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Messages - Nonmomentus Brain

#1
Off-Topic / Re: magnetically driven engines
April 04, 2017, 10:35:58 AM
Quote from: mumblemumble on April 03, 2017, 09:32:04 PMYou know, I do find the ideas of "laws" brought into debate of this rather ironic : laws must ALWAYS match what actually happens, and if the law is broken, the law is no more : In science, these "laws" are there to show the CURRENT understood limits of physics and the world, but this never means they cannot be broken, just our current UNDERSTANDING is broken.

And due to this, these laws should always be challenged : Because sometimes laws are incorrect, or done wrong, or whathaveyou. It happens, and is forgivable, but true physics exist because they can happen over and over again.

And if the laws hold up to countless experiments and no experiment disproves them, they are clearly useful enough to describe the world. If you do have an experiment which disproves them, go ahead and publish. If you pass peer review and independent scientists can recreate your findings, have fun with your Nobel prize. Of course, that is a very, very big if.

Quote from: mumblemumble on April 03, 2017, 09:32:04 PMBeyond that, it should always be based on TANGIBLE ideas : Some of the laws of energy seem a bit silly to me, particularly when calling the mass of something "potential energy" : The energy is not from the mass itself, but the weight of it being pulled downward. The downward force is the energy force, not the object itself, potential energy is not actual energy which exists inside the object, it is the energy of the downward force, which can only be observed when a pathway down is given.

This is where it becomes clear that you don't have even a basic understanding of terminology in physics. Energy is, in fact, defined in terms of force; specifically, the integral of force over distance (basically, it's a measure of how much force you have to apply over a distance to get your object from one point to another).
If you had a 10 kilogram rock on Earth's surface and wanted to lift it up by 2 meters, you'd have to exert a force of about 100 Newtons, which is the weight of the rock, to compensate for gravity. Lifting it 2 m, then, would require about 200 Joules (say, in terms of electrical energy for an engine, minus inefficiencies), and the potential energy of the rock would increase by those 200 J. If you want to get those 200 J back "out of the rock", you could drop it and use it to spin a generator, which would then (minus inefficiencies) get you 200 J of electrical energy. That's how energy works. If you want to use weird terminology nobody else uses (say, "energy force"), that changes nothing about what happens and just makes it harder for others to understand you.

Regardless, requiring laws of physics to be "tangible ideas" would be foolish. Potential differences driving current in a circuit, oscillations being described by complex numbers, quantum mechanics – none of that is tangible, but it works – otherwise a lot of modern technology couldn't have been built. If you were to restrict physics to those things a layperson finds intuitively understandable and "tangible", you won't get very far.



I've heard the following said: "You can never hope to defeat your enemies without knowing how they think. And by the time you know enough science to fight the scientific, it's already too late – you're already a scientist." Basically, if you knew what you're talking about, you wouldn't be talking about it because you'd know why you're wrong.

I'm sure there's a book out there written to give an introduction to physics to people with your level of understanding. If you're in this discussion because you're interested in facts, reading something like that would be a good idea. If you're interested in winning the argument, reading up on the matter would be a good idea as well, because at least you'd learn how concepts are expressed in physics and what specific terms mean, so you could express your ideas in a way we can actually understand. If you're just here because you like arguing, well, I guess it's your right to waste your time like that.



Quote from: Thyme on March 19, 2017, 01:00:34 PMI intentionally avoided mathematics. I doubt it will help the case, as "conservative force field" and similar arguments did not convince mumble2.

While it is unlikely to convince him, other people will probably be interested in it. And even if not, practicing the basics doesn't hurt. (And yes, I'm still planning to do the math, I just haven't gotten around to it yet.)
#2
Off-Topic / Re: magnetically driven engines
March 14, 2017, 12:52:08 PM
mumblemumble, I'd like to ask you about what math you're acquainted with, so I know how detailed my explanations can be and where I have to start. Are you acquainted with functions? Vectors? Differentiation and integration? Different coordinate systems? (I don't know very much about education systems except for the one I'm in and I don't know how that compares to others, so I'm afraid I have to ask.)
It won't be necessary to test magnets to determine their properties, that's already been done and expressed in some very useful equations, which we only need to apply to our particular situation. And those equations do work – otherwise, it would hardly be possibly to calculate what one needs for a generator, or a particle accelerator, or a CRT screen, or any other magnet-based technology.

