Help with some Space Math (sorta)

[Anaraxes]

There's a Java ChView that sounds like it's meant to be an update. I didn't actually launch it since modern Java won't let it run. (The ChView web pages explain that they ask for unrestricted file system access in order to save changs to the data files to your disk drive. I wasn't sufficiently interested in computer archaeology to pursue it further.)

If you want to see if this version serves your needs, here's what Google gave me:
http://www.solstation.com/index.html
http://www.ocean-of-storms.com/jaymin/software/chview/

[Anthony]

They made it a java servlet rather than a standalone, which is why it won't run.

[Varyon]

ChView looks to be a dead end for me. Any ideas on the other questions? Here's what I'm thinking of going with so far.

For the arrangement of the core stars, it occurs to me that any platonic solid would allow for a number of core stars that, while not all equidistant from each other, would all have the same number of other core stars adjacent to them. However, I like the distinction of Inner Core vs Outer Core, so I think I'll go with a regular tetrahedron as the Inner Core, and each face of the Inner Core tetrahedron serving as a face of another regular tetrahedron, with the Outer Core systems located at the tips. This gives an arrangement where each Inner Core system is 50 light years from each of the other Inner Core systems, and is also 50 light years from 3 of the 4 Outer Core systems. The Outer Core systems, meanwhile, are each 50 light years from 3 of the 4 Inner Core systems. 3-dimensional math is a bit beyond me (hence wanting a modeling program of some sort), so I'm not certain how far the Outer Core systems are from each other, or from the non-adjacent Inner Core system.

Unless there are objections, I'm going to assume using the Pythagorean Theorem for calculating the time needed to travel is appropriate, and am going to go with the path being in the form of an arc (unless, of course, the vessel can simply fly straight toward or away from the sun).

How far a vessel can go on in a set amount of time is still beyond me to calculate quickly; if I ever do manage to run a game in the setting, I'll probably keep a spreadsheet open that I can plug some numbers into to check each of the possible destinations, although unless the players really press their luck in terms of energy in their ship's capacitors, it's not terribly likely to come up.

For the stuttering-gravity idea, unless someone wants to give some input, I think I'm going to abandon it. As it currently stands, if we assume the intensity of felt gravity is proportional to the time spent at it (so spending 50% of the time at 1G would result in a felt gravity of 0.5G), we're looking at something like taking twice as long to reach a destination (1:1 alternation of on/off) in exchange for being able to experience all of 0.1G acceleration during the trip. This either means no gravity during travel, or some sort of spin gravity - but all vessels in Harpyias are what GURPS calls streamlined (with nautical lines for the larger, more efficient ones), which is normally forbidden from having spin gravity. Considering a streamlined vessel is targeted from the front as though it were 1 SM smaller, would it be appropriate to allow for spin gravity for such, but treat them as though they were 1 SM smaller for determining maximum gravity? For example, an SM+10 streamlined vessel would have a maximum 0.15G for gravity. Does that sound about right?

[Isikyus]

Considering this, would it be appropriate to simply use the Pythagorean Theorem here? Say your destination is in an orbit 1 AU away from your current location, and is located 45 degrees away from where you'd intersect its orbit flying straight out; for simplicity, assume it isn't moving. The long way to travel to it would be to follow your own orbit for 0.5*pi days, reaching the 45-degree mark, then boosting out for 1 day to reach your destination. Boosting out first, then making the 45-degree trip, would similarly take pi/2 + 1 = ~2.57 days. Going at a diagonal, we use the Pythagorean Theorem to determine it would take ((pi/2)^2 + 1^2)^0.5 = ~1.86 days instead. Does this seem correct?

This works as far as I'm able to tell.
If you use equatorial coordinates like Anthony's solar system maps (newer thread), the (distance)/2 term drops out and you end up with clean Pythagorean distances.


