To calculate the frame buffer size in bytes, you need to know the resolution of the system (width x height) and the bits per pixel. The formula for the frame buffer size in bytes is:
[ \text{Frame Buffer Size (bytes)} = \frac{\text{Width} \times \text{Height} \times \text{Bits per Pixel}}{8} ]
For a system with a resolution of, for example, 1920x1080 (Full HD), the frame buffer size would be ( \frac{1920 \times 1080 \times 12}{8} = 27,648,000 ) bytes, or approximately 26.4 MB. Adjust the width and height as needed for different systems to find their frame buffer sizes.
how pixel screen positions are stored and retrieved from frame buffer?
to store 12 bits per pixel 1.for system with resolution 640 by 480 frame buffer size=(640*480*12)/8=0.46Mbyte 2.for system with resolution 1280 by 1024 frame buffer sizs=(1280*1024*12)/8=1.96Mbyte 3.for system with resolution 2560 by 2048 frame buffer sizs=(2560*2048*12)/8=7.86Mbyte to store 24 bits per pixel 1.for system with resolution 640 by 480 frame buffer size=(640*480*24)/8=0.92Mbyte 2.for system with resolution 1280 by 1024 frame buffer sizs=(1280*1024*24)/8=3.93Mbyte 3.for system with resolution 2560 by 2048 frame buffer sizs=(2560*2048*24)/8=15.72Mbyte
A frame buffer temporarily stores an entire image frame for image or video capture applications. Here, 'buffer' implies 'temporary memory'. 'Random access' means that an interfacing microprocessor or other electronic component can read from ('access') arbitrary ('random') memory locations. Given that this is in the context of a frame buffer, that means you could read the first pixel, then the pixel on the 10th row and 9th col, then the last pixel, or any order desired. This is in contrast to 'sequential access', which only allows reading consecutive memory locations (ie read pixel1, then pixel 2, pixel3, ...). Hopefully it is clear that random access allows for more control and is necessary for image processing operations. On the other hand, sequential access gives less control, but is sufficient for transferring an image frame to memory or from camera to LCD display. To be clear, 'random' here means 'arbitrary', and is also used to describe the general-purpose random access memory (RAM) in PCs.
On a Windows Vista system, the size of one pixel is 631 bytes, however, the size on disk is 4 KB, or 4,096 bytes. Hope this helps.
A 200x100 image with an 8-bit color depth means each pixel is represented by 1 byte. To calculate the total size, you multiply the width by the height and then by the number of bytes per pixel: 200 pixels * 100 pixels * 1 byte/pixel = 20,000 bytes. Therefore, the image size is 20,000 bytes.
To convert bytes to pixels, you need to know the color depth of the image. For example, if you're using a common color depth of 24 bits per pixel (8 bits for each of the RGB channels), then each pixel would require 3 bytes (since 24 bits ÷ 8 bits/byte = 3 bytes). Therefore, 10,000,000 bytes would equal approximately 3,333,333 pixels (10,000,000 bytes ÷ 3 bytes/pixel). If the color depth changes, the number of pixels would need to be recalculated accordingly.
The Z-buffer algorithm is a convenient algorithm for rendering images properly according to depth. To begin with, a buffer containing the closest depth at each pixel location is created parallel to the image buffer. Each location in this depth buffer is initialized to negative infinity. Now, the zIntersect and dzPerScanline fields are added to each edge record in the polyfill algorithm. For each point that is to be rendered, the depth of the point against the depth of the point at the desired pixel location. If the point depth is greater than the depth at the current pixel, the pixel is colored with the new color and the depth buffer is updated. Otherwise, the point is not rendered because it is behind another object.
To convert pixels to bytes, you need to know the color depth of the image, which is usually measured in bits per pixel (bpp). The formula to calculate the size in bytes is: (width × height × color depth) / 8. For example, for a standard 24-bit color image, the calculation would be (width × height × 24) / 8. This gives you the total size in bytes for the image.
Taken from http://bmrc.berkeley.edu/frame/research/mpeg/mpeg_overview.html The typical data rate of an I-frame is 1 bit per pixel while that of a P-frame is 0.1 bit per pixel and for a B-frame, 0.015 bit per pixel.
The number of pixels in a 2MB image depends on the color depth and format of the image. For example, a standard 24-bit color image (which uses 3 bytes per pixel) would contain approximately 682,666 pixels in 2MB (2,000,000 bytes divided by 3 bytes per pixel). However, if the image has a different color depth or compression, the pixel count would vary.
The data from the framebuffer determines which of the colors in the palette are used for the current pixel it is rendering. This output data provides primary-color data from the lookup table.
The number of pixels in 2000 KB depends on the color depth and format of the image. For example, a standard image with a color depth of 24 bits (8 bits for each of the RGB channels) would have about 3 bytes per pixel. Therefore, 2000 KB (or 2,000,000 bytes) would contain approximately 666,667 pixels (2,000,000 bytes ÷ 3 bytes/pixel). However, this is a rough estimate; actual pixel counts can vary based on compression and image format.