Can someone smarter than me take a look at this. I'm trying to implement a Bezier curve algorithm I found here in objective-c.
The output is way wrong. I think I converted the code correctly so either the original was wrong or was not ment to be used like this... If I use the built in NSBezierPath the curve looks great but I can't use the built in NSBezierPath.
NSBezierPath example
NSBezierPath *bezierPath = [[NSBezierPath alloc] init];
[bezierPath setLineWidth:1.0f];
[bezierPath moveToPoint:NSMakePoint(x1, y1)];
[bezierPath curveToPoint:NSMakePoint(x4, y4) controlPoint1:NSMakePoint(x2, y2) controlPoint2:NSMakePoint(x3, y3)];
My code trying to draw a bezier curve
- (void)drawBezierFrom:(NSPoint)from to:(NSPoint)to controlA:(NSPoint)a controlB:(NSPoint)b color:(NSUInteger)color
{
float qx, qy;
float q1, q2, q3, q4;
int plotx, ploty;
float t = 0.0;
while (t <= 1)
{
q1 = t*t*t*-1 + t*t*3 + t*-3 + 1;
q2 = t*t*t*3 + t*t*-6 + t*3;
q3 = t*t*t*-3 + t*t*3;
q4 = t*t*t;
qx = q1*from.x + q2*to.x * q3*a.x + q4*b.x;
qy = q1*from.y + q2*to.y * q3*a.y + q4*b.y;
plotx = round(qx);
ploty = round(qy);
[self drawPixelColor:color atX:plotx y:ploty];
t = t + 0.003;
}
}
Edit
See Bezier curve algorithm in objective-c needs a tweak for a completed functional Bezier Curve method.
I used this Bezier function with my xy plotter and found a small error with the 'to'.
The to.x to.y and b.x b.y need to be switched so that the pen starts at from and ends at to.
qx = q1*from.x + q2*a.x + q3*b.x + q4*to.x;
qy = q1*from.y + q2*a.y + q3*b.y + q4*to.y;
It looks to me like you have the wrong coefficients with each point and that one of your adds became a multiply. I think what you want is this:
qx = q1*from.x + q2*a.x + q3*to.x + q4*b.x;
qy = q1*from.y + q2*a.y + q3*to.y + q4*b.y;
Related
gif
Creating View, Porjection Matrix
void CameraSystem::CreateMatrix()
{
camera->aspect = viewport->Width / viewport->Height;
projection_matrix = XMMatrixPerspectiveFovLH(camera->fov, camera->aspect, camera->near_z, camera->far_z);
XMVECTOR s, o, q, t;
XMFLOAT3 position(camera->position.m128_f32[0], camera->position.m128_f32[1], camera->position.m128_f32[2]);
s = XMVectorReplicate(1.0f);
o = XMVectorSet(0.0f, 0.0f, 0.0f, 1.0f);
q = XMQuaternionRotationRollPitchYaw(camera->pitch, camera->yaw, camera->roll);
t = XMLoadFloat3(&position);
world_matrix = XMMatrixAffineTransformation(s, o, q, t);
view_matrix = XMMatrixInverse(0, world_matrix);
camera->look = XMVector3Normalize(XMMatrixTranspose(view_matrix).r[2]);
camera->right = XMVector3Normalize(XMMatrixTranspose(view_matrix).r[0]);
camera->up = XMVector3Normalize(XMMatrixTranspose(view_matrix).r[1]);
camera->position = world_matrix.r[3];
cb_viewproj.data.view_matrix = XMMatrixTranspose(view_matrix);
cb_viewproj.data.projection_matrix = XMMatrixTranspose(projection_matrix);
}
this code creating projection matrix with aspect, fov, near, far.
and create world, view with camera transform. and these work perfect for rendering but maybe not for create ray.
