GregTech 5/Electricity/zh-cn

从5.0版开始（适用于Minecraft 1.7.2），格雷科技开始使用自己的能源网系统，因为GregoriusT并不满足于的能源网系统.

电压与电流
格雷科技使用“电压”（V）和电流（A）两个术语来描述新能源网. 电流为1A等价于一个IC2的EU电力包；电压则是指代这个电力包的大小.

EU/刻（tick, t）用于描述每刻的EU接收总量. 举例说明：如果一个机器接收了一个32V的电力包和一个24V的电力包，那么总接收电力就是32 + 24 = 56 EU/t.

和IC2能源网不同，所有格雷科技的电力相关设备都有电压和电流限制.

不同机器所能输入或输出的电流不同.
 * 格雷科技的在升压模式输出1A，在降压模式则为4A.
 * 中每有一个电池就输出1A.
 * 可接受的机器要么能接受2A，要么能接受3A.
 * 发电机输出为1A.

为机器供能时必须格外小心：
 * 电压若超出机器承受范围会导致爆炸. 
 * 电流超出机器承受范围则不会有任何问题，只要保证电压在范围内即可. 

机器以及合成表也有电压等级. 多方块机器的电压等级由其决定. 一定要注意机器和合成表的电压等级之间的联系：

单个合成的超频次数可以叠加. 换句话说，如果机器的电压等级比合成的电压等级高出数个级别，那么超频的倍数可以叠加.
 * 若合成的电压等级高于当前机器的电压等级，则机器不能执行该合成.
 * 若合成的电压等级等于当前机器的电压等级，则机器可以正常执行该合成.
 * 若合成的电压等级低于当前机器的电压等级，则机器会处于超频模式. 此时，机器执行该合成的耗时为原来的二分之一，能耗则为原来的两倍，也就是说，机器功率变为原来的4倍.

在5.0版本中，格雷科技有10个电压等级.

备注：ULV算作0级（T0）.

线缆与损耗
因为格雷科技使用自建的能源系统，您现在需要使用格雷的线缆来为机器供能. 需要注意的是，格雷科技里唯一能接受IC2的EU的机器是（不要和IC2的变压器弄混）.

所有格雷科技的线缆都有电压上限、电流上限以及损耗： Do note that packets can rebound. Even if the logical path that a packet dictates that EU should not travel in that direction, you should not take for granted that your cables will not have some stray EU packets travelling through them. For example a 32V package is sent trough a Tin Cable which has a loss of 1EU per block to a machine 8 blocks away. After 8 blocks of cables the 32V Package is down to 24V when it arrives at the machine. Should the machine need for example 30EU/t to operate. A second package sent in the same tick is needed every 4 Ticks. Thus a 2A supply is needed for the machine with this setup. Cable losses are applied to each EU Package, netting you a 2x power loss.
 * 线缆承载的电压若超出上限，则会起火并熔毁. 
 * 线缆承载的电流若超出上限，则会起火并熔毁. 
 * The loss of a cable is per Block a EU package travels.

每种材质的线缆都有对应的1x、2x、4x、8x、12x以及16x的未绝缘导线和1x、2x、4x、8x、和12x的绝缘导线.

'''特别要注意的是，未绝缘导线的损耗是绝缘导线的2倍. '''

举例说明：
 * 一根绝缘的1x锡线可以承载1A和32V的电力，损耗为1V/m. 这意味着在32格之后电力才会损耗殆尽.
 * 一根未绝缘的1x锡线可以承载1A和32V的电力，损耗则为2V/m. 这意味着只需要16格电力就会损耗殆尽.

下表给出了当前格雷科技众多导线的属性：

备注：[1]无绝缘版本；[2]暂无合成

Also any GT Block and Battery outputting Energy has an energy loss on output. This means there is no such thing as lossless cables in GregTech.

A power outputting machine will take (8 * 4 ^ Tier) + (2 ^ Tier) EU from its storage and output only (8 * 4 ^ Tier) EU. The energy lost is therefore (2 ^ Tier).

An example: Say a turbine is supposed to output 32V.

output = 32 = (8 * 4 ^ Tier).

Solving for Tier gives you 1. The energy loss will then be (2 ^ Tier). In this case it is 2.

So the turbine takes 34 EU from its storage, outputs 32 and then voids 2 EU per packet output.

Here is a table documenting some of the cable properties in GregTech:

Optimal Cable length between Batteries for maximum efficiency.

The EU loss of GregTech Cables and Batteries scales linearly with the number of sequential Cables and the number of Batteries, but since voltage is topped up at every battery there will be a loss that is increasing exponentially for every identical segment of a Battery and x-number of Cable links. This exponential loss from more batteries also reduces the impact of the linear loss, but this ofc comes at the cost of more exponential loss. This means that there must exist a sweet spot, because with short segments the extra exponential loss of more segments will be more detrimental to the efficiency than the linear loss from making each segment longer, for long segments this will be reversed. So lets do the math!

Lets first define our terms, a segment is the length of a Battery plus a number of sequential Cables. The efficiency of such a segment will be (8 * 4^T - (D - 1)L) / (8 * 4^T + 2^T). T is the tier (ULV is tier 0, LV is tier 1 and so on). L is the loss of the cable in voltage/meter/ampere. D is the distance of the segment, so the length of the Cables plus the battery.

But this is no good since we want to figure out the optimal length when there is an element of exponential decline that we haven't accounted for. We do this by making an expression of how much efficiency we get in each single block if there was a uniform exponential decline over the whole segment. This turns out to be ((8 * 4^T - (D - 1)L) / (8 * 4^T + 2^T))^(1 / D).

We now take the derivative of that expression with respect to D to get how the efficiency changes when we change the length of the segments, when we do this we get such a ghastly monstrosity that not even WolframAlpha can deal with it algebraically. But this wont stop us on our quest for efficiency! Lets solve it numerically!

Step 1: go to http://www.wolframalpha.com/ because we are lazy. Step 2: Enter "(d/dD) ((8 * 4^T - (D - 1)L) / (8 * 4^T + 2^T))^(1 / D) = 0, T=, L=". It will solve the problem numerically for each separate case. So if you want to know the optimal length of Annealed Copper Cable between your MV Batteries, you enter T=2, L=1 and it will give you the optimal length of each segment (This includes the battery!). In the case of Annealed Copper Cable this turns out to be about 24.1, so 23 cables between each battery is optimal. For more information on other cables, see the table above.

机器爆炸
Using GregTech machines without thought and care can be fairly unsafe. If a machine gets contact with rain on any of the 6 sides of the block, it can catch fire. If a machine gets lit on fire, it can explode.

能量转换
格雷科技并不接受由IndustrialCraft²的线缆传输的EU，同时也有部分mod的EU驱动的设备也不支持由格雷科技的线缆传输的EU. 因此，这种情况下需要对两种EU进行相互转换.

要将IC2的转换为GT5的EU，只需将IC2能源方块的输出面紧贴GT变压器的输入面即可. 换句话说，将IC2的变压器或存电箱带点的一面紧贴在格雷科技变压器的输入面上.

要将GT5的EU转换为IC2的EU，只需将GT的导线连接在IC2的方块上即可.

示范截图：