Author Topic: Semi-Official 7.x Suggestion Thread  (Read 169044 times)

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Offline Paul M

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Re: Semi-Official 7.x Suggestion Thread
« Reply #315 on: May 22, 2016, 10:21:36 AM »
I would suggest that instead of a terra-former producing x atm of gas that the formula should be:

(tech_rating)*(12756/(diameter of current body))**3  [12756 is the diameter of Earth]

So using the NCN for Mars a single terraformer should produce:

(0.002 atm/year)*(12756/6800)**3 = 0.0132 atm/year

So smaller bodies terraform faster but larger bodies go even slower.  Basically this takes into effect the volume of gas the engine has to produce.  The cube is geometrically correct but other factors could be used to balance it for game purposes.  Even a linear increase would speed up terraforming smaller planets and moons.  Correspondingly large bodies will terraform much slower.
 
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Offline QuakeIV

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Re: Semi-Official 7.x Suggestion Thread
« Reply #316 on: May 23, 2016, 02:34:48 AM »
Fund it.
 

Offline sloanjh

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Re: Semi-Official 7.x Suggestion Thread
« Reply #317 on: May 23, 2016, 07:04:52 AM »
The cube is geometrically correct but other factors could be used to balance it for game purposes. 

Why do you say that cubed is correct when the surface area goes like R**2?

John
 

Offline Paul M

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Re: Semi-Official 7.x Suggestion Thread
« Reply #318 on: May 23, 2016, 08:32:06 AM »
Why do you say that cubed is correct when the surface area goes like R**2?

John

I was being lazy.  I was working on the assumption that the volume of the gas can be related to the volume of the planet itself to avoid a messy formula for the ~10 km of dense gas above the planet that is the atmosphere.  But using the square is correct as the full formula can be approximately given by (Re/Rtf)^2 .  I just didn't follow the math all the way through as I was avoiding writing out the formula and doing the algebra.

But on the other hand my feeling is this is something that should be adjusted for game balance more than just out of geometric considerations.
 

Offline swarm_sadist

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Re: Semi-Official 7.x Suggestion Thread
« Reply #319 on: May 23, 2016, 10:22:38 AM »
Just use the surface area of the planet. The atmosphere is mostly within 20km of the planet so there is no need for fancy calculations.

KISS (Keep It Simple Stupid). Words to live by.
 

Offline Paul M

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Re: Semi-Official 7.x Suggestion Thread
« Reply #320 on: May 24, 2016, 02:56:07 AM »
I am continually informed that my definition of "simple" doesn't alway match up with other peoples meaning of the word.  But the culprit was lazy thinking on my part more than anything else.
 

Offline swarm_sadist

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Re: Semi-Official 7.x Suggestion Thread
« Reply #321 on: May 24, 2016, 09:58:26 AM »
I am continually informed that my definition of "simple" doesn't alway match up with other peoples meaning of the word.  But the culprit was lazy thinking on my part more than anything else.

Nothing wrong with being complex.

Surface Area of a Sphere = 4*pi*r^2
 

Offline bean

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Re: Semi-Official 7.x Suggestion Thread
« Reply #322 on: May 24, 2016, 10:00:28 AM »
If we're fixing that formula, we should probably also take into account planetary gravity, both for realism and to avoid making large planets too hard to terraform.  Adding a factor of (gPlanet/gEarth) would work.
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Offline swarm_sadist

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Re: Semi-Official 7.x Suggestion Thread
« Reply #323 on: May 24, 2016, 11:23:15 AM »
I was thinking about that as well, but the surface area goes up at a pretty steady pace with planet mass. Surface Area < Volume, usually.
 

Offline bean

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Re: Semi-Official 7.x Suggestion Thread
« Reply #324 on: May 24, 2016, 12:11:48 PM »
I was thinking about that as well, but the surface area goes up at a pretty steady pace with planet mass. Surface Area < Volume, usually.
Well, surface area is going to scale with the square of radius, while mass scales with the cube of radius.  But gravity scales with the inverse square of radius, so the surface gravity scales linearly with radius.  So in theory, and assuming constant planet density, the required time will scale inversely with the radius of the planet. 
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Offline iceball3

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Re: Semi-Official 7.x Suggestion Thread
« Reply #325 on: May 24, 2016, 05:03:41 PM »
Well, surface area is going to scale with the square of radius, while mass scales with the cube of radius.  But gravity scales with the inverse square of radius, so the surface gravity scales linearly with radius.  So in theory, and assuming constant planet density, the required time will scale inversely with the radius of the planet.
Doesn't the inverse square law only apply to the diminishing effects of gravity over range, with the radius in question being said range?
Volume and mass are almost directly linearly correlated (granted, given the same density). With mass being directly tied to gravitational pull on a linear basis...
So I think in this case you might just be using the wrong maths. Surface area of the planet would be a lot better value to terraform by. Feel free to correct me if I'm wrong, though.
 

