Hi Guys,
One interesting implication of a space-time universe is that we cannot calculate the mass of the universe at any one time. Lemma can see the cover of our four dimensional ball, but we cannot. We can see other planets or galaxies at some earlier point in time than we are at. Remember as I pointed out every thing we call a planet is just the “point” on a vector where ore universe crosses the vector. We assume the vector is linear (eg θ, φ, and ξ are constant and time—“r”– grows). If not time and distance interact.
From what I have read about calculating the mass of the universe those doing the measurement assume a roughly spherical known universe. In a sense they collapse over time. Their result is a mass that is far greater than would be predicted by the known ordinary matter. The discrepancy has led to a model that has in it dark energy and dark matter. However, they are not measuring the ‘Now’ universe. They are at best measuring the points where the spiral intersects the 3-d vectors in 4-space. Recall that there is more to the 4-d ball than our 3-d vector. Is it possible that part of the mass they measure is the mass of the 4-d ball not in our spiral? That’s just a thought. It would certainly explain why dark matter is so hard to find. What would the mass of an information ball look like to a five-dimensional Lemma?
One thing to note is that as time passes the spiral gets longer. Does that mean that the density of the ‘Now’ universe is less or that at an earlier ‘Now’ it would be greater? Would that affect our measure? Would it allow for faster expansion due to weaker gravity?
I looked at your conclusion, Artemy. Even though the equation:
L = a / 2 * [ φ * √ 1 + φ² + ln( φ + √ 1 + φ² )
has higher powers. I don’t see automatic acceleration of the lengthening of the spiral with time. If the circle itself is growing at a steady pace, the spiral length will continue to increase, but I can see no acceleration. The difference between the lengths of the spiral when the circle (at 2* j) has radius X to when the circle has radius Y is a constant. It is a/2. See the figure below. In this case X is 4.074 (the length for the spiral to have a length of 13.8), and Y is 3.260 (four fifths of 4.074. The length of the spiral at radius=3.260 is 11.04, which is four fifths of 13.8. All of the ten segments bear the same ratio to each other.

Let us suppose though that the expansion is indeed accelerating. That might imply that we should be looking at a logarithmic spiral. The equation for that is:
R=a*(e^ k)℘
There is also a set of spirals that belong to the Archimedes family Those have equations that can be written generally as:
R=a*(℘^p) ,
where the value of p determines the acceleration of the spiral.
For instance if p=1/2, the spiral is called a Fermat’s spiral and decelerates; if p=1, it is an Archimedes spiral and is a linear relationship. It is what I’ve discussed up to now; if p= -1/2, the spiral is called a lituus spiral. The lituus spiral accelerated and then leaves the spiral as does the hyperbolic spiral in which p= -1. These are the named spirals, but other values are quite possible. The Archimedes family it more promising since using the logarithmic spiral does no permit values of zero unless either a aproaches zero or k is negative. I have tried setting p to 1.25 while keeping the spiral length at 13.8. I am not sure of the segment lengths I have calculalted for the segments, but the radius growth with steady growth of ℘ is for real. The table below shows that result. In this case the value of a is 0.40956. Thus the equation is:
R=0.40956*℘^1.25 , as opposed to:
R=0.6484*℘,
where in both cases ℘ takes values from 0 to 2*π.
π
Circle & exp 1.0 diametres | Circle segment lengths | Exp=1.0 Length of spiral segments | Ezp=1.0 Length of Spiral | Exp=1.25 Length of spiral segments | Exp=1.25 diameter of circles | Exp=1.25 Length of spiral | Exp=1.25Circle diametres | Exp=1.25 Circle diametres |
8.148 | 2.56 | 2.468 | 13.793 | 2.776 | 9.9474 | 13.8 | 8.148 | 1.006 |
7.335 | 2.56 | 2.219 | 11.325 | 2.413 | 8.7300 | 11.023 | 7.142 | 0.986 |
6.519 | 2.56 | 1.966 | 9.106 | 2.o72 | 7.5261 | 8.610 | 6.164 | 0.947 |
5.691 | 2.56 | 1.715 | 7.140 | 1.755 | 6.3691 | 6.538 | 5.217 | 0.915 |
4.890 | 2.56 | 1.468 | 5.425 | 1.434 | 5.2529 | 4.765 | 4.302 | 0.877 |
4.075 | 2.56 | 1.224 | 3.957 | 1.150 | 4.1825 | 3.3306 | 3.425 | 0.832 |
3.295 | 2.56 | 0.978 | 2.733 | 0.843 | 3.1653 | 2.180 | 2.593 | 0.784 |
2.445 | 2.56 | 0.759 | 1.755 | 0.673 | 2.2086 | 1.338 | 1.809 | 0.720 |
1.630 | 2.56 | 0.563 | 0.996 | 0.422 | 1.3304 | 0.6653 | 1.089 | 0.645 |
0.815 | 2.56 | 0.433 | 0.433 | .0.245 | 0.5544 | 0.2432 | 0.454 | 0.454 |
Totals | 25.6 | 13.8 | 13.8 |
This sheds a new light on what signifies time. When R and j are linearly related it is easy to see R as time, but when they are not, it is clear that time is tied to ℘ even though ℘ is related to circumference rather than radius. The derivative of radius (R’) is expansion.
Speaking of exponents higher than one, there is also a spiral called a Galilean spiral with the equation:R=a*℘² –l, which when the l is zero looks much like the one below

If it looks familiar, it should because it is similar one that combined two Archimedes spirals showing what would happen if the viewer looked both ways. It is different in that the internal-nodal point (where the lines cross) is in a different distance from the origin, but always at ℘=π. For the Galileo spiral they cross at 1.02, where as for the earlier double spiral they cross at a radius of 2.037. When the exponent is 1.25, they cross at a radius of 1.7125. The Galilean spiral shows much more expansion than the Archimedes spiral or the spiral with an exponent of 1.25. If one were to use 5 as an exponent, then when ℘=π, the spiral crosses at a radius of 0.127

Although, not necessary for the theory, it might be fun to speculate if something is different in the evolution of the universe before the two spirals cross.
Ciao,
Bruce
PS Really glad your visit was before Calgary had seen covid-19