Creating Mouse Ray
MouseRay CameraSystem::CreateMouseRay()
{
MouseRay mouse_ray;
POINT cursor_pos;
GetCursorPos(&cursor_pos);
ScreenToClient(ENGINE->GetWindowHandle(), &cursor_pos);
// Convert the mouse position to a direction in world space
float mouse_x = static_cast<float>(cursor_pos.x);
float mouse_y = static_cast<float>(cursor_pos.y);
float ndc_x = 2.0f * mouse_x / (float)ENGINE->GetWindowSize().x - 1.0f;
float ndc_y = (2.0f * mouse_y / (float)ENGINE->GetWindowSize().y - 1.0f) * -1.0f;
float ndc_z = 1.0f;
ndc.x = ndc_x;
ndc.y = ndc_y;
XMMATRIX inv_view = XMMatrixInverse(nullptr, view_matrix);
XMMATRIX inv_world = XMMatrixInverse(nullptr, XMMatrixIdentity());
XMVECTOR ray_dir;
XMVECTOR ray_origin;
ndc_x /= projection_matrix.r[0].m128_f32[0];
ndc_y /= projection_matrix.r[1].m128_f32[1];
ray_dir.m128_f32[0] = (ndc_x * inv_view.r[0].m128_f32[0]) + (ndc_y * inv_view.r[1].m128_f32[0]) + (ndc_z * inv_view.r[2].m128_f32[0]) + inv_view.r[0].m128_f32[3];
ray_dir.m128_f32[1] = (ndc_x * inv_view.r[0].m128_f32[1]) + (ndc_y * inv_view.r[1].m128_f32[1]) + (ndc_z * inv_view.r[2].m128_f32[1]) + inv_view.r[1].m128_f32[3];
ray_dir.m128_f32[2] = (ndc_x * inv_view.r[0].m128_f32[2]) + (ndc_y * inv_view.r[1].m128_f32[2]) + (ndc_z * inv_view.r[2].m128_f32[2]) + inv_view.r[2].m128_f32[3];
ray_origin = XMVector3TransformCoord(camera->position, inv_world);
ray_dir = XMVector3TransformNormal(ray_dir, inv_world);
ray_dir = XMVector3Normalize(ray_dir);
XMtoRP(ray_origin, mouse_ray.start_point);
XMtoRP(ray_dir * 1000.f, mouse_ray.end_point);
return mouse_ray;
}
get cursor pos, and converto ndc, it's perpect, i guess 'ndc, porjection calulate' has some propelm, but i couldn't find any other code.
inversed view matrix to ray direction, and i guess this code also has problem.
As you can see, the mouse does not seem to generate a ray in the exact direction of the pointed mouse. The frustum feels narrower than it really is, and the further away from the map, the worse it gets.
It seems as if the camera position is fixed, but the camera position is being updated from the world matrix as shown in the properties window.
The fact that views and projection matrices are also honestly generated can be inferred from the fact that the rest except for raycasting are rendered correctly.
I assume that the ray direction vector is miscalculated.
I don't know what more calculations can be made in the next code.
raycast test by reactphysics3d library.
I have a problem with collision detection of a circle and a rectangle. I have tried to solve the problem with the Pythagorean Theorem. But none of the queries works. The rectangle collides with the rectangular bounding box of the circle.
if (CGRectIntersectsRect(player.frame, visibleEnemy.frame)) {
if (([visibleEnemy spriteTyp] == jumper || [visibleEnemy spriteTyp] == wobble )) {
if ((visibleEnemy.center.x - player.frame.origin.x) * (visibleEnemy.center.x - player.frame.origin.x) +
(visibleEnemy.center.y - player.frame.origin.y) * (visibleEnemy.center.y - player.frame.origin.y) <=
(visibleEnemy.bounds.size.width/2 * visibleEnemy.bounds.size.width/2)) {
NSLog(#"Check 1");
normalAction = NO;
}
if ((visibleEnemy.center.x - (player.frame.origin.x + player.bounds.size.width)) *
(visibleEnemy.center.x - (player.frame.origin.x + player.bounds.size.width)) +
(visibleEnemy.center.y - player.frame.origin.y) * (visibleEnemy.center.y - player.frame.origin.y) <=
(visibleEnemy.bounds.size.width/2 * visibleEnemy.bounds.size.width/2)) {
NSLog(#"Check 2");
normalAction = NO;
}
else {
NSLog(#"Check 3");
normalAction = NO;
}
}
}
Here is how I did it in one of my small gaming projects. It gave me best results and it's simple. My code detects if there is a collision between circle and the line. So you can easily adopt it to circle - rectangle collision detection by checking all 4 edges of the rectangle.