Offline bean

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Re: Semi-Official 7.x Suggestion Thread
« Reply #326 on: May 24, 2016, 05:27:47 PM »
Doesn't the inverse square law only apply to the diminishing effects of gravity over range, with the radius in question being said range?
Volume and mass are almost directly linearly correlated (granted, given the same density). With mass being directly tied to gravitational pull on a linear basis...So I think in this case you might just be using the wrong maths. Surface area of the planet would be a lot better value to terraform by. Feel free to correct me if I'm wrong, though.
I'm not using the wrong math, but we're into some moderately esoteric physics, so I'm going to have to show it to explain.  If I'm misunderstanding you, then I'm preemptively sorry.
The shell theorem states that a spherically symmetric body will behave as if it is a point mass concentrated at its geometrical center, and that it doesn't matter if it is solid or hollow.  So we can calculate the gravitational force on someone standing on the surface of the Earth using the mass and radius of the Earth, just as we do to work out the pull on the moon.  It also states that anything inside a spherically symmetrical shell will not be affected by that shell's gravity at all.  The parts closer to you are counterbalanced by the parts farther away.  This means that someone digging down into a spherical planet will see exactly what someone would see if they were peeling away the planet instead. 
(The second part isn't particularly relevant, but it is interesting, so I'm not going to take it out.)
Now for the math:
1. The planet's mass is equal to rho*4/3*pi*r^3, and we'll assume that rho (density) is constant.  The acceleration due to gravity on a planet's surface gPlanet (assuming that the second object is small relative to the first one) is G*mplanet/r^2.  If we substitute in the equation for the planet's mass we get G*rho*4/3*pi*r^3/r^2, which simplifies to G*rho*4/3*pi*r.
2. The pressure on a planet's surface P is (slightly simplified) equal to (MassAtm/4*pi*r^2)*gPlanet.  If this doesn't make sense, consider conservation of momentum.  The pressure on the piece of air just above the ground must be equal to the force of gravity on the column of air above that piece.  Otherwise, the air would be accelerating upwards or downwards.  Either would be bad if sustained. 
3. Let's assume that each terraforming platform produces so many tons of atmosphere each year, dMassAtm.  Using this we can work out:
dP = (dMassAtm*gPlanet)/(4*pi*r^2) = (dMassAtm*G*rho*4/3*pi*r)/(4*pi*r^2).
Cancelling terms, we get:
dP = (dMassAtm*G*rho)/(3*r)
In other words, the change in pressure is proportional to density (smaller planet for a given mass means less surface area to spread the weight out across and higher gravity) and inversely proportional to radius (bigger planet means that you have to pump out more mass).  In aurora terms, we'd spec the terraforming machines on Earth, and then multiply the rate by the relative density and by the inverse of the relative radius.
Make sense?
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Offline sloanjh

