How is it possible object can pass through ring spirte like in the image below?
Please can you help me I have no idea how can i do that.
I think you posted a incorrect image. To get the image you posted you just have to draw the red bar on top of the black ring.
I guess you want the ring on the left side to be on top and the right side to be over so it visually goes through. Well this is simply not so easy in 2D since draw order.
I have a couple of suggestion you can explore.
Always draw the ring on top of the bar but when a collision is happening you calculate where the bar overlaps and don't draw the pixels in that place. You can use a Pixmap for calculations like this. Depending on the size of your images this could be very expensive to calculate each frame.
A faster but slightly more hacky way could be to split red bar in multiple images and if a certain part of it should be overlapped by the ring draw it first otherwise draw it after the ring. Depending on what the red bar is going to look in your end product and how much possible angles the bar could have I can imagine this can be very tricky to get right.
Use 3D for this. You could have a billboard with a slight angle for the ring and have the bar locked on the distance axis at the rings center. However, on certain angles of entrance and exit you will get Z fighting since the pixels will be at the same distance from the camera. This might or might not be noticable and I have no idea how LibGDX would handle Z fighting.
I wanna add this solution :
if the object gonna pass through the ring horizontally i propose to devise sprite ring in to to sprite (sprite 1 & sprite 2)
you just have to draw sprites in that order :
Sprite1
Sprite Object
Sprite2
You can do the same if the object is gonna pass through ring vertically
PS : this solution don't work if the object is going to passs through ring both Vertically and Horizontally
Hope this was helpfull
Good luck
Related
Currently I am using canvas to draw the gold circle, then snip off the areas half way to adjacent stars with globalCompositeOperation : destination-out, then paste the result into the main canvas.
I am contemplating changing to svg for this.
The closest method I have found is clip masks. I would have to create a set of clip-masks for each circle (star), this seems excessive if there were 2000 stars.
Another way would be to create polygons, but the math required to calculate the path may be beyond my capabilities.
So before I go down either of these routes, I would like to know if there is a better way or which of the above methods are recommended.
I am trying to draw multiple circles on top of each other (with different radii) using PIXI.Graphics; it looks like this:
declare var PIXI:any;
const particle = new PIXI.Graphics();
particle.beginFill(0xFFFF00).drawCircle(0,0,5).endFill(); //yellow
particle.beginFill(0xFFA500).drawCircle(0,0,3).endFill(); //orange
The yellow circle is always drawn on top of the orange
However, after some fiddling around it seems like that the order of both drawing commands does not determine which of both graphics gets rendered in the foreground (more due to the color?!). This might stem from the GPU implementation -> I dont know the details;
Any ideas how to enforce within a graphics the right z order?
I am not really sure what's exactly your problem because for me they do render in the correct order, you can check this fiddle online:
https://www.pixiplayground.com/#/edit/2f5I0uuKBDq1i7HOMUoXh
Orange circle with a smaller radius is drawn on top of the yellow circle with a bigger radius.
I created my own little 2D-Engine with DirectX (okey, should be more like a GUI in the end) and tried to create rounded edges for a simple Rectangle. Since I never done this with a graphics framework before I had no idea how to supply this.
For now, I just overlapped 5 Rectangles and 4 circles (the circles are used for the rounded edges). It does work with opaque colors but if I add alpha into the rectangles the circles are making problems. (Shown in the image below - i should have choose another colors...)
<# Open Image #>
I can't find a solution myself (I googled and whondered I found nothing about rounded edges in DirectX) and I do believe there is a much powerful and faster method doing this. So my final question is, what are the common algorythm to create a rectangle with rounded edges in Direct3D9 ?
The common way to draw rounded quads is the use of textures with an alphachannel. It's very easy and the most of the gui's uses images to achieve a specific look. If you draw only single-colored boxes it may look very generic after a while (even if they have fancy rounded corners ;) ).
But if you want to draw rounded quads directly, I would suppose to generate a custom geometry, which fits the desired area directly without overlapping (need for alphablending). In you case it would be something like this:
The more triangles you're using for the corner the smoother it will look.
I'm using DirectX10 to simulate a water surface, and I'm now with a height map,which is a 2D array of the heights(y) at the points (x,z). But to draw it on the screen, I must turn it into a mesh or have a index to draw triangle topology.
But the data is too large to do it manually. Are there any methods for me to draw it on the screen. I hope it's easy to implement. If there is function included in DirectX10 which can make it, the it's the best one for me.
Create a mesh that format a grid of squares (each made of two triangles) and set all vertices y = 0. In the vertex shader sample the heightmap and add the value stored in the heightmap to the y of the vertice.
This might help you.
P.S: If the area you want it to cover is too big you should take a look at terrain LOD techniques (should work the same for water).
I'm sure you can make a mesh out of it. I doubt you can generate the heightmap for a water surface that is too large to "meshify".
Why are you looking at Diamond square. For a 512x512 heightmap all you need to do is define a set of point and then generate the triangles for it. Its really very simple.
I know this is more high school math(wow been a long time since I was there) but I am trying to solve this programatically so I am reaching out to the collective knowledge of stackoverflow
Given this layout:
Midpoint is my reference point and in an array I have the vector points of all other points (P)
I can get to this state with code of having the light blue area by breaking it into four quadrants and doing a lame bubble sort to find largest(y) or lowest(x) value in each quadrant.
I need to find only the quadrants that outer border fully hits red no white space. For example the lower left and the up right dont have any white space hitting the light blue rectangle.
I am sure my terminology is all off here and im not looking for any specific code but if someone could point me to a more optimized solution for this problem or the next step in what I already have.
Thank you
I might do some BFI solution first, then perhaps look to generalize it or at least reduce it to a table-drive loop.
So, if it's exactly these shapes, and not a general solution, I think you should proceed sort of like this:
Derive the coordinates of the blue rectangle. I suspect one thing that's confusing you is that you have each individual x and y for the blue rect but you can't easily loop through them.
Derive the coordinates of the midpoint of each rectangle edge. You are going to need this because you care about quadrants. It will be trivial to do this once you have done 1.
Write different code for each 1/2 rectangle edge. There is no doubt a more clever way but this will get working code.
Make it more elegant now if you care. I betg you can reduce the rules to an 8-row
table full of things like 1, -1, or something like that.
First, you can't define red area by a single vector, since it's disjoint. You need the same number of vectors as the number of distant red regions.
Second, do we assume that different red figures neither intersect nor share a border? In the next clause I do.
Third, under assumption in point 2, the quadrant will have all red outer side iff there exists a contiguous red figure that intersects both its axes (i.e. rays). To determine this for all quadrants, you should only traverse all (P) points in the order they're given. This takes linear time and solves the problem.