How does a 3D model handled unit wise ?
When i have a random model that i want to fit in my view port i dunno if it is too big or not, if i need to translate it to be in the middle...
I think a 3d object might have it's own origine.
You need to find a bounding volume, a shape that encloses all the object's vertices, for your object that is easier to work with than the object itself. Spheres are often used for this. Either the artist can define the sphere as part of the model information or you can work it out at run time. Calculating the optimal sphere is very hard, but you can get a good approximation using the following:
determine the min and max value of each point's x, y and z
for each vertex
min_x = min (min_x, vertex.x)
max_x = max (max_x, vertex.x)
min_y = min (min_y, vertex.y)
max_y = max (max_y, vertex.y)
min_z = min (min_z, vertex.z)
max_z = max (max_z, vertex.z)
sphere centre = (max_x + min_x) / 2, (max_y + min_y) / 2, (max_z + min_z) / 2
sphere radius = distance from centre to (max_x, max_y, max_z)
Using this sphere, determine the a world position that allows the sphere to be viewed in full - simple geometry will determine this.
Sorry, your question is very unclear. I suppose you want to center a 3D model to a viewport. You can achieve this by calculating the model's bounding box. To do this, traverse all polygons and get the minimum/maximum X/Y/Z coordinates. The bounding box given by the points (min_x,min_y,min_z) and (max_x,max_y,max_z) will contain the whole model. Now you can center the model by looking at the center of this box. With some further calculations (depending on your FOV) you can also get the left/right/upper/lower borders inside your viewport.
"so i tried to scale it down"
The best thing to do in this situation is not to transform your model at all! Leave it be. What you want to change is your camera.
First calculate the bounding box of your model somewhere in 3D space.
Next calculate the radius of it by taking the max( aabb.max.x-aabb.min.x, aabb.max.y-aabb.min.y, aabb.max.z-aabb.min.z ). It's crude but it gets the job done.
To center the object in the viewport place the camera at the object position. If Y is your forward axis subtract the radius from Y. If Z is the forward axis then subtract radius from it instead. Subtract a fudge factor to get you past the pesky near plane so your model doesn't clip out. I use quaternions in my engine with a nice lookat() method. So call lookat() and pass in the center of the bounding box. Voila! You're object is centered in the viewport regardless of where it is in the world.
This always places the camera axis aligned so you might want to get fancy and transform the camera into model space instead, subtract off the radius, then lookat() the center again. Then you're always looking at the back of the model. The key is always the lookat().
Here's some example code from my engine. It checks to see if we're trying to frame a chunk of static terrain, if so look down from a height, or a light or a static mesh. A visual is anything that draws in the scene and there are dozens of different types. A Visual::Instance is a copy of the visual, or where to draw it.
void EnvironmentView::frameSelected(){
if( m_tSelection.toInstance() ){
Visual::Instance& I = m_tSelection.toInstance().cast();
Visual* pVisual = I.toVisual();
if( pVisual->isa( StaticTerrain::classid )){
toEditorCamera().toL2W().setPosition( pt3( 0, 0, 50000 ));
toEditorCamera().lookat( pt3( 0 ));
}else if( I.toFlags()->bIsLight ){
Visual::LightInstance& L = static_cast<Visual::LightInstance&>( I );
qst3& L2W = L.toL2W();
const sphere s( L2W.toPosition(), L2W.toScale() );
const f32 y =-(s.toCenter()+s.toRadius()).y();
const f32 z = (s.toCenter()+s.toRadius()).y();
qst3& camL2W = toEditorCamera().toL2W();
camL2W.setPosition(s.toCenter()+pt3( 0, y, z ));//45 deg above
toEditorCamera().lookat( s.toCenter() );
}else{
Mesh::handle hMesh = pVisual->getMesh();
if( hMesh ){
qst3& L2W = m_tSelection.toInstance()->toL2W();
vec4x4 M;
L2W.getMatrix( M );
aabb3 b0 = hMesh->toBounds();
b0.min = M * b0.min;
b0.max = M * b0.max;
aabb3 b1;
b1 += b0.min;
b1 += b0.max;
const sphere s( b1.toSphere() );
const f32 y =-(s.toCenter()+s.toRadius()*2.5f).y();
const f32 z = (s.toCenter()+s.toRadius()*2.5f).y();
qst3& camL2W = toEditorCamera().toL2W();
camL2W.setPosition( L2W.toPosition()+pt3( 0, y, z ));//45 deg above
toEditorCamera().lookat( b1.toOrigin() );
}
}
}
}
Related
I'm making a sniper shooter arcade style game in Gamemaker Studio 2 and I want the position of targets outside of the viewport to be pointed to by chevrons that move along the circumference of the scope when it moves. I am using trig techniques to determine the coordinates but the chevron is jumping around and doesn't seem to be pointing to the target. I have the code broken into two: the code to determine the coordinates in the step event of the enemies class (the objects that will be pointed to) and a draw event in the same class. Additionally, when I try to rotate the chevron so it also points to the enemy, it doesn't draw at all.
