I am currently working on simulating a cloth like material and then displaying the results via Open Scene Graph.
I've gotten the setup to display something cloth like, by just dumping all the vertices into 1 Vec3Array and then displaying them with a standard Point based DrawArrays. However I am looking into adding the faces between the vertices so that a further part of my application can visually see the cloth.
This is currently what I am attempting as for the PrimitiveSet
// create and add a DrawArray Primitive (see include/osg/Primitive). The first
// parameter passed to the DrawArrays constructor is the Primitive::Mode which
// in this case is POINTS (which has the same value GL_POINTS), the second
// parameter is the index position into the vertex array of the first point
// to draw, and the third parameter is the number of points to draw.
unsigned int k = CLOTH_SIZE_X;
unsigned int n = CLOTH_SIZE_Y;
osg::ref_ptr<osg::DrawElementsUInt> indices = new osg::DrawElementsUInt(GL_QUADS, (k) * (n));
for (uint y_i = 0; y_i < n - 1; y_i++) {
for (uint x_i = 0; x_i < k - 1; x_i++) {
(*indices)[y_i * k + x_i] = y_i * k + x_i;
(*indices)[y_i * (k + 1) + x_i] = y_i * (k + 1) + x_i;
(*indices)[y_i * (k + 1) + x_i + 1] = y_i * (k + 1) + x_i + 1;
(*indices)[y_i * k + x_i] = y_i * k + x_i + 1;
}
}
geom->addPrimitiveSet(indices.get());
This does however cause memory corruption when running, and I am not fluent enough in Assembly code to decipher what it is trying to do wrong when CLion gives me the disassembled code.
My thought was that I would iterate over each of the faces of my cloth and then select the 4 indices of the vertices that belong to it. The vertices are inputted from top left to bottom right in order. So:
0 1 2 3 ... k-1
k k+1 k+2 k+3 ... 2k-1
2k 2k+1 2k+2 2k+3 ... 3k-1
...
Has anyone come across this specific use-case before and does he/she perhaps have a solution for my problem? Any help would be greatly appreciated.
You might want to look into using DrawArrays with QUAD_STRIP (or TRIANGLE_STRIP because quads are frowned upon these days). There's an example here:
http://openscenegraph.sourceforge.net/documentation/OpenSceneGraph/examples/osggeometry/osggeometry.cpp
It's slightly less efficient than Elements/indices, but it's also less complicated to manage the relationship between the two related containers (the vertices and the indices).
If you really want to do the Elements/indices route, we'd probably need to see more repro code to see what's going on.
My problem is simple.
I am searching a mathematical function to distribute number over an interval.
For example I have this list :
[2; 4; 9; 14]
And in my case I wish
2 -> 1 = f(2)
14 -> 20 = f(14)
4 -> f(4) = ?
9 -> f(9) = ?
This is just an example I am searching for f(x).
Someone would have any idea ?
Thanks for advance ! :)
If you want a linear function, then:
f(x) = lowerFunc + (x - lowerX) * (upperFunc - lowerFunc) / (upperX - lowerX),
where:
lowerFunc: function value at the lower end
upperFunc: function value at the upper end
lowerX: x parameter at the lower end
upperX: x parameter at the upper end.
For your example:
f(x) = 1 + (x - 2) * (20 - 1) / (14 - 2)
= 1 + (x - 2) * 19/12
f(2) = 1
f(4) = 4.1666
f(9) = 12.08333
f(14) = 20
I have following alghoritm to find line intersection of two planes:
public static function getIntersectOf2Planes ( self $P1 , self $P2 )
{
/* Line equation in parametric form:
x = x0 + t*a
y = y0 + t*b
z = z0 + t*c
*/
$x0 = ( $P1->B * $P2->D - $P2->B * $P1->D ) / ( $P1->A * $P2->B - $P2->A * $P1->B ) ;
$a = ( $P1->B * $P2->C - $P2->B * $P1->C );
$y0 = ( $P2->A * $P1->D - $P1->A * $P2->D ) / ( $P1->A * $P2->B - $P2->A * $P1->B ) ;
$b = ( $P2->A * $P1->C - $P1->A * $P2->C );
$z0 = 0;
$c = ( $P1->A * $P2->B - $P2->A * $P1->B );
$IntersectionLine = new Line3D( $x0, $a, $y0, $b, $z0, $c );
return $IntersectionLine;
}
and it works fine, but to compute $x0 and $y0 i have to divide by:
( $P1->A * $P2->B - $P2->A * $P1->B )
and in some cases, the value of this expression is equal to zero, so I get an "dividing by zero" error :(
What should I do in this case?
