When I click on the top left corner it gives me 0,0 but I want the opposite result. I want to have a 0,0 coordinate when I click on the bottom left corner.
Basically when I go down the y coordinate is higher and when I go up the y coordinate is lower.It only happens when I do Gdx.input.getY();
This is true for almost any computer language or function.
The reason for this goes back. In earlier computers Cathode Ray Tubes (CRTs) would "draw" the image with a cathode ray from the upper left corner to the lower right.
To ease the interface between the graphics card memory and the CRT, the memory was read from the beginning and the image was drawn from the top left (with the lowest memory address) to the lower right (with the highest memory address).
The reason for this order likely originates from the writing style in countries where the computer was invented.
The tradition of computing from left to right and top to bottom persists today, thus the origin for the plane field in in the upper left corner.
Use the Camera's unproject method, which takes a Vector3 holding screen coordinates and converts them to camera (or world) coordinates. Try something like:
camera.unproject(screenCoords.set(screenX, screenY, 0);
Note, this will modify screenCoords to hold world coordinates (in your case... flipping the y, and taking into account any translation of the camera).
Related
How to invert Y axis? When I touch on bottom or top of the screen, the Y value is opposite I want
You can't invert an axis per se. You see, in computer graphics, the 2D coordinate system is a bit different from the canonical one taught at school in maths. The difference is that in computer graphics the y-axis is in the opposite direction, that is, from the origin to the bottom it has positive values and from the origin to the top it has negative values. Also, the origin is at the top left corner of the screen. If you can't get used to it then you can always take the opposite value to what you get, for this, asume ycoord holds the value obtained then you can do ycoord = -ycoord and that will get you the value as you're used to. Also, if you want the origin to be in the bottom left corner then you should check your y-coordinate, if it's positive then substract the vertical resolution to it, and if it's negative then add the vertical resolution to it.
But keep mind that you're going against the standard definition for coordinate systems in computer graphics.
I would say this is a duplicate questions of this one:
Move a shape to place where my finger is touched
Check on my answer there, so I won't repeat my self.
Or in short - use camera.unproject() method to get world coordinates from screen coordinates.
I have a little tech game I am messing around with and I can't figure out the formula to position 1 object given another objects origin.
So I have a Spaceship and a Cannon. I have the game setup to use units, so 1 unit = 16 pixels (pixel art).
Basically my cannon should be placed 0.5625 units on the X and 0 on the Y relative to the origin of the Spaceship, which is located at 0, 0 (bottom left corner).
The cannon should is independent on the angle of the spaceship, it can aim in different directions rather than being fixed to aim the way of the spaceship.
I have it constantly following the cursor, which works fine. Now when I rotate the Spaceship, obviously the origin of the Spaceship is changing in world coordinates, so my formula to place the cannon is all messed up, like so:
protected Vector2 weaponMount = new Vector2();
weaponMount.set(getBody().getPosition().x + 0.5625f, getBody()
.getPosition().y);
Obviously if I position the ship at a 90° angle, X is going to be different and the cannon would be waaaayyy off the ship. Here is a screenshot example of what I mean:
What would be the formula for this? I have tried using cos/sin but that does not work.
Any ideas?
weaponMount.set(0.5625f,0).setAngle(SpaceshipAngle).add(getBody().getPosition());
Where SpaceshipAngle is the angle of your Spaceship.
The origin of the spaceship is the point, arround which the spaceship will rotate and scale (the Texture of it). The position instead is always the lower left corner of the Texture and does not depend on the rotation.
Your problem is, that your offset does not depent on the rotation of your spaceship.
To take care about this rotation you should store a Vector2 offset, which describes your weapons offset (in your case it is a Vector2(0.5625f, 0)).
Next store a float angle describing your spaceships rotation.
Then you can rotate the offset by using: offset.setAngle(rotation).
The last thing is to set the weapons position. The code for this did not change so much:
weaponMount.set(getBody().getPosition().x + offset.x, getBody()
.getPosition().y + offset.y);
I'm experimenting with LibGDX and 3D in a projection view. Right now I'm looking at how to determine the outermost bounds of my viewport in world space at z=0.0, in order to draw coordinate grid no larger than necessary. However, I seem to have outpaced my education in that I haven't taken a formal linear algebra class and am still a little fuzzy on matrix math.
Is there a way to determine where I should start and stop drawing lines without resorting to using picking and drawing a transparent plane to intersect with?
LibGDX's unproject function takes screen coordinates in a Vector3 and returns a Vector3 in world space from the near clipping plane to the far, given the provided z. However, given that I have a translated and rotated Camera (an encapsulation of the viewprojection matrix and a slew of convenience methods), it occurs to me that I can't pick an arbitrary z to put in the window coordinate vector and just set it to 0.0 after unprojection, as that point probably won't be the furthest viewable point in the viewport. So how do I know what z value to use in the window coordinate that will give the the x and y I need in world space that's at z=0.0?
EDIT (UPDATE):
So apparently it looks like the problem I'm looking at is plane intersection, which would require ray tracing. So now I suppose my question is this: is ray tracing 4 times per render loop (or, I suppose whenever the camera has moved) worth the payoff of being able to dynamically draw a worldspace coordinate grid no larger than the viewport? If not, is there a cheaper algorithm I can use to estimate where I should start and stop drawing lines?
