spaceship damage from solar radiation

[Drifter]

I'm trying to figure out how close a ship or spacestation can get to a sun.

I am getting my temperatures from Traveller: First In, which I believe uses the same formulae as SPACE. B=(278*fourth root of L)/square root of R to find B, blackbody temp, L is luminosity of the star in solar units, R is the orbital radius. The temperature is from T=B*(fourth root of (1-A))*(1+G); B is the blackbody number, A is albedo and G is greenhouse factor (zero in this case since its a ship in space).

What I'm trying to wrap my brain around is what the damage would be. Say the ship is subject to 800 degrees F/400 C - what does that do? Where do I find the damage for that sort of thing?

Does armor help? How about ice armor? I know that increasing the albedo helps but does painting the ship silver help more than adding layers of steel?

Note that I don't own the current 4e versions of Space or Spaceships, so I'll have to purchase those if the information is in there. So I'd like advice on what to get.

Also, I'm having trouble figuring out how to determine how LONG the ship is subject to those temps. I know the orbit and the period - highly elliptic, total period of 132 days around an M2 sunk, 0.05AU at closest approach. But how do I determine how long the ship is in a dangerous area?

[Anthony]

Short answer: heat is not damage; 800F will never ever melt the hull. However, it will eventually increase the entire body of the spacecraft to 800F, which will cause problems for any components not capable of operating at those temperatures.

[Ulzgoroth]

Originally Posted by Drifter

I'm trying to figure out how close a ship or spacestation can get to a sun.

I am getting my temperatures from Traveller: First In, which I believe uses the same formulae as SPACE. B=(278*fourth root of L)/square root of R to find B, blackbody temp, L is luminosity of the star in solar units, R is the orbital radius. The temperature is from T=B*(fourth root of (1-A))*(1+G); B is the blackbody number, A is albedo and G is greenhouse factor (zero in this case since its a ship in space).

What I'm trying to wrap my brain around is what the damage would be. Say the ship is subject to 800 degrees F/400 C - what does that do? Where do I find the damage for that sort of thing?

Does armor help? How about ice armor? I know that increasing the albedo helps but does painting the ship silver help more than adding layers of steel?

First of all, I'd note that the blackbody model there is oversimplified for a spacecraft. It assumes a uniform absorption coefficient over the surface, where one would actually want a reflective surface facing the heat sources and a black surface on the other side. Not to mention that most real spaceships will have to have active heat radiators as well (allowing them to emit some extra heat at a higher temperature.)

But moving past that, there's at least three equilibrium temperature ranges to think about. If the equilibrium temperature is comfortably low, there's no problem of course. There's an intermediate range where it will cause reduced functioning as crew or components become stressed and fail while operating outside their tolerated range. And then there's a catastrophic range where equilibrium temperature is immediately destructive on its own (melting, igniting the atmosphere, evaporating the crew, that sort of thing) and the only way to survive is to not equilibrate.

The catastrophic range might be represented as burning damage, and the intermediate range as heat exhaustion.

Armor might help slow temperature equilibration of the inner parts of the ship, dependent on qualities which have little to do with its use as armor, but won't change the equilibrium.

Originally Posted by Drifter

Also, I'm having trouble figuring out how to determine how LONG the ship is subject to those temps. I know the orbit and the period - highly elliptic, total period of 132 days around an M2 sunk, 0.05AU at closest approach. But how do I determine how long the ship is in a dangerous area?

This might be of some use.

Or you could of course do the math in full. I haven't looked at that in a while.

[Drifter]

Originally Posted by Anthony

Short answer: heat is not damage; 800F will never ever melt the hull. However, it will eventually increase the entire body of the spacecraft to 800F, which will cause problems for any components not capable of operating at those temperatures.

Thanks. I was looking at the problem wrong. I'm looking at component failure, say a 3d6 roll every X time units while the ship is over a certain temp.

Then again that means breaking the ship down into components. I guess I could go with the parts you build a ship with, such as communications, weapons, turrets, etc. Things directly attached to the hull would be effected first/most, as they heat up.

I wonder how much a ship can take? Say, 1000F for a week?

[Drifter]

Originally Posted by Ulzgoroth

First of all, I'd note that the blackbody model there is oversimplified for a spacecraft. It assumes a uniform absorption coefficient over the surface, where one would actually want a reflective surface facing the heat sources and a black surface on the other side. Not to mention that most real spaceships will have to have active heat radiators as well (allowing them to emit some extra heat at a higher temperature.)

But moving past that, there's at least three equilibrium temperature ranges to think about. If the equilibrium temperature is comfortably low, there's no problem of course. There's an intermediate range where it will cause reduced functioning as crew or components become stressed and fail while operating outside their tolerated range. And then there's a catastrophic range where equilibrium temperature is immediately destructive on its own (melting, igniting the atmosphere, evaporating the crew, that sort of thing) and the only way to survive is to not equilibrate.

The catastrophic range might be represented as burning damage, and the intermediate range as heat exhaustion.

Armor might help slow temperature equilibration of the inner parts of the ship, dependent on qualities which have little to do with its use as armor, but won't change the equilibrium.

This might be of some use.

Or you could of course do the math in full. I haven't looked at that in a while.