Here's the coordinate algorithm and the code to draw the chevrons, respectively
//determine the angle the target makes with the player
delta_x = abs(ObjectPlayer.x - x); //x axis displacement
delta_y = abs(ObjectPlayer.y - y); //y axis displacement
angle = arctan2(delta_y,delta_x); //angle in radians
angle *= 180/pi //angle in radians
//Determine the direction based on the larger dimension and
largest_distance = max(x,y);
plusOrMinus = (largest_distance == x)?
sign(ObjectPlayer.x-x) : sign(ObjectPlayer.y-y);
//define the chevron coordinates
chevron_x = ObjectPlayer.x + plusOrMinus*(cos(angle) + 20);
chevron_y = ObjectPlayer.y + plusOrMinus*(sign(angle) + 20);
The drawing code
if(object_exists(ObjectEnemy)){
draw_text(ObjectPlayer.x, ObjectPlayer.y-10,string(angle));
draw_sprite(Spr_Chevron,-1,chevron_x,chevron_y);
//sSpr_Chevron.image_angle = angle;
}
Your current code is slightly more complex that it needs to be for this, if you want to draw chevrons pointing towards all enemies, you might as well do that on spot in Draw. And use degree-based functions if you're going to need degrees for drawing anyway
var px = ObjectPlayer.x;
var py = ObjectPlayer.y;
with (ObjectEnemy) {
var angle = point_direction(px, py, x, y);
var chevron_x = px + lengthdir_x(20, angle);
var chevron_y = py + lengthdir_y(20, angle);
draw_sprite_ext(Spr_Chevron, -1, chevron_x, chevron_y, 1, 1, angle, c_white, 1);
}
(also see: an almost-decade old blog post of mine about doing this while clamping to screen edges instead)
Specific problems with your existing code are:
Using a single-axis plusOrMinus with two axes
Adding 20 to sine/cosine instead of multiplying them by it
Trying to apply an angle to sSpr_Chevron (?) instead of using draw_sprite_ext to draw a rotated sprite.
Calculating largest_distance based on executing instance's X/Y instead of delta X/Y.
I'm trying to solve an problem where I cannot find the Relative Offset of a Point inside a Box that exists inside of a space that can be arbitrarily rotated and translated.
I know the WorldSpace Location of the Box (and its 4 Corners, the Coordinates on the Image are Relative) as well as its Rotation. These can be arbitrary (its actually a 3D Trigger Volume within a game, but we are only concerned with it in a 2D plane from top down).
Looking at it Aligned to an Axis the Red Point Relative position would be
0.25, 0.25
If the Box was to be Rotated arbitrarily I cannot seem to figure out how to maintain that given we sample the same Point (its World Location will have changed) its Relative Position doesnt change even though the World Rotation of the Box has.