I'm creating a 3D renderer in Java, which currently can render the wireframe of a cube using Points and lines and rotate the cube, the question is, what should Z be? And what should be set to Z? I'm guessing that the size of the cube should be set to Z?
Thanks for your time! Any answers would be much appreciated.
Z usually means the out-of-plane direction if the current viewport lies in the x-y plane.
Your 3D world has its own coordinate system. You'll transform from 3D world coordinates to viewport coordinates when you render.
I think you might be missing some basic school math/geometry here. However, it's actually not that hard to understand.
Imagine a flat plane, e.g. a sheet of paper.
The first coordinate axis will go straight from left to right and we'll call it X. So X = 0 means your point is on the left border. X = 10 might mean your point is on the right border (really depends on how big you define a unit of 1; this could be in centimeters, inches, etc.). This is already enough to describe some point in one dimension (from left to right).
Now, we need a second axis. Let's call it Y. It's running from the top border (Y = 0) to the bottom (Y = 10). Now you're able to describe any point on the plane as you've got two positions. For example, (0, 0) would be the top left corner. (10, 10) would be the bottom right corner. (5, 0) would be the center point of the top border, etc.
What happens if we add yet another dimension? Call it Z. This will essentially be the height of your point above the sheet. For example, Z = 0 could mean your point is sitting on the sheet of painter, while Z = 10 means your point is sitting 10 cm above the paper. Now you use three coordinates to describe a point: (5, 5, 0) is the center of the paper. (5,5,5) is the center of the cube sitting on your paper filling it and being 10 cm high.
When programming in 3D, you can use the same terminology. The only real difference is, that you're using a so called projection/view matrix to determine how to display this 3d positions on screen. The easiest transform could be the following matrix:
1 0 0
0 1 0
Multiplying this with your 3d coordinates you'll get the following two terms:
2d-x = 3d-x
2d-y = 3d-y
This results in you viewing the cube (or whatever you're trying to display) from straight above essentially ignoring the Z axis again (you can't render something sticking out of your display, unless using some kind of 3d glasses or similar technology).
Overall, it's up to you how you use the coordinates and interpret them. Usually x and y refer to the plane (position on the ground or position inside a 2D world) while z might be the height or the depth (front or back). It really depends on the specific case. But in generic, it's really just another dimension like x and y.
3D means 3 "Dimensions". One dimension is "X", the other "Y", the third "Z". None have a sepcific direction, though it's convenient to conventionally assign a direction, for example "Forward", "Left", and "Up".
Something whose X, Y, and Z values are all equal to 0 resides at the origin, or center of the space. You can write this as (0,0,0) where the order of the parameters are (x,y,z).
A point or vertex at the location (1,0,0) is one unit in the X direction from the origin. So if you moved from (0,0,0) to (1,0,0), you would be moving purely in the X direction.
(0,1,0) is one unit in the Y direction away from the origin.
(0,0,1) is one unit in the Z direction away from the origin.
(1,1,0) is one unit in the X direction and one unit in the Y direction. So if X means "Forward", and Y means "Left", then (1,1,0) is forward-and-left of the origin.
So a basic cube can be defined by the following vertices:
(1,1,-1)
(1,-1,-1)
(-1,1,-1)
(-1,-1,-1)
(1,1,1)
(1,-1,1)
(-1,1,1)
(-1,-1,1)
I have an assignment to implement ray tracing in Java.
I'm not asking for much, just to have some information on how to construct the rays from the camera through a pixel given its x and y. I've found over the Internet a lot of sources that explain that but in 2D, and I need how to do that in 3D.
Thanks in advance
The question is how to find the coordinates in space of a point on the screen whose position is given by (x,y) in screen coordinates.
I don't know what coordinates system you're using for the screen, so I'll make some educated guesses and you can adjust accordingly.
The center of the screen has known location [X,Y,Z]center in space. I'll presume the origin of the screen coordinate system is there. We have a "direction" vector d which is normal (perpendicular) to the screen, and an "up" vector u. I'll presume that the +y direction on the screen is u. We can take the cross-product of these vectors, r = dxu, which I will take to be the +x direction on the screen. So the location of a point on the screen whose screen coordinates are (x, y) will be [X, Y, Z]center + xr + yu, and we're done.
The basic idea is this: you have a camera at a given position (x,y,z) with a given resolution. You have a set of light sources. You have an orientation of the camera (an angle, think of how you'd tilt/rotate your head to look up/down etc...). Now what you want to do is essentially for each camera pixel, "extend" perpendiculars (rays) until they touch geometry. Then you know "what to render" (namely the bit of geometry you've just found). Next up is to determine whether or not the object is shadowed, which you do by "extending" rays towards the light sources until they either touch geometry (your spot is shadowed by the other bit of geometry, for the given the light source) or until they reach the light source (your spot is lit by the given light source).
That's the basics, it gets a lot more difficult when you consider things like reflection, diffusion of light, and so on.