milon, that doesn't mean that magnetism "doesn't exist", just that the phenomenon expresses itself in different ways in different frames of reference. Coriolis force, for example, certainly exists to an observer in a rotating system – not to one in an inertial frame, but that won't stop that hurricane from spinning. "Fictitious force" is kind of a misnomer.
#3
Off-Topic / Re: magnetically driven engines
March 13, 2017, 02:45:40 PM
This thread is full of words and neither side has, so far, been able to convince the other with them. Unless I didn't see some posts, it seems in this discussion about physics, one thing is notably absent: mathematics, without which you don't have serious physics. There are ways in which this discussion might benefit from a more mathematical look:

Firstly, you, mumblemumble, might show us mathematically why it has to work. If you're insufficiently well-versed in mathematics and/or physics, you can just do a little research and link us to someone else's calculations. If this machine works, surely someone must have done the math and put their work on the internet. We can then check if there are any errors. If no significant ones are found, we will have to admit defeat.

Secondly, if you present us with specifications for a perpetual motion device, we could use mathematics to find whether it does what you say it would. Some of the methods we would use might exceed your current mathematical abilities, but we'll be glad to help understand (as long as you admit that you don't understand, instead of immediately accusing us of making errors where there are none – of course, if we actually did make an error and you point out that it doesn't make sense, we'll correct it) and surely, infinite energy is worth doing a little bit of reading on something so interesting! Of course, this will require you to accept, if not the current results, then at least the methods of physics. But if you're unwilling to do that, then our time is wasted anyway. Preferably, the specifications you'd provide would be such that if this is shown to not work, you would admit defeat rather than endlessly providing a slightly different design that we'd have to examine. Ideally, we'd look at the general case, but that might be a bit too abstract for you to accept.

Of course, both of these approaches carry the risk of spawning another discussion about what is and what isn't an error here. It would be good to remember, then, that some of us here have decided to make this their career, which means considerable training and, therefore, considerable skill. Of course, mistakes can still happen, but if we're told that there's a mistake somewhere, we check and there isn't, it's very unlikely that we missed it twice (especially since we're several people who'll check each other and who are, like you, interested in figuring out what is actually true, rather than just winning the argument) and much more likely that there just isn't a mistake. As I said, we'll be happy to help understand, but only if you actually make an effort.

(Note that when I say "we" here, I mean something along the lines of "I when I have time, and possibly someone else if they feel like it"; I obviously don't have any kind of authority here, but the plural sounds nicer, I think.)

Of course, I can't make a post about mathematics and perpetual motion without mentioning that conservation of energy is a result of Noether's Theorem (and time invariance). Noether's Theorem is a mathematical result and therefore definitely true (unless the very principles of mathematics are wrong, but if that were the case, we'd have bigger problems), but it's some pretty serious math and not easily understood and I definitely wouldn't expect you to just accept it without actually understanding the proof.

So, what does everyone think? Would a mathematical approach help conclude this discussion or just make everything worse? Are you willing to try it? Have I overstepped my boundaries by using the first person plural?
#4
Off-Topic / Re: magnetically driven engines
January 09, 2017, 04:20:52 PM
The movement observed isn't due to energy in the magnet, but due to the potential energy of the object that's moved. It's similar to gravity: If something falls down, that's not because the Earth has energy that it gives to that thing, but because the object has potential energy that can be released by falling down. And, obviously, you can't generate free energy from Earth's gravity because once something has fallen down, you have to use at least as much energy as was released in the fall to get it to its prior height again (in practice, inefficiencies like friction mean it'll take more).

In the case of the magnet, if something is attracted to it, you'll have to exert at least as much energy to move it away from the magnet as was released when it was accelerated towards it. For magnetic repulsion, it's the same, just the other way around. Of course, you could just have a big magnet, let it attract a lot of stuff and not bother with removing it again. In that case, of course, you'd gain energy, but only the energy that was already in the objects. Imagine asteroids falling towards Earth: They definitely do release a lot of energy, but they already contained it as potential energy because they were so high above the Earth.

