It is my understanding that PVR textures, made with texturetool, are simply compressed images. Therefore the difference lies in the file size.
Frankly, the file size doesn't interest me. What I want to know is, can a PVR texture consume less RAM than a normal .PNG texture? Or does this depend entirely on the texture format (like RGBA8888 etc)?
The essential question would be:
Given X.png and X.pvr, if I display both with texture format RGBA8888, will one consume less RAM than the other?
Yes, the PVR will consume less RAM at all stages — it's unpacked live by the GPU as it's accessed. There's no intermediate decompression.
A PVR-like approach used in digital video is that instead of storing RGB at every pixel, convert to YUV, then store Y at every pixel and U and V only twice per four-pixel block. So you go from 128 bits for the block to 64 bits. To get back to RGB the outputter reads the exactly correct Y and interpolates or accesses the most nearby U and V as necessary.
Schemes like PVR do a similar thing of not storing the full value at every pixel but inferring parts of it from nearby context. What counts as nearby is picked directly corresponding to however the caching is arranged on that GPU. It's usually more in-depth that just scaling down the sampling resolution of some of the channels, e.g. specifying a base offset for samples and then using a tiny precision for each is also common.
So the GPU can always get a value for pixel X by reading only values in a very small, local region of the data.
This contrasts with traditional schemes like PNG where having to know every pixel in the stream prior to X is acceptable if it improves the compression. Processing such things live would flood the GPU's memory bandwidth and hence be completely impractical, so such textures are decompressed from disk and then uploaded.
So schemes like PVR tend to lead to poorer compression and lower per-pixel quality but the win is that they can sit in VRAM compressed. A game will often increase the resolution of its textures if using PVR to try to find a comfortable balance.
Uncompressed textures are:
16 bit per pixel (RGB565, RGBA4444),
24 bit per pixel (RGB888)
PVRTC textures are either 4bpp or even 2bpp. So yes they do use less memory.
Also they perform better because need less memory bandwidth to fetch textures.
Related
I am comparing jpeg to jpeg in a constant 'video-stream'. i am using EMGU/OpenCV to compare each pixels at the byte level. There are 3 channels to each image (RGB). I had heard that it is common practice to store only the pixels that have changed between frames as a way of conserving memory space. But, if for instance/example I say EVERY pixel has changed (pls note i am using an exaggerated example to make my point and i would normally discard such large changes) then the resultant bytes saved is 3 times larger than the original jpeg.
How can I store such motion changes efficiently?
thanks
While taking the consecutive images the camera might also move or not. If the camera is fixed, only the items on the view move and some portion of the image changes every time. If the camera also moves, even if the objects stand still, the image changes significantly. There are some algorithms to discard the effect of the motion of the camera. So the main idea is when compared with the sampling frequency of the camera (e.g. 25 frames per second) most of the objects nearly standing still.
Because most of the image is unchanged between the frames, it becomes feasible to use difference of the images. It provides some compression ratios. However after some amount of time the newly received image shows big difference with the reference image, so it becomes better to get a new image reference. Which is named a "reference frame".
In fact, modern video compression algorithms uses advanced techniques to detect the objects and follow them, which results better compression ratios.
Wikipedia - Different compression techniques
Check This - OpenCV should handle the storing of consecutive images in different video formats.
i have just downloaded the latest win32 jpegtran.exe from http://jpegclub.org/jpegtran/ and observed the following:
i have prepared a 24 BPP jpeg test image with 14500 x 10000 pixels.
compressed size in file system is around 7.5 MB.
decompressing into memory (with some image viewer) inflates to around 450 MB.
monitoring the jpegtran.exe command line tool's memory consumption during lossless rotation (180) i can see the process consuming up to 900 MB memory!
i would have assumed that such jpeg lossless transformations don't require decoding the image file into memory and instead would just perform some mathematical transformations on the encoded file itself - keeping the memory footprint very low.
so which of the following is true?
some bug in this particular tool's implementation
some configuration switch i have missed
some misunderstanding at my end (i.e. jpeg lossless transformations also need to decode the image into memory?)
the "mathematical operations" consuming even more memory than "decoding the image into memory"
edit:
according to the answer by JasonD the reason seems to be the latter one. so i'll extend my question:
are there any implementations that can do those operations in small chunks (to avoid high memory usage)? or does it always need to be done on the whole and there's no way around it?
PS:
i'm not planning to implement my own codec / algorithm. instead i'm asking if there are any implementations out there that meet my requirements. or if there could be in theory, at least.