I know, that the case when this expression is equal to zero, doesn't mean that there is no intersection, because it's happen when I have planes perpendicular to one of the axies.
For example for:
Plane1:
A = 0
B = 0
C = 100
D = 0
Plane2:
A = 50
B = 0
C = 0
D = -250
so the equation of line should exists.
PS I wrote this code with a view to:
https://math.stackexchange.com/questions/2766615/line-by-two-planes-intersection?noredirect=1#comment5706281_2766615
In short, you have to implement the intersection algorithm for the case when (a1*b2 - a2*b1) = 0 (ie. when the planes are not independent when you set z=0).
To expand on that, first we need to understand how you got this far. First let us write down the equation of two planes:
a1x + b1y + c1z + d1 = 0
and
a2x + b2y + c2z + d2 = 0
When two planes intersect, the intersection is a line. So, the most usual way to solve that is by finding a point first on such a line and then figuring out its orientation (a, b, c) in your case. The orientation is a straight forward cross product. The intersection point is typically calculated by setting one of the co-ordinates to 0 and then solving the 2 linear equations that you get. In your code, this is done by setting:
z = 0
But this only works when the equations
a1x + b1y + d1 = 0 and a2x + b2y + d2 = 0
are capable of giving a solution for x and y, which is not the case when a1b2-a2b1=0. So In such cases, you can solve the same by setting either x or y to 0 which again gives two linear equations that you can solve to obtain a point on the line. Then you can compute the parametric form much like how you did. For example (setting y to 0):
x0 = (c1d2 - c2d1)/(a1c2 - a2c1)
y0 = 0
z0 = (a2d1 - a1d2)/(a1c2 - a2c1)
But for this to be a defined value you need to have (a1c2 - a2c1) to be non-zero. Does this help?
Here's the link to the question..
http://www.codechef.com/problems/J7
I figured out that 2 edges have to be equal in order to give the maximum volume, and then used x, x, a*x as the lengths of the three edges to write the equations -
4*x + 4*x + 4*a*x = P (perimeter) and,
2*x^2 + 4*(a*x *x) = S (total area of the box)
so from the first equation I got x in terms of P and a, and then substituted it in the second equation and then got a quadratic equation with the unknown being a. and then I used the greater root of a and got x.
But this method seems to be giving the wrong answer! :|
I know that there isn't any logical error in this. Maybe some formatting error?
Here's the main code that I've written :
{
public static void main(String[] args)
{
TheBestBox box = new TheBestBox();
reader = box.new InputReader(System.in);
writer = box.new OutputWriter(System.out);
getAttributes();
writer.flush();
reader.close();
writer.close();
}
public static void getAttributes()
{
t = reader.nextInt(); // t is the number of test cases in the question
for (int i = 0; i < t; i++)
{
p = reader.nextInt(); // p is the perimeter given as input
area = reader.nextInt(); // area of the whole sheet, given as input
a = findRoot(); // the fraction by which the third side differs by the first two
side = (double) p / (4 * (2 + a)); // length of the first and the second sides (equal)
height = a * side; // assuming that the base is a square, the height has to be the side which differs from the other two
// writer.println(side * side * height);
// System.out.printf("%.2f\n", (side * side * height));
writer.println(String.format("%.2f", (side * side * height))); // just printing out the final answer
}
}
public static double findRoot() // the method to find the 2 possible fractions by which the height can differ from the other two sides and return the bigger one of them
{
double a32, b, discriminant, root1, root2;
a32 = 32 * area - p * p;
b = 32 * area - 2 * p * p;
discriminant = Math.sqrt(b * b - 4 * 8 * area * a32);
double temp;
temp = 2 * 8 * area;
root1 = (- b + discriminant) / temp;
root2 = (- b - discriminant) / temp;
return Math.max(root1, root2);
}
}
could someone please help me out with this? Thank You. :)
I also got stuck in this question and realized that can be done by making equation of V(volume) in terms of one side say 'l' and using differentiation to find maximum volume in terms of any one side 'l'.
