The Dyson sphere temperature.

[Omeganian]

I am not sure where to put it now (here or pure ST, please transfer if wrong), after the other topic I put this there was locked for being dead, but since it concerns the discussion about the SW vs. ST shields, the old topic is here, and they told me to start a new topic if I want to (which I do), I put it here.

Mr. Wong wrote on Stardestroyer.net, that the star inside the sphere produces about "less than 45%" Sunpower.

He didn't write however, that it is a lot less. The sphere is not a planet, it has eternal day not to mention a few problems with heat dissipation. Simply speaking, the outer surface must dissipate all the energy produced by the star (getting all of it would be the entire point of building the sphere), while following the Stefan–Boltzmann law

And, according to the Second law of thermodynamics, the temperature of the outer surface can't be higher than that of the inside. Now, if you apply the formula to 45% Sunpower (SD.net has the constants), you get an outer surface temperature of, approximately:

T= ((3.827E26 W*44%)/(1E22m²*4*Pi)/(5.67E-8 W/(m²·K4))^(1/4) = 394 Kelvin = 121 Celsius.

In order to get 300 Kelvin - And I don't think the temperature could be higher - You need to lower the power to about:

E= 300 K^4*5.67E-8 W/(m²·K4)*1E22m²*4*Pi = 5.77E25 W = 15% Sunpower.

To get 15 Celsius (the average for Earth), you need to drop it to:

E= 288 K^4*5.67E-8 W/(m²·K4)*1E22m²*4*Pi = 5.9E25 W = 12.8% Sunpower.

Of course, it doesn't take into account the insulation provided by the sphere, which may make a considerable temperature difference between the outer and inner surface. Perhaps we can ignore the metal looking parts - a dense, strong material is commonly a good heat conductor - but the soil is lousy in this respect, and, having to support life and being constantly recycled by it, it can't be tampered with much. A mere fifty degrees difference will lower the power to 6-7% Sunpower.

The interesting part is, the stars with the appropriate power shown on the Hertzsprung-Russell diagram are noticeably colder than the Sun - which, of course, is even less flattering for the ST shields.

[Junghalli]

I haven't watched the episode, but wasn't the star in the middle of the sphere supposed to be dying? If so, its luminousity might have increased considerably from when the sphere was inhabited.

Was the sphere's inner surface still habitable?

[Omeganian]

The star was somewhat unstable, and the sentient creatures have long since abandoned the sphere. But the plant life was still there, the surface was green.

[Darth Wong]

That's a very interesting angle of analysis. I must admit that I just took the lazy man's way out and neglected to look at several factors which could further reduce the energy estimate.

[Wyrm]

This confused me at first since, if taken with the sun's actual output (L_{\sun} = 3.827e26 W), then at 1 AU = 1.495e11 m, the effective temperature would be:

L = σAT^4; A = 4πR²
T = L/(σ4πR²)^(1/4)
T = 393.67547242 K = 120 °C

Ker-sizle.

But then I realized that the total surface area of the earth is four times its cross-section to the solar input. Most of the sun's energy does not hit it from zenith, which it does on a Dyson sphere.

(By the way, in the lingo, Sunpower is called the solar luminosity.)

So, yea, the luminosity of the star does have to be reduced further, as you say. Also, if we built a Dyson sphere around the sun, it would have to be 279.33987328e6 km in diameter, or 1.87 AU.

[Enola Straight]

Do we have any canon dialogue from the episode? Someone stating outer/inner surface temps, mass, radiant output?

IIRC, Picard ordered the Enterprise-D to drop out of warp, and shuddered when it encountered unexpected gravitational shear. Perhaps there is some technology that enables inner surface's heat energy to be converted to gravitational energy?

[Wyrm]

Something that big is going to have a lot of gravity, period. I calculated the damn thing to be on the order of a few solar masses in and of itself. However, I note that in this post I made a boo-boo: I took the 200 million km diameter to be the radius., thus my calculated mass of 6.333 solar masses to be... somewhat less than that — about 4 times less, or 1.58325 solar masses. Still a heavy bugger, and it seems, much heavier than the star it contains. See the recent Dyson Sphere temperature thread — the star inside the sphere has to be colder (and therefore, less massive) than the sun in order to keep the temperature to Earth normal.

[seanrobertson]

*snip*

Of course, it doesn't take into account the insulation provided by the sphere, which may make a considerable temperature difference between the outer and inner surface. Perhaps we can ignore the metal looking parts - a dense, strong material is commonly a good heat conductor - but the soil is lousy in this respect, and, having to support life and being constantly recycled by it, it can't be tampered with much. A mere fifty degrees difference will lower the power to 6-7% Sunpower.

The interesting part is, the stars with the appropriate power shown on the Hertzsprung-Russell diagram are noticeably colder than the Sun - which, of course, is even less flattering for the ST shields.

