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[wierd]

Ok, I decided to cheat, because in reality, I actually dont like math much. (Grin, Just what math can tell me, and what I can do with it.)

As such, I found a little calculator on the internet for calculating surface gravity of a celestial body, based on its density and sphere radius.  Plugging in the amended numbers (For the 2m square value given for 1 play tile, instead of the 1 meter square value I had originally estimated) gives us approximately 210 km sphere radius. (Rounded up.) This means we need 167g/cc density of the planet to achieve 1g surface gravity. The normal earth has a density of 5.5g/cc. This planet is SUPER DENSE compared to the earth. (It may only be 1/30 the volume, but it throws a LOT of weight around!)

The calculator also has a nifty feature, and gives the rate of gravitation acceleration at the surface, which it says is 9.8m/sec. (Anyone want to cross reference that with the large body of "Falling object" science we've done?) We might be able to derive an empirical value for surface gravity instead of the assumed 1g if we use the ACTUAL drop rate.

More interesting facts: Slade, at a density of 200000 kg/m^3,  is more dense than the core of our sun, and 15x more dense than the estimated density of the material at the earth's core.

[itg]

Since 200000 kg/m^3 is 200 g/cc, These numbers are actually very feasible. One would assume that the core of the DF planet is mostly if not entirely slade, and the eerie glowing pits easily compensate for the fact that a solid slade planet would be too heavy.

For the record, I'll explain how the math is done. This stuff should be pretty familiar to anyone who took freshman physics, but I'll do my best to explain it at as basic a level as I can.

First some standard physics notation:

F=force, G=Gravitational constant, about 6.67×10^-11 N m^2/kg^2
M=Mass of planet, m=mass of test particle
r=radius, a=acceleration, d=density, V=volume


F = GMm/r^2     (Newton's law of gravitation)

ma = GMm/r^2     (applying Newton's 2nd law, F = ma)

a = GM/r^2     (canceling out the mass of the test particle)

a = G(dV)/r^2     (applying mass = density*volume)

a = G(d*4π/3*r^3)/r^2     (applying the formula for the volume of a sphere, V = 4π/3r^3)

a = (4π/3)Gdr     (rearranging and simplifying)

r = 3a/(4πGd)     (rearranging to solve for r)

d = 3a/(4πGr)     (rearranging to solve for d)

The last three equations are easy to use to solve for gravital acceleration at the surface (a), radius of the planet (r), and average density of the planet (d), given that you know the other two parameters.

[CaptainArchmage]

Nice job you people have done here.

The interior of the DF planet is not entirely solid, it is full of holes, but what is the exact ratio you get between open space and filled space in the HFS?

You have a planet with a mass of 7.7584772 * 10^21 kg based on wierd's information, and the orbital year is exactly 336 days.

Edit:

Volume is 3.879 * 10 ^ 16 m^3 based on 210km radius
Density is 167,000kg/m^3
Mass is 6.48 * 10 ^ 21 kg.

The synodic month should be longer than the Moon's sidereal month, and the moon's synodic and sidereal months happen an integer number of times per year, so there are either exactly 12 (retrograde orbit) or exactly 14 (prograde orbit) sidereal months in a year, depending on the direction the moon goes around the planet. Putting my visualisation into words there is a bit difficult.

From adventure mode, the Sun rises in the East and sets in the west as normal. The following assumes that the planet orbits around the sun in the same direction as the Earth does. To determine which direction the planet rotates in relative to the orbit around the sun, you need to know whether the planet's sidereal day is longer or shorter than the synodic day. If the synodic day is longer, then the planet rotates prograde, if the synodic day is shorter, then the planet rotates retrograde.

The sidereal period is therefor either exactly 24 or 28 days. If the sidereal period is 28 days, it is because the Moon goes around the planet in a retrograde direction, and assuming the Moon's orbit is not perfectly circular (the science people have brought up says it is not), this means tidal forces will eventually cause it to de-orbit and crash into the planet.

This is the equation for semi-major axis, assuming the Planet's mass is significantly larger than the Moon's:

(T^2) / (4*pi^2 / (G*M)) = a^3 where a is the semi-major axis
M = 7.758 * 10 ^ 21 kg
M = 6.48 * 10 ^ 21 kg

The denominator is equal to 9.134 * 10 ^ -11 somethings

If prograde orbit of 24 days:

T = 2,073,600 seconds
The semi-major axis is 38,340,280 meters = about 38,340 km 35873293.02 meters = about 35,900 km


If retrograde orbit of 28 days:

T = 2,419,200 seconds
The semi-major axis is 42,489,974.79 meters = about 42,500km 40015432.33 meters = about 40,000 km

It should be possible to determine how long until the Moon crashes into the Planet in this case.