For reference, the Red Point represents an Object that exists in the scene that the Box is encompassing.
bool UPGMapWidget::GetMapMarkerRelativePosition(UPGMapMarkerComponent* MapMarker, FVector2D& OutPosition)
{
bool bResult = false;
if (MapMarker)
{
const FVector MapMarkerLocation = MapMarker->GetOwner()->GetActorLocation();
float RelativeX = FMath::GetMappedRangeValueClamped(
-FVector2D(FMath::Min(GetMapVolume()->GetCornerTopLeftLocation().X, GetMapVolume()->GetCornerBottomRightLocation().X), FMath::Max(GetMapVolume()->GetCornerTopLeftLocation().X, GetMapVolume()->GetCornerBottomRightLocation().X)),
FVector2D(0.f, 1.f),
MapMarkerLocation.X
);
float RelativeY = FMath::GetMappedRangeValueClamped(
-FVector2D(FMath::Min(GetMapVolume()->GetCornerTopLeftLocation().Y, GetMapVolume()->GetCornerBottomRightLocation().Y), FMath::Max(GetMapVolume()->GetCornerTopLeftLocation().Y, GetMapVolume()->GetCornerBottomRightLocation().Y)),
FVector2D(0.f, 1.f),
MapMarkerLocation.Y
);
OutPosition.X = FMath::Abs(RelativeX);
OutPosition.Y = FMath::Abs(RelativeY);
bResult = true;
}
return bResult;
}
Currently, you can see with the above code that im only using the Top Left and Bottom Right corners of the Box to try and calculate the offset, I know this is not a sufficient solution as doing this does not allow for Rotation (Id need to use the other 2 corners as well) however I cannot for the life of me work out what I need to do to reach the solution.
FMath::GetMappedRangeValueClamped
This converts one range onto another. (20 - 50) becomes (0 - 1) for example.
Any assistance/advice on how to approach this problem would be much appreciated.
Thanks.
UPDATE
#Voo's comment helped me realize that the solution was much simpler than anticipated.
By knowing the Location of 3 of the Corners of the Box, I'm able to find the points on the 2 lines these 3 Locations create, then simply mapping those points into a 0-1 range gives the appropriate value regardless of how the Box is Translated.
bool UPGMapWidget::GetMapMarkerRelativePosition(UPGMapMarkerComponent* MapMarker, FVector2D& OutPosition)
{
bool bResult = false;
if (MapMarker && GetMapVolume())
{
const FVector MapMarkerLocation = MapMarker->GetOwner()->GetActorLocation();
const FVector TopLeftLocation = GetMapVolume()->GetCornerTopLeftLocation();
const FVector TopRightLocation = GetMapVolume()->GetCornerTopRightLocation();
const FVector BottomLeftLocation = GetMapVolume()->GetCornerBottomLeftLocation();
FVector XPlane = FMath::ClosestPointOnLine(TopLeftLocation, TopRightLocation, MapMarkerLocation);
FVector YPlane = FMath::ClosestPointOnLine(TopLeftLocation, BottomLeftLocation, MapMarkerLocation);
// Convert the X axis into a 0-1 range.
float RelativeX = FMath::GetMappedRangeValueUnclamped(
FVector2D(GetMapVolume()->GetCornerTopLeftLocation().X, GetMapVolume()->GetCornerTopRightLocation().X),
FVector2D(0.f, 1.f),
XPlane.X
);
// Convert the Y axis into a 0-1 range.
float RelativeY = FMath::GetMappedRangeValueUnclamped(
FVector2D(GetMapVolume()->GetCornerTopLeftLocation().Y, GetMapVolume()->GetCornerBottomLeftLocation().Y),
FVector2D(0.f, 1.f),
YPlane.Y
);
OutPosition.X = RelativeX;
OutPosition.Y = RelativeY;
bResult = true;
}
return bResult;
}
The above code is the amended code from the original question with the correct solution.
assume the origin is at (x0, y0), the other three are at (x_x_axis, y_x_axis), (x_y_axis, y_y_axis), (x1, y1), the object is at (x_obj, y_obj)
do these operations to all five points:
(1)translate all five points by (-x0, -y0), to make the origin moved to (0, 0) (after that (x_x_axis, y_x_axis) is moved to (x_x_axis - x0, y_x_axis - y0));
(2)rotate all five points around (0, 0) by -arctan((y_x_axis - y0)/(x_x_axis - x0)), to make the (x_x_axis - x0, y_x_axis - y0) moved to x_axis;
(3)assume the new coordinates are (0, 0), (x_x_axis', 0), (0, y_y_axis'), (x_x_axis', y_y_axis'), (x_obj', y_obj'), then the object's zero-one coordinate is (x_obj'/x_x_axis', y_obj'/y_y_axis');
rotate formula:(x_new, y_new)=(x_old * cos(theta) - y_old * sin(theta), x_old * sin(theta) + y_old * cos(theta))
Update:
Note:
If you use the distance method, you have to take care of the sign of the coordinate if the object might go out of the scene in the future;
If there will be other transformations on the scene in the future (like symmetry transformation if you have mirror magic in the game, or transvection transformation if you have shockwaves, heatwaves or gravitational waves in the game), then the distance method no longer applies and you still have to reverse all the transformations your scene has in order to get the object's coordinate.
