I tested with the temp addon and got the following info:
Hull heat: 29 (Water evaporate: 6050 / melting: 10285)
Yepp, that's what it's supposed to look like 
[and yeah there will be just minimal heat in the hull.]
Posts by AkhkharuXul
-
-
So could you tell me does that have to be covered in water from the outside? If so would I just dump source blocks from the roof down or does it even matter? I wanted to seal it up a bit but was told if you do that it might also cause issues. I really wanted to make a nice design for the reactors.
Yes, it needs to be surrounded by water; most reactors do. [Water only counts in a 3x3 cube around the reactor core itself]
The water source block 2+ blocks above a reactor is for ones that have too much heating and evaporate water, this reactor is not like that.
So you could basically make a 3x3 cube of water then place the reactor + chambers in it (I think that works - placing in the water I mean) or just place the reactor then fill the rest of the 3x3x3 cube. Or place the reactor + chambers then place the water source block above so it'll pour over the unit.
[basically as long as there are water blocks [source/flowing, doesn't matter] in the remaining 24 blocks of the 3x3x3 cube (reactor taking up 3 blocks) you're all safe.] -
Could someone help me design a Reactor with the highest effiency and no cooldown that only uses IC2?
I seen a few I'm still learning myself and trying understand the best designs. My main problem is not even the placement of cells there looks to be 100's on these forums alone but actually building the chamber and where to place things seems to be the biggest downfall. Any suggestions or visual images would be wonderful
No cooldown = Mark 1
Highest efficiency Mark 1? 2.33Design: http://test.vendaria.net/index…CXXXXXXXXXXXXXXXXXXXXXXXX
There are Mark 1 reactors that produce more EU/t but none (so far) with a higher efficiency (and many believe it isn't possible).
-
Quote
This would actually be a more simple model if all TFs had a buffer (inductive capacity) of their high side and would only transfer if it were empty/full if stepping down/up respectively
I'm pretty sure they do, but that it doesn't matter (I could be wrong though). I just know I've blown up a couple items when the only possible explanation was the TF (LVT) was holding a packet of MV and sent it to a device. [Well I guess another explanation could be that a packet was both sent and received from/to the exploding device during the same tick... ] -
what are breeders for? D:
Turning nearly-depleted uranium cells into actual uranium cells. [You can craft 8x nearly-depleted uranium cells with 1 uranium]
Add coal dust to the nearly-depleted uranium to get isotope cells [Which are then "recharged" in a nuclear reactor; reactors for this specific purpose are called breeders]So basically, they're to increase your uranium yields.
-
Its impossible to make endless uranium with a breeder even if the depleted cells would instantly turn into uranium cells. Only 25% will turn into depleted cells so it will always decrease. Nice way to increase your uranium income by alot through :).
My CASUC reactor will need alot of MSFU to store the enormous amount of energy it makes in 1 cycle. Also with a breeder it will have a effective efficiency of like 8 :D. In the end eff is most important. Getting 8 mil eu from a single uranium ore is nice :).
This theoretical CASUC Breeder: http://test.vendaria.net/index…IXIUIUIXIUIUIXIUIUIXXXXXX
(Which would need to be supplied with 1.5 filled water buckets every second.)Would require 16 isotope cells per cycle (2 uranium, 4 tin, 32 coal dust) [ya, I know 16 coal dust, but the other 16 is used to turn the re-enriched uranium back into uranium cells]
And would recharge those 16 isotope cells and produce 44 million EU.Uranium efficiency per uranium ore? 22.
So if efficiency is the greatest thing, I win?
-
Yaeah, sorry for that. As I mentioned above, it was a typo, and where I stated half a Diamond more for doing the long run without IC² was just fail in math on my side, too
Right, it's a whole diamond more, even without IC
[So with IC it's what, twice as many? Even more cos you can macerate the iron first?]
-
How about:
16 obsidian = 12 iron = 3 gold
.... 64 obsidian = 12 gold
12 gold = 1.5 diamonds ??? [nvm still 3 diamonds; EE is just plain silly -but you did skip a step somewhere ... nvm you confused me with the 4 gold ore = 8 dust = 8 ingots = 1 diamond (making me think 8 gold = 1 diamond)]Edit: Well the math's right, EE is broken.
[Maybe they're figuring on the coal it'd take to smelt the gold, since you could convert it to diamond ore, but then need to use 1 coal per 7 diamond ore to make 7 diamonds :)]
*won't do anymore EE calculations* [They don't even have the numbers section anymore it's all picture nonsense] -
Display More
Well, I just started using EE in conjunction with IC² and did some Math.
