LATIN SMALL LETTER SIDEWAYS O WITH STROKE·U+1D13

Character Information

Code Point
U+1D13
HEX
1D13
Unicode Plane
Basic Multilingual Plane
Category
Lowercase Letter

Character Representations

Click elements to copy
EncodingHexBinary
UTF8
E1 B4 93
11100001 10110100 10010011
UTF16 (big Endian)
1D 13
00011101 00010011
UTF16 (little Endian)
13 1D
00010011 00011101
UTF32 (big Endian)
00 00 1D 13
00000000 00000000 00011101 00010011
UTF32 (little Endian)
13 1D 00 00
00010011 00011101 00000000 00000000
HTML Entity
ᴓ
URI Encoded
%E1%B4%93

Description

U+1D13, or the Latin Small Letter Sideways O with Stroke, is a lesser-known character in the Unicode standard used in digital text. Its primary usage can be found within typographical contexts, where it serves as an alternative representation of the letter 'o'. This unique character showcases the versatility and creativity of typeface design, offering a distinct style that deviates from traditional Latin script. Although its application may be limited due to its uncommon status, it demonstrates how Unicode's expansive library can cater to diverse typographic needs and preferences. The Latin Small Letter Sideways O with Stroke is a testament to the evolving nature of written communication and the continuous exploration of visual expression in digital text.

How to type the symbol on Windows

Hold Alt and type 7443 on the numpad. Or use Character Map.

  1. Step 1: Determine the UTF-8 encoding bit layout

    The character has the Unicode code point U+1D13. In UTF-8, it is encoded using 3 bytes because its codepoint is in the range of 0x0800 to 0xffff.

    Therefore we know that the UTF-8 encoding will be done over 16 bits within the final 24 bits and that it will have the format: 1110xxxx 10xxxxxx 10xxxxxx
    Where the x are the payload bits.

    UTF-8 Encoding bit layout by codepoint range
    Codepoint RangeBytesBit patternPayload length
    U+0000 - U+007F10xxxxxxx7 bits
    U+0080 - U+07FF2110xxxxx 10xxxxxx11 bits
    U+0800 - U+FFFF31110xxxx 10xxxxxx 10xxxxxx16 bits
    U+10000 - U+10FFFF411110xxx 10xxxxxx 10xxxxxx 10xxxxxx21 bits
  2. Step 2: Obtain the payload bits:

    Convert the hexadecimal code point U+1D13 to binary: 00011101 00010011. Those are the payload bits.

  3. Step 3: Fill in the bits to match the bit pattern:

    Obtain the final bytes by arranging the paylod bits to match the bit layout:
    11100001 10110100 10010011