The blackbody temp only gives the initial numbers, the actual temp formula gives the albedo variable to account for reflectivity. But that is a good start, combined with what Anthony said - different parts of the hull would probably be different colors, and absorb heat at different rates. What might be 800 degrees if painted white might be 1500 degrees if black.

I guess I'm concerned with the intermediate range. What the crew can do to mitigate damage and system failures when subject to those conditions. Once it reaches the last stage the game is up, I'd imagine.

I've not wanted to try to fit Kepler's 2nd law into Excel, but I guess I'll have to buckle down and try :)

[Anthony]

Originally Posted by Drifter

I wonder how much a ship can take? Say, 1000F for a week?

If it has enough passive and/or active cooling, it can withstand heat indefinitely; otherwise less than a day.

[Drifter]

Originally Posted by Anthony

If it has enough passive and/or active cooling, it can withstand heat indefinitely; otherwise less than a day.

Well, by indefinitely maybe we can say 'as long as the system is operational'. And it remains operational indefinitely for the load it was meant to bear, while anything beyond that would strain the system. Eventually it would give out.

But thats not in my scenario. I"m looking more at a situation with some forewarning. Indefinite is probably the best answer here. Thanks for input.

[malloyd]

Originally Posted by Drifter

Also, I'm having trouble figuring out how to determine how LONG the ship is subject to those temps. I know the orbit and the period - highly elliptic, total period of 132 days around an M2 sunk, 0.05AU at closest approach. But how do I determine how long the ship is in a dangerous area?

Are you sure you have the right formula? I think you may have the parentheses in the wrong place. An M2 star should have a luminosity of about 0.023 Lsun, at 0.05 AU energy intercepted should be 0.023/(0.05)^2, or 9.2 times that at Earth orbit. Temperature goes as the fourth root of energy, so 1.74 times. Blackbody temperature at Earth orbit is about freezing, so that's probably where the 278 is coming from, peak temperature is about 484K, or about 412 F. I don't see where 800 or 1000 could be coming from.

This is a pretty eccentric orbit (e = 0.87 for the M = 0.44 Msun expected for an M2 star). I'm not up for the integration at this time of night, but given that you drop to freezing temperatures at 0.15 AU and only go over Earthlike temperatures at 0.11 AU just by inspection of equal areas it looks like you spend considerably less than 1/10th of the orbit above them, so yeah a week of serious stress per orbit looks about right, certainly no more than 2 weeks.

[Anthony]

An M2 dwarf has a typical luminosity of about 0.03, and at 0.05 AU that's multiplied by 400 for a total of 12 x sunlight, or about the same as 0.28 AU from the sun. That should actually be survivable indefinitely with a basic parasol design, as long as the parasol can withstand the temperature; it's too hot for something like aluminum, but there's no shortage of materials that can handle 600K or so (because it's a thin sheet, temperature is somewhat higher), and as long as the parasol is properly sized, only a quite small fraction of the heat will actually reach the ship.

[Drifter]

Originally Posted by malloyd

Are you sure you have the right formula? I think you may have the parentheses in the wrong place. An M2 star should have a luminosity of about 0.023 Lsun, at 0.05 AU energy intercepted should be 0.023/(0.05)^2, or 9.2 times that at Earth orbit. Temperature goes as the fourth root of energy, so 1.74 times. Blackbody temperature at Earth orbit is about freezing, so that's probably where the 278 is coming from, peak temperature is about 484K, or about 412 F. I don't see where 800 or 1000 could be coming from.

This is a pretty eccentric orbit (e = 0.87 for the M = 0.44 Msun expected for an M2 star). I'm not up for the integration at this time of night, but given that you drop to freezing temperatures at 0.15 AU and only go over Earthlike temperatures at 0.11 AU just by inspection of equal areas it looks like you spend considerably less than 1/10th of the orbit above them, so yeah a week of serious stress per orbit looks about right, certainly no more than 2 weeks.

Well, I plugged in the wrong number for one, typing in 0.422 instead of 0.0422. So I'm only dealing with not much over a hundred degrees at an approach of 0.05AU. So I can drop the craft lower, to about 0.015 to get that 800 degree range.
Thats still way above the numbers you get. Eventually I found that the key is albedo. If I leave it at 1, perfectly reflective, temp is -460, but for anything less the temp goes up dramatically. I'll have to research it to find real values, but if I plug in 0.75 (just a wild guess) I get that 800 degree range.
And there is the other issue. I'm using luminosity values from the First In table on page 50. M0 is given as 0.5, M5 is given as 0.011. Search the internet I've found matching data, and sites that have differing numbers. And none seem to match each other. And nothing matches the corresponding mass rates given. I'm going with the First In numbers, which I assume was playtested pretty well, which gives me a value of 0.0422 lum for an M2 V star.

As to period, I figured out Kepler after all. T^2=(A^3)/M, T being time, A being the semimajor axis, M is star mass. I hadn't taken mass into account before, so thanks for the reminder. At a min distance of 0.015AU, max of 0.478, with a stellar mass of 0.384 (again from First In), I get a total period of 72 days. The ship still spends only about 10 days at its closest approach. I'm not up to figuring out a chart for temp variations, I'm all out of brain on this.