I hope this clears things up. If you have any questions, just ask!
#5
General Discussion / Re: Steel burns on Rimworld?
August 01, 2016, 08:06:42 PM
I don't deserve such a good opinion – all I did was expand on your great work! I won't deny that I appreciate it, though.

Thank you for giving the source! It was an interesting read. I hadn't heard about Thornton's rule before.

Now that we've ruled out ignition of the walls in a closed room, let's look at rooms with ventilation. I will assume that the required oxygen is brought into the room and the gaseous combustion product are expelled, but the insulation of the walls, ceiling and floor is still perfect.

Since the released energy is now, again, determined by the amount of wood rather than the amount of oxygen, it is necessary to determine how much wood we can store within this room. Since I know next to nothing about fluid dynamics, I will simply assume that if 50% of the volume is taken up by wood, that leaves enough space for the air to flow. This gives us 100 m3 of wood. If we assume red oak (which is the type of wood used in your first post, so we already have the released energy per kg), that gives us a density of 0.63 g/cm3.(1) Therefore, we have 63000 kg of wood, which is equivalent to about 650 GJ, according to the value of 10.3 MJ/kg given by you.

Of course, not all of this heat is used to heat the walls. Some of it is used to heat up the oxygen that enters the room, while some of it is lost because hot CO2 and H2O leave the room.
To find these values, it is necessary to know how much of those gasses must be heated. Assuming, again, that the wood is entirely composed of cellulose, the previously established reaction equation C6H10O5 + 6 O2 --> 6 CO2 + 5 H2O is useful again.
If the mass of the wood is 63000 kg and the molar mass of cellulose is 162 g/mol, that gives us 390 kmol of cellulose. With the factors from the reaction equation, it is easy to calculate that the molar amounts of oxygen and carbon dioxide are both 2300 kmol, while the molar amount of water vapor is 1900 kmol. One can already see that the amount of oxygen inside the room (1.8 kmol, according to my previous calculations) is little more than a rounding error here.
Knowing the molar heat capacities of oxygen, carbon dioxide and water vapor (29.4 J/mol*K, 37.1 J/mol*K and 33.6 J/mol*K, respectively)(2) and assuming a temperature increase of 1200 °C, one can calculate that heating the oxygen takes 81 GJ of energy, while the heat loss caused by carbon dioxide is 102 GJ and that caused by the water is 77 GJ. The combined loss is 260 GJ, still far less than the energy released through the combustion of the wood.

However, we cannot filter the oxygen from the outside air and only pump that in. A large amount of (for this purpose) inert nitrogen will be brought in as well, and it well be heated as well. If, by molar amount, 20% of the air is oxygen and 80% nitrogen, that means that 9200 kmol of nitrogen will have to be heated as well. With the heat capacity of nitrogen being 29.1 J/mol*K(2), this means that heating it to 1200 °C will need 320 GJ.

This leaves 70 GJ (or 70000 MJ) to heat the walls. According to our previous numbers, that should be more than enough to heat the entirety of the walls to the required temperature. When I, however, decided to calculate what the increase in temperature would be, I noticed that it was far lower than 1200 °C. This lead me to check your initial value for the energy required to heat one wall. I then noticed that you were off by a factor of 10 when converting from joules to megajoules, resulting in a value of 5220 MJ instead of the correct 52200 MJ. Fortunately, this error did not cause our conclusions to be wrong, since it only reinforces our stance that it is not possible. Furthermore, when I calculated the same value using Wolfram|Alpha, the result was a mere 41000 MJ(3).

After having discovered this error, I decided that instead of calculating the temperature change of the entire wall, I would once again calculate the depth to which a temperature change of 1200 °C could be caused, using the same method as previously. The result was 20 cm, which should be sufficient to ignite the wall.

The next question to be answered is whether the exothermic oxidation of iron is sufficiently exothermic to be self-sustaining. In other words, does the combustion of a certain amount of iron provide sufficient energy to ignite at least the same amount of iron?

