设图像的长宽比为4:3 600万像素的数码相机的空间分辨率是多少? 它拍的一副彩色图像需多少个字节来储存

2025-03-14 23:29:55
推荐回答(3个)
回答1:

600W=2863*2147像素,至于多少字节,那是随机的,因为JPEG图像是压缩的,和被拍摄物体的空间频率分布有关系,如果是拍摄一张白纸,细节非常少,文件就很小。如果拍一个西瓜,有很多纹路细节很多文件就大。如果是存为BMP图像,则要看是采用什么样的编码方式,16位彩色BMP图像需要600*(16/8)=12兆,24位彩色图像需要600*(24/8)=18兆,32位真彩色需要600*(32/8)=24兆字节

回答2:

“空间分辨率”通常是医学影像学诊断设备的评价指标。要考虑断层的厚度。并不适合平面影像的评价。
在相机说明书的提供的参数中,有最大分辨率下不同宽高比图片的宽/高边的像素数参考值。数码卡片机的传感器一般都是4:3幅面的,600万像素的4:3幅面的传感器理论上最大分辨率可以达到2828X2121,选用3:2幅面时传感器需要裁切,只能达到2828X1414(400万像素)了。
在最大分辨率下的照片占用的存储空间要根据设定的图像存储格式和压缩率,以及图像的复杂程度来确定,并不是一个固定数值。通常RAW无损存储格式占用存储空间较大,高压缩率的JPG格式占用存储空间较小。说明书中一般也提供参考数据。

回答3:

600万像素=2848*2136,理论上600万像素需要6M的存储空间,但实际上差别很大,色彩、细节少的有的不到1M,细节丰富的会有3M多,但由于一般都是直出JPG格式,都不会超过6M这个极限的。

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