Could someone comment if I just miscalculated something? Because it strikes me as odd you would get half a Diamond extra per Obsidian Stack by going the long path even without macerating the Gold Ore.Quote64 Obsidian = 12 Gold Ore = 24 Gold Ingots = 6 Diamonds
You macerated the gold ore. 64 obsidian = 12 gold ore = 12 gold ingots = 2 diamonds.
vs 64 obsidian = 8 lapis ore = 2 diamonds.
(though I don't use EE myself so I was just going off your numbers. It just struck me as odd cos from what I saw of EE everything had a "diamond value" basically, or ## needed to make diamonds)
Erg, I failed at maths? 64 obsidian = 12 gold ore = 12 gold ingots = 3 diamonds (according to your calculations... ) but you did macerate on the line I quoted -
I ran a test with 41 glass instead of 1 and the same stack of 64 cobble. The result:
Packets of 32 were sent from the MFS unit over the glass; loss was one unit per packet; 1200 packets later the operation was complete.
Conclusions: The Wiki and current documentation are wrong/misleading on various points.
Curious... what was the total energy used to macerate the 64 cobble? 39600? (cos it seems like it should be 2 loss per 128 packet not 1 per 32 packet... maybe it's having a problem with fractions/decimals) [which would be 39000 EU for the same 64 cobble] either way both those numbers are still cheaper then the advertised 625 EU / operation.
-
Thanks for the info. Yeah the teleporting went to hell. My players are really sad today. We were starting build a teleport network.
Ah, it is for SMP. You could have players bring their own lapotron crystals to teleport themselves
Or let each teleporter have its own energy supply (probably nuclear power) and then you could just figure it out in trips per minute [or hour(s)] based off the energy generation capabilities. -
Actually you wouldn't get 50/50 then, but even with 4 uran rows and 3 iso ones you'll only need to supply a few urans each cycle...
Rows not columns, isn't a reactor only 6 rows high? *shrugs* and I was trying to work with their -CASUC models anyway (and obviously just theorycrafting all this)
I don't think it'll be possible to have a reactor that would have a positive flow of uranium, as even my last model would have required some isotopes (even if you figure 1/4 uranium will turn into an isotope cell on it's own)
[Looking like 14 isotopes per cycle, using the 4 chamber reactor given... 2/3 uran/iso ratio]
Hey, the tool is back up: http://test.vendaria.net/index…IXIUIUIXIUIUIXXXXXXXXXXXX
650 heat so 1.3 buckets / second would be required?
And you would charge as much uranium as you burn. (4 recharged rows / 2 cycles = 2 rows / cycle) [assuming 4k heat and the math is solid]-- I actually already had redpower2 installed... so I could probably build this in game... at least it'd bring back the chance for explosions if the math is off (or the system hiccups in any way)
-
If you have 50/50 uran and iso, with each iso connected to 2 uran you should be able to roll with just iso cells ...
Now, if only I hadn't randomly lost access to the reactor planner site...
Haha I had the same idea, horizontal rows of Uran/Iso/Uran/Iso/Uran would (or well should) completely recharge the iso cells during 1 cycle. [And you'd get like 2.714 efficiency] (and it would only require 7 Uranium Cells [and 14 Iso cells] per cycle once it's constantly breeding.) ... Obviously I was running with the assumption that you were using a 4 chamber reactor and could fill all but the top row with uranium/iso cells. [I should figure out the heat it'd generate, but I don't feel like it], should be far less then all uran so, I'm going to guess it's a maybe?
Edit: Oh I see what Desuman's up to *counters* if you alternate rows of iso/uran/iso/uran/iso it'd take 2 cycles for the top/bottom rows of Isotopes to recharge but the middle row would recharge every cycle. Once it's up and running it'd be free energy as you wouldn't have to add any uranium to the system (just coal dust). [Erg, you'd still have to add isotope cells ... not sure how many though]
-
Quote
How is that not having any energy loss?
It is, but only between the batboxes and the MFE (and it's only a little bit of loss). I was going to suggest LVTs between there instead of gold cables, but I figured it was his loss so I didn't care
[Looks like TMI anyway so he could have just used some more batboxes :)] -
*adds another problem to the list*
Unless the chunk with the receiving teleporter (and possibly all the chunks in the middle) the receiving MFSU(s) won't even get power [cos it, or part of the power network would be in unloaded chunks]. (Guess that'd only be a problem for the return-trip though, but it did sound like you wanted to power the second teleporter with the MFSUs you have there.) -
Display More
I ran my own experiment with an MFS, MVTF, LVTF, 1x glass 'wire', 2x gold wire (was out of copper wire at that moment).
The first run I got odd values, so I retested with a full stack of 64 cobblestone.
