I am trying to understand those reference points in Corona. It is said that the default reference point is the center of the object, however the x position of center of the textObject isn't at 100 but 150 when I run this code:
local textObject = display.newText( "Hello World!", 100, 100, nil , 20 )
I displayed those position with touch event on the screen and printing the x and y position of touched places:
local function onScreenTouch(event)
print("x: "..event.x.." y: "..event.y);
end
Runtime:addEventListener("tap", onScreenTouch)
Very good question and it's something that drove me nuts for a long time.
By default the reference point is display.CenterReferencePoint unless you are passing the X, Y as parameters to an API call. In most cases (for rectangles), passing an X, Y becomes Top Left reference point only for the initial draw of the object. Lines of course you're passing end points and for circles your X, Y is the center.
myRect = display.rect(X, Y, width, height) -- X, Y = Top, Left
in that case, X and Y represent the Top Left corner of the box however afterwards if you do:
myRect.x = X -- X = center of object
myRect.y = Y -- Y = center of object
then you are now using a center reference point unless you explicitly change it back to something else. Now because this is confusing, I tend to do this:
myRect = display.rect(0,0, width, height)
myRect.x = 100
myRect.y = 100
just as a way that I'm really only positioning things using the center point.
Hope that explains it.
Rob
Related
I am trying to use vtkImageReSlicer to extract a 2d slice from a 3d
vtkImageData object. But I can't seem to get the recipe right. Am I doing it right?
I am also a bit confused about ResliceAxes Matrix. Does it represent a cutting plane? If
I move the ReSliceAxes origin will it also move the cutting plane? When I
call Update on the vtkImageReSlicer, the program crashes. But when I don't
call it, the output is empty.
Here's what I have so far.
#my input is any vtkactor that contains a closed curve of type vtkPolyData
ShapePolyData = actor.GetMapper().GetInput()
boundingBox = ShapePolyData.GetBounds()
for i in range(0,6,2):
delta = boundingBox[i+1]-boundingBox[i]
newBoundingBox.append(boundingBox[i]-0.5*delta)
newBoundingBox.append(boundingBox[i+1]+0.5*delta)
voxelizer = vtk.vtkVoxelModeller()
voxelizer.SetInputData(ShapePolyData)
voxelizer.SetModelBounds(newBoundingBox)
voxelizer.SetScalarTypeToBit()
voxelizer.SetForegroundValue(1)
voxelizer.SetBackgroundValue(0)
voxelizer.Update()
VoxelModel =voxelizer.GetOutput()
ImageOrigin = VoxelModel.GetOrigin()
slicer = vtk.vtkImageReslice()
#Am I setting the cutting axis here. x axis set at 1,0,0 , y axis at 0,1,0 and z axis at 0,0,1
slicer.SetResliceAxesDirectionCosines(1,0,0,0,1,0,0,0,1)
#if I increase the z value, will the cutting plane move up?
slicer.SetResliceAxesOrigin(ImageOrigin[0],ImageOrigin[1],ImageOrigin[2])
slicer.SetInputData(VoxelModel)
slicer.SetInterpolationModeToLinear()
slicer.SetOutputDimensionality(2)
slicer.Update() #this makes the code crash
voxelSurface = vtk.vtkContourFilter()
voxelSurface.SetInputConnection(slicer.GetOutputPort())
voxelSurface.SetValue(0, .999)
voxelMapper = vtk.vtkPolyDataMapper()
voxelMapper.SetInputConnection(voxelSurface.GetOutputPort())
voxelActor = vtk.vtkActor()
voxelActor.SetMapper(voxelMapper)
Renderer.AddActor(voxelActor)
I have never used vtkImageReslice, but I have used vtkExtractVOI for vtkImageData, which allows you to achieve a similar result, I think. Here is your example modified with the latter, instead:
ImageOrigin = VoxelModel.GetOrigin()
slicer = vtk.vtkExtractVOI()
slicer.SetInputData(VoxelModel)
#With the setVOI method you can define which slice you want to extract
slicer.SetVOI(xmin, xmax, ymin, ymax, zslice, zslice)
slicer.SetSampleRate(1, 1, 1)
slicer.Update()
voxelSurface = vtk.vtkContourFilter()
voxelSurface.SetInputConnection(slicer.GetOutputPort())
voxelSurface.SetValue(0, .999)
voxelMapper = vtk.vtkPolyDataMapper()
voxelMapper.SetInputConnection(voxelSurface.GetOutputPort())
voxelActor = vtk.vtkActor()
voxelActor.SetMapper(voxelMapper)
Renderer.AddActor(voxelActor)
In my app I am currently able to work out whether the user's tap is within a rectangular area simply by checking all of the following conditions are true:
Finger X > rectangle X
Finger Y > rectangle Y
Finger X < rectangle X + rectangle Width
Finger Y < rectangle Y + rectangle Height
However, I now have to determine if the user taps within a circular area. Currently I have a circular shape on screen and have resorted to just checking it's bounding rectangle, which works but obviously isn't great.
