Yes it is trapped for a long time but that doesn't change the fact that the amount of it is determined by the volume of the core, not the surface area of the star.
Why are Class L brown dwarves (dwarfs) just as hot as Class O blue Giants?
- Thread starter GraXXoR
- Start date
Yes it is trapped for a long time but that doesn't change the fact that the amount of it is determined by the volume of the core, not the surface area of the star.
In case of fringe stars, millions of years.
For others, core temperature of stars is many magnitudes higher than what their surface temperature is. Our sun is a measly 5730K surface, but it goes to 15 million or so under.
My guess would be that, for simplicity, the game simply uses a 1/r^2 relation between received heat and distance to the star, modified by the surface[1] temperature of the star.
However, that 1/r^2 only holds true for a point-like source (or far-field). For an extended source (like a star that fills half your cockpit screen), this isn't true any more - essentially, you'd have to integrate over the visible area.
So I would side with GraXXoR - yep, if you fill the same visible area with a O class compared to a L class, it should be _much_ hotter.
Then again, since my FAS's heat exchangers are topside, I should never be able to scoop with my top to the star (which I usually do) since the heat exchangers then would have to run hotter than the star in order to be able to shed some heat. Just don't ask any compicated questions, otherwise the whole building mya come crashing down on us. And do a reality check if you ever get into the situation that you actually would need to use ED to simulate a space probe scooping fuel from a star [alien].
[1]which gives us another two degrees of freedom in that equation - where did the devs put the "surface temperature"? Actual visible surface? Exclusion zone? Center? And where from do they start the 1/r^2 calculation - center, surface, exclusion zone?
However, that 1/r^2 only holds true for a point-like source (or far-field). For an extended source (like a star that fills half your cockpit screen), this isn't true any more - essentially, you'd have to integrate over the visible area.
So I would side with GraXXoR - yep, if you fill the same visible area with a O class compared to a L class, it should be _much_ hotter.
Then again, since my FAS's heat exchangers are topside, I should never be able to scoop with my top to the star (which I usually do) since the heat exchangers then would have to run hotter than the star in order to be able to shed some heat. Just don't ask any compicated questions, otherwise the whole building mya come crashing down on us. And do a reality check if you ever get into the situation that you actually would need to use ED to simulate a space probe scooping fuel from a star [alien].
[1]which gives us another two degrees of freedom in that equation - where did the devs put the "surface temperature"? Actual visible surface? Exclusion zone? Center? And where from do they start the 1/r^2 calculation - center, surface, exclusion zone?
You're not understanding the physics...
Energy MUST PASS THROUGH the surface of the star to escape...
The energy escaping from a surface is *proportional* to A*T^4
where A is the surface area and T is the temperature in Kelvin.
That really is all there is to it.
Forget it - you're clearly stuck with simple black body physics - stars are more complex.
If you multiply the amount of energy generated at the core which subequently escapes through the surface of the star by 16, the surface temperature will double. If the star stays the same size.
Alternatively, multiply core energy by 16 and maintaining surface temperature requires 16 times the surface area to radiate said energy
And the star will expand to 4 times its diameter.
If more energy is escaping from the star either it gets bigger or its surface heats up.
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Suit yourself.
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*Sigh* I hate having to resort to this, seemingly unfair tactic.
Unsheaths wikipedia article:
Stellar luminosity:
A star's luminosity can be determined from two stellar characteristics: size and effective temperature.[3] The former is typically represented in terms of solar radii, R⊙, while the latter is represented in kelvins, but in most cases neither can be measured directly. To determine a star's radius, two other metrics are needed: the star's angular diameter and its distance from Earth, often calculated using parallax. Both can be measured with great accuracy in certain cases, with cool supergiants often having large angular diameters, and some cool evolved stars having masers in their atmospheres that can be used to measure the parallax using VLBI. However for most stars the angular diameter or parallax, or both, are far below our ability to measure with any certainty. Since the effective temperature is merely a number that represents the temperature of a black body that would reproduce the luminosity, it obviously cannot be measured directly, but it can be estimated from the spectrum.
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There's only one way to sort this, slide rules at 10 paces ladies and gentlemen, nominate a second.
Have already repped you this thread...
There's only one way to sort this, slide rules at 10 paces ladies and gentlemen, nominate a second.
You're showing your age
.I somehow doubt that half of the people here know what a slide rule is, let alone know how to use one...
Ok, admittedly, I don't know any more, either. When I was at school, we still had to get one (large one, 30 cm or so) and learned some of the basics, but soon after we also were among the first pupils to get (or be allowed to use) electronic calculators instead.
GraXXoR, I suggest you start just flat pressing 'ignore' button on a select number of people
Thanks for the suggestion but I've tried using ignore in other forums and have often ended up not following what the Hell was going on.
As a once lecturer at a university I made it a point to try and illuminate even the dullest 10W, 1700K lightbulbs in the room. And my God, there were a lot of them.
The lack of basic mathematics and fundamental scientific understanding -let alone logicical reasoning and scientific rigor- was so severe that at times I marvelled that some of them could even operate the vending machine in the corridor outside my office using coins rather than beating it with a club until a Snickers bar fell out. ;-)
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Quoting wiki is easy
From the same article:
From the same article:
Exactly....
As close to controlled conditions I could manage:
Steady state cabin temp 62%
9000K + Star...
Steady state cabin temp: 61%
900K+ Star
Top star is radiating 10^4 i.e. 10,000 times the radiation per unit area!
Distance Ratio: 0.69 : 4.5 = ~6.5x
Inverse square law: 1/(6.5*6.5) = 0.024 (only 2.4% energy from class A reaching me cf Class T)
10,000 * 0.024 = 24!!!
The radiation from the Class A should be 24 times stronger than from the Class T assuming that they were point sources, which they obviously are not. But it's a close enough approximation given that both stars are about the same RELATIVE ANGULAR DIAMETER.
Just for fun:
Radius ratio of the stars: 0.17 : 1.4
Surface areas ratio: 0.17^2 : 1.4^2 (This Class A star has 67.8x surface area)
67.8 x 10,000 = 678,000 times the energy is being released every second by the Class A than the Class T!
*Note: It's midnight, I have been drinking. I do not take responsibility for houses repurchased or spouses angered due to inaccuracies in this post.
;-)
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Quoting wiki is easy
From the same article:
Please explain what relevance this has to ANYTHING I have talked about specifically in this thread.
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They aren't. You are VASTLY closer to the dwarf.
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900k 'star' I think....
that looks safe right?
Over 1000K!!! DEADLY as a star apparently. and look how far out that is - many closer planets get far far hotter
I was replying to the specific thing I quoted.
All I can say is, we're in good companyYou're showing your age
I somehow doubt that half of the people here know what a slide rule is, let alone know how to use one...
Okay, let's take it to absurd extremes: you can find Y-class brown dwarf stars with surface temperatures of 300K or less (I found one as low as 30K, no typo) - i.e. cooler than your pilot. They still overheat your ship if you get close to them, while a metal-rich planet basking at over 1000K is absolutely fine to land on.
Sure, you're much closer than you are to an O-class star, but it shouldn't be possible for them to overheat you at all!
Sure, you're much closer than you are to an O-class star, but it shouldn't be possible for them to overheat you at all!
Well pointed out. Wish I'd made that observation myself and saved the forum 4 pages of drama. 笑