((2+2)+2) is twice that of (2+2+2), so one of them is an imposter.
The bad news is, to find the volume we have to find the jacobian, which is a 6*6 matrix, work out its determinant, and then integrate it.
(which made no distinction between azimuthal and polar angle)
Users browsing this forum: No registered users and 1 guest