To answer this question, one first must find the processes that take place in the combustion of iron in our scenario. I assume that at first, a certain (oxide-free) volume of iron is at the required temperature. Then, it oxidizes, with the required oxygen being at the required temperature. The energy released in the oxidization melts the iron oxide to allow for the oxygen to reach the iron underneath, which is heated by the oxidization of the first volume. Furthermore, sufficient oxygen to oxidize the iron underneath is heated as well. Of course, I know that in reality, these are not discrete steps. It is, however, a useful simplification.

For my calculations, I assume that a volume of 1 cm3 oxidizes and must provide enough energy to melt the resulting iron oxide and heat the 1 cm3 of iron underneath as well as the required oxygen by 1200 °C. As has been shown in my previous post, the oxidization of 1 cm3 of iron into iron(II) oxide releases 39 kJ. This volume of iron has a mass of 8 g and, due to the molar mass of 56 g/mol, contains 0.14 mol of iron.
Due to the coefficients in the reaction equation 2 Fe + O2 --> 2 FeO, 0.14 mol of iron(II) oxide are formed. The molar heat of fusion of iron(II) oxide is 24.1 kJ/mol4, which gives us 3.4 kJ of energy required to melt the iron(II) oxide.
The heating of the next volume requires 4.2 kJ of energy, as can be calculated from the temperature difference of 1200 °C, the molar amount of 0.14 mol and the molar heat capacity of 25.1 kJ/mol(2). 0.14 mol of iron require 0.07 mol of oxygen for oxidization. This amount of oxygen requires 2.5 kJ to be heated.
Overall, of the 39 kJ released, 10 kJ are required to keep the reaction going; therefore, the net release of energy is 29 kJ/cm3. If all walls (80 m3 of steel) combust completely, this would mean a release of 2.3 TJ of energy, which seems too high, since 29 kJ/cm3 is only slightly higher than the 28 kJ/cm3 I have calculated for wood in my previous post and the volume of the steel is 80% of that of the wood. This would indicate that there would not be a factor 4 between the two values. However, in this post, I have used values for red oak instead of values for cellulose, as I have done in the previous post. I shall calculate the energy released by the combustion of 100 m3 of cellulose.

100 m3 of cellulose have, given the previously established density of 1.6 g/cm3, a mass of 160000 kg. Dividing by the molar mass of 162 g/mol, this gives us 1000 kmol of cellulose. Knowing that the heat of combustion of cellulose is -2800 kJ/mol, it can easily be calculated that the energy released in the combustion of 1000 kmol of cellulose is 2.8 TJ, which agrees with the value of energy received for the combustion of the steel.

This also shows that cellulose and wood are not the same material and should not be treated as such. It is necessary to carefully consider whether they are sufficiently similar to be seen as equivalent for any particular situation. However, I do not think that the previous results are to be considered wrong, since the amount of oxygen required to burn a certain amount of cellulose should be similar to that required to burn a certain amount of wood.

In conclusion, if the necessary temperatures can be reached, it seems that a storage room for wood can ignite steel if it catches fire and the ventilation systems are sufficient. Furthermore, it seems that fire can spread through steel objects, if sufficient oxygen is present.

1http://www.wolframalpha.com/input/?i=density+of+red+oak
2http://www.wolframalpha.com/input/?i=molar+heat+capacity+of+O_2,+CO_2,+H_2O,+N_2,+Fe
3http://www.wolframalpha.com/input/?i=%28specific+heat+capacity+of+iron%29*%28density+of+iron%29*%2820m^3%29*%28580+%C2%B0C%29&rawformassumption=%22UnitClash%22+-%3E+{%22%C2%B0C%22,+{%22DegreesCelsiusDifference%22}}
4http://www.wolframalpha.com/input/?i=molar+heat+of+fusion+of+iron%28II%29+oxide
#6
General Discussion / Re: Steel burns on Rimworld?
July 31, 2016, 08:52:18 PM
Quote from: brcruchairman on July 31, 2016, 03:28:51 PM
First off, *applause* Well played, sir; your work was really well thought out, well-sourced, and basically awesome. That's what I was hoping for when asking for a physicist. :) [snip]
Your high praise makes me very happy! Though I am not yet one, I do, in fact, plan to become a physicist eventually.

