Good evening. I am trying to smoothly move rectangle shape across the screen with glTranslatef. The concept is Windows 10 notifications. You know that they appear from right bottom of the screen, ease in-out animation and so on. I am trying to recreate it. But I dont know how can I achieve it.
Here is a basic requirement:
Move slowly, not in one frame. For example Windows 10 notifications. They move slowly, ease in-out (if possible).
I would appreciate any help, any tip, so thanks in advance!
You could start by finding the height (or width, as needed) define and initialize a variable (say, x) at half the height (again, or width) and move the notif x pixels up, divide x by 2, and raise the notif by 2 again, and repeat a number of times so that the notif doesn't take to long to appear, or look awkward, and finally move to the final desired position. (If you don't like how it looks, you can subtract x by 1 or some other number instead.)
Related
I'm making a 2D platformer game. I have created a texture for the platform, that is meant to be repeated over and over to fill the entire platform, without going over. My first attempt was to draw all the pixels from the bitmap manually, but this caused the background to flicker through while moving the platform (the movement and drawing threads are seperate, so the movement can run at a specific speed, while the FPS doesn't need to suffer). I found this technique worked better:
// Init
bitmap = new BitmapDrawable(res, Texture.PLATFORM.getBitmap());
bitmap.setTileModeXY(Shader.TileMode.REPEAT, Shader.TileMode.REPEAT);
// Drawing loop
int x = getX() + (isStill() ? 0 : (int)MainActivity.offsetX);
int y = getY() + (isStill() ? 0 : (int)MainActivity.offsetY);
bitmap.setBounds(x, y, x + getWidth(), y + getHeight());
bitmap.draw(canvas);
However, the bitmap appears to be staying static while the platform is acting as a "view hole" to see through to the bitmap. The only work around I can think of is to somehow "offset" the static bitmap:
bitmap.offset(x, y);
Obviously, that isn't a function. I couldn't find one that would do what I want when looking through the docs.
To summon things up, the BitmapDrawable is causing the background to not move with the platform, making it look super weird.
Thanks in advance!
Try these tips in your code:(I assumed the game moves forward in the horizontal direction)
The GUY should only move up and down(with the appropriate touch input) and not forward and backward as you want the focus(or camera alternatively) solely on the GUY.I noticed that the WALL was moving up in your video when the GUY moved from initial higher position of the wall to little bit lower position later, rectify this because the GUY should move down(try to implement Gravity effect).
The WALL should only move forward(mostly) and backward(less often I guess).The WALL shouldn't move up and down normally. Do not apply Gravity effect to it. You can create at least 2 BitmapDrawable instance of WALL for a screen. They are going to be reused sequencially(for eg: If the 1st one goes totally outside of the screen, reshow it in the desired position using setBounds() method) and continue same for others the whole game.
The currently BLUE BACKGROUND, if it is a part of a larger map, then it needs to be appropriately offsetted.
One of the obstacles that I can think of at the time of writing this is to move the WALL down until it goes out of the screen which results in the death of the GUY.
At those places, where I have used the word move, you need to use the setBounds(a, b, c, d) method to make necessary position based changes as I didn't find other way to update the position of a BitmapDrawable instance. I think, you need to use game framework like libGdx to get method of luxury like setOffset(x, y) or of similar sort.
Sorry that I could only present you the ideas without specific code as I do not have past experience working in a project like this. Hope, it helps you in anyway possible.
I want to launch projectiles from the bottom-right corner of the screen towards the left side of the screen. Now, I want the projectiles to fly with random velocities and angles according to the screen dimensions, just like that. Now, I know this is very simple but from some reason I can't manage to make this work.
Here is what I have tried so far:
My first try - Launch function
private void launchProjectile() {
projectiles.peek().getBody().applyForce(projectiles.peek().getBody().getWorldVector(new Vector2(MathUtils.random(-20,-1*SCALAR_HEIGHT),
MathUtils.random(2*SCALAR_HEIGHT,8*SCALAR_HEIGHT)).scl(MathUtils.random(3*SCALAR_HEIGHT,5*SCALAR_HEIGHT))),
projectiles.peek().getBody().getWorldCenter(), true);
Gdx.app.log("System", String.valueOf(SCALAR_HEIGHT));
}
Here is my second try - Launch function
private void launchProjectile() {
float xVelocity;
float yVelocity;
xVelocity = (float) MathUtils.random(0,0)*SCALAR_WIDTH/2;
yVelocity = (float) MathUtils.random(20,20)*SCALAR_HEIGHT;
velocityProjectile.set(xVelocity,yVelocity); // Sets the velocity vector to the above values
velocityProjectile.sub(projectiles.peek().getBody().getPosition());
velocityProjectile.nor(); // Normalize the vector - Now it's fine and ready!