So, equations are like this :-
P = 4(l+b+h);
S = 2(l*b+b*h+l*h);
V = l*b*h;
so equation in l for V = (l^3) - (l^2)P/4 + lS/2 -------equ(1)
After differentiation we get:-
d(V)/d(l) = 3*(l^2) - l*P/2 + S/2;
to get max V we need to equate above equation to zero(0) and get the value of l.
So, solutions to a quadratic equation will be:-
l = ( P + sqrt((P^2)-24S) ) / 24;
so substitute this l in equation(1) to get max volume.
Having searched the web, I see various people in various forums alluding to approximating a cubic curve with a quadratic one. But I can't find the formula.
What I want is this:
input: startX, startY, control1X, control1Y, control2X, control2Y, endX, endY
output: startX, startY, controlX, controlY, endX, endY
Actually, since the starting and ending points will be the same, all I really need is...
input: startX, startY, control1X, control1Y, control2X, control2Y, endX, endY
output: controlX, controlY
As mentioned, going from 4 control points to 3 is normally going to be an approximation. There's only one case where it will be exact - when the cubic bezier curve is actually a degree-elevated quadratic bezier curve.
You can use the degree elevation equations to come up with an approximation. It's simple, and the results are usually pretty good.
Let's call the control points of the cubic Q0..Q3 and the control points of the quadratic P0..P2. Then for degree elevation, the equations are:
Q0 = P0
Q1 = 1/3 P0 + 2/3 P1
Q2 = 2/3 P1 + 1/3 P2
Q3 = P2
In your case you have Q0..Q3 and you're solving for P0..P2. There are two ways to compute P1 from the equations above:
P1 = 3/2 Q1 - 1/2 Q0
P1 = 3/2 Q2 - 1/2 Q3
If this is a degree-elevated cubic, then both equations will give the same answer for P1. Since it's likely not, your best bet is to average them. So,
P1 = -1/4 Q0 + 3/4 Q1 + 3/4 Q2 - 1/4 Q3
To translate to your terms:
controlX = -0.25*startX + .75*control1X + .75*control2X -0.25*endX
Y is computed similarly - the dimensions are independent, so this works for 3d (or n-d).
This will be an approximation. If you need a better approximation, one way to get it is by subdividing the initial cubic using the deCastlejau algorithm, and then degree-reduce each segment. If you need better continuity, there are other approximation methods that are less quick and dirty.
The cubic can have loops and cusps, which quadratic cannot have. This means that there are not simple solutions nearly never. If cubic is already a quadratic, then the simple solution exists. Normally you have to divide cubic to parts that are quadratics. And you have to decide what are the critical points for subdividing.
http://fontforge.org/bezier.html#ps2ttf says:
"Other sources I have read on the net suggest checking the cubic spline for points of inflection (which quadratic splines cannot have) and forcing breaks there. To my eye this actually makes the result worse, it uses more points and the approximation does not look as close as it does when ignoring the points of inflection. So I ignore them."
This is true, the inflection points (second derivatives of cubic) are not enough. But if you take into account also local extremes (min, max) which are the first derivatives of cubic function, and force breaks on those all, then the sub curves are all quadratic and can be presented by quadratics.
I tested the below functions, they work as expected (find all critical points of cubic and divides the cubic to down-elevated cubics). When those sub curves are drawn, the curve is exactly the same as original cubic, but for some reason, when sub curves are drawn as quadratics, the result is nearly right, but not exactly.
So this answer is not for strict help for the problem, but those functions provide a starting point for cubic to quadratic conversion.
