I don't know which numbers do what in the coordinates example here. I imagine they mean things like place the top left corner at this position and the bottom right corner at this position, but I don't know which number corresponds to which position.
I've been trying to fool around with the numbers to get a small green rectangle but keep getting weird results like the following, and don't know which numbers need to be what is order to make the rectangle symmetrical and at the bottom
This is what the rectangle should look like
The height of the rectangle is 50, the height of the screen is 1000, and the width of the screen is 1700.
Here's my draw function
function love.draw()
love.graphics.setColor(0.28, 0.63, 0.05) -- set the drawing color to green for the ground
love.graphics.polygon("fill", objects.ground.body:getWorldPoints(objects.ground.shape:getPoints())) -- draw a "filled in" polygon using the ground's coordinates
-- These are the grounds coordinates. -11650 950 13350 950 13350 1000 -11650 1000
love.graphics.setColor(0.76, 0.18, 0.05) --set the drawing color to red for the ball
love.graphics.circle("fill", objects.ball.body:getX(), objects.ball.body:getY(), objects.ball.shape:getRadius())
love.graphics.setColor(0.20, 0.20, 0.20) -- set the drawing color to grey for the blocks
love.graphics.polygon("fill", objects.block1.body:getWorldPoints(objects.block1.shape:getPoints()))
love.graphics.polygon("fill", objects.block2.body:getWorldPoints(objects.block2.shape:getPoints()))
print(objects.block1.body:getWorldPoints(objects.block1.shape:getPoints()))
end
As described at https://love2d.org/wiki/love.graphics, Löve's coordinate system has (0, 0) at the upper left corner of the screen. X values increase to the right, Y values increase down.
The polygon function expects the drawing mode as it's first parameter, and the the remaining (variable) parameters are the coordinates of the vertices of the polygon you wish to draw. Since you want to draw a rectangle you need four vertices/eight numbers. You do not have to list the upper left corner of the rectangle first, but that's probably the easiest thing to do.
So in your case, you want something like:
love.graphics.polygon('fill', 0, 950, 0, 1000, 1700, 1000, 1700, 950)
I've not worked with the physics system, so I'm not quite sure how it's coordinate system relates to "screen" coordinates. The values you show in the comment in your code listing seem like they should give a rectangle (although x = -11650 wouldn't be on screen). You might try experimenting without the physics system first.
Also, since the physics system in Löve is just a binding to Box2D, you might want to read its documentation (http://box2d.org/about/). Not really sure what you're trying to do with feeding shape:getPoints into body:getWorldPoints.
Related
By default, 0,0 coordinates are in the top, left corner. However, for my scene, it would be very helpful if it was in my bottom, left corner.
Is that even possible? If so, how can I set that?
This is a mathematical approach which I think is a lot easier.
Simply define a function which converts your local Y coordinate to what Corona uses:
function localY ( y )
return 600-y --Assuming your screen size is 600 pixels
end
Then simply use
x, localY(y) instead of x,y
You can do the same for x if decided to change it.
I have an application built with SceneKit that is currently displaying several nodes. I can figure out which node is pressed and want to use that to make a label appear below the Node that was touched. Now, when I set the label's center the following way...
nameLabel.center = CGPointMake(CGFloat(result.node.position.x), CGFloat(result.node.position.y+20)
…it appears in the upper left corner since the node is on (1, 0). What I figured is that the sceneView's (0, 0) is in the center of the screen while the actual physical display's (0, 0) is in the top left corner.
Is there a way to convert the two different numbers into each other? (I could hardcode since I know where the Node's are or create a separate label for each Node but that is not really a perfect solution.)
Thanks in advance :)
You can use the projectPoint: method:
var projected = view.projectPoint(result.node.position))
//projected is an SCNVector3
//projected.x and y are the node's position in screen coordinates
//projected.z is its depth relative to the near and far clipping planes
nameLabel.center = CGPointMake(CGFloat(projected.x), CGFloat(projected.y+20)
The picture shows a simple UIView after applying the following transform:
- (CATransform3D) transformForOpenedMenu
{
CATransform3D transform = CATransform3DIdentity;
transform.m34 = -1.0 /450;
transform = CATransform3DRotate(transform, D2R(40), 0, 1, 0);
transform = CATransform3DTranslate(transform, 210, 150, -500);
return transform;
}
I'm trying to make the distances highlighted with black to have equal length. Could you please help me understand the logic behind the values and calculations?
Cheers
UPD Sept 13
Looks like removing 3DTranslate keeps distances equal. I see I can use layer's frame property to reposition rotated view to the bottom left of the screen. Not yet sure, but this might actually work.
The .m34 value you are setting is best set on the sublayerTransform of the containing view rather than the view you are transforming.
I don't fully understand the maths behind affine transforms so I made this project which allows me to play around with the transform values to achieve the effect I want. You can plug in the values from your code above and see what it looks like, though note that there is already a perspective value applied using the sublayerTransform property mentioned above.
For your specific case, I think you want to adjust the anchor point of the layer to (0.0,0.5) and apply the rotation transform only. This assumes you want the menu to swing back like a door, with the hinges on the left edge.
