Mail Archives: djgpp/1998/10/17/20:13:30.1
Message-ID: | <007c01bdfa2c$81be9da0$827d14cb@dragon>
|
From: | "James Takac" <pdragon AT jtn DOT net DOT au>
|
To: | <djgpp AT delorie DOT com>
|
Subject: | Re: 3d sphere
|
Date: | Sun, 18 Oct 1998 10:10:43 +1000
|
MIME-Version: | 1.0
|
X-Priority: | 3
|
X-MSMail-Priority: | Normal
|
X-Mailer: | Microsoft Outlook Express 4.72.3110.5
|
X-MimeOLE: | Produced By Microsoft MimeOLE V4.72.3110.3
|
X-MIME-Autoconverted: | from 8bit to quoted-printable by icc.net.au id KAA06640
|
X-MIME-Autoconverted: | from quoted-printable to 8bit by delorie.com id UAA29274
|
Reply-To: | djgpp AT delorie DOT com
|
actually you only need to calculate 1/8th of the circle and plot both
positive and negative values of the coords
-----Original Message-----
From: Arthur <arfa AT clara DOT net>
To: djgpp AT delorie DOT com <djgpp AT delorie DOT com>
Date: Sunday, October 18, 1998 5:28 AM
Subject: RE: 3d sphere
>> If you are not speed dependen, make two loops:
>> loop b=0° to 180°
>> loop a=0° to 180°
>> ribbon=cos(b)
>> offset in ribbon=cos(a)
>> screenpos= (a,b)
>>
>> offcourse you have to multiply a and b with different konstants,
>> bependin on how many ribbons you have, how big they are and the size of
>> the sphere on screen.
>>
>> cos(a) and cos(b) witt ossilate between -1 and 1 so if you want a screen
>> x-size of 100, make screen_x=cos(a)*50+50;
>>
>> Hop it helped(and hope it works, BTW, ordinary sin() and cos() works
>> with radians, radians=degrees*pi/180 so use sin(a*3.1415/180) etc).
>
>Remember that for a circle, you only need to calculate one quarter of it,
>since the other four quarters are simply mirror images of the first quater.
>
>James Arthur
>jaa AT arfa DOT clara DOT net
>ICQ#15054819
>
- Raw text -