Author Topic: Need a volunteer!  (Read 1409 times)

Legacy_ElgarL

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Need a volunteer!
« on: September 22, 2010, 04:03:56 am »


               I'm seeking someone who knows the basics of editing 2da files. But, mostly has knowledge of the PRC classes and is willing to put in a lot of work creating and modifying 2da's. Having an understanding of counting in hex would be an advantage too.

I'm so busy coding a new client (patched) that I've really no time to do the 2da work thats needed to implement it. I'll do it myself in the end, but it just extends deadlines doing everything solo '<img'>.

Please don't ask if your time is severly limited. I'm a retired person so most of my day is spent programming and working on this project.
               
               

               
            

Legacy_Calvinthesneak

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« Reply #1 on: September 22, 2010, 05:26:29 am »


               I know more than basics, I've made custom prestige classes, spells, feats, as well as clothing and such.



This would be to my benefit as the PRC is in use on my world.  I however am a full time programming student.  I can get things done but it's not going to be something I can do all day every day.
               
               

               
            

Legacy_ElgarL

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« Reply #2 on: September 22, 2010, 06:28:48 pm »


               If you are willing look me up in the PRC IRC chat - http://www.nwnprc.co...e.php?page_id=2



I'm always in there. AFK at times, but always reading past messages.
               
               

               
            

Legacy_Calvinthesneak

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« Reply #3 on: September 22, 2010, 09:55:46 pm »


               Wilco.  Not sure where you are in terms of Time Zone, I'm in GMT -8 (Pacific Daylight Savings).  I'll pop in after I finish with my SQL labs for the day.
               
               

               
            

Legacy_ElgarL

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« Reply #4 on: September 22, 2010, 10:25:19 pm »


               GMT here (UK). I'll likely be asleep before I see you today, but I'm normally a night owl '<img'>.
               
               

               
            

Legacy_B_Harrison

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« Reply #5 on: September 23, 2010, 01:24:20 am »


               I'm in the UK also and know 2das rather well, but have too many other current projects to commit much time (and I'm not familiar with hex at all). If you ever need a second helper or backup 2da person though, I'd still be happy to help out.
               
               

               
            

Legacy_Calvinthesneak

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« Reply #6 on: September 23, 2010, 01:43:37 am »


               If you can count in binary Ben, you can do Hex. It's just breaking your binary into four bit groupings and converting.


Lets see if I can damage my brain further today. We'll pick a number.. say 208.  Convert to Binary.  *steam comes out of ears*

1101 0000  (calculator with windows can help with conversion) Take each group of four and convert to a hex number.


Leftmost group is 1101, which converts to 13.  13 Hex is D. 

Rightmost group is 0000 which obviously converts to 0 still...0.

So your Hex humber is D0.
               
               

               


                     Modifié par Calvinthesneak, 23 septembre 2010 - 12:54 .
                     
                  


            

Legacy_Baragg

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« Reply #7 on: September 26, 2010, 10:07:13 pm »


               

Calvinthesneak wrote...

If you can count in binary Ben, you can do Hex. It's just breaking your binary into four bit groupings and converting.


Lets see if I can damage my brain further today. We'll pick a number.. say 208.  Convert to Binary.  *steam comes out of ears*

1101 0000  (calculator with windows can help with conversion) Take each group of four and convert to a hex number.


Leftmost group is 1101, which converts to 13.  13 Hex is D. 

Rightmost group is 0000 which obviously converts to 0 still...0.

So your Hex humber is D0.


Was that my mind that just blew out?...............
               
               

               
            

Legacy_kalbaern

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« Reply #8 on: September 28, 2010, 03:15:55 am »


               Here's a link to a converter I use. I've too few brain cells left to risk an explosion of my own.

http://www.cleaveboo...ol/calnumba.htm

Just type the value under the base 10 block and click calculate. The answer will appear next to the base 16 block.
               
               

               


                     Modifié par kalbaern, 28 septembre 2010 - 02:18 .