I don't know about the library in question, but in order to perform a lossless rotation on a jpeg image, you would at least have to decompress the DCT coefficients in order to rotate them, and then re-compress.
The DCT coefficients, fully expanded, will be the same size or larger than the original image data, as they have more bits of information.
It's lossless, because the loss in a jpeg is caused by quantization of the DCT coefficients. So long as you don't decode/re-encode/re-quantize these, no loss will be incurred.
But it will be memory intensive.
jpeg compression works very roughly as follows:
Convert image into YCbCr colour space.
Optionally downsample some of the channels (colour error is less perceptible than luminance error, so it is typical to 2x downsample the chroma channels). This is obviously lossy, but very predictably/stably so.
Transform 8x8 blocks of the image by a discrete cosine transform (DCT), moving the image into frequency space. The DCT coefficients are also in an 8x8 block, and use more bits for storage than the 8-bit image data did.
Quantize the DCT coefficients by a variable amount (this is the quality setting in most packages). The aim is to produce as many small and especially zero coefficients as possible. The is the main "lossy" aspect of jpeg compression.
Zig-zag through the 2D data to turn it into a 1D stream of coefficients which is roughly in frequency order. High frequencies are more likely to be zero'd out, so many packets will ideally end in a stream of zeros which can be truncated.
Compress (non-lossily) the (now quite compressible) data using huffman encoding.
So a 'non-lossy' transformation would want to avoid doing as much as possible of that - especially anything beyond the DCT quantization, but that does not avoid expanding the data.
I'm making a Worms-style bitmap destructible terrain game using OpenGL. I'd like to know where the limitiations in terms of video memory are for the size of the worlds.
Currently, I use blocks of 512*512 RGBA textures for the terrain.
How much memory, very roughly, can I expect such a 512*512 RGBA texture to take up?
Is there any internal, automatic compression going on?
How much video memory can I expect most user's computers to have free?
How much memory, very roughly, can I expect such a 512*512 RGBA texture to take up?
Not enough information. You should always use sized OpenGL image formats (GL_RGBA8, GL_RGBA16).
GL_RGBA8 takes up 32-bits per pixel, which is 4 bytes. Therefore, 512*512*4 = 1MB.
Is there any internal, automatic compression going on?
No.
How much video memory can I expect most user's computers to have free?
How much are you using currently?
OpenGL will page image data in and out according to the available space. If you run out of GPU memory, OpenGL will happily allocate system memory and upload the images as needed.
But to be honest, your little Worms game isn't going to actually cost anything in terms of memory size. Maybe 64MB when you're done, tops. It's nothing you need to be concerned about.
I would not worry about that very much. Even with 8192*2048 world (4 screens wide and 2 screens tall, which is very big for Worms-style game) you would require only 8*2*4=64Mb (add mipmaps, other textures, framebuffer) you should fit into 128MB bounds. As far as I know even older GPUs have that kind of memory (we don't speak about GeForce4 cards, right?).
Older GPUs may have limitation on how big each texture could be, but since you already split your world into 512x512 chunks it won't be a problem.
If video memory becomes an issue you could allow users to use half-sized textures (i.e. downsample the world to 4096*1024 and 256x256 chinks) and fetch new / discard unused regions on demand.
With 32-bpp (4 bytes) you get 4*512*512 = 1 MB
See this regarding texture compression: http://www.oldunreal.com/editing/s3tc/ARB_texture_compression.pdf
Again, this depends on your engine, but if I were you I would do this:
Since your terrain texture will probably be reusing some mosaic-like textures, and you need to know whether a pixel is present, or destroyed, then given you are using mosaic textures no larger than 256x256 you could definitely get away with an GL_RG16 internal format (where each component would be a texture coordinate that you would need to map from [0, 255] -> [0.0, 1.0] and you would reserve some special value to indicate that the terrain is destroyed) for your terrain texture, making every 512x512 block take up 0.5MB.
Although it's temping to add an extra byte to indicate terrain presence, but a 3 byte format wouldn't cache too well
I am loading a RGBA texture which is 1024 x 1024. I expected the on-memory texture size would be 1024 x 1024 x 4 => 4 MB . But when I try to print the memory consumption I can see that the texture is taking around 7 - 8 MB, almost double. I was just wondering whether IPad is converting every channel from byte to half-float,
So is there any way to specify that every pixel should take 4 bytes and not 8 bytes.
The easiest way to specify it is using a sized internal format (like GL_RGBA8 instead of GL_RGBA), although I'm not sure if these are supported in ES. But I would be surprised if an ES device would store a standard RGBA texture with more than 8 bits per channel.