Let's say that a ball has a ballRadius, and location (xBall, yBall). The line is defined with two points (xStart, yStart) and (xEnd, yEnd).
Implementation of a simple collision detection:
float ballRadius = ...;
float x1 = xStart - xBall;
float y1 = yStart - yBall;
float x2 = xEnd - xBall;
float y2 = yEnd - yBall;
float dx = x2 - x1;
float dy = y2 - y1;
float dr = sqrtf(powf(dx, 2) + powf(dy, 2));
float D = x1*y2 - x2*y1;
float delta = powf(ballRadius*0.9,2)*powf(dr,2) - powf(D,2);
if (delta >= 0)
{
// Collision detected
}
If delta is greater than zero there are two intersections between ball (circle) and line. If delta is equal to zero there is one intersection – perfect collision.
I hope it will help you.
I've got two points between which im drawing a line (x1,y1 and x2,y2) but i need to know the coordinates of x3,y3 which is gapSize away from point x2,y2. Any ideas on how to solve this problem (the program is written in objective-c if that is helpful at all)?
You can simply calculate the angle in radians as
double rads = atan2(y2 - y1, x2 - x1);
Then you get the coordinates as follows:
double x3 = x2 + gapSize * cos(rads);
double y3 = y2 + gapSize * sin(rads);
Is this what you meant?
Compute the distance between P1 and P2: d=sqrt( (y2-y1)^2 + (x2-x1)^2)
Then x2 = (d*x1 + gapSize*x3) / (d+gapSize)
So x3 = (x2 * (d+gapSize) - d*x1) / gapSize
Similarly, y3 = (y2 * (d+gapSize) - d*y1) / gapSize
Sorry for the math. I didn't try to code it but it sounds right. I hope this helps.
There are many ways to do this. Simplest (to me) is the following. I'll write it in terms of mathematics since I can't even spell C.
Thus, we wish to find the point C = {x3,y3}, given points A = {x1,y1} and B = {x2,y2}.
The distance between the points is
d = ||B-A|| = sqrt((x2-x1)^2 + (y2-y1)^2)
A unit vector that points along the line is given by
V = (B - A)/d = {(x2 - x1)/d, (y2-y1)/d}
A new point that lies a distance of gapSize away from B, in the direction of that unit vector is
C = B + V*gapSize = {x2 + gapSize*(x2 - x1)/d, y2 + gapSize*(y2 - y1)/d}
I've got two points between which im drawing a line (x1,y1 and x2,y2) but i need to know the coordinates of x3,y3 which is gapSize away from point x2,y2. Any ideas on how to solve this problem (the program is written in objective-c if that is helpful at all)?
You can simply calculate the angle in radians as
double rads = atan2(y2 - y1, x2 - x1);
Then you get the coordinates as follows:
double x3 = x2 + gapSize * cos(rads);
double y3 = y2 + gapSize * sin(rads);
Is this what you meant?
Compute the distance between P1 and P2: d=sqrt( (y2-y1)^2 + (x2-x1)^2)
Then x2 = (d*x1 + gapSize*x3) / (d+gapSize)
So x3 = (x2 * (d+gapSize) - d*x1) / gapSize
Similarly, y3 = (y2 * (d+gapSize) - d*y1) / gapSize
Sorry for the math. I didn't try to code it but it sounds right. I hope this helps.
There are many ways to do this. Simplest (to me) is the following. I'll write it in terms of mathematics since I can't even spell C.