I am writing most of this past midnight, so there might be a number of errors. Please check anything that seems peculiar and point it out to me.

You seem to have forgotten to provide a reference for your source number 3 – would it be possible for you to correct that? I'm interested in how they got that number.

Another problem regarding your calculations of the energy potential from the oxygen is that you seem to have multiplied the given value with the entirety of the air instead of just the oxygen content.
Because of these problems, I shall try and determine myself how much energy we can get out of the air.
Firstly, I will have to determine what burns in order to find how much oxygen is needed and how much energy is released. Cellulose seems to make up roughly 50% of wood1. In order to simplify things, I will pretend that wood is pure cellulose. The molar heat of combustion of cellulose is about -2800 kJ/mol2 (the value is negative because energy is released). In case it seems peculiar to have a molar value for a polysaccharide such as cellulose, it seemed so to me as well. The paper gives the formula as C6H10O5, which appears to be the smallest unit in the structure. Regardless, we can work with this. It is easy to find that the combustion can be described by the following reaction equation:

C6H10O5 + 6 O2 --> 6 CO2 + 5 H2O

We can see that for each mol of cellulose (as the source number 2 seems to define it), 6 mol of oxygen are required. Therefore, for each mol of oxygen, 470 kJ of energy are released. Assuming oxygen is an ideal gas, one mol takes up 22.4 L of volume. If 20% of the air is oxygen, the amount of oxygen is given by 40 m3/22.4 m3/kmol, which is about 1800 mol. Multiplying this with the previously established value for the energy release per mol of oxygen, we get 1800 mol * 470 kJ/mol = 850 MJ of energy released if all the oxygen is used up.

Another way to get to this value is to take the mass of the air (240 kg, assuming 1.2 kg/m3 of density and 200 m3 of volume), multiply it by 20% to get an approximation of the oxygen mass and divide by the molar mass (32 g/mol for molecular oxygen). This gives us about 1500 mol of oxygen, which is close enough to the other value. The difference is probably due to rounding errors and the fact that there is a difference between mass percentage and volume percentage. Simply multiply by the 470 kJ/mol and we get a value of 705 MJ.

For comparison with your source, we can also calculate the energy released per kilogram of oxygen: 470 kJ/mol / 32 g/mol = 15 kJ/g or 15 MJ/kg. This greater than your source's value of 2.75 MJ/kg by a factor of more than 5 – again, I would be very interested to read that article.

Of course, what you have already noticed is that my value for the energy released is lower than yours by a factor of almost 80. I was quite confused by this until I checked your calculations as well – you appear to have dropped a decimal point when calculating the energy released from the 200 m3. It should be 660 MJ rather than 66000. As you can see, this is far less than the 41000 MJ required to heat the walls up.

Of course, these values are still too high, since the air itself must also be heated (we'll ignore the fuel).
Assuming 240 kg of air and a specific heat capacity of 1 kJ/kgK3, we need about 290 MJ of energy to heat the air by 1200 °C. If we go with the most optimistic result, that leaves about 560 MJ to heat the walls.

It is obvious that this is insufficient to heat the entire wall. However, because we are assuming a wood-air mixture, the energy will be released very rapidly due to the massive surface area. Since the assumption that the entire wall would have to be heated was based on the premise that the heat would be conducted away otherwise, we might relax this requirement since there might simply not be sufficient time in order for that to happen.
So, if we don't have to heat the entire wall, how much of the wall can we heat? The maximum depth up to which we can heat the steel by 1200 °C is given by 550 MJ/(1200 °C * 8 g/cm3*80 m2*0.5J/(g*°C)), in other words the energy we put into it divided by the product of the temperature change, the density of the material, the area and the specific heat capacity. This gives us a depth of about 1.4 mm. This may not seem like much, but due to the exothermic nature of combustion, the burning steel might keep the reaction going.

In addition, if the heating is sufficiently rapid to prevent the formation of an oxide layer before ignition, only about 900 °C of heating are required (see my previous post), which would allow the steel to be heated sufficiently to a depth of 1.9 mm.


