It's talking about information capacity. You can store roughly the same amount of data in 9.5 bits as you can 6 trits because 29.5 (724) is approximately 36 (729).
I suppose, a trit is a "bit with three states" (0, 1 and 2).
Since you can form 3^6=729 different states with 6 trits (every trit can take three different states and you have 6 trits), the above calculation considers how many bits one needs to form 729 different states.
In fact (i.e. reality) you would need ten bits to form 729 different states, but a common calculation to derive how many bits are needed to form a specified number of states is to take the logarithm base 2 of the number. This yields log_2(3^6) = log_2(729) ~ 9.5.
Note: If you want to use the logarithm-base-2 calculation to determine how many bits are needed, you have to take the ceiling of the result.
this might all be true, but as computers still works binary (high or low, on or off, open or closed) for a trit you would still need two binary positions. That is not really a good basis to talk about better information density. So as I see it, it takes 100% more space but gives 58% more info? unless ofcourse I completely misunderstand the concept
A trits would be based off of its own discrete transistor with three states rather than two. So it would occupy (hypothetically) the exact same amount of space, but would hold much more data.
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这是在谈论信息容量。您可以在 9.5 位中存储与 6 个 trit 大致相同的数据量，因为 29.5 (724) 大约为 36 (729)。
It's talking about information capacity. You can store roughly the same amount of data in 9.5 bits as you can 6 trits because 29.5 (724) is approximately 36 (729).
因为 3^6 (729) 大约等于 2^9.5 (724)。
您可以阅读三进制数字系统了解更多详细信息。
Because 3^6 (729) is approximately 2^9.5 (724).
You can read about the ternary number system for more details.
我认为，trit 是“具有三种状态的位”（
0、1和2）。由于您可以用 6 个 trit 形成 3^6=729 种不同的状态（每个 trit 可以采用三种不同的状态，并且您有 6 个 trit），因此上述计算考虑了形成 729 个不同状态需要多少位。
事实上（即现实），您需要 10 位才能形成 729 个不同的状态，但要得出形成指定数量的状态需要多少位，常见的计算方法是取该数字的以 2 为底的对数。得出
log_2(3^6) = log_2(729) ~ 9.5。注意：如果要使用以2为底的对数计算来确定需要多少位，则必须取结果的上限。
I suppose, a trit is a "bit with three states" (
0,1and2).Since you can form
3^6=729different states with 6 trits (every trit can take three different states and you have 6 trits), the above calculation considers how many bits one needs to form 729 different states.In fact (i.e. reality) you would need ten bits to form 729 different states, but a common calculation to derive how many bits are needed to form a specified number of states is to take the logarithm base 2 of the number. This yields
log_2(3^6) = log_2(729) ~ 9.5.Note: If you want to use the logarithm-base-2 calculation to determine how many bits are needed, you have to take the ceiling of the result.
http://en.wikipedia.org/wiki/Ternary_numeral_system
“一个 trit 包含 log23 （大约1.58496) 信息位。”
1.58496*6～9.5
http://en.wikipedia.org/wiki/Ternary_numeral_system
"One trit contains log23 (about 1.58496) bits of information."
1.58496 * 6 ~ 9.5
日志(3**6)/日志(2) ~ 9.5097750043269382
log(3**6)/log(2) ~ 9.5097750043269382
二进制是离散值。你只有整数值，所以我猜 9.5 位实际上是 10 位。
Binary are discrete values. You have only whole values, so I guess 9.5 bits are in actuality 10 Bits.
这可能都是正确的，但由于计算机仍然以二进制方式工作（高或低、开或关、开或关），因此您仍然需要两个二进制位置。这并不是谈论更好的信息密度的良好基础。所以在我看来，它占用了 100% 的空间，但提供了 58% 的信息？
当然，除非我完全误解了这个概念
this might all be true, but as computers still works binary (high or low, on or off, open or closed) for a trit you would still need two binary positions. That is not really a good basis to talk about better information density. So as I see it, it takes 100% more space but gives 58% more info?
unless ofcourse I completely misunderstand the concept
一个
（约 1.58496）
trit包含信息位。因此
6 trit = 9.5 位（大约）Wikipedia< 中有一些更有趣的信息/a>
One
(about 1.58496)
tritcontainsbitsof information.and hence
6 trit = 9.5 bits (approx)Some more interesting information in Wikipedia
trits 将基于其自己的具有三种状态而不是两种状态的分立晶体管。因此，它会占用（假设的）完全相同的空间量，但会容纳更多的数据。
A trits would be based off of its own discrete transistor with three states rather than two. So it would occupy (hypothetically) the exact same amount of space, but would hold much more data.