Any help would be appreciated.
The distance between two points in two dimensions is defined as
dist = sqrt((x2-x1)^2 + (y2-y1)^2)
To check if your tap point is inside a circle, take the centre of your circle as (x1,y1), and the 'tap location' as (x2,y2), and check if
sqrt((x2-x1)^2 + (y2-y1)^2) < R
With R being the radius of your circle.
Edit:
As John mentioned, from a computational point of view it is more interesting to compare vs R^2, to avoid the sqrt for every tap. So the condition becomes:
(x2-x1)^2 + (y2-y1)^2 < R^2
I have the x, y co-ordinates of a point on a rotated image by certain angle. I want to find the co-ordinates of the same point in the original, non-rotated image.
Please check the first image which is simpler:
UPDATED image, SIMPLIFIED:
OLD image:
Let's say the first point is A, the second is B and the last is C. I assume you have the rotation matrice R (see Wikipedia Rotation Matrix if not) et the translation vector t, so that B = R*A and C = B+t.
It comes C = R*A + t, and so A = R^1*(C-t).
Edit: If you only need the non rotated new point, simply do D = R^-1*C.
First thing to do is defining the reference system (how "where the points lies with respect to each image" will be translated into numbers). I guess that you want to rely on a basic 2D reference system, given by a single point (a couple of X/Y values). For example: left/lower corner (min. X and min. Y).
The algorithm is pretty straightforward:
Getting the new defining reference point associated with the
rotated shape (min. X and min. Y), that is, determining RefX_new and
RefY_new.
Applying a basic conversion between reference systems:
X_old = X_new + (RefX_new - RefX_old)
Y_old = Y_new + (RefY_new -
RefY_old)
----------------- UPDATE TO RELATE FORMULAE TO NEW CAR PIC
RefX_old = min X value of the CarFrame before being rotated.
RefY_old = max Y value of the CarFrame before being rotated.
RefX_new = min X value of the CarFrame after being rotated.
RefY_new = max Y value of the CarFrame after being rotated.
X_new = X of the point with respect to the CarFrame after being rotated. For example: if RefX_new = 5 with respect to absolute frame (0,0) and X of the point with respect to this absolute frame is 8, X_new would be 3.
Y_new = Y of the point with respect to CarFrame after being rotated (equivalently to point above)
X_old_C = X_new_C(respect to CarFrame) + (RefX_new(CarFrame_C) - RefX_old(CarFrame_A))
Y_old_C = Y_new_C(respect to CarFrame) + (RefY_new(CarFrame_C) - RefY_old(CarFrame_A))
These coordinates are respect to the CarFrame and thus you might have to update them with respect to the absolute frame (0,0, I guess), as explained above, that is:
X_old_D_absolute_frame = X_old_C + (RefX_new(CarFrame_C) + RefX_global(i.e., 0))
Y_old_D_absolute_frame = Y_old_C + (RefY_new(CarFrame_C) + RefY_global(i.e., 0))
(Although you should do that once the CarFrame is in its "definitive position" with respect to the global frame, that is, on picture D (the point has the same coordinates with respect to the CarFrame in both picture C and D, but different ones with respect to the global frame).)
It might seem a bit complex put in this way; but it is really simple. You have just to think carefully about one case and create the algorithm performing all the actions. The idea is extremely simple: if I am on 8 inside something which starts in 5; I am on 3 with respect to the container.
------------ UPDATE IN THE METHODOLOGY
As said in the comment, these last pictures prove that the originally-proposed calculation of reference (max. Y/min. X) is not right: it shouldn't be the max./min. values of the carFrame but the minimum distances to the closer sides (= perpendicular line from the left/bottom side to the point).
------------ TRIGONOMETRIC CALCS FOR THE SPECIFIC EXAMPLE
The algorithm proposed is the one you should apply in any situation. Although in this specific case, the most difficult part is not moving from one reference system to the other, but defining the reference point in the rotated system. Once this is done, the application to the non-rotated case is immediate.
Here you have some calcs to perform this action (I have done it pretty quickly, thus better take it as an orientation and do it by your own); also I have only considered the case in the pictures, that is, rotation over the left/bottom point:
X_rotated = dx * Cos(alpha)
where dx = X_orig - (max_Y_CarFrame - Y_Orig) * Tan(alpha)
Y_rotated = dy * Cos(alpha)
where dy = Y_orig - X_orig * Tan(alpha)
NOTE: (max_Y_CarFrame - Y_Orig) in dx and X_orig in dy expect that the basic reference system is 0,0 (min. X and min. Y). If this is not the case, you would have to change this variables.