1Pettersen, R.C., 1984. The chemical composition of wood. The chemistry of solid wood, 207, pp.57-126.
2Colbert, J.C., Xiheng, H. and Kirklin, D.R., 1981. Enthalpy of combustion of microcrystalline cellulose. JOURNAL OF RESEARCH of the National Bureau of Standards, 86(6), pp.655-660.
3http://www.engineeringtoolbox.com/air-specific-heat-capacity-d_705.html




Quote from: Shurp on July 31, 2016, 06:56:22 PM
OMG, you're right, we're looking at this all wrong.  The issue isn't the steel, it's the oxygen.  Steel is 8+ times denser than wood.  The volume of oxygen you need to bring in to keep that surface burning compared to an equal surface of wood is stupendous.  Unless you have it seriously compressed or your spraying pure oxygen at it you'll never keep it lit. 

And I imagine steel burns *hotter* than wood (ie: releases more energy per molecule burned), which makes it even harder to keep enough oxygen present to keep the fire going (higher temperature combustion products pushing away fresh oxygen)

I'll see if I can back your assumptions up with a few calculations. Again, I might make mistakes, so be careful and if something looks wrong, just check and ask.

I will look at the volume of the different substances that can be oxidized by the same amount of oxygen. While I am aware that I thereby fail to take surface conditions etc. into account, it should be sufficient to shed some light on whether the assumption in your first paragraph is true.

If we assume that when steel combusts, iron(II) oxide is formed (this is the oxide that would require the least amount of oxygen), then for each atom of iron, one atom of oxygen is required. Since in an atmosphere, only molecular oxygen O2 and not atomic oxygen is present in any meaningful amount4, this means that each molecule of oxygen can oxidize two atoms of iron. As a reaction equation:
2 Fe + O2 --> 2 FeO
The molar mass of iron is about 56 g/mol. This means that each mol of oxygen can oxidize 112 g of iron. The density of steel has previously been established as 8 g/cm3. A simple division shows that a mol of oxygen can oxidize 14 cm3 of iron.

As discussed above, each mole of oxygen can oxidize on sixth of C6H10O5, the substance that makes up a large percentage of wood. Its molar mass is 162 g/mol. Therefore, one mol of oxygen can oxidize 27 g of cellulose. The density of wood varies widely, so I will use the density of cellulose instead (which is a good idea, since that's the substance for which I've actually calculated anything so far). It seems to be around 1.6 g/cm3.5
Using the same method as before, it can be calculated that each mole of oxygen can oxidize about 17 cm3 of cellulose. This is higher than the value for steel, which supports your assumption. However, the difference is quite small and cannot account for the difference in flammability observed.

In response to your second paragraph, I will examine how much energy is released by the combustion of cellulose and iron.
As can be seen above, the molar heat of combustion for cellulose is about -2800 kJ/mol.
The molar heat of combustion for iron is easily calculated if one has knowledge of the reaction equation (which is given above) and the heat of formation of every involved substance. Since the heat of formation is defined to be 0 for the most stable form of an element, the only substance that must be taken into account here is iron(II) oxide, the heat of combustion of which is -272 kJ/mol.6 Since the heat of combustion is given by the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants and the heats of formation of the reactants are 0 in this case, the heat of combustion is -272 kJ/mol as well.

As you can see, this contradicts your assumption that steel "releases more energy per molecule burned" (technically, steel doesn't consist of molecules, actually). However, it is clear that this is not because you what you meant is wrong, but because your wording is. Since organic molecules consist of many atoms, they cannot be compared directly to the single metal atoms in this case.

Instead, we might look at the energy per volume, which is more relevant here.
Using the values given above, we can easily calculate that each of cellulose produces about 28 kJ per cm3.
We can equally easily calculate that the corresponding value for steel is 39 kJ per cm3.

Again, it can be seen that the assumption is confirmed, but the difference is not as great as expected. However, there is a significant difference in the temperature of combustion of the two materials. While wood burns at 600 °C (as has been established previously), pure iron ignites at temperatures beyond 930 °C (as previously established). This is not because of the heat released in combustion, but because of the activation energy required to start the reaction.

