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引用: 我们都知道,当多张连续图像以一定的速率(比如每秒 24 张)播放,即可形成视频。但视频传输却并不是单纯地发送连续图像这么简单。假设一张图像分辨率为 1280x720,且每个 RGB 像素点大小为 24bit,则单张图像大小约为 21.1Mb;如果每秒需要传输 24 张,则大约需要占用约 506.4Mb [^1] 的带宽。显然这样的带宽占用是不现实的。
修正: 21.1Mb/8 = 2.64MB
The text was updated successfully, but these errors were encountered:
引用: 我们都知道,当多张连续图像以一定的速率(比如每秒 24 张)播放,即可形成视频。但视频传输却并不是单纯地发送连续图像这么简单。假设一张图像分辨率为 1280x720,且每个 RGB 像素点大小为 24bit,则单张图像大小约为 21.1Mb;如果每秒需要传输 24 张,则大约需要占用约 506.4Mb [^1] 的带宽。显然这样的带宽占用是不现实的。 修正: 21.1Mb/8 = 2.64MB
@jiangzhibin151 改成了下面这段话:
我们都知道,当多张连续图像以一定的速率(比如每秒 24 张)播放,即可形成视频。但视频传输却并不是单纯地发送连续图像这么简单。假设一张图像分辨率为 1280x720,且每个 ARGB 像素点大小为 32bit,则单张图像大小约为 29.5Mbit;如果每秒需要传输 24 张,则大约需要占用约 737.5Mbit/s 的带宽。显然这样的带宽占用是不现实的。
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不是 29.5 x 24 = 708 吗?
的确是 708(按错计算器了
mthli
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引用:
我们都知道,当多张连续图像以一定的速率(比如每秒 24 张)播放,即可形成视频。但视频传输却并不是单纯地发送连续图像这么简单。假设一张图像分辨率为 1280x720,且每个 RGB 像素点大小为 24bit,则单张图像大小约为 21.1Mb;如果每秒需要传输 24 张,则大约需要占用约 506.4Mb [^1] 的带宽。显然这样的带宽占用是不现实的。
修正:
21.1Mb/8 = 2.64MB
The text was updated successfully, but these errors were encountered: