My new Canon Powershot came with a 32MB memory card which indicated the storage of 12 photos at the high m-pixel setting. With those figures in hand, my math calculation indicates 2.7MB per photo are being used at that high m-pixel setting.

After purchasing a larger memory card (2GB) my calculation (2GB divided by 2.7MB) indicated that there should be enough memory for approx 740 photos at that high m-pixel setting. Instead, my camera now indicates the new 2GB memory card has enough storage for 940 photos at the high m-pixel setting.

What am I missing here? Help needed for simple math. Thanks. :confused:

Well first 2G is actually 2048M, but another issue might be minimum size for a file may be higher on the smaller card.

Scott - Thanks for the feedback. Looks like my math may be intact but my knowledge of memory cards might be suspect. Given the possible variations you described regarding file size everything must be in order with this new camera.

Yeah, it's the way the storage on SD cards works and your assumptions on file size that are messing you up!

Because it's a storage media, your 32MB is actually 32,000,000 bytes (yep, the manufacturers are telling you a clever lie!) so if you divide that by 1024 you will actually get your proper kilobytes;

32,000,000 / 1024 = 31250KB

So, I reckon your photos are actually 2.4MB... but that's the PROPER 2.4MBs

Anyway, your 2.4MB pictures are about 2400KB.

NOOOOOOW...
If you divide your storage by your photos you'll see it's just over 13 (your camera will round it down for obvious reasons!) and with a small amount taken up for TOC and the like, I can see why your camera would display 12!

But here's where it all falls down! If you go on the above calculations, to get 940 pictures at this size you'd need a 2.2GB card!

What I think it may be doing is creating an 'average' file size for that setting... your file size will depend upon the compression and the colour scale in the photo. So your camera will 'know' that not ALL the photos will be of the maximum size, and average it out... with your small card this won't matter as it would only shift it by maybe one photo... but on a larger disk, the shift is a lot higher!

Saying that... a shift of 100 photos? That does seem a BIT much! Ha ha!

Bits, Bytes, Nybbles, marketing and overhead.

1 bit = 1 on/off state
1 Nybble is 4 bits
1 byte is 8 bits

Difference between kbps and KBps.
One is Killobits/s and the other is KiloBytes/s.
Sometimes it takes 10 bits to transmit 1 byte of information because of oberhead.

b = bits and B = bytes. K = 1000 or 1024 depending on context.

2 GB bemory is a power of 2 number of bits. Just like 512 MB 1GB. 512 is right, 1GB assumes 1 kilobyte = 1024 bits.

Disk drives they actually like to sell as capacity in bytes, but overhead reduces capacity.

example:
It could be that a file containing 1 bit, is actually 2048 bits long + overhead where as on a larger sd card a 1 bit file could occupy 8192 bits + overhead.
Compressin reduces the file size.

So, it's just an estimate. Disclaimer, if you will.

Remember to format and new SD cards with the camera even though the camera may be able to store onto the card. The PC may not be able to read it directly.

See, I was thinking along the same vein as you KISS. But with storage medium they actually assume that 1GB = 1,000,000,000 bytes (8,000,000,000 bits).

Also, I was thinking about the overheads, OS and TOS of the SD card... but that would reduce the capacity, not boost it! Ha ha!

Interesting stuff. I almost forgot my initial question. Anyway, it's all good info. Thanks for the replies.

B is bit and B is Byte or 8 bits.

Yep, I totally accept that... but the notation of bits is usually reserved for data rates and read/write speeds (e.g. 2Mb/s) and nothing here is referring to bits.

As this is storage question it is all to do with bytes... unless we multiply all our figures by 8 of course! Ha ha!