perspectiveCamera = new PerspectiveCamera(90, 80, 48);
perspectiveCamera.position.set(0,0, 10f);
perspectiveCamera.lookAt(0,0,0);
perspectiveCamera.near = .01f;
perspectiveCamera.far = 300f;
My ScreenWidth x ScreenHeight are 800 x 480;
pCamera.unproject(mytouchPoint) shall suppose to give results between
x = 0 to 80
y = 0 to 48
but I m getting 0.000xyz for both x and y axis
Don't use such a small value for your camera's near member, it will cause floating point errors and/or z-fighting.
The width and height values you provided to PerspectiveCamera constructor, are used to calculate the aspect ratio. There is no single 2D resolution (the size of the screen-plane in world coordinates) in a 3D perspective.
You cannot simply unproject a 2D screen coordinate to a single 3D coordinate. For each 2D screen coordinate, there are an "infinite" amount of 3D coordinates possible. Therefor the unproject method of the camera will use the z-coordinate of the provided screen coordinate to decide which of those 3D coordinates to return. If z is zero, it will give the coordinate on the near-plane. If z is one, it will give the coordinate on the far-plane.
Assuming you used z=0 for myTouchPoint and given you have a very small near-plane (since you near value is very small), the unprojected value will be vary small and therefor (almost) equal to zero.
For more information, you might want to have a look at: http://blog.xoppa.com/interacting-with-3d-objects/
I found a way to easily do it. Its fast too.
Just create a plane at required depth z and find intersection of ray on it.
float zDepth=-10;//your decision or and object z position
public boolean touchDown(int screenX, int screenY, int pointer, int button) {
Ray ray = camera.getPickRay(screenX,screenY);
Plane plane=new Plane();
plane.set(0,0,1,0);// the xy plane with direction z facing screen
plane.d=zDepth;//***** the depth in 3d for the coordinates
Vector3 yourVector3Position=new Vector3();
Intersector.intersectRayPlane(ray, plane, yourVector3Position);
}
In a project of mine (VC++2010, MFC), I want to draw a circle using the CDC::Ellipse. I set two points: the first one is the center of the circle, the second one is a point I want it to be on the circumference.
I pass to the CDC::Ellipse( int x1, int y1, int x2, int y2 ) the coordinates of the upper-left corner and lower-right one.
Briefly: with Pitagora Theorem I calculate the distance between the two points ( radius ), then I subtract this value from the coordinates of the center to obtain the upper-left corner and add to obtain the lower-right one.
When I draw the cirlce and the points, and I zoom in, I see that the second one isn't on the circumference as expected, it is slightly inside unless you set it at 0°, 45°, 90° and so on with respect to the absolute sistem of coordinates.
Then I tried to draw the same circle using CDC::Polyline, I gave to this method the points obtained rotating another point around the center, at the distance equal to the radius. In this case the point is on the circumference every where I set it.
The overlap of these two circles has shown that they perfectly overlap at 0°, 45°, 90° and so on, but the gap is maximum at 22.5°, 67.5° and so on.
Has anyone ever noticed a similar behavior?
Thanks to everybody that can help me!
