Computing desk
< September 3	<< Aug \| September \| Oct >>	September 5 >

Welcome to the Wikipedia Computing Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

September 4[edit]

Recovery of Deleted File on SD Card[edit]

Help!

I accidentally deleted a .zip file from a Secure Digital card. The file is about 275KB in size, and was deleted before a backup could be made. In order to stop the deleted data from being overwritten, I have made the SD card read-only by using the write-protect switch. The SD card's filesystem is NTFS, and my operating-system is Microsoft Windows 7 Professional Edition. I need to recover the deleted file ASAP - is there any free software or freeware I can use to recover my deleted file?

Any and all help is appreciated, and thank you to everyone.

I don't know offhand what the windows software is but there is Linux software to do the same thing. There are Live Discs that do the same thing. It's likely your file's recoverable. Write protecting it is the right idea. I'm sure others will know of some windows software. Shadowjams (talk) 07:13, 4 September 2010 (UTC)[reply]

PhotoRec and Recuva both come highly recommended and can undelete basically anything as long as it hasn't been over-written. Good luck! Zunaid 07:17, 4 September 2010 (UTC)[reply]

What is the cellular 3G base station price range?[edit]

If someone want to cover around 5,000 km2 with 3G coverage - what amount of investment into cellular hardware he will be looking at? Assuming wired data network is already exist and installation costs for towers are out of scope of this question(and spectrum licenses out of scope too). One million users inside 5.000 km2. Basically, how much one base station roughly cost and how big would be the cell it will cover? Was trying to search Google, but no luck so far. —Preceding unsigned comment added by 70.52.186.74 (talk) 07:31, 4 September 2010 (UTC)[reply]

I don't know about the cost, but the number of base stations would depend on the specific mobile technology and frequency used, the local geography and the amount of bandwidth required. The table in this article suggests a CDMA-2000 network operating at 1800 MHz has cells 14 km in radius (each covering an area of ~600 km²). However, considering how close the mobile masts are in my area of the UK, I think that might be an idealised radius. Indeed, OFCOM's Mobile phone base station database indicates that a city of 100,000 people has well over 50 base stations and a city the size of Manchester has nearly a 1,000 base stations. Astronaut (talk) 08:46, 4 September 2010 (UTC)[reply]

Thank you! You actually shifted my keywords selection toward right direction and I found an article that mention number of base stations to cover Germany: they have more than 20.000 GSM base station covering 99% of the population and 13.000 UMTS base stations covering 80% of the population. That is perfect for my purposes. Still wondering about rough price estimation that cellular companies pay for base stations(preferably 3G)... 70.52.186.74 (talk) 11:15, 4 September 2010 (UTC)[reply]

Some more searches - and I found an answer! Thanks a bunch! If not you, I would still be wondering. I also think that number of base stations for Manchester is that high because there is different cell companies operate there while some stations are shares, most are not or support different types of network(like GSM vs CDMA and such). And another interesting number - total base station in UK is ~53 000... 70.52.186.74 (talk) 11:39, 4 September 2010 (UTC)[reply]

AFAIK all mobile phone networks in the UK use GSM or it's 3G decendents (UMTS etc) Nil Einne (talk) 13:00, 4 September 2010 (UTC)[reply]

The numbers I mentioned above are from a database that is aimed at people concerned about radiation exposure from mobile comms. Therefore it lists the number of masts in an area and lets you see the frequency, signal strength and company/companies on each mast (interesting because I found out why there is a dead spot round my sister's house :-) If only one operator is in an area of flat geography, then maybe fewer masts will be required, but you would still need sufficient capacity to support the bandwidth your 1 million population might use. Astronaut (talk) 23:47, 4 September 2010 (UTC)[reply]

Scanning over the network[edit]

The charity where I volunteer recently bought a new printer/copier/fax/scanner (a Xerox Workcenter 4118). It has a builtin network card which is connected to the rest of the network using a standard ethernet cable, and a standard telephone connection is used so faxes can be sent and received. However, reading the user manual it seems the scanner function can only be used if it is connected a USB or parallel cable. Is there a way to get scanning to work across the network using any PC on the network, perhaps using a third-party piece of software? Being an underfunded charity, there is a strong preference for a low price, or better still, free solution. Thanks. Astronaut (talk) 08:12, 4 September 2010 (UTC)[reply]

Go to Start --> Control Panel --> Scanners and Cameras --> Add Device --> Xerox --> Next. Then, right-click on the scanner and choose Properties --> Xerox Settings --> Add and then enter the IP address of the scanner. After scanning, double-click on the scanner, and WIA should reveal any scanned documents.--Best Dog Ever (talk) 08:24, 4 September 2010 (UTC)[reply]

Thanks, I'll try that. How can I verify whether the Xerox machin supports WIA? And will I need to install the TWAIN scan driver, or does WIA replace that functionality? Astronaut (talk) 08:52, 4 September 2010 (UTC)[reply]

Not a problem. WIA is a replacement for TWAIN. TWAIN gives you more control over the scanner, but always requires the installation of software from the manufacturer to function. TWAIN is also more tempermental. So, I always use WIA, simply because it's easier to set up. By the way, if the generic Xerox driver mentioned above doesn't work, click the "Have disk" button instead of "Next" and then point it to drivers on any disks that came with the scanner.--Best Dog Ever (talk) 09:04, 4 September 2010 (UTC)[reply]

I also work for a charity with a Xerox scanner/printer :-). In the case of our printer, the scanner does not work unless you pay a Xerox agent to enable it. At that point it will scan to a network - usually to a drive on the server. --Phil Holmes (talk) 11:19, 4 September 2010 (UTC)[reply]

Check the documentation at http://www.office.xerox.com/support/dctips/dctips.html. Looks like you need to run [[Internet Information Services|IIS]. ---— Gadget850 (Ed) ^talk 11:35, 4 September 2010 (UTC)[reply]

Extracting a date from a serial number in MS Excel[edit]

The number is in the form YYMMDDXXXXXXX. How do I extract the first 6 digits and reformat into an actual date? Roger (talk) 18:08, 4 September 2010 (UTC)[reply]

You can use LEFT(...,6) to extract the first six characters from an input string. You can similarly use MID(...) to extract any substring. Use these to extract the 2-character year, month, and date; and then feed them into one of the date and time functions to produce a date in the form of your choice (as a numeric value, or as a formatted string, or as a DATE object, and so on.

=CONCATENATE(MID($A$1, 3,2), "/", MID($A$1, 5,2), "/" MID($A$1, 1,2))

... might do the trick for you (where $A$1 is the source cell). Nimur (talk) 18:35, 4 September 2010 (UTC)[reply]

Perfect solution! Thanks. Roger (talk) 19:15, 4 September 2010 (UTC)[reply]

Just wanted to add a thanks from my side, I was thinking about this exact problem just earlier this week. I needed to extract dates of birth from I.D. numbers in our membership database. Thanks! Zunaid 08:31, 5 September 2010 (UTC)[reply]

This assumes the US-style mm/dd/yy format for the date and could go horribly wrong if the user's regional settings have dd/mm/yy. A more culture-independent version is =DATE(LEFT(A1,2)+2000, MID(A1,3,2), MID(A1,5,2)). If the date could be earlier than 2000 then the first parameter should be something like LEFT(A1,2)+ IF(VALUE(LEFT(A1,2))>50,1900,2000). AndrewWTaylor (talk) 11:01, 5 September 2010 (UTC)[reply]