Furthermore, there's a problem with your assumption that the hot combustion product would "push away" fresh oxygen: the combustion products of steel, various iron oxides, aren't gaseous. In fact, I presume that the higher temperature would only serve to increase convective airflow, thereby bringing in more oxygen. I admit, I didn't think of that either while calculating the numbers above.

The lower density of magnesium is largely because of the lower mass of the magnesium atoms themselves. The actual difference in oxygen required is much lower than you might think.

I hope I could help.

4You are alive.
5Sun, C.C., 2005. True density of microcrystalline cellulose. Journal of pharmaceutical sciences, 94(10), pp.2132-2134.
6http://www.wolframalpha.com/input/?i=heat+of+formation+of+iron%28II%29+oxide




Good night. I hope my sleepiness did not introduce too much error (yes, I know I should have waited until I was well-rested with something like this, but it was too much fun).
#7
General Discussion / Re: Steel burns on Rimworld?
July 31, 2016, 02:51:56 PM
Quote from: Shurp on July 31, 2016, 02:41:24 PM
The issue with real steel fires is not so much temperature as vaporization.  When you heat up wood hot enough the surface molecules break free and can mix with the oxygen in the air.  The boiling point of iron is 2862'C (thanks google!) so even if you have it hot enough to burn you're only going to torch the surface.

(Steel wool burns so nicely because it's *all* surface)

Wouldn't it be sufficient to merely melt the iron oxide, causing it to flow away and allow the (still solid) iron underneath to oxidize? There wouldn't be a flame, but there isn't one either when burning steel wool and according to your own post, that counts as burning. The temperatures would be quite a bit lower: 1360 °C in the case of iron(II)oxide and even lower if there are impurities.
#8
General Discussion / Re: Steel burns on Rimworld?
July 31, 2016, 06:17:54 AM
Quote from: brcruchairman on July 30, 2016, 03:53:25 PM
[work by the quoted author; too long to be quoted here]
While your post is an interesting read and certainly well-thought-out, there is a major problem with it.


You assume that at 600 °C, steel burns. This, however, is based solely on the temperature of a burning match, which might be an appropriate figure for a wood fire, but tells us nothing about steel. I have found a source indicating that the temperature of ignition of iron is somewhat higher, at 930 °C in an atmosphere of pure oxygen.1 At a lower partial pressure of oxygen, the temperature of ignition is likely to be higher.

Under the assumption that the temperature of a match is equal to that of burning oak (according to your findings, 600 °C), it becomes clear that it is impossible to heat steel to the required temperature simply by burning wood, since the second law of thermodynamics prevents the flow of heat from a colder body to a hotter one on its own. However, the properties of oak trees and matches may be sufficiently dissimilar to allow for sufficient temperatures under certain conditions.


As an additional factor to be considered, in the oxidation of steel, iron oxide forms. Unlike the combustion products in more usual fires (mainly CO2 and H2O for wood), iron oxide is not gaseous at these temperatures. Therefore, it remains at the surface and blocks additional oxygen from reaching the unoxidized steel unless it is removed somehow.

According to Wolfram|Alpha2, the melting points of various iron oxides appear to be too high in order for them to flow away from the point of oxidation at the given temperature. Not only would this prevent further combustion, but as a layer of iron oxide would form while it was being heated (since the greater heat would accelerate the rusting reaction, according to the Arrhenius equation), combustion might not start at all. On the other hand, in another study3, it was found that if exposed to oxygen during the heating process, iron ignites at slightly above 1400 K (or about 1150 °C). This indicates that oxides form during the heating process and inhibit ignition, but that reaching the melting points of the oxides as given by Wolfram|Alpha is not necessary for ignition. It is possible that this is because the melting point is lower, which can be caused by impurities (for example, different oxides mixed with each other, as is likely to be the case here).


In conclusion, though the previous findings indicate that it is possible to heat steel to a temperature of 600 °C, this temperature is insufficient for the ignition of iron, even under optimal conditions. However, it might be possible to reach sufficiently high temperatures anyway. Further examination is warranted.

Again, I would like to stress that I do not mean to attack your work, but merely add to it. I hope that my post was as interesting to read as yours.