To find both local extremes and inflection points, the following get_t_values_of_critical_points() should provide them. The
function compare_num(a,b) {
if (a < b) return -1;
if (a > b) return 1;
return 0;
}
function find_inflection_points(p1x,p1y,p2x,p2y,p3x,p3y,p4x,p4y)
{
var ax = -p1x + 3*p2x - 3*p3x + p4x;
var bx = 3*p1x - 6*p2x + 3*p3x;
var cx = -3*p1x + 3*p2x;
var ay = -p1y + 3*p2y - 3*p3y + p4y;
var by = 3*p1y - 6*p2y + 3*p3y;
var cy = -3*p1y + 3*p2y;
var a = 3*(ay*bx-ax*by);
var b = 3*(ay*cx-ax*cy);
var c = by*cx-bx*cy;
var r2 = b*b - 4*a*c;
var firstIfp = 0;
var secondIfp = 0;
if (r2>=0 && a!==0)
{
var r = Math.sqrt(r2);
firstIfp = (-b + r) / (2*a);
secondIfp = (-b - r) / (2*a);
if ((firstIfp>0 && firstIfp<1) && (secondIfp>0 && secondIfp<1))
{
if (firstIfp>secondIfp)
{
var tmp = firstIfp;
firstIfp = secondIfp;
secondIfp = tmp;
}
if (secondIfp-firstIfp >0.00001)
return [firstIfp, secondIfp];
else return [firstIfp];
}
else if (firstIfp>0 && firstIfp<1)
return [firstIfp];
else if (secondIfp>0 && secondIfp<1)
{
firstIfp = secondIfp;
return [firstIfp];
}
return [];
}
else return [];
}
function get_t_values_of_critical_points(p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y) {
var a = (c2x - 2 * c1x + p1x) - (p2x - 2 * c2x + c1x),
b = 2 * (c1x - p1x) - 2 * (c2x - c1x),
c = p1x - c1x,
t1 = (-b + Math.sqrt(b * b - 4 * a * c)) / 2 / a,
t2 = (-b - Math.sqrt(b * b - 4 * a * c)) / 2 / a,
tvalues=[];
Math.abs(t1) > "1e12" && (t1 = 0.5);
Math.abs(t2) > "1e12" && (t2 = 0.5);
if (t1 >= 0 && t1 <= 1 && tvalues.indexOf(t1)==-1) tvalues.push(t1)
if (t2 >= 0 && t2 <= 1 && tvalues.indexOf(t2)==-1) tvalues.push(t2);
a = (c2y - 2 * c1y + p1y) - (p2y - 2 * c2y + c1y);
b = 2 * (c1y - p1y) - 2 * (c2y - c1y);
c = p1y - c1y;
t1 = (-b + Math.sqrt(b * b - 4 * a * c)) / 2 / a;
t2 = (-b - Math.sqrt(b * b - 4 * a * c)) / 2 / a;
Math.abs(t1) > "1e12" && (t1 = 0.5);
Math.abs(t2) > "1e12" && (t2 = 0.5);
if (t1 >= 0 && t1 <= 1 && tvalues.indexOf(t1)==-1) tvalues.push(t1);
if (t2 >= 0 && t2 <= 1 && tvalues.indexOf(t2)==-1) tvalues.push(t2);
var inflectionpoints = find_inflection_points(p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y);
if (inflectionpoints[0]) tvalues.push(inflectionpoints[0]);
if (inflectionpoints[1]) tvalues.push(inflectionpoints[1]);
tvalues.sort(compare_num);
return tvalues;
};
And when you have those critical t values (which are from range 0-1), you can divide the cubic to parts:
function CPoint()
{
var arg = arguments;
if (arg.length==1)
{
this.X = arg[0].X;
this.Y = arg[0].Y;
}
else if (arg.length==2)
{
this.X = arg[0];
this.Y = arg[1];
}
}
function subdivide_cubic_to_cubics()
{
var arg = arguments;
if (arg.length!=9) return [];
var m_p1 = {X:arg[0], Y:arg[1]};
var m_p2 = {X:arg[2], Y:arg[3]};
var m_p3 = {X:arg[4], Y:arg[5]};
var m_p4 = {X:arg[6], Y:arg[7]};
var t = arg[8];
var p1p = new CPoint(m_p1.X + (m_p2.X - m_p1.X) * t,
m_p1.Y + (m_p2.Y - m_p1.Y) * t);
var p2p = new CPoint(m_p2.X + (m_p3.X - m_p2.X) * t,
m_p2.Y + (m_p3.Y - m_p2.Y) * t);
var p3p = new CPoint(m_p3.X + (m_p4.X - m_p3.X) * t,
m_p3.Y + (m_p4.Y - m_p3.Y) * t);
var p1d = new CPoint(p1p.X + (p2p.X - p1p.X) * t,
p1p.Y + (p2p.Y - p1p.Y) * t);
var p2d = new CPoint(p2p.X + (p3p.X - p2p.X) * t,
p2p.Y + (p3p.Y - p2p.Y) * t);
var p1t = new CPoint(p1d.X + (p2d.X - p1d.X) * t,
p1d.Y + (p2d.Y - p1d.Y) * t);
return [[m_p1.X, m_p1.Y, p1p.X, p1p.Y, p1d.X, p1d.Y, p1t.X, p1t.Y],
[p1t.X, p1t.Y, p2d.X, p2d.Y, p3p.X, p3p.Y, m_p4.X, m_p4.Y]];
}
subdivide_cubic_to_cubics() in above code divides an original cubic curve to two parts by the value t. Because get_t_values_of_critical_points() returns t values as an array sorted by t value, you can easily traverse all t values and get the corresponding sub curve. When you have those divided curves, you have to divide the 2nd sub curve by the next t value.
