-
Yes£¬u are right!
It is not a good way to solve the problem. I will try to find a better method.
-
well done!
the follow is my friend's work,it's welldone,just waiting for his final goods. the author is 8J from http://flash.ting365.com/bbs/showthr...threadid=17309
8J's goods
[Edited by qingdiao on 10-09-2000 at 07:48 AM]
-
try this
Hey guys, I've been watching your posts & have been trying to modify these files you've posted... To speed up the most recently posted file, you can
1) Toggle high quality (either medium or low)
2)set array to (4) it does make a difference on my machine
3)and make the object a flash object/drawing not a bmp till the code is ironed out.
I'm working on cleaning up the code... I'll post the files when I do.
peace.
x2fr
-
Double Mask?
Next build is getting there! I've got chineseflag displaying in the (0-3) polygon, although it's skewed a bit despite sharing coordinates with the line MCs in the array. I haven't tried any graphics optimizing yet, but I thought I should post this to show where I'm going with it.
I've also created 'flagb' by duplicating flag and setting canflag in place of chineseflag in a new mask MC, then aligning it with array (4-7) coordinates.
For some reason flagb doesn't get recognized by the interpreter ... is it that you can only display one mask at a time perhaps?
http://members.home.net/wright.r/Flash5/line3d_4a16.swf
And the .fla:
http://members.home.net/wright.r/Flash5/line3d_4a16.fla
Here's the file with flag MC unmasked for a wild 3d-skewing bitmap:
http://members.home.net/wright.r/Flash5/line3d_4a19.swf
Richard
[Edited by Dickee on 10-10-2000 at 09:26 AM]
-
cool!
My friend eastecho had find a better way, Please see this:
http://www.flashempire.com/flash/3d.html
That is what we need, I think.
-
how hard that I can login here these days,the net have something wrong.
myfriend eastecho worked well,that is what we need.
see my new thread
http://flash.ting365.com/bbs/showthr...threadid=17690
and this:eastecho's thread
http://flash.ting365.com/bbs/showthr...threadid=17581
-
I managed to get the download for eastecho's new file ... from ... ±ß³ÇÀË×Ó (Jimbob's) post. But I can't view the pictest4.swf or download the .fla for yours ... all I get is text of the file with right or left click! Aaargh! Oh well, I have the graphic and code from the thread ... thankyou!
If you can, please post the links directly ... it's a little difficult to follow Chinaren-speak ... although I downloaded the fonts when prompted from the 1st posting on this thread! Could you point me to a good tut (in English) to learn your language, and also an English link to PRC history, from the Chinaren point-of-view?
Richard
-
fla
hey,Dickee,I guess your mean is to let me give u away to
learn chinese?I have no idea,
please try this ,this software can translate chinese words into english,also english to chinese.
http://download.sina.com.cn/scgi/detail.pl?s_id=4
u can chose a link to download it.
-
please help dickee!!!!!!!!
Could you tell me where I can download the chines text encoding plug in for IE 5 . Everytime install on demand starts it just hangs and doesn't download anything for me!!
It is getting very frustrating and I would also like to know how well that util that eastjim or quingdau posted works. I can't even tell where to click for the downloads without the chinese encoding!!!!!!!
-
virtually
Den's triangle code with a right-triangle bitmap displayed (similar to my original post, only sized to fit ... hmmm, I'm not sure where I got this from now ... is this yours qingdiao?!):
http://ulome.myetang.com/triangles5.swf
..edit..
Very strange ... click on the above link, you get 403 - access forbidden ... copy and paste to address bar without clicking a different link, still 403 ... but click a different link, then copy & paste to address bar, and you'll get the .swf!
qingdiao's download if you missed it:
http://kacaflash.home.chinaren.com/pictest4.fla
eastecho's surface solution:
http://flash.ting365.com/flash/texture_2.zip
And the view:
http://www.flashempire.com/flash/3d.html
As you can see by these screenshots, I'm having the same problem with the forum ... only ascii code for the fonts:
http://members.home.net/wright.r/Fla...mpireForum.JPG
However, I display the font package when I go to his site for the example:
http://members.home.net/wright.r/Fla...pireForum2.JPG
After seeing this prompt many times upon entering other Asian sites, I finally accepted the download for the font package (which is a short form package to keep the size down ... or so it said when downloading). I can't tell you more about your freeze-up problem, other than going to MS Windows downloads for the complete language package for the particular fonts required. I just removed and re-downloaded the Chinese (Simplified) Text Support with no problems.
However, font or no, in future just use your cursor to highlight the links on the page, and find the correct link by reading your taskbar display.
Richard
[Edited by Dickee on 10-11-2000 at 09:11 PM]
-
The man named ±ß³ÇÀË×Ó is eastecho. My name just jimbob. The file was made by eastecho. What I made was Flying Flag.
We had been discussed these question in flashempire forum. And will soon get the answer I think.
-
eastjim
I asked a Chinese co-worker of mine if he could translate for me and he said it was very difficult to translate technical information.
Maybe there is someone in my Flash users group.
regards,
spindoctor
Originally posted by eastjim
The man named ±ß3ÇÀËxÓ is eastecho. My name just jimbob. The file was made by eastecho. What I made was Flying Flag.
We had been discussed these question in flashempire forum. And will soon get the answer I think.
-
Well, I know that Chinese is very hard to learn. I will discuss with eastecho, we will made English version after we find the answer.
Well, I will try to explain some theory about this later, but my English is poor. I hope u can understand me.
-
:)
-
eastjim
I am able to understand you fine. Maybe it is harder to explain the details of programming.
spindoctor
Originally posted by eastjim
Well, I know that Chinese is very hard to learn. I will discuss with eastecho, we will made English version after we find the answer.