Code snippet:
this is how I calculate the radius given 2 points:
centerPX = vvFPoint( 1380, 845 );
secondPointPX = vvFPoint( 654,654 );
double radiusPX = (sqrt( (secondPointPX.x - centerPX.x) * (secondPointPX.x - centerPX.x) + (secondPointPX.y - centerPX.y) * (secondPointPX.y - centerPX.y) ));
( vvFPoint is a custom type derived from CPoint )
this is how I draw the "circle" with the CDC::Ellipse:
int up = (int)(((double)(m_p1.y-(double)originY - m_radius) / zoom) + 0.5) + offY;
int left = (int)(((double)(m_p1.x-(double)originX - m_radius) / zoom) + 0.5) + offX;
int down = (int)(((double)(m_p1.y-(double)originY + m_radius) / zoom) + 0.5) + offY;
int right = (int)(((double)(m_p1.x-(double)originX + m_radius) / zoom) + 0.5) + offX;
pDC->Ellipse( left, up, right, down);
(m_p1 is the center of the circle, originX/Y is the origin of the image, m_radius is the radius of the circle, zoom is the scale factor, offX/Y is an offset in the client area of my SW)
this is how I draw the circle "manually" (and quite trivial method) using a custom polyline class:
1) create the array of points:
point.x = centerPX.x + radiusPX;
point.y = centerPX.y;
for ( i=0; i < 3600; i++ )
{
pt1.RotateDeg ( centerPX, (double)0.1 );
poly->AddPoint( pt1 );
}
(RotateDeg is a custom method to rotate a point using first argument as a pivot and second argument as angle value in degrees, AddPoint is a custom method to create the array of points, poly is my custom polyline object).
2) draw it:
When I call the Draw( CDC* pDC ) I use the previous array to draw the polyline:
pDC->MoveTo(p);
I hope this can help you to reproduce my weird observations!
code snippet 2:
void vvPoint<Tipo>::RotateDeg(const vvPoint<Tipo> ¢er, double angle)
{
vvPoint<Tipo> ptB;
angle *= -(M_PI / 180);
*this -= center;
ptB.x = ((this->x * cos(angle)) - (this->y * sin(angle)));
ptB.y = ((this->x * sin(angle)) + (this->y * cos(angle)));
*this = ptB + center;
}
But to let you better understand my observations I would like to add a few images so you can see where my whole question started from... The problem is: I can't add images since I need to have 10 reputation. I uploaded a .zip file on dropbox and if you want I can send you the URL of this file. Let me know if this is the correct (and safe..) way to bypass this problem.
Thanks!
This might be a possible explanation. As MSDN says about CDC::Ellipse (with my emphasis):
The center of the ellipse is the center of the bounding rectangle
specified by x1, y1, x2, and y2, or lpRect. The ellipse is drawn with
the current pen, and its interior is filled with the current brush.
The figure drawn by this function extends up to, but does not include,
the right and bottom coordinates. This means that the height of the
figure is y2 – y1 and the width of the figure is x2 – x1.
The way you described how you calculate the bounding rectangle is not entirely clear (some source code would have helped) but, given the second paragraph quoted above, you possibly need to add 1 to your x2 and y2 values, to make sure you have a circle with the desired radius.
It's also worth noting that there may be slight rounding differences between your two drawing methods where you have an odd-sized bounding box (i.e. so the centre point falls logically on a half-pixel).
UPDATE
Using your code snippets (thanks), and assuming no zoom and zero offsets etc., I get a radius of 750.704 pixels and the following parameters for the ellipse:
pDC->Ellipse(629, 94, 2131, 1596);
According to MSDN, this means that the ellipse will be drawn in a figure of the following dimensions:
width = (2131 - 629) = 1502
height = (1596 - 94) = 1502
So as far as I can see, this should produce a circle rather than an ellipse.
The next thing to do is to find out how you're drawing the polygon - for that we need to see the implementation of RotateDeg - can you post that code? I'm suspecting some simple rounding error here, that maybe gets magnified when you zoom.
UPDATE 2
Just looking at this code:
for ( i=0; i < 3600; i++ )
{
pt1.RotateDeg ( centerPX, (double)0.1 );
poly->AddPoint( pt1 );
}
You are rotating your polygon points incrementally by 0.1 degrees each time. This will possibly accumulate some errors, so it may be worth doing it something like this instead:
for ( i=0; i < 3600; i++ )
{
vvFPoint ptNew = pt1;
ptNew.RotateDeg ( centerPX, (double)i * 0.1 );
poly->AddPoint( ptNew );
}
Maybe this will mean you have to change your RotateDeg function to take care of the correct quadrants.