A fuller proof of this (hopefully all the special characters show up):

Measure your position in normal polar coordinates, as a distance r from the central star, and an angle φ from some arbitrary zero — the fixed stars, say, or the main habitable planet.
Write this as a complex number in polar form, z = re^(iφ)

Now, suppose our ship runs the drive for a short time 2*Δ_t, at an angle θ to the radial direction. (θ=0 indicates travel directly away from the star, θ=π/2 orbiting at a constant distance, θ=π travel towards the star, and so on.)
Call our starting point z_0, and the new location z_1.

We can approximate the new r-coordinate r_1 as r_1 = r_0 + (r_0 * Δ_t) cos(θ).
Similarly, take φ_1 = φ_0 + Δ_t sin(θ).
As usual for calculus, assume Δ_t is small enough, and we can treat these as exact.

r_1 = r_0(1 + Δ_t) cos(θ), so
z_1 = r_0 * (1 + Δ_t cos(θ)) * e^(i*φ_0) * e^(i * Δ_t sin(θ))

ln(re^(iφ)) = ln(r) + iφ
This gives us equatorial coordinates u = ln(r), v = φ.

So taking ln(z_0) = u_0 + iv_0, and similarly for z_1, we have:
u_1 + iv_1 = u_0 + ln(1 + Δ_t cos(θ)) + iv_0 + i*Δ_t sin(θ)

For a longer travel time t, we can calculate the new position by adding up sufficiently-small steps:
u_t + iv_t = u_0 + iv_0 + lim_{n→∞} n( ln(1 + t/n cos(θ)) + it/n sin(θ))

Simplifying the limit gives lim_{n→∞} ln( (1 + t/n cos(θ))^n ) + it sin(θ))
= ln( lim_{n→∞} (1 + t/n cos(θ))^n ) + it sin(θ))
= ln( e^(t cos(θ)) ) + i sin(θ))
= t cos(θ) + it sin(θ))

So the distance from (u_0, v_0) to (u_t, v_t) is exactly t sqrt(cos^2(θ) + sin^2(θ)), which is Pythagorean.

[Varyon]

Originally Posted by Isikyus

This works as far as I'm able to tell.

Thank you. It's been too long since I did Calculus and the like, so I can't really follow the proof, but I'll take your word for it. Would you agree that the path would be an arc rather than a straight line?

[Anthony]

Wow, my old logarithmic solar system maps. It's a way of looking at orbits that I haven't seen other people do (my original purpose was to allow doing orbital mechanics using paper and transparent overlays, which is probably a pointless thing to do).

[Isikyus]

Originally Posted by Varyon

Thank you. It's been too long since I did Calculus and the like, so I can't really follow the proof, but I'll take your word for it. Would you agree that the path would be an arc rather than a straight line?

You're welcome!

Yes, the paths are arcs, specifically arcs of a logarithmic spiral.

The exceptions are boosting straight in or out (as you mentioned), and travelling between two points on the same circular orbit — in that case the fastest path is along the orbit itself.

[Varyon]

Originally Posted by Isikyus

You're welcome!

Yes, the paths are arcs, specifically arcs of a logarithmic spiral.

The exceptions are boosting straight in or out (as you mentioned), and travelling between two points on the same circular orbit — in that case the fastest path is along the orbit itself.

Excellent. Yes, that's the shape I was assuming for the arcs; sounds like I wasn't too off on that part of things.

Thinking further on how far a ship can go on a charge, it occurs to me this would be an egg-shaped area, which probably prevents using any sort of simple equation to predict where one can reach. Looks like I'll have to stick with "determine places nearby, calculate time to reach each to see if any can be reached."

It looks like most of my questions have been answered (or abandoned, like that weird stutter-gravity bit), so thanks everyone for that. I think I'm actually going to stick with tetrahedrons for the relationships of all the aetheric stars - the core (inner and outer) will be as described previously, with the outer cores serving as a point of another tetrahedron, another tetrahedron on the opposite face, and so forth, possibly with some of the "side" faces spawning new tetrahedrons of their own. Certainly, the strangely-regular nature of the spacing of aetheric stars will raise some questions in-setting, but that's perfectly acceptable. I'd love to find something that would let me visualize this; the closest I've found is this, but that only lets me add one additional tetrahedron.

