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Do radiators get rid of heat while it's still an asset?


THX1138
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It probably doesn't matter much since it's only during the start-up of mining but the radiator panels report a percentage of cooling while the equipment is still warming up and below optimum-efficiency temperature. Does this mean the radiators are slowing down the accumulation of heat in a way that's detrimental to the start-up period of mining?

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I screwed up my screenshot. It was supposed to show that the radiator had a percentage of cooling being carried out and the drill and ISRU were below optimum temperature. My thread is ruined!

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53 minutes ago, THX1138 said:

Does this mean the radiators are slowing down the accumulation of heat in a way that's detrimental to the start-up period of mining?

Could well be.  I've always thought they lacked a radiator valve myself.

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I'm pretty sure they only bleed off unwanted heat, i.e. when you go over the optimum temperature.  Otherwise, you'd expect the heat-up rate of drills and ISRU to slow down a lot as they climb close to optimum, and they wouldn't end up right at optimum temperature when they stabilize.

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I've tested that myself, and I'm pretty sure the answer is no. Until your unit is up to full operating temperature, it doesn't shed any heat energy. Once it's there, it suddenly tries to shed all of it.

All the heat that your radiators shed before then is just small amounts of warmth (from the launch and landing) from the skin and interiors of other parts of your rocket.

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6 hours ago, THX1138 said:

I screwed up my screenshot. It was supposed to show that the radiator had a percentage of cooling being carried out and the drill and ISRU were below optimum temperature. My thread is ruined!

You can still Edit your original post.

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This need testing with writing down facts...kind of like science, but in real life. What I am wondering, is how does the heat generation occur? I mean, does a drill generate an amount of heat related to how much ore it is drilling through, or is it just a simple on/off state?

How do we calculate the maximum amount of cooling a base will need? (when drilling, refining and doing all the hot stuff at once...except sundiving of course)

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32 minutes ago, THX1138 said:

I know but the people who were gonna respond have already responded and won't see the updated picture so there's no point.

Usually people come back to the thread where they post a replie to see other's people reaction. At least it's what I do.

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1 hour ago, Blaarkies said:

This need testing with writing down facts...kind of like science, but in real life. What I am wondering, is how does the heat generation occur? I mean, does a drill generate an amount of heat related to how much ore it is drilling through, or is it just a simple on/off state?

It generates heat based on its maximum times its thermal efficiency. The ore rate is irrelevant.

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How do we calculate the maximum amount of cooling a base will need? (when drilling, refining and doing all the hot stuff at once...except sundiving of course)

Right click on the part in the VAB or SPH, and each part tells you the maximum it needs. If you have a mix of fixed and extendable radiators, it gets complicated -- I think the extendable ones always take heat first.

Edited by bewing
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2 hours ago, Blaarkies said:

This need testing with writing down facts...kind of like science, but in real life. What I am wondering, is how does the heat generation occur? I mean, does a drill generate an amount of heat related to how much ore it is drilling through, or is it just a simple on/off state?

How do we calculate the maximum amount of cooling a base will need? (when drilling, refining and doing all the hot stuff at once...except sundiving of course)

Well this seems like pretty simple math to me.

You need to find the optimum and maximum temperatures of every heat-generating object in your base.   You're going to want to subtract the optimum temperature from the maximum, and then you'll add those all together and get the amount required for heat.

Then you'll divide this number by the radiator heat expelling amount, and then you should get the amount of required radiators.

In Mathematical Terms:

(Maximum - Optimum) / Radiator_Amount = Radiator Amount

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35 minutes ago, Netherman555 said:

Well this seems like pretty simple math to me.

You need to find the optimum and maximum temperatures of every heat-generating object in your base.   You're going to want to subtract the optimum temperature from the maximum, and then you'll add those all together and get the amount required for heat.

Then you'll divide this number by the radiator heat expelling amount, and then you should get the amount of required radiators.

In Mathematical Terms:

(Maximum - Optimum) / Radiator_Amount = Radiator Amount

I calculated it basically like that a few savegames ago, but the ISRU kept blowing up as my lander reached the physics bubble(I was landing the lander next to it, but at 2.3km the ISRU blew up)...I was thinking I completely misunderstood the heating mechanics, now it seems like it was just a bug

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14 hours ago, Snark said:

I'm pretty sure they only bleed off unwanted heat, i.e. when you go over the optimum temperature.  Otherwise, you'd expect the heat-up rate of drills and ISRU to slow down a lot as they climb close to optimum, and they wouldn't end up right at optimum temperature when they stabilize.

The hotter something is, the better a radiator works.  A radiator will shed more heat when it reaches optimum temp than it will be when its cold.

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6 minutes ago, Corona688 said:

The hotter something is, the better a radiator works.  A radiator will shed more heat when it reaches optimum temp than it will be when its cold.

Yes, but it's not going to just coincidentally hit thermal equilibrium right precisely at five hundred point zero degrees for a mining drill.

If we imagine a hypothetical radiator cooling down some component, where there isn't any special don't-pull-unwanted-heat logic present, we'd expect to see temperature settle down at some equilibrium temperature, which it would asymtotically approach (that is, the closer it gets to that temperature, the slower its temperature changes).  The equilibrium temperature will be some oddball number whose exact value depends on a bunch of numbers such as heat generation rate, radiative efficiency, ambient temperature, yadda yadda.  It might be higher than 500, it might be lower than 500, but it's almost certainly not going to be exactly 500.  And the graph of temperature over time is going to be a smooth curve with no sudden changes in slope.

Whereas what we do in fact observe with the radiators is:  the drill heats up really fast up to exactly 500 degrees, and then suddenly stops dead like it hit a wall.  That's the behavior of a radiator that's been coded to care about the drill's optimum temperature.

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