I'm not sure I understood your post. Anyway, what about (10  x)*25.5 ? It is equal to 255 when x is zero and equal to 0 when x is 10.
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apatriarca
Member Since 03 Jul 2006Online Last Active Today, 07:43 AM
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 Group Crossbones+
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 Age 29 years old
 Birthday February 11, 1985

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Torino, Italy
#5140688 Inverse Proportion and Zero
Posted by apatriarca on 20 March 2014  09:57 AM
#5134720 This is what atan2() does if given two Vector2D points on 2D plane?
Posted by apatriarca on 26 February 2014  05:07 AM
There is no need to convert these vectors to angles to compute a rotation matrix. All you need to compute a rotation matrix in 2D are the cosine and sine of the angle and you already have them encoded in the direction of the difference vector. You indeed simply have to normalize the difference vector:
(cos(angle), sin(angle) ) = normalize(P2  P1).
If you think you have to compute the angle to compare them between different fingers positions, then it is not true again. If P3 and P4 are different positions of your fingers, then you can compute the cosine and sine of the rotation angle between the two vectors by simply using a complex division. If do not know complex numbers I suggest to learn about them, but you can implement the division using the following formula:
(a, b) / (c, d) = ( a*c + b*d, b*c  a*d ) / (c*c + d*d).
#5125823 .obj file loader
Posted by apatriarca on 23 January 2014  03:30 AM
Is there any reason you are using (*model).GetIndices() instead of model>GetIndices() ?
#5125637 Barycentric coordinates of circumcenter of tetrahedron
Posted by apatriarca on 22 January 2014  08:10 AM
The steps in the derivation are very similar to the steps in the triangle case and you should get:
Dot(BA,BA)*s + Dot(BA,CA)*t + Dot(BA,DA)*u = (1/2)*Dot(BA,BA)
Dot(CA,BA)*s + Dot(CA,CA)*t + Dot(CA,DA)*u = (1/2)*Dot(CA,CA)
Dot(DA,BA)*s + Dot(DA,CA)*t + Dot(DA,DA)*u = (1/2)*Dot(DA,DA)
which you can then solve using Cramer's rule (or something else).
#5125616 Mathematics for 3D programming  Exercises
Posted by apatriarca on 22 January 2014  05:01 AM
Making a simple 3D game/tech demo or a ray tracer or .. is IMHO probably one of the best way to exercise those math skills.
#5123310 Need a design advice on a Mesh Class
Posted by apatriarca on 13 January 2014  10:53 AM
If Context::CreateBuffer calls Mesh::CreateBuffer you have an infinite loop. This is in my opinion the main reason to avoid doing that.
#5123269 shadows in 2d
Posted by apatriarca on 13 January 2014  07:18 AM
I guess it depends on the type of game you are working on (i.e. where is the camera and the relative positions of lights and objects) but it can be done. For example, this article on gamedev explains a technique to implement soft shadows in 2D.
#5117517 Genetic algorithm in a game
Posted by apatriarca on 17 December 2013  02:44 AM
#5115643 OpenGl as a blitter
Posted by apatriarca on 09 December 2013  07:43 AM
P.S. If you really want to learn about OpenGL, I suggest to start using the modern APIs. The programmable pipeline is now common in the mobile and Web world too.
#5114300 Using one instance of Behaviour Tree for many agents?
Posted by apatriarca on 04 December 2013  07:07 AM
#5114262 Quaternions and matrices
Posted by apatriarca on 04 December 2013  03:14 AM
#5110747 question about quadratic bezier patch
Posted by apatriarca on 20 November 2013  07:12 AM
#5110713 question about quadratic bezier patch
Posted by apatriarca on 20 November 2013  03:02 AM
#5110503 question about quadratic bezier patch
Posted by apatriarca on 19 November 2013  10:51 AM
Let P(i, j) be the control points of the quadratic Bézier surface patches, C(P0, P1, P2, t) represents a quadratic Bézier curve with control points P0, P1 and P2 with parameter t and S(u, v) the quadratic Bézier surface patch equation. We are trying to explains to you that
S(u, 0.7) = C( C(P(0,0), P(0, 1), P(0,2), 0.7) , C(P(1,0), P(1, 1), P(1,2), 0.7) , C(P(2,0), P(2, 1), P(2,2), 0.7) , u).
#5110435 question about quadratic bezier patch
Posted by apatriarca on 19 November 2013  05:45 AM
That's latex code. It look like it is less readable than I assumed. I don't know how to write equations in this forum anymore (the eqn tag does not seem to work and I can't upload images from 3rd party services).
If you fix one of the two coordinates (for example v = 0.7 as in your example), the corresponding basis polynomials B_j(v) becomes constants. You can thus sum the terms P_ij B_j(v) for 0 <= j <= 2 to obtains the control points for a quadratic curve depending only on the parameter u.