[Isikyus]

Originally Posted by Varyon

Thinking further on how far a ship can go on a charge, it occurs to me this would be an egg-shaped area, which probably prevents using any sort of simple equation to predict where one can reach. Looks like I'll have to stick with "determine places nearby, calculate time to reach each to see if any can be reached."

While it's an egg-shape in real space, on a map like Anthonys' the area should be an ellipse — or a circle, with with right scales for angle and distance-from-sun.

Since it's a logarithmic scale, the gap between any two adjacent orbits is the same vertical distance in "map space". Suppose this is 1 cm; then 1 cm vertically = 1 day of travel.
Horizontal lines on Anthony's map are circular orbits in real space. If the full map (360 degrees of orbit) is 4pi cm wide, you have 4pi cm horizontally = 4pi days for a full orbit, so 1 cm = 1 day horizontally as well.

Because travel time is Pythagorean, this applies to every direction: x cm on the map at any angle is x days of travel. If you have, say, 1/2 day of charge left, you can reach anywhere within an 0.5 cm radius circle in map space.

Originally Posted by Varyon

I'd love to find something that would let me visualize this; the closest I've found is this, but that only lets me add one additional tetrahedron.

I'm not sure what software there is for this, but I can see some ways to get a physical model. For instance, there's this model of the tetrahedron-with-tetrahedrons-on-the-faces arrangement you're using for the core stars — that page also lets you rotate and view the 3D model online, but as a shop page it isn't editable.

You might also be able to make a stick-and-ball model with something like toothpicks and gumdrops, or maybe paper (though that might make it hard to see the inner stars).

[Varyon]

Originally Posted by Isikyus

While it's an egg-shape in real space, on a map like Anthonys' the area should be an ellipse — or a circle, with with right scales for angle and distance-from-sun.

Since it's a logarithmic scale, the gap between any two adjacent orbits is the same vertical distance in "map space". Suppose this is 1 cm; then 1 cm vertically = 1 day of travel.
Horizontal lines on Anthony's map are circular orbits in real space. If the full map (360 degrees of orbit) is 4pi cm wide, you have 4pi cm horizontally = 4pi days for a full orbit, so 1 cm = 1 day horizontally as well.

Because travel time is Pythagorean, this applies to every direction: x cm on the map at any angle is x days of travel. If you have, say, 1/2 day of charge left, you can reach anywhere within an 0.5 cm radius circle in map space.

I may wish to study Anthony's process, then, and generate some maps for my systems.

Originally Posted by Isikyus

I'm not sure what software there is for this, but I can see some ways to get a physical model. For instance, there's this model of the tetrahedron-with-tetrahedrons-on-the-faces arrangement you're using for the core stars — that page also lets you rotate and view the 3D model online, but as a shop page it isn't editable.

You might also be able to make a stick-and-ball model with something like toothpicks and gumdrops, or maybe paper (though that might make it hard to see the inner stars).

If I had a bundle of d4's, I'd totally start sticking them together to see what sort of shapes I got. Indeed, I may end up doing just that. Stick-and-ball may be doable as well, but I'm not exactly artistically inclined, so I'd probably end up making lopsided rather than regular tetrahedrons. From your comments, however, it occurs to me that the old chemistry stick-and-ball models may well actually work well for my purposes, although it looks like a big bundle of d4's would be cheaper.

EDIT: Looking at that 3d model you linked, it would appear you could place a single "double tetrahedron" (a tetrahedron with another tetrahedron growing out of one face) between each outer tetrahedron, adding an additional star. Indeed, it looks like you could basically stack tetrahedrons indefinitely, and I suspect this will be the model I'll go with just as soon as I find a good way to visualize it...