Thus, we wish to find the point C = {x3,y3}, given points A = {x1,y1} and B = {x2,y2}.
The distance between the points is
d = ||B-A|| = sqrt((x2-x1)^2 + (y2-y1)^2)
A unit vector that points along the line is given by
V = (B - A)/d = {(x2 - x1)/d, (y2-y1)/d}
A new point that lies a distance of gapSize away from B, in the direction of that unit vector is
C = B + V*gapSize = {x2 + gapSize*(x2 - x1)/d, y2 + gapSize*(y2 - y1)/d}
Essentially I'm trying to calculate the bisector line between two points. I've got two methods, one works the other doesn't. I can't quite figure out why the other one doesn't work. The one that works is a little more computationally intensive and since this routine is run a lot, I'd like to use the simpler one... except it doesn't work. I'm probably missing something simple but I find this amusing since I seem to have a better grasp of trig than I do of high school algebra.
Note: the function is passed the end points (endPoint1, endPoint2).
Here's the one that works (using trig to calculate the bisector):
CGPoint midPoint = CGPointMake((endPoint1.x + endPoint2.x) / 2, (endPoint1.y + endPoint2.y) / 2);
//Normalize an end point
CGPoint nPoint = CGPointMake(endPoint1.x - endPoint2.x, endPoint1.y - endPoint2.y);
//Find theta and rotate 90°
CGFloat theta = atanf(nPoint.y / nPoint.x);
if (nPoint.x < 0.0f) theta += M_PI;
else if (nPoint.x > 0.0f && nPoint.y < 0.0f) theta += (M_PI * 2);
theta += M_PI_2;
//Calculate another point along new theta and de-normalize the point
CGPoint centerPoint = CGPointMake(cosf(theta) * 10, sinf(theta) * 10);
centerPoint.x += midPoint.x;
centerPoint.y += midPoint.y;
//Create the line definition
LineDef def = LineDefForPoints(midPoint, centerPoint);
Here's the one that doesn't, but I'd like it to:
CGPoint midPoint = CGPointMake((endPoint1.x + endPoint2.x) / 2, (endPoint1.y + endPoint2.y) / 2);
//Calculate the slope and invert
CGFloat m = (endPoint1.y - endPoint2.y) / (endPoint1.x - endPoint2.x);
//Take the negative reciprocal
m = -1/m;
//Calculate another point on the line
CGPoint centerPoint = CGPointMake(midPoint.x + 10, midPoint.y + (m * 10));
//Create the line definition
LineDef def = LineDefForPoints(midPoint, centerPoint);
So I'd swear this should work. The change in Y is equal to m times the change in x. I've calculated the mid point, figured out the slope of the perpendicular line and calculated another point on that line. However the line definitions created aren't equivalent when given the same end points, so I'm missing something.
By the way, LindeDef is a simple struct with three CGFloat variables for the a, b & c components of a straight line. And creating a LineDef from two points is trivial (I happen to be using a block to do this):
LineDef (^LineDefForPoints)(CGPoint, CGPoint) = ^LineDef(CGPoint p1, CGPoint p2){
LineDef line = {0,0,0};
line.a = p2.y - p1.y;
line.b = p1.x - p2.x;
line.c = line.a*p1.x + line.b*p1.y;
return line;
};
Slope-intercept form is fragile for this; use vectors.
〈V〉 = B - A
midpoint = 〈V〉/2 + A
⟂〈V〉 = 〈Vy, -Vx〉
Dammit, it was something simple. I reformatted my actual production code so I could put it on Stack Exchange and in the process edited out my mistake. The code I posted in the question actually works perfectly. Originally the line: m = -1 / m; was folded into the original assignment like so: CGFloat m = -1 / (endPoint1.y - endPoint2.y) / (endPoint1.x - endPoint2.x);. Of course, now the problem is obvious...I forgot parenthesis. I separated the line into two in stack exchange so I could explain my 'reasoning' in the comments, hoping to find the problem.
Sorry, for all the trouble.