LATIN SMALL LETTER BOTTOM HALF O·U+1D17

Character Information

Code Point
U+1D17
HEX
1D17
Unicode Plane
Basic Multilingual Plane
Category
Lowercase Letter

Character Representations

Click elements to copy
EncodingHexBinary
UTF8
E1 B4 97
11100001 10110100 10010111
UTF16 (big Endian)
1D 17
00011101 00010111
UTF16 (little Endian)
17 1D
00010111 00011101
UTF32 (big Endian)
00 00 1D 17
00000000 00000000 00011101 00010111
UTF32 (little Endian)
17 1D 00 00
00010111 00011101 00000000 00000000
HTML Entity
ᴗ
URI Encoded
%E1%B4%97

Description

U+1D17, or Latin Small Letter Bottom Half O, is a rare and distinct Unicode character that holds a significant position within the world of typography and digital text representation. As part of the Unicode 5.1 standard, it was introduced to represent an alternative form of the lowercase letter "o," with a unique bottom half resembling a backwards "C." Its usage is often seen in specialized texts such as linguistic studies, typographic design, or cultural explorations, where its unusual visual appeal and distinct character may be desirable. While it does not have widespread application due to its unconventional appearance, the Latin Small Letter Bottom Half O offers a creative alternative to the conventional lowercase "o," providing an opportunity for designers and writers to experiment with different visual elements in digital text.

How to type the symbol on Windows

Hold Alt and type 7447 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+1D17. 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+1D17 to binary: 00011101 00010111. 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 10010111