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Re: Semi-Official 7.x Suggestion Thread
« Reply #327 on: May 25, 2016, 07:49:49 AM »
I'm not using the wrong math, but we're into some moderately esoteric physics, so I'm going to have to show it to explain.  If I'm misunderstanding you, then I'm preemptively sorry.
The shell theorem states that a spherically symmetric body will behave as if it is a point mass concentrated at its geometrical center, and that it doesn't matter if it is solid or hollow.  So we can calculate the gravitational force on someone standing on the surface of the Earth using the mass and radius of the Earth, just as we do to work out the pull on the moon.  It also states that anything inside a spherically symmetrical shell will not be affected by that shell's gravity at all.  The parts closer to you are counterbalanced by the parts farther away.  This means that someone digging down into a spherical planet will see exactly what someone would see if they were peeling away the planet instead. 
(The second part isn't particularly relevant, but it is interesting, so I'm not going to take it out.)
Now for the math:
1. The planet's mass is equal to rho*4/3*pi*r^3, and we'll assume that rho (density) is constant.  The acceleration due to gravity on a planet's surface gPlanet (assuming that the second object is small relative to the first one) is G*mplanet/r^2.  If we substitute in the equation for the planet's mass we get G*rho*4/3*pi*r^3/r^2, which simplifies to G*rho*4/3*pi*r.
2. The pressure on a planet's surface P is (slightly simplified) equal to (MassAtm/4*pi*r^2)*gPlanet.  If this doesn't make sense, consider conservation of momentum.  The pressure on the piece of air just above the ground must be equal to the force of gravity on the column of air above that piece.  Otherwise, the air would be accelerating upwards or downwards.  Either would be bad if sustained. 
3. Let's assume that each terraforming platform produces so many tons of atmosphere each year, dMassAtm.  Using this we can work out:
dP = (dMassAtm*gPlanet)/(4*pi*r^2) = (dMassAtm*G*rho*4/3*pi*r)/(4*pi*r^2).
Cancelling terms, we get:
dP = (dMassAtm*G*rho)/(3*r)
In other words, the change in pressure is proportional to density (smaller planet for a given mass means less surface area to spread the weight out across and higher gravity) and inversely proportional to radius (bigger planet means that you have to pump out more mass).  In aurora terms, we'd spec the terraforming machines on Earth, and then multiply the rate by the relative density and by the inverse of the relative radius.
Make sense?

The important part here is your number 2 - the pressure on the surface is (assuming atmosphere depth is much smaller than radius) the weight of the mass of air above you (i.e. mass per unit surface area) times the planet's surface gravity.  The total mass of the entire atmosphere is then simply that times the surface area, which goes like radius squared.  I considered folding in surface gravity when I pinged Paul about R^2, but since the range of habitable surface gravities isn't very big (5x-ish at most) I didn't think it was worth bringing up that complexity.  The reason I'm going here is to make sure people understand that your original point (scale by relative surface gravities) is the real thing they should be paying attention to, both since that's the thing with a selection effect to not vary much and because that's what Aurora tracks, not the dependency on density.  They're both correct formulae, it's just the surface gravity part is a lot simpler to grok IMO.

John
 

Offline bean

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Re: Semi-Official 7.x Suggestion Thread
« Reply #328 on: May 25, 2016, 10:35:34 AM »
The important part here is your number 2 - the pressure on the surface is (assuming atmosphere depth is much smaller than radius) the weight of the mass of air above you (i.e. mass per unit surface area) times the planet's surface gravity.
Exactly.  If you're doing really serious atmospheric modeling, you take variations in gravity into account, but it isn't worth doing here. 

Quote
The total mass of the entire atmosphere is then simply that times the surface area, which goes like radius squared.  I considered folding in surface gravity when I pinged Paul about R^2, but since the range of habitable surface gravities isn't very big (5x-ish at most) I didn't think it was worth bringing up that complexity.
I think it's very worth bringing up, to reduce how strongly planetary scaling cuts the effectiveness of terraforming stations.  If we assume that density is pretty much constant, the range of habitable planetary radii is also going to be 5-ish, too.  With varying density, it might be 50% larger.  This means that the ratio between large and small planets is going to be in the double digits, and large worlds will take forever to terraform.  Folding in gravity takes the scaling from square to linear (approximately).
Quote
The reason I'm going here is to make sure people understand that your original point (scale by relative surface gravities) is the real thing they should be paying attention to, both since that's the thing with a selection effect to not vary much and because that's what Aurora tracks, not the dependency on density.  They're both correct formulae, it's just the surface gravity part is a lot simpler to grok IMO.

John
Fair enough.  I did suggest that originally, then went into the full math to explain why it was important, but it's mathematically easier to use density when you do that.
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Offline NihilRex

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Re: Semi-Official 7.x Suggestion Thread
« Reply #329 on: June 03, 2016, 10:30:35 AM »
<Snip>
 and large worlds will take forever to terraform.
<Snip>

I see no problem with this part.  Reshaping a world should be the work of generations, and not "Sister Jenny found a planet on her lunch break, it should be ready for the Amish Colonists by Tuesday."
 
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