1Grosse, A.V. and Conway, J.B., 1958. Combustion of metals in oxygen. Industrial & Engineering Chemistry, 50(4), pp.663-672.
2http://www.wolframalpha.com/input/?i=melting+points+of+Fe_2O_3,+Fe_3_O_4,+FeO
3Nguyen, K. and Branch, M.C., 1987. Ignition Temperature of Bulk 6061 Aluminum, 302 Stainless Steel and 1018 Carbon Steel in Oxygen. Combustion Science and Technology, 53(4-6), pp.277-288.
#9
Ideas / Re: Holidays
July 05, 2016, 05:13:32 PM
It might be possible for different factions to have a different list of generated holidays. Have a list of possible, generic names and randomize the date. Tribals might have a high probability of having holidays on a day of season change. Characters from the same faction would share the same holidays. Spacers wouldn't have any connection there. Even if recruited, a character would retain the old holidays.

Different holidays would be celebrated in different ways. Some might be celebrated by piling a lot of wood on a big heap and setting it on fire, others by setting off fireworks, others again by having an extraordinary feast. If a day is considered a holiday by a pawn, these interactions would be made available during that day and there would be a message informing the player. The player has the option of ordering the celebration to be held by placing something similar to the party spot where the celebration should be held, but doesn't have to. If this is done, the characters who consider the day a holiday celebrate there and do whatever they do to celebrate and get a large mood boost (but of course don't get any actual work done). If not, then they get a smaller mood boost.

Several new traits might be introduced. One might allow a pawn to take part in other faction's celebrations, another might worsen the mood if another faction's holiday is properly celebrated or if the pawn's own holiday is not properly celebrated.

By generating a multitude of different holidays, the concerns expressed by Mossy piglet could be averted and more diverse stories might be created. By allowing a variety of types of celebrations, holidays would be more than a simple mood buff and, again, would create more stories.
#10
In case you missed it: I've finished both versions of the texture. Here they are again.

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#11
I'm happy that you like it so much!
I've finished the side view. I only found the side view and back view for the vanilla assembly bench in the game's files, so I presume those two are all that are needed. I hope you like this much as one as the other!

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#12
I've finished it in one orientation (should be attached to this post). As you can see, I modified the assembly bench by doing three things: Covering tools that look like they would be used by a human operator (the alternative would have been finding a way to animate little robotic arms), connecting the "boxes" (because there's no human to move the item from one to another) and adding a hot furnace (to make it look more industrial; a "lore explanation" might be that while a human can work with irregularities in the supplied steel, the machine can't adapt like that and has to melt it down to create a homogeneous material with which to work).

If you find this acceptable, I'll get to work on the other required orientation. Else, please tell me what you'd like changed (though, as you can probably tell, I'm not very good at this myself). If you want me to, I can provide the .psd file with the different layers and such, including some "cut content" that turned out not to look as good as I had imagined. Is .png an acceptable file type?

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#13
For the CR patch (which I presume is still planned to be released, considering that it has been stated that work has been done on it), will there be multiple ammunition types for a weapon type? Will , for example, be different types of ammunition be used for a laser precision rifle and a laser pistol? And will there be different ammunition options, similar to how CR has AP, FMJ and HP rounds?
#14
Seems like an interesting idea. Would you be interested in me trying to modify the current texture a bit so it's distinguishable from the vanilla assembly bench at a glance?
#15
Ideas / Re: Your Cheapest Ideas
April 11, 2016, 10:50:28 AM
Idea: Railgun turret.

Works similarly to mortars. Building of a large size, requires a colonist to operate, takes long to research, requires ammunition (possibly, but not necessarily, the existing artillery shells), long cooldown/warm-up (capacitors need to charge and systems need to cool down, possibly somewhat longer than the mortar's cooldown), extreme range.

Main differences to mortars: Requires a lot of electricity. Incapable of indirect fire, so it requires a line of sight. Extremely accurate and very fast projectile travel time (most of the time, it will hit what it tries to hit). No AoE damage – only a single target is hit (it does not pierce targets because the projectiles are designed to release their entire kinetic energy against a single target). However, it should be able to hit multiple body parts and do massive damage, taking out most humanoids in a single shot.

It would probably be used against single dangerous targets (centipedes, for example). It would be mostly useless against large groups of enemies (like tribes).