Awesome, awesome work, Omeganian. If I'm not the first to say so, let me at least be among the many who should welcome you whole-heartedly to SD.net.

If memory serves, the Enterprise's shields were expected to withstand three hours in a 150,000 km altitude at "23 percent" maximum strength.

6% of Sol's output is ... 2.296E25W? (I've had two glasses of wine and don't feel entirely myself; subjectively speaking, that seems high. Is my math right?)

Assuming the star's as large as ours in spite of its much lesser luminosity, and further assuming the Enterprise's ellipsoid shield profile is roughly 100,000 m^2, she would absorb solar energy at a rate of 373 GW.

In three hours, then, Enterprise would have to cope with 4.028E15W -- a bit less than one megaton over time (4.18E15J).

With full shields, the Enterprise could therefore expect handle about 17,500 terajoules -- the equivalent of four megatons.

This begs the question (not the logical fallacy, mind): how do the shields' instantaneous capacity compare to the ability to cope with energy over half a day? Logic dictates it's far easier to cope with, as Lord M. Wong once noted, a thousand ruler slaps than one (or a few) solid hits with a sledgehammer. Of course, logic and Star Trek in any form rarely coincide

[seanrobertson]

That's a very interesting angle of analysis. I must admit that I just took the lazy man's way out and neglected to look at several factors which could further reduce the energy estimate.

Your work was still most useful, m'lord

After all, when you crunched those numbers, the idea was to be rather generous to Trekkies.

I post in other, only semi-related forums, and I find that generosity remains most useful: When one grants TNG-loving Trekkies weapons and shields that, if not quite reach, but at least approach, the low megaton or high kiloton range, many of their arguments are cut off at what Southern U.S. people call the "stump."

That is, rather than try to undercut Trek's offensive technological capabilities, I try to present it in a somewhat realistic light. That makes the work of demonstrating the Galactic Empire's technological superiority that much easier; my workload is cut 50% or better.

[Count Dooku]

This is a bit off topic, but does anyone know what the minimum mass of a star [of pure hydrogen] would be? The less massive the star, the less energy it outputs, and the longer it lives (I remember that much from intro to astronomy). It's also reasonable to assume you could make a smaller, and more practical, sphere. The inhabitants of that sphere would be able to live there for hundreds of billions, of years, rather than just a few billion.

Having said all that, I think it goes without saying that the Ent D's shields must have been terrible (or the usual bad writing was in place). I mean, it's not like the inside of the sphere was a a few million kilometers from the star - it was probably a hundred million kilometers away (at the very least). Unless the Ent D was thrust into the sphere at near relativistic speeds, it would have received the equivalent EM radiation Venus does. Massive solar flares or not, there couldn't have been that much energy there.

[Wyrm]

Depends on the metallicity of the star, but a good rule of thumb is 75 times Jupiter's mass (ie, 0.0715 solar masses). Although it is true that a less massive star will live longer, but that's because it's power output is much less. This, of course, limits the amount of energy available to the civilization creating it.

Also, any plants used have to be adapted to use the different spectrum of the less-massive star. In particular, it means that the blue part of the spectrum will be deficient, which means that the most energetic catalytic reactions available to plants might be impossible.

[Omeganian]

I thought of something else.

A sphere has no gravity on its inner surface. I don't remember the exact proof, but I think it is centered around the very fact that no work is being performed. Doesn't it mean that if you want gravity, you will actually need to spend energy? But that would mean megawatts per square meter.

[Wyrm]

I thought of something else.

A sphere has no gravity on its inner surface. I don't remember the exact proof, but I think it is centered around the very fact that no work is being performed. Doesn't it mean that if you want gravity, you will actually need to spend energy? But that would mean megawatts per square meter.

No. Although it's true that there's no gravity inside a spherical shell (which is what a Dyson sphere is), it's not because no work is being done. Sitting in your chair, gravity does no work on you either, but that doesn't mean gravity disappears. Rather, there's no gravity on the inside of a sphere because the forces of all the little pieces of the shell cancel out.

If you want 'gravity' holding you to the inside of the sphere, you need to spin up the Dyson sphere. Of course, only the equator of the sphere will have gravity, as the effect vanishes at the poles.

[Omeganian]

That way, the atmosphere will fly away at the poles. Do you need work if you want to prevent that?

[Darth Wong]

I thought of something else.

A sphere has no gravity on its inner surface. I don't remember the exact proof, but I think it is centered around the very fact that no work is being performed.

Actually, it is based on the fact that if you integrate all the way around the sphere, the accumulated gravitational forces will cancel out.

Doesn't it mean that if you want gravity, you will actually need to spend energy? But that would mean megawatts per square meter.

Energy is not measured in megawatts, or megawatts per square metre.

If you want 'gravity' holding you to the inside of the sphere, you need to spin up the Dyson sphere. Of course, only the equator of the sphere will have gravity, as the effect vanishes at the poles.

