Apparently this is verlet integration:
[source lang="cpp"]void Particle::integrate(float timestep){ Vec2 Temp = P.Position; P.Position += P.Position  P.OldPosition + P.Acceleration*Timestep*Timestep; P.OldPosition = Temp;}[/source]
I got it from this article,
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#4949362 Numerical integration methods
Posted by thecoast47 on 14 June 2012  06:07 PM
#4824860 I give up.. Generating indices for a cylinder?
Posted by thecoast47 on 18 June 2011  01:09 PM
I have a theory ....
here is code for a circle in 2D...
If i were you i would probably try using this to to place points around the Z axis and then increment Y Axis .
If you do this a couple times and then somehow connect the vertices the right way... then you should get a cylinder.
Or if you want to brute force it... you can just connect the subdivisions together by drawing lines....If you draw enough 3D circles translated on the Y axis enough times then you also get a cylinder...
here is code for a circle in 2D...
If i were you i would probably try using this to to place points around the Z axis and then increment Y Axis .
If you do this a couple times and then somehow connect the vertices the right way... then you should get a cylinder.
void RenderCircle(){ float numSubdivisions = 24; glBegin(GL_TRIANGLE_FAN); glColor4f(R, G, B,0); glVertex3d(CENTER.X,CENTER.Y,Depth); // Set edge colour for rest of shape glColor4f(0, 0, 0,0); for (float angle=0; angle<=PI*2; angle+=((PI*2)/numSubdivisions) ) { glVertex3d( radius*(float)cos(angle) + CENTER.X, radius*(float)sin(angle) + CENTER.Y, Depth); } glVertex3d(CENTER.X+radius,CENTER.Y,Depth); glEnd(); }
Or if you want to brute force it... you can just connect the subdivisions together by drawing lines....If you draw enough 3D circles translated on the Y axis enough times then you also get a cylinder...
#4803554 Where do I start?
Posted by thecoast47 on 27 April 2011  08:27 AM
Here are the languages i started with:
visual basic 3
visual basic 6
JavaScript
java (with openGL)
c++(with SDL/OPENGL) <industry standard
Also you need to learn:
MATH:
linear algebra
vectors
Matrix math
trigonometry
trig identities
Programming CONCEPTS:
sorting algorithms
Multi threading
Object orientation
dynamic memory
blitting
buffers
I recommend learning java then c++.
Or if you want to start with c++,just buy a book on C++ and another on SDL
visual basic 3
visual basic 6
JavaScript
java (with openGL)
c++(with SDL/OPENGL) <industry standard
Also you need to learn:
MATH:
linear algebra
vectors
Matrix math
trigonometry
trig identities
Programming CONCEPTS:
sorting algorithms
Multi threading
Object orientation
dynamic memory
blitting
buffers
I recommend learning java then c++.
Or if you want to start with c++,just buy a book on C++ and another on SDL
#4767220 The Specifics of a good Platformer's Physics
Posted by thecoast47 on 30 January 2011  05:45 PM
I've done Line segment collision recently.
My collision footage is here: http://www.youtube.c...h?v=ARYgb_kQqI
All you you have to do is find the slope of the X and Y velocities of your shape and the wrap and invisible line along the tiles that have a slope. That way you can find the point of intersection and check if the player is close or equal to the point of intersection.
I use slope intercept for the equation of the line but you could also use the other form( know its something Like Ax +By +C = 0).
Let the position of the player be (30,60)
Players Line Equation(Y = Mx+
So slope of the player = M
M = Yvel/Xvel
Now we need to find B so all we need to do now is plug in any point that belongs to the slope into the equation.
So 60 = M(30) + B therefore B = Y  (M(X))
SO you have the Y intersection
And now you need X;
Basically Repeat these steps for the invisible Line and then Substitute the Y of the player's line equation into the Equation of the invisible line
SO your going to get Mx + b = Mx + B
and Solve For X
and there you go you have a point of intersection to check collision!
Hope this helped
k.
My collision footage is here: http://www.youtube.c...h?v=ARYgb_kQqI
All you you have to do is find the slope of the X and Y velocities of your shape and the wrap and invisible line along the tiles that have a slope. That way you can find the point of intersection and check if the player is close or equal to the point of intersection.
I use slope intercept for the equation of the line but you could also use the other form( know its something Like Ax +By +C = 0).
Let the position of the player be (30,60)
Players Line Equation(Y = Mx+
So slope of the player = M
M = Yvel/Xvel
Now we need to find B so all we need to do now is plug in any point that belongs to the slope into the equation.
So 60 = M(30) + B therefore B = Y  (M(X))
SO you have the Y intersection
And now you need X;
Basically Repeat these steps for the invisible Line and then Substitute the Y of the player's line equation into the Equation of the invisible line
SO your going to get Mx + b = Mx + B
and Solve For X
and there you go you have a point of intersection to check collision!
Hope this helped
k.