How do you determine the GPU memory consumption? I would rather guess the additional memory is due to other important GPU resources, like VBOs and not to forget the framebuffer itself (the memory you render into), that takes a reasonable amount of memory. And remember, when using mip-maps these additionally require around 33% of the base texture's memory.
And if you're talking about the size of the CPU data you create the texture from, then this doesn't have anything to do with the texture's size anyway and only depends on the size of your own data.
You have to specify the type and internal format of your texture when you create it using glTexImage2D.
Yours is probably set to GL_FLOAT or something .
Lookup the documentation here : http://www.opengl.org/sdk/docs/man/xhtml/glTexImage2D.xml
For iPhone game development, I switched from PNG format to PVRTC format for the sake of performance. But PVRTC compression is creating files that are much bigger than the PNG files.. So a PNG of 140 KB (1024x1024) gets bloated to 512 KB or more in the PVRTC format.. I read somewhere that a PNG file of 50KB got compressed to some 10KB and all, in my case, its the other way around..
Any reason why it happens this way and how I can avoid this.. If PVRTC compression is blindly doing 4bpp conversion (1024x1024x0.5) irrespective of the transparencies in the PNG, then whats the compression we are achieving here..
I have 100s of these 1024x1024 images in my game as there are numerous characters each doing some complex animations.. so in this rate of 512KB per image, my app would get more than 50MB.. which is unacceptable for my customer.. ( with PNG, I could have got my app to 10MB)..
In general, uncompressed image data is either 24bpp (RGB) or 32bpp (RGBA) flatrate. PVRTC is 4bpp (or 2bpp) flatrate so there is a compression of 6 or 8 (12 or 16) times compared to this.
A requirement for graphics hardware to use textures natively is that the format of the texture must be random accessible for the hardware. PVRTC is this kind of format, PNG is not and this is why PNG can achieve greater compression ratios. PVRTC is a runtime, deployment format; PNG is a storage format.
PVRTC compression is carried out on 4x4 blocks of pixels at a time and at a flat bit rate so it is easy to calculate where in memory to retrieve the data required to derive a particular texel's value from and there is only one access to memory required. There is dedicated circuitry in the graphics core which will decode this 4x4 block and give the texel value to your shader/texture combiner etc.
PNG compression does not work at a flat bitrate and is more complicated to retrieve specific values from; memory needs to be accessed from multiple locations in order to retrieve a single colour value and far more memory and processing would be required every single time a texture read occurs. So it's not suitable for use as a native texture format and this is why your textures must be decompressed before the graphics hardware will use them. This increases bandwidth use when compared to PVRTC, which requires no decompression for use.
So for offline storage (the size of your application on disk), PNG is smaller than PVRTC which is smaller than completely uncompressed. For runtime memory footprint and performance, PVRTC is smaller and faster than PNG which, because it must be decompressed, is just as large and slow as uncompressed textures. You might gain some advantage with PNG at initialisation for disk access, but then you'd lose time for decompression.
If you want to reduce the storage footprint of PVRTC you could try zip-style compression on the texture files and expand these when you load from disk.
PVRTC (PowerVR Texture Compression) is a texture compression format. On devices using PowerVR e.g. most higher end mobile phones including the iPhone and other ARM-based gadgets like the iPod it is very fast to draw since drawing it is hardware accelerated. It also uses much less memory since images are represented in their compressed form and decoded each draw, whereas a PNG needs to be decompressed before being drawn.
PNG is lossless compression.
PVRTC is lossy compression meaning it approximates the image. It has a completely different design criteria.
PVRTC will 'compress' (by approximating) any type of artwork, giving a fixed bits per texel, including photographic images.
PNG does not approximate the image, so if the image contains little redundancy it will hardly compress at all. On the other hand, a uniform image e.g. an illustration will compress best with PNG.
Its apples and oranges.
Place more than one frame tiled onto a single image and blit the subrectangles of the texture. This will dramatically reduce your memory consumption.
If you images are, say, 64x64, then you can place 256 of them on a 1024x1024 texture in a 16x16 arrangement.
With a little effort, images do not need to be all the same size, just so long as you keep track in the code of the rectangle in the texture that each image is at.
This is how iPhone game developers do it.
I agree with Will. There is no point in the question. I read the question 3 times, but I still don't know what Sankar want to know. It's just a complain, no question.
The only thing I can advice, don't use PVRTC if you mind to use it. It offers performance gain and saves VRAM, but it won't help you in this case. Because what you want is just reducing game volume, not a consideration about trade-off between performance and quality.