The problem you're seeing is caused by your CATransform3DTranslate call. You're essentially setting the Y Axis off center, and hence seeing a different perspective view of the frame.
Think of it this way;
You're standing in the center of a long narrow field stretching off into the horizon. The edge of the field appears as if it is converges to a center point somewhere off in the distance. The angle of each edge to the converging point will appear equal if you are at the center of the field. If, on the other hand, you move either to the left or the right, the angles change and one will seem greater than the other (inversely opposite of course).
This is essentially what is happening with your view; As your converging points are to the right, changing the Y axis away from 0 will have the same effect as moving to the left or right in my example above. You're no longer looking at the parallel lines from the center.
so in your code above Setting the ty in CATransform3DTranslate to 0 Should fix your problem I.E.
transform = CATransform3DTranslate(transform, 210, 0, -500);
You may also need to alter the tz and tx Value to make it fit.
OK, so what eventually solved my question is this:
3D transform on Y axis to swing the view like a door transform = CATransform3DRotate(transform, D2R(40), 0, 1, 0);
set Z anchor point on a layer, to move it back targetView.layer.anchorPointZ = 850;
adjust layer position so that the view is located slightly to the bottom left of the parent view:
newPosition.x += 135 * positionDirection;
newPosition.y += 70 * positionDirection;
This sequence adjusts position without CATransform3DTranslate and keeps the 'swinged' effect not distorted.
Thanks everybody!
I think that this might be a bug in InDesign but I thought that it might be handy to know.
Here's how to recreate: Create a rectangle 200px x 200px, with a 4pt stroke. Set Align stroke to INSIDE. Set the top left xy pos of the rectangle to 100px,100px so that you can easily see changes, and leave the Reference Point in the top left. Now when you change stroke widths, everything is fine, and change corner types, everything is fine, EXCEPT the Fancy Corners. So if you choose Fancy Corners, the x/y becomes 96,96 and total width/height changes to 208 x 208 which is incorrect. Now change the Align Stroke to "OUTSIDE", and the x/y changes back to 100,100 and the rectangle size changes back to 200 x 200. Completely backwards, but just for the Fancy Corners.
Please comment if this is an expected response.
It is stated, that to rotate a line by a certain angle, you multiply its end point coordinates by the matrix ({Cos(a), Sin(a)} {-Sin(a) Cos(a)}), where a is rotation angle. The resulting two numbers in matrix will be x and y coordinates of rotated line's end point. Rotation goes around line's start point.
Simplifying it, new coordinates will be {x*Cos(a) - y*Sin(a)} for x and {x*Sin(a) + y*Cos(a)} for y.
Task is to rotate a triangle, using this method. But the following code that uses this method, is giving out some crap instead of rotated image (twisted form of original triangle, rotated by "random" angle):
x0:=200;
y0:=200;
bx:=StrToInt(Edit1.Text);
by:=StrToInt(Edit2.Text);
cx:=StrToInt(Edit4.Text);
cy:=StrToInt(Edit5.Text);
a:=StrToInt(Edit3.Text);
//Original triangle
Form1.Canvas.Pen.Color:=clBlue;
Form1.Canvas.MoveTo(x0,y0);
Form1.Canvas.LineTo(bx,by);
Form1.Canvas.LineTo(cx,cy);
Form1.Canvas.LineTo(x0,y0);
//New triangle
Form1.Canvas.Pen.Color:=clGreen;
Form1.Canvas.MoveTo(x0,y0);
b1x:=Round(bx*cos(a*pi/180)-by*sin(a*pi/180));
b1y:=Round(bx*sin(a*pi/180)+by*cos(a*pi/180));
c1x:=Round(cx*cos(a*pi/180)-cy*sin(a*pi/180));
c1y:=Round(cx*sin(a*pi/180)+cy*cos(a*pi/180));
Form1.Canvas.LineTo(b1x,b1y);
Form1.Canvas.MoveTo(x0,y0);
Form1.Canvas.LineTo(c1x,c1y);
Form1.Canvas.LineTo(b1x,b1y);
end;
Well, I'm out of ideas. What am I doing wrong?
Thanks for your time.
The formula you are using rotates a point around (0, 0). To achieve the required result change your calculation to:
b1x:=x0 + Round((bx-x0)*cos(a*pi/180)-(by-y0)*sin(a*pi/180));
b1y:=y0 + Round((bx-x0)*sin(a*pi/180)+(by-y0)*cos(a*pi/180));
c1x:=x0 + Round((cx-x0)*cos(a*pi/180)-(cy-y0)*sin(a*pi/180));
c1y:=y0 + Round((cx-x0)*sin(a*pi/180)+(cy-y0)*cos(a*pi/180));
You appear to be rotating each individual line round its initial start point coordinates. So line 1 will get rotated about its start point (x0,y0); then line 2 will get rotated about bx,by; then line 3 will get rotated round cx. This will result in a twisted triangle. Instead you will need to rotate all three lines round the start point of the first line.