When all splitting is proceeded, you have the control points of all sub curves. Now there are left only the cubic control point conversion to quadratic. Because all sub curves are now down-elevated cubics, the corresponding quadratic control points are easy to calculate. The first and last of quadratic control points are the same as cubic's (sub curve) first and last control point and the middle one is found in the point, where lines P1-P2 and P4-P3 crosses.
Conventions/terminology
Cubic defined by: P1/2 - anchor points, C1/C2 control points
|x| is the euclidean norm of x
mid-point approx of cubic: a quad that shares the same anchors with the cubic and has the control point at C = (3·C2 - P2 + 3·C1 - P1)/4
Algorithm
pick an absolute precision (prec)
Compute the Tdiv as the root of (cubic) equation sqrt(3)/18 · |P2 - 3·C2 + 3·C1 - P1|/2 · Tdiv ^ 3 = prec
if Tdiv < 0.5 divide the cubic at Tdiv. First segment [0..Tdiv] can be approximated with by a quadratic, with a defect less than prec, by the mid-point approximation. Repeat from step 2 with the second resulted segment (corresponding to 1-Tdiv)
0.5<=Tdiv<1 - simply divide the cubic in two. The two halves can be approximated by the mid-point approximation
Tdiv>=1 - the entire cubic can be approximated by the mid-point approximation
The "magic formula" at step 2 is demonstrated (with interactive examples) on this page.
Another derivation of tfinniga's answer:
First see Wikipedia Bezier curve
for the formulas for quadratic and cubic Bezier curves (also nice animations):
Q(t) = (1-t)^2 P0 + 2 (1-t) t Q + t^2 P3
P(t) + (1-t)^3 P0 + 3 (1-t)^2 t P1 + 3 (1-t) t^2 P2 + t^3 P3
Require these to match at the middle, t = 1/2:
(P0 + 2 Q + P3) / 4 = (P0 + 3 P1 + 3 P2 + P3) / 8
=> Q = P1 + P2 - (P0 + P1 + P2 + P3) / 4
(Q written like this has a geometric interpretation:
Pmid = middle of P0 P1 P2 P3
P12mid = midway between P1 and P2
draw a line from Pmid to P12mid, and that far again: you're at Q.
Hope this makes sense -- draw a couple of examples.)
In general, you'll have to use multiple quadratic curves - many cases of cubic curves can't be even vaguely approximated with a single quadratic curve.
There is a good article discussing the problem, and a number of ways to solve it, at http://www.timotheegroleau.com/Flash/articles/cubic_bezier_in_flash.htm (including interactive demonstrations).
I should note that Adrian's solution is great for single cubics, but when the cubics are segments of a smooth cubic spline, then using his midpoint approximation method causes slope continuity at the nodes of the segments to be lost. So the method described at http://fontforge.org/bezier.html#ps2ttf is much better if you are working with font glyphs or for any other reason you want to retain the smoothness of the curve (which is most probably the case).
Even though this is an old question, many people like me will see it in search results, so I'm posting this here.
I would probably draw a series of curves instead of trying to draw one curve using a different alg. Sort of like drawing two half circles to make up a whole circle.
Try looking for opensource Postcript font to Truetype font converters. I'm sure they have it. Postscript uses cubic bezier curves, whereas Truetype uses quadratic bezier curves. Good luck.