Well, I will try to explain some theory about this later, but my English is poor. I hope u can understand me.
-
Let me try :D
First, I must point out that this algorithm was made by 8j, Thanks for his great work.
Well, Let me try to explain it, here we go:
our purpose is to transform a right-angled triangle into any triangle we need. Please look at the following pic.
In the pic, we have the MC --- right-angled triangle A1A2A3 and we must transform into target triangle ABC . we already know that coordinate: A(X1,Y1)B(X2,Y2)C(X3,Y3)
to do this, we have two step work:
the first: transform right-angled triangle A1A2A3 into right-angled triangle AB'C'. That's easy ,just change the _xcalse and _yscale of A1A2A3 then rotate triangle A1A2A3 .
the second: transform right-angled triangle AB'C' into target triangle ABC . we just need to change the _yscale of right-angled triangle AB'C'.
so, We already know the MC A1A2A3 and the coordinate: A(X1,Y1)B(X2,Y2)C(X3,Y3), all we need to compute out is the length of AB' (we can define to a) ,the length of B'C' (we can define to b) and the angle AB'E.
Because right-angled triangle AEB' is similar to AB'D , we get AE/AB'=AB'/AD ,
so a=AB'=SQRT(AE*AD)
we can draw a line BH , make angle BHF a right-angle. so right-angled triangle CBH is similar to right-angled triangle CDF , then FD/BH=CF/CH
==> FD=(CF*BH)/CH
==> AD=FD-FA=(CF*BH)/CH-FA=((Y3-Y1)(X3-X2))/(Y3-Y2)-(X3-X1)
so we get a.
we draw another line B'H' make angle B'H'F a right-angle.
we know angle EAB' = angle B'C'H'
COS(EAB')=AE/AB'
SIN(B'C'H')=B'H'/B'C'
so (AE/AB')^2=1-(B'H'/B'C')^2
so b=B'C'=SQRT((X3-X2)^2/(1-((X2-X1)/a)^2))
Now we get a and b , It is easy to compute out angle AB'E.
That's all, I hope u can understand.
Best regards,
Jimbob
[Edited by eastjim on 10-12-2000 at 12:01 AM]
-
Re: Let me try :D
I appreciate your help very much. Thanks to you and 8j. I can understand what you have written.
Continued success and I look forward to following your other tutorials.
spinDoctor
Originally posted by eastjim
First, I must point out that this algorithm was made by 8j, Thanks for his great work.
Well, Let me try to explain it, here we go:
our purpose is to transform a right-angled triangle into any triangle we need. Please look at the following pic.
In the pic, we have the MC --- right-angled triangle A1A2A3 and we must transform into target triangle ABC . we already know that coordinate: A(X1,Y1)B(X2,Y2)C(X3,Y3)
to do this, we have two step work:
the first: transform right-angled triangle A1A2A3 into right-angled triangle AB'C'. That's easy ,just change the _xcalse and _yscale of A1A2A3 then rotate triangle A1A2A3 .
the second: transform right-angled triangle AB'C' into target triangle ABC . we just need to change the _yscale of right-angled triangle AB'C'.
so, We already know the MC A1A2A3 and the coordinate: A(X1,Y1)B(X2,Y2)C(X3,Y3), all we need to compute out is the length of AB' (we can define to a) ,the length of B'C' (we can define to b) and the angle AB'E.
Because right-angled triangle AEB' is similar to AB'D , we get AE/AB'=AB'/AD ,
so a=AB'=SQRT(AE*AD)
we can draw a line BH , make angle BHF a right-angle. so right-angled triangle CBH is similar to right-angled triangle CDF , then FD/BH=CF/CH
==> FD=(CF*BH)/CH
==> AD=FD-FA=(CF*BH)/CH-FA=((Y3-Y1)(X3-X2))/(Y3-Y2)-(X3-X1)
so we get a.
we draw another line B'H' make angle B'H'F a right-angle.
we know angle EAB' = angle B'C'H'
COS(EAB')=AE/AB'
SIN(B'C'H')=B'H'/B'C'
so (AE/AB')^2=1-(B'H'/B'C')^2
so b=B'C'=SQR((X3-X2)^2/(1-((X2-X1)/a)^2))
Now we get a and b , It is easy to compute out angle AB'E.
That's all, I hope u can understand.
Best regards,
Jimbob
-
Thank you jimbob!
for your great tranlate!
I would like to say that,this is the first part of the arithmetic(hope I used the right word,forgive my poor english),In other case, such as A is on the left of B or on the right of C,you will need to use other ways to compute.
I'm still working on this,I'll write a complete tutorial about the arithmetic as soon as I finish it.
Thank you.
(I use my new name SouthPig here ,haha )
by the way:
Here is the demo of this arithmetic
http://www.flashkit.com/board/showth...threadid=37117
and source code is here:
http://movfx.51.net/show/triangle2.zip
enjoy!
[Edited by SouthPig on 10-11-2000 at 11:44 PM]
-
Welcome 8j !!!
Hehe..... I am waiting for you next work
-
If you can do that...
About a week ago, I came up with a circular nav system which I was hoping to build on so that it would be totally manipulatable. However, I'm having all sorts of problems doing certain things.
At present, I can adjust the x and y axis' and I can adjust the number of balls used, but I have problems with including a z attribute, and with resizing the balls to match. Plus, if it's possible with the way I've approached the problem, I'd like to be able to move the circle in any possible direction, as opposed to just the x and y locked positions.
The URL for the SWF is http://www.design4web.net/nav.html and the code is at http://www.design4web.net/code.html
I hope someone can help!
Posting Permissions
- You may not post new threads
- You may not post replies
- You may not post attachments
- You may not edit your posts
-
Forum Rules
|
Click Here to Expand Forum to Full Width
|