[itg]

The interior of the DF planet is not entirely solid, it is full of holes, but what is the exact ratio you get between open space and filled space in the HFS?

There's plenty of wiggle room here, since there's no evidence of how deep the eerie glowing pits are. I figure we can require that the numbers make sense in the end, then use that constraint to fix the proportion of holes to slade.

The synodic month should be longer than the Moon's sidereal month, and the moon's synodic and sidereal months happen an integer number of times per year, so there are either exactly 12 (retrograde orbit) or exactly 14 (prograde orbit) sidereal months in a year, depending on the direction the moon goes around the planet. Putting my visualisation into words there is a bit difficult.

Good catch on the integer number of sidereal months. I think I can visualize it. Wouldn't that be something if Toady actually deliberately based the game calendar on the sidereal month of a moon in retrograde orbit? I wonder if there's any in-game way to determine the direction of the moon's orbit...

The sidereal period is therefor either exactly 24 or 28 days. If the sidereal period is 28 days, it is because the Moon goes around the planet in a retrograde direction, and assuming the Moon's orbit is not perfectly circular (the science people have brought up says it is not), this means tidal forces will eventually cause it to de-orbit and crash into the planet.

I'm not convinced there's any evidence the moon's orbit is not circular. As I recall, that idea was brought up to explain why the length of synodic months vary, but the all the months appear to me to be exactly the same length. Granted, I haven't done to-the-hour measurements, so there's still the possibility I'm wrong.

[Snaake]

The interior of the DF planet is not entirely solid, it is full of holes, but what is the exact ratio you get between open space and filled space in the HFS?

There's plenty of wiggle room here, since there's no evidence of how deep the eerie glowing pits are. I figure we can require that the numbers make sense in the end, then use that constraint to fix the proportion of holes to slade.

The synodic month should be longer than the Moon's sidereal month, and the moon's synodic and sidereal months happen an integer number of times per year, so there are either exactly 12 (retrograde orbit) or exactly 14 (prograde orbit) sidereal months in a year, depending on the direction the moon goes around the planet. Putting my visualisation into words there is a bit difficult.

Good catch on the integer number of sidereal months. I think I can visualize it. Wouldn't that be something if Toady actually deliberately based the game calendar on the sidereal month of a moon in retrograde orbit? I wonder if there's any in-game way to determine the direction of the moon's orbit...

The sidereal period is therefor either exactly 24 or 28 days. If the sidereal period is 28 days, it is because the Moon goes around the planet in a retrograde direction, and assuming the Moon's orbit is not perfectly circular (the science people have brought up says it is not), this means tidal forces will eventually cause it to de-orbit and crash into the planet.

I'm not convinced there's any evidence the moon's orbit is not circular. As I recall, that idea was brought up to explain why the length of synodic months vary, but the all the months appear to me to be exactly the same length. Granted, I haven't done to-the-hour measurements, so there's still the possibility I'm wrong.

- Indeed, there's wiggle room, while at the same time some data to base the proportion off (you could count how many/how many tiles of eerie glowing pits there are per embark square, on average; similar estimates have been made for curious structures and candy spires in the past)

- The calendar being based on sidereal months is very compelling. So compelling, that I'd say that the 12-month calendar is in itself probably the best evidence for the retrograde orbit. I think that because there are no tides(?), the only thing in-game based on the lunar phases are werebeasts. This also means that dwarves/at least one of the other civilized races is at least decent when it comes to astronomy (looks at the moon's location relative to fixed stars, instead of just at it's phases). An invention made in antiquity in the real world, but still.

- Lastly, What little I do know/remember about orbital dynamics is that no system with more than 2 orbiting masses is ever really perfectly circular; ellipses or anomalies of some kind are the norm. So at the very least, the phrase you should be using is "(not) effectively circular", or something to that effect.

[CaptainArchmage]

The orbit of the moon would not be circular at least because of the Sun's gravitational influence.

As I pointed out, if the Dwarf Fortress planet orbits the sun retrograde, the moon could be orbiting the planet prograde. We would need to have some way to determine whether the planets sidereal day is longer or shorter than the synodic day.

I think the 28-day lunar orbit fits better with the dwarven months.

The moon crashing into the planet might be a good apocalypse scenario later on in the game. It may have been planned when the moon was put in. Has the moon been around since 23a?

[itg]

- Lastly, What little I do know/remember about orbital dynamics is that no system with more than 2 orbiting masses is ever really perfectly circular; ellipses or anomalies of some kind are the norm. So at the very least, the phrase you should be using is "(not) effectively circular", or something to that effect.

I thought it was understood that when I said "circular orbit" I was making an idealization. That's standard practice for physics problems. Considering that wierd's calculation for the radius of the planet relies heavily on some huge assumptions (for instance, that the world is round, despite all appearances) and is at best an order of magnitude estimate, the tiny effect of the Sun's gravity on the moon's orbit will make up an infinitesimal portion of our total error. None of this is meant to take away from wierd's work, by the way. He's done a great job, but there's no clean way to turn the flat world of DF into a sphere.