The X_rotated and Y_rotated give the perpendicular distance from the point to the closest side of the carFrame (respectively, left and bottom side). By applying these formulae (I insist: analyse them carefully), you get the X_old_D_absolute_frame/Y_old_D_absolute_frame that is, you have just to add the lef/bottom values from the carFrame (if it is located in 0,0, these would be the final values).
EDIT - Thanks for all the answers everyone. I think I accidentally led you slightly wrong as the square in the picture below should be a rectangle (I see most of you are referencing squares which seems like it would make my life a lot easier). Also, the x/y lines could go in any direction, so the red dot won't always be at the top y boundary. I was originally going for a y = mx + b solution, but then I got stuck trying to figure out how I know whether to plug in the x or the y (one of them has to be known, obviously).
I have a very simple question (I think) that I'm currently struggling with for some reason. I'm trying to have a type of minimap in my game which shows symbols around the perimeter of the view, pointing towards objectives off-screen.
Anyway, I'm trying to find the value of the red point (while the black borders and everything in green is known):
It seems like simple trigonometry, but for some reason I can't wrap my head around it. I just need to find the "new" x value from the green point to the red point, then I can utilize basic math to get the red point, but how I go about finding that new x is puzzling me.
Thanks in advance!
scale = max(abs(x), abs(y))
x = x / scale
y = y / scale
This is the simple case, for a square from (-1, -1) to (1, 1). If you want a different sized square, multiply the coordinates by sidelen / 2.
If you want a rectangle instead of a square, use the following formula. (This is another solution to the arbitrarily-sized square version)
scale = max(abs(x) / (width / 2), abs(y) / (height / 2))
x = x / scale
y = y / scale
Let's call the length of one side of the square l. The slope of the line is -y/x. That means, if you move along the line and rise a distance y toward the top of the square, then you'll move a distance x to the left. But since the green point is at the center of the square, you can rise only l/2. You can express this as a ratio:
-y -l/2
——— = ———
x d
Where d is the distance you'll move to the left. Solving for d, we have
d = xl/2y
So if the green dot is at (0, 0), the red dot is at (-l/2, xl/2y).
All you need is the angle and the width of the square w.
If the green dot is at (0,0), then the angle is a = atan(y/x), the y-coordinate of the dot is w/2, and therefore the x-coordinate of the dot is tan(1/a) * (w/2). Note that tan(1/a) == pi/2 - tan(a), or in other words the angle you really want to plug into tan is the one outside the box.
Edit: yes, this can be done without trig, too. All you need is to interpolate the x-coordinate of the dot on the line. So you know the y-coordinate is w/2, then the x-coordinate is (w/2) * x/y. But, be careful which quadrant of the square you're working with. That formula is only valid for -y<x<y, otherwise you want to reverse x and y.
Imagine a circle. Imagine a pie. Imagine trying to return a bool that determines whether the provided parameters of X, Y are contained within one of those pie pieces.
What I know about the arc:
I have the CenterX, CenterY, Radius, StartingAngle, EndingAngle, StartingPoint (point on circumference), EndingPoint (point on circumference).
Given a coordinate of X,Y, I'd like to determine if this coordinate is contained anywhere within the pie slide.
Check:
The angle from the centerX,centerY through X,Y should be between start&endangle.
The distance from centerX,centerY to X,Y should be less then the Radius
And you'll have your answer.
I know this question is old but none of the answers consider the placement of the arc on the circle.
This algorithm considers that all angles are between 0 and 360, and the arcs are drawn in positive mathematical direction (counter-clockwise)
First you can transform to polar coordinates: radius (R) and angle (A). Note: use Atan2 function if available. wiki
R = sqrt ((X - CenterX)^2 + (Y - CenterY)^2)
A = atan2 (Y - CenterY, X - CenterX)
Now if R < Radius the point is inside the circle.
To check if the angle is between StartingAngle (S) and EndingAngle (E) you need to consider two possibilities:
1) if S < E then if S < A < E the point lies inside the slice
2) if S > E then there are 2 possible scenarios
if A > S
then the point lies inside the slice
if A < E
then the point lies inside the slice
In all other cases the point lies outside the slice.
Convert X,Y to polar coordinates using this:
Angle = arctan(y/x);
Radius = sqrt(x * x + y * y);
Then Angle must be between StartingAngle and EndingAngle, and Radius between 0 and your Radius.
You have to convert atan2() to into 0-360 before making comparisons with starting and ending angles.
(A > 0 ? A : (2PI + A)) * 360 / (2PI)