100312 - 99096 = 1216
1216/2 = 608
625*64 = 40000
98488 - 60088 = 38400
(98488 - 60088) / 64 = 600Ignoring the initial charge (which apparently cost exactly 608 for wrench and replace), I observed a sporadic draw of 32eU/pulse that vaguely matches the 2eU/tick specification of the unit.
Strangely I also see that an operation costs 600 and not 625 as predicted.
Ran a similar test with just a batbox + macerator (and a set amount of EU). Same results:
Placing macerator consumed 608 EU, cycle of 64 items consumed 38400 EU (600 EU / process).
And the batbox definitely was only sending packets of 32 EU, but I wouldn't call them sporadic.
Watched it for a minute, it sent 75 packets. 75/60 = 1.25 / sec x 32 EU = 40 EU/s
Alblaka has already said that there's 20 ticks per second so... 40/20 = 2 EU/t (as advertised)*runs a couple tests at the end* mixed results, but average EU / process was 600.... 2@608 and 2@592
[Makes me wonder how the whole "need + loss" thing works, or if it always sends packet size and then just
subtracts the loss from storage. And it seems like machines don't "call for" energy unless they can receive
a full packet size. (maybe that's why the 32 EU ticks seemed sporadic for you cos it would have sent 4
quickly then waited a second or two before sending another set of 4, since you were using a MFE) *shrugs* ] -
all this talk about marks and heat/cool surpluses makes me faint. could someone give me a simple 2-3 chamber reactor that I can run NON-STOP without having an explosion or having to carry round 5 buckets of water with me?
pleeeease!?
http://test.vendaria.net/index…CXXXXXXXXXXXXXXXXXXXXXXXX
(obviously needs to be surrounded with water when set-up, but after that you'd only need to carry 3 uranium cells to re-fill it)
[There definitely needs to be a "good reactor designs" sticky :)]
Also, any mark 1 (assuming it's no -CASUC) would be able to be run non-stop as they do not build-up heat / melt parts. [Though now some people are trying to claim their Mark 3 reactors are MK 1 when using timers *boggles*] -
not nitpicking , but personally i hate parts getting melted due to over heating. I do my best to get as much out of the uranium as i can. To simplify your Reactor a bit with no consumed parts.
http://test.vendaria.net/index…XXXXXXXXXXXXXXXXXXXXXXXXX
(the parts getting melted bit was directed at plating dont care much for them lol rather enjoy having HD's, not at your design)
I think you missed the bit where his is a mark 1-1 so you can run it anywhere (including completely surrounded by lava), and his design didn't burn up parts. (nvm it looks like 1 cooling cell next to the 2 Plating on the right will melt... eventually >1 cycle so still mk 1 as you could just move it when you refill the uranium.)
Personally I prefer surrounding reactors with water, cos external water cooling is OP.
[I was against it at first, but seeing how much it actually does it's a no-brainer, and a "just deal with it"] -
Doesn't tell me how long it takes to re-enrich uranium, but it gives me a good guess :
Assumption : it takes a full cycle baseline to re-enrich uranium at 0 heat. So at 4000 heat, it would take half a cycle. I set a timer to 80 minutes upon reactor start and that should be when it is time to change out the breeding fuel.
I know it takes more then 1 cycle (at 0 heat) to re-enrich depleted uranium (well unless maybe you surrounded 1 isotope cell with 4 uranium cells).
According to Demork's calculations:
It would take 4 cycles for 1 uranium cell to re-enrich 1 isotope (at 0 heat).
at 6000 to 9000 (8999) heat, 1 cycle would be enough for 1 uranium to re-enrich 1 isotope.
at 9000 or more (aka "over 9000!") heat, only 1/2 a cycle would be required.So basically, 10000 seconds = reactor cycle (C), (H) heat index = 4, 2, 1, 0.5 (for <3000, <6000,<9000, >9000 heat), (N) Number of uranium cells around the isotope cell.
Timer = CH/N -
The machine "buffer" or storage is pretty well documented in the wiki for most machines. Usually it's equal to the cost of 1 operation. Induction furnaces are a bit different though. But for a Macerator, one operation costs 625 EU and it has 625 EU storage. So when one operation completes it would dump the storage to the progress bar then "call for" 625 EU. (which is slightly more then 512) [some machines work like that, others gradually use the power, constantly filling the progress bar.]
But yes, it's been documented that transformers send multiple packets of LV or MV when sending. [So it's technically possible to get 512 EU/t out of a 32 EU/t LV transformer (I'm pretty sure)]
Alblaka also mentioned that the EU senders check for "needed power" and send "need + loss energy" so in your example the MFSU probably sends 1 packet of 512 [which gets converted to 16 packets of 32] then it'd send 113 EU which would be converted to 3 packets of 32 and 1 packet of 17 [or 4 packets of 28.25]; assuming your machine needed 625 EU.
