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I was looking at a chart of hiragana and trying to find the characters to form the name of a song Inochi no Namae. I started with i, then no, but there's no chi. I then started wondering where chu and cho are, among others. I'm guessing it has to do with the diacritic marks. Then I got "na," and the "mae" threw me.

Any help appreciated.

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2 Answers 2

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You will find "chi" in the "t" row and "i" column, hence "ti". There are various ways to transcribe Japanese into Latin script. Whether you spell it chi or ti, it is the same Japanese sound: ち.

For cha, chu, and cho, it is chi + ya, chi + yu, and chi + yo. You could also spell it tya, tyu, and tyo.

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  • Added details for cha, chu, and cho.
    – Dono
    Commented Jul 12, 2012 at 14:50
  • Thanks! Makes sense. And "Mae"? And you ignored my edit. :[
    – temporary_user_name
    Commented Jul 12, 2012 at 14:50
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    What is problematic about mae? It is ma + e. You can find ma in the m-row with a-column and e is along with the rest of the vowels, the fourth from the left.
    – Dono
    Commented Jul 12, 2012 at 14:52
  • Oh. It really is that simple. Coming from an English perspective that's a very unusual way of breaking up a word, but now I see the principle.
    – temporary_user_name
    Commented Jul 12, 2012 at 14:56
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    Just in case anybody is wondering, "to" and "cho" are two completely separate sounds (besides the fact that they end with an "o" sound) in all romanization methods that I've seen. "to" is と and "cho" is ちょ.
    – atlantiza
    Commented Jul 12, 2012 at 15:02
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As @Dono answered, "chi" is the "t+i" sound in the chart. Some others that maybe are not obvious are "shi" (s+i), "tsu" (t+u), and "fu" (h+u).

Also, the singular "n" can be difficult in some instances. Since you are a beginner, you are probably mostly seeing words in romaji (Roman letters). A romaji "n" will be associated with a vowel immediately following it unless a dash (-) or apostrophe (') is used. For example, "tanin" (other people) is "ta+ni+n". However, "man'in" (many people) is "ma+n+i+n".

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