One other point, you mentioned that you see the problem when you zoom into the image. If this means you are using you zoom variable, it is worth checking in this line ...:
pDC->Ellipse( left, up, right, down);
... that the parameters still form a square shape, so (right - left) == (down - up).
UPDATE 3
I just ran your RotateDeg function, in its current form, to see how the error accumulates (by feeding in the previous result to the next iteration). At each step, I calculated the distance between the new point and the centre and compared this with the required radius.
The chart below shows the result, where you can see an error of 4 pixels by the time the points have been calculated.
I think that this at least explains part of the difference (i.e. your polygon drawing is flawed) and - depending on zoom - you may introduce asymmetry into the ellipse parameters, which you can debug by comparing the width to the height as I described above.
So I decided to write a ray tracer the other day, but I got stuck because I forgot all my vector math.
I've got a point behind the screen (the eye/camera, 400,300,-1000) and then a point on the screen (a plane, from 0,0,0 to 800,600,0), which I'm getting just by using the x and y values of the current pixel I'm looking for (using SFML for rendering, so it's something like 267,409,0)
Problem is, I have no idea how to cast the ray correctly. I'm using this for testing sphere intersection(C++):
bool SphereCheck(Ray& ray, Sphere& sphere, float& t)
{ //operator * between 2 vec3s is a dot product
Vec3 dist = ray.start - sphere.pos; //both vec3s
float B = -1 * (ray.dir * dist);
float D = B*B - dist * dist + sphere.radius * sphere.radius; //radius is float
if(D < 0.0f)
return false;
float t0 = B - sqrtf(D);
float t1 = B + sqrtf(D);
bool ret = false;
if((t0 > 0.1f) && (t0 < t))
{
t = t0;
ret = true;
}
if((t1 > 0.1f) && (t1 < t))
{
t = t1;
ret = true;
}
return ret;
}
So I get that the start of the ray would be the eye position, but what is the direction?
Or, failing that, is there a better way of doing this? I've heard of some people using the ray start as (x, y, -1000) and the direction as (0,0,1) but I don't know how that would work.
On a side note, how would you do transformations? I'm assuming that to change the camera angle you just adjust the x and y of the camera (or the screen if you need a drastic change)
The parameter "ray" in the function,
bool SphereCheck(Ray& ray, Sphere& sphere, float& t)
{
...
}
should already contain the direction information and with this direction you need to check if the ray intersects the sphere or not. (The incoming "ray" parameter is the vector between the camera point and the pixel the ray is sent.)
Therefore the local "dist" variable seems obsolete.
One thing I can see is that when you create your rays you are not using the center of each pixel in the screen as the point for building the direction vector. You do not want to use just the (x, y) coordinates on the grid for building those vectors.
I've taken a look at your sample code and the calculation is indeed incorrect. This is what you want.
http://www.csee.umbc.edu/~olano/435f02/ray-sphere.html (I took this course in college, this guy knows his stuff)
Essentially it means you have this ray, which has an origin and direction. You have a sphere with a point and a radius. You use the ray equation and plug it into the sphere equation and solve for t. That t is the distance between the ray origin and the intersection point on the spheres surface. I do not think your code does this.
So I get that the start of the ray would be the eye position, but what is the direction?
You have camera defined by vectors front, up, and right (perpendicular to each other and normalized) and "position" (eye position).
You also have width and height of viewport (pixels), vertical field of view (vfov) and horizontal field of view (hfov) in degrees or radians.
There are also 2D x and y coordinates of pixel. X axis (2D) points to the right, Y axis (2D) points down.
For a flat screen ray can be calculated like this:
startVector = eyePos;
endVector = startVector
+ front
+ right * tan(hfov/2) * (((x + 0.5)/width)*2.0 - 1.0)
+ up * tan(vfov/2) * (1.0 - ((y + 0.5f)/height)*2.0);
rayStart = startVector;
rayDir = normalize(endVector - startVector);
That assumes that screen plane is flat. For extreme field of view angles (fov >= 180 degreess) you might want to make screen plane spherical, and use different formulas.
how would you do transformations
Matrices.