One more thing: We can in principle measure the stability of the moon's orbit from year to year by keeping tabs on the dates of the full moons. If the moon spirals inward or outward at all, the length of the month will show it. I bet the months stay exactly the same length, year after year, even if you somehow make it to year 2 billion. I guess my point here is that we should assume the physics of the DF universe are simplified (In fact, we know this), so it's entirely possible and in fact likely that the moon's orbit is exactly circular.

The orbit of the moon would not be circular at least because of the Sun's gravitational influence.

As I pointed out, if the Dwarf Fortress planet orbits the sun retrograde, the moon could be orbiting the planet prograde. We would need to have some way to determine whether the planets sidereal day is longer or shorter than the synodic day.

I think the 28-day lunar orbit fits better with the dwarven months.

The moon crashing into the planet might be a good apocalypse scenario later on in the game. It may have been planned when the moon was put in. Has the moon been around since 23a?

I agree that would make a good scenario (Majora's Mask, anyone?), but I seriously, seriously doubt it was planned. It's just possible that Toady intended the calendar to be based on sidereal months, then calculated the dates of the full moon based on that (the other likely possibility being that he arbitrarily put 13 full moons into the year because the full moons aren't "supposed to" line up perfectly with the calendar months), but the rest of this is just an entertaining exercise.

[wierd]

Agreed.

If I could get accurate "at equator" temperature measurements over time, I could better resolve the diameter of the equator, and possibly solve for other possible topologies, such as elipsoids or toroids.

However, the temp map given by legends mode is "timeless", and appears to be "average daily" temperatures. I need a plot showing temperatures over a 24/hr period exactly on the equator, for several days, to constrain the topology for such calculations. (The current method only gives longitudinal curvature. Getting say, a week of "hourly measurements at equator" would let me compute in the other direction as well, refining the results.

This would let me know if we are dealing with a "niven style" ringworld, or a "halo style" one, etc.

A toroidal "planet" would lay down much better than a spherical one, just that I lack enough data to constrain its dimensions.

That we could even solve for a sphere, given the very limited data available, is pretty spectacular.

[Snaake]

Does temperature change based off the day/night cycle in adventure mode, or is it just constant regardless of the time of day (or night)? Because if it does change, I'd guess it's a fairly straightforward model, eg. regular or triangular sine wave, that's the same all across the world. But that would have to be verified too, of course. The point is, that would be a possible source for data.

[wierd]

Unknown. I don't usually play adv mode, and suck at it so horribly that I have no real interest in it.

I do know that temperature calculations are active, etc, but I don't know if DF computes a change in temp over a 24hr period or not in adv mode.

[itg]

I just did a quick test, and I can confirm temperature does change with time of day in adventure mode. It seems to work as you'd expect, getting warmer in the afternoon and cooler at night. Hour-by hour measurements will have to wait until later, though.

[CaptainArchmage]

Unknown. I don't usually play adv mode, and suck at it so horribly that I have no real interest in it.

I do know that temperature calculations are active, etc, but I don't know if DF computes a change in temp over a 24hr period or not in adv mode.

It does, based on what I've heard. People report rivers freezing at night and thawing at dawn.

[wierd]

Then that is very exciting. We could conduct 1 week observations on several worlds (to flatten rainshadow and weather biases) to get a threshold for insolation falloff over time, and thus derive equatorial curvature as well as the previously derived longitudinal curvature. We could then determine if the planet is an elipsoid, a sphere (with certainty), or the inside of a torus (and how big.)

A torus solution would introduce the least distortion at the "polar regions" on the map, because it would resolve as a line, and not as a point. It would however, make orbital calculations..... "quite interesting".

[itg]

I think it's also worth considering, depending on how the temperature data turns out, alternative ways to project the flat map onto the sphere. For example, instead of squishing the map near the north pole to make a triangle-shaped region, you could leave it square-shaped (think barcode sticker on a golf ball). You'd lose tileability, but that's merely a convenience for rendering.

[wierd]

That would be a "cube map" projected onto a sphere, and would be 6 identically sized regions per planet. Sadly, the curvature data from the first iteration says this can't happen, otherwise the insolation falloff zone won't be right. (At least for a sphere anyway!)

That may not be the case for an elipsoid, where equatorial curvature is less than longitudinal curvature. (Sphere that's been "mooshed" down at the poles.) That would result in a longitudinal curvature that is a parabola, and not a circle, which would place the 30 degree insolation latitude at a different position on the solid.

Again, I need data for equatorial curvature, and I can only get that from manual adv measurement sets.