The larger problem is that the effect will resolve to a tangential force and a plane-normal one everywhere but the equator, so everything will be pushed toward the equator.

[Omeganian]

Doesn't it mean that if you want gravity, you will actually need to spend energy? But that would mean megawatts per square meter.

Energy is not measured in megawatts, or megawatts per square metre.

I just estimated the force needed to pull the objects down with enough power to simulate normal gravity. Just the air is a kg/cm^2, pulled by a force of 1g. If you will have to supply power all the time... Well, I don't think the star will be enough for that, and there are considerable amounts of water and land as well

[Darth Wong]

Doesn't it mean that if you want gravity, you will actually need to spend energy? But that would mean megawatts per square meter.

Energy is not measured in megawatts, or megawatts per square metre.

I just estimated the force needed to pull the objects down with enough power to simulate normal gravity. Just the air is a kg/cm^2, pulled by a force of 1g. If you will have to supply power all the time... Well, I don't think the star will be enough for that, and there are considerable amounts of water and land as well

Force is not energy either. Work is only done when the hypothetical device actually induces movement through application of force.

[Omeganian]

Force is not energy either. Work is only done when the hypothetical device actually induces movement through application of force.

Well, that's what I'm asking. You need to make the air apply pressure on the inner surface of the globe, will you need to perform work if you don't want it to fly away?

[Connor MacLeod]

Doesn't it mean that if you want gravity, you will actually need to spend energy? But that would mean megawatts per square meter.

Energy is not measured in megawatts, or megawatts per square metre.

I just estimated the force needed to pull the objects down with enough power to simulate normal gravity. Just the air is a kg/cm^2, pulled by a force of 1g. If you will have to supply power all the time... Well, I don't think the star will be enough for that, and there are considerable amounts of water and land as well

dude, I dont think you're getting this. You seem to be mixing terminology here, and its muddling up what you are saying. Power is measured in watts, intensity is "energy per unit area", and Mike explained force.

Are you trying to ask whether the Sphere would have to constaintly expend energy (from say, a reactor or something) to keep it's internal gravity active or something?

[Connor MacLeod]

Force is not energy either. Work is only done when the hypothetical device actually induces movement through application of force.

Well, that's what I'm asking. You need to make the air apply pressure on the inner surface of the globe, will you need to perform work if you don't want it to fly away?

Er, no, why? Planets seem to do that fine without needing a constant input of energy, and the sphere itself acts like a wall, which would doubtless contain the atmosphere (the openings like the one the Enterprise went through may have some sort of "atmospheric containment" fields over it as well, since as far as I can recall we didn't see any evidence of it being evacuated into space when the E-D got sucked in)

[Omeganian]

Force is not energy either. Work is only done when the hypothetical device actually induces movement through application of force.

Well, that's what I'm asking. You need to make the air apply pressure on the inner surface of the globe, will you need to perform work if you don't want it to fly away?

Er, no, why? Planets seem to do that fine without needing a constant input of energy, and the sphere itself acts like a wall, which would doubtless contain the atmosphere (the openings like the one the Enterprise went through may have some sort of "atmospheric containment" fields over it as well, since as far as I can recall we didn't see any evidence of it being evacuated into space when the E-D got sucked in)

What about the atmosphere falling on the star?

[Connor MacLeod]

What about the atmosphere falling on the star?

What about it? And how is the star relevant to the atmosphere "escaping" or needing work to keep it in? Isn't the star inside the Dyson Sphere?

[Omeganian]

What about the atmosphere falling on the star?

What about it? And how is the star relevant to the atmosphere "escaping" or needing work to keep it in? Isn't the star inside the Dyson Sphere?

We don't just need to keep it inside, we need to keep it on the inner surface of the sphere. I am merely asking by what measures can you achieve sufficient gravity that no air will be sucked into the star from any spot of the sphere. The sphere is not a planet, its inner surface has no natural gravity.

[Connor MacLeod]

We don't just need to keep it inside, we need to keep it on the inner surface of the sphere. I am merely asking by what measures can you achieve sufficient gravity that no air will be sucked into the star from any spot of the sphere.

Why would it be "sucked into the star?" For one thing, I doubt the star is anything like a real life star in terms of mass or gravity. Secondly, I don't see what the hell you're basing your conclusions on, since we know the Dyson Sphere has gravity (as I recall, we see rivers and oceans and green stuff on the surface - something HAS to be keeping it all in place. Likewise, something has to be keeping the atmosphere in place.)

The sphere is not a planet, its inner surface has no natural gravity.

But, as I recall, that Dyson sphere DID have gravity. Which brings us back to the same fucking poitn I already asked. Are you assuming gravity somehow needs a "constant" input of energy to somehow remain active?

[Darth Wong]

I think he's assuming that if it's not "natural", then the applicable laws of physics must be totally different, so it must require a constant input of energy. Pretty common reasoning, actually. My falling-out with Robert Brown was over this exact issue, as it related to landspeeders in SW.