View Full Version : help in drawing object's face (new)

I wish to draw this polygon with 6 coodinates (front face) which are (-x,y,1),(-x,-y,1),(0,-y,1),(0,y-d,1),(-w,y-d,1),

(-w-d,y,1). The arc is between the last 2 coodinates with an angle of 90deg. And d<w.

I have problems with the arc part, below is what i did:

glBegin(GL_POLYGON);

glNormal3f( 0.0f, 0.0f, 1.0f);

glVertex3f(-x, y, 1.0f);

glVertex3f(-x,-y, 1.0f);

glVertex3f(0,-y, 1.0f);

glVertex3f(0,y-d,1.0f);

float arcx = d*cos(i);

float arcy = d*sin(i);

for (i=0;i<= pi/2; i++){

glVertex3f(arcx,arcy,1.0f);

}

glVertex3f(-w-d, y, 1.0f);

glEnd();

think this is nt very correct. Can someone help?

[This message has been edited by coda (edited 01-10-2004).]

ZbuffeR

01-10-2004, 02:43 AM

float arcx = d*cos(i);

float arcy = d*sin(i);

for (i=0;i<= pi/2; i++){

glVertex3f(arcx,arcy,1.0f);

}

Did you really expect that arcx would change at all during the loop ??? Learn C before OpenGL...

Do this :

float x=1.0f;

float y=1.0f;

float d=0.5f;

int SUBDIVS=10;

glBegin(GL_POLYGON);

glNormal3f( 0.0f, 0.0f, 1.0f);

glVertex3f(-x+d, d, 1.0f);

glVertex3f(-x+d,-y-d, 1.0f);

glVertex3f(d,-y-d, 1.0f);

glVertex3f(d,0,1.0f);

float arcx;

float arcy;

for (i=0;i<= SUBDIVS; i++) {

arcx = d*cos(i*M_PI_2/SUBDIVS);

arcy = d*sin(i*M_PI_2/SUBDIVS);

glVertex3f(arcx,arcy,1.0f);

}

glEnd();

thanx for your help. The arc is drawn but my original intention is to draw the arc in the inverted way i.e mirror image along the line y=x such that the rect look "chipped" off.. i tried changing the angle but turned out funny. Also, d is radius of arc but I still cannot figure out why you use those coodinates in the vertices (before the for loop)

[This message has been edited by coda (edited 01-10-2004).]

ZbuffeR

01-10-2004, 05:19 PM

>>>>The arc is drawn but my original intention is to draw the arc in the inverted way i.e mirror image along the line y=x such that the rect look "chipped" off.. i tried changing the angle but turned out funny.<<<<

Not very clear. Mirror along y=x would change nothing for the circular part. If you meant something else, you can try it out by offsetting the angle by M_PI and moving the origin.

>>>>Also, d is radius of arc but I still cannot figure out why you use those coodinates in the vertices (before the for loop)<<<<

Come on, it is not so hard. Plot the coords by hand, on a paper, and you will understand http://www.opengl.org/discussion_boards/ubb/tongue.gif

okie i think i figured it out, your origin is at the rounded corner, distance d from the edge? hmm now my problem is how to get the arc to look inverted- much like a arc in the 3rd quadrant of a circle.

ZbuffeR

01-11-2004, 03:21 AM

float x=1.0f;

float y=1.3f;

float d=0.5f;

int SUBDIVS=10;

// glBegin(GL_POLYGON);

glBegin(GL_TRIANGLE_FAN);

// glBegin(GL_LINE_LOOP); // use this to debug your drawing

glNormal3f( 0.0f, 0.0f, 1.0f);

glVertex3f(-x,-y, 1.0f);

glVertex3f(-x, 0, 1.0f);

float arcx;

float arcy;

for (i=0;i<= SUBDIVS; i++) {

arcx = d*cos(i*M_PI_2/SUBDIVS+M_PI);

arcy = d*sin(i*M_PI_2/SUBDIVS+M_PI);

glVertex3f(arcx,arcy,1.0f);

}

glVertex3f(0,-y, 1.0f);

glEnd();

I have nothing to do, so...

I believe that you got 'funny results' because the polygon is no more convex. As it is ultimately tesselated into triangles by the GL renderer, it is better to use 'TRIANGLE_FAN' (read the docs), as behavior is specified. Start at the lower left corner to avoid concavity problems.

And read the docs ! The red book is avalable online, check the opengl main page.

I finally can draw the arc in the inverted way by changing the angle and using for (i=10;i>=0;i--)....etc . thanx for your help!

Powered by vBulletin® Version 4.2.5 Copyright © 2018 vBulletin Solutions Inc. All rights reserved.