From f81bc0a1ceeb39f8177b6b64fcb375c9c6ce8985 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Robert=20J=C3=A4schke?= Date: Sun, 21 Jul 2024 14:18:52 +0200 Subject: [PATCH] fixed grammar --- .../2024-07-21-fragmenting-the-tile-data.markdown | 15 ++++++++------- 1 file changed, 8 insertions(+), 7 deletions(-) diff --git a/_posts/2024-07-21-fragmenting-the-tile-data.markdown b/_posts/2024-07-21-fragmenting-the-tile-data.markdown index 02d16e1..ea8513a 100644 --- a/_posts/2024-07-21-fragmenting-the-tile-data.markdown +++ b/_posts/2024-07-21-fragmenting-the-tile-data.markdown @@ -86,13 +86,14 @@ For comparison, here's the plot for the 2240 colour tiles of size ![sub-band sizes for the 2240 colour tiles of size 500x500](/img/subband_sizes_tiles500.png) -Here the most frequent number of part (36) does not fit to the -abovementioned formula but we also have colour images, thus three -components (probably [YCbCr](https://en.wikipedia.org/wiki/YCbCr)). I -suppose that the Y component (luma) is encoded in more detail (=higher -depth) than the colour components. For example, having depth 5 for -luma and depth 3 for Cb and Cr each, would result in 36 sub-bands (3 * -5 + 1 + (2 * (3 * 3 + 1)). +Most of these tiles have 36 parts that represent encoded sub-bands +which does not fit to the abovementioned formula. However, we also +have colour images, thus three components (probably +[YCbCr](https://en.wikipedia.org/wiki/YCbCr)). I suppose that the Y +component (luma) is encoded in more detail (=higher depth) than the +colour components. For example, having depth 5 for luma and depth 3 +for Cb and Cr each, would result in 36 sub-bands (3 * 5 + 1 + (2 * +(3 * 3 + 1)). A good starting point for further reading is Chapter 16 "Wavelet-Based Image Compression" of the book [Introduction to Data