// Sets the start velocity of the projectile Trajectory to the current velocity
projectiles.peek().getBody().setLinearVelocity(velocityProjectile.scl(18+SCALAR_HEIGHT));
}
In both tries, the projectile flies way more than I need and it doens't take in consideration the screen size like it should.
Can you guys please tell me what is the right way to do this?
Thanks!!
Start with this page: http://www.iforce2d.net/b2dtut/projected-trajectory
In the "How fast should it be launched to reach a desired height?" section, you can see how much vertical velocity will be required to make the projectile reach the top of the screen. So you would pick a random number less than that, to make sure it doesn't go off the top of the screen.
Next, in the "How high will it go?" section, you can see the formula to find out how many time steps it will take for the projectile to reach maximum height. It will then take the same amount of time to come back down to the starting height. For example, let's say it would take 60 time steps to reach maximum height. That means it would take 120 time steps to fall down again to the same height as it started. Then you can set the horizontal part of the launch velocity so that it cannot go outside the screen in 120 time steps.
I am trying to use the Java Robot class to create a bot to automate some tedious tasks for me, I have never used the Robot class. I have looked up the Class in the Java docs, usage seems straightforward but I have an issue of finding a certain image(I say image, I mean a certain part of the screen) effectively. Is there any other way other than loading 'x' ammount of pixels, checking them, checking the next ammount etc until I find the image I am looking for? Also is there any list of the Button and MouseButton identifiers needed for the Java Robot class as I cna not find any.
For the mouse button identifiers, you are supposed to use BUTTON1_MASK and other button mask constants from java.awt.event.MouseEvent. For example, to click the mouse you would do something like:
Robot r = new Robot();
r.mousePress(MouseEvent.BUTTON1_MASK);
r.mouseRelease(MouseEvent.BUTTON1_MASK);
I believe BUTTON1_MASK is the left mouse button, BUTTON2_MASK is the middle mouse button, and BUTTON3_MASK is the right mouse button, but it has been a month or so since I have used Robot.
As for the checking for an image, I have no idea how that is normally done. But the way you specified in your question where you just check every group of pixels shouldn't be too computationally expensive because you can get the screen image as an array of primitives, then just access the desired pixel with a bit of math. So when checking the "rectangle" of pixels that you are searching for your image in, only keep checking the pixels as long as the pixels keep matching. The moment you find a pixel that does not match, move onto the next "rectangle" of pixels. The probability that you will find a bunch of pixels that match the image that end up not being the image is extremely low, meaning that each rectangle will only need to check about 5 or fewer pixels on average. Any software that performs this task would have to check every pixel on the screen at least once (unless it makes a few shortcuts/assumptions based on probabilities of image variations occurring), and the algorithm I described would check each pixel about 5 times, so it is not that bad to implement, unless you have a huge image to check.
Hope this helps!
I am developing java game using box2d for my physics, I have got helicopter, ex:
I reduced gravity by setting:
body.setGravityScale(0.03f);
So it acts bit realistic (is affected by gravity only little bit, floating in the air)
To move it, down/up left/right I have controller, thats how I control my helicopter:
body.applyLinearImpulse(new Vector2(pValueX * 3, pValueY * 3), mainBody.getWorldCenter());
Where pValueX and pValueY are 1 or -1 (directions up/down left or right)
It works good, but now I am trying to achieve more realistic effect, when moving helicopter left/right I wanted to tilt it little bit so it works like real helicopter, but could not find proper way how to do it, I have tried applying force in different part of the body, but it makes my helicopter rotating 360 degrees if keep pressing left or right.
This question is old, but in case it's still relevant, I created a helicopter using JBox2D (which pretty much maps directly to Box2D). For tilting left/right (i.e. forwards/backwards relative to the pilot):-
heli.applyTorque(TURN_TORQUE);
or
heli.applyTorque(-TURN_TORQUE);
This rotates the heli, and then if the player wants lift:
Vec2 force = new Vec2();
force.y = (float)Math.cos(chopper.getAngle()) * -1;
force.x = (float)Math.sin(chopper.getAngle());
force.mulLocal(ROTOR_FORCE);
heli.applyForceToCenter(force);
What you can do is, just define two constants as maxForceLeft and maxForceRight. When you press left apply some force on the cockpit part of the helicopter and keep comparing it with the maxForceLeft,once it reaches that value stop applying the force.Do the same for the right button by applying the force on the tail rotor part of the helicopter.In this way you can avoid rotating it 360 degrees.Depending upon the kind of effect you want for your helicopter you can apply the forces in either upward or downward direction.
http://www.iforce2d.net/b2dtut/rotate-to-angle
What you need is rotating the body to a desired angle..
This is a great tutorial to achieve this.
I hope this would help.
I have an app where the user draws pictures and then these pictures are converted to pdf. I need to be able to crop out the whitespace before conversion. Originally I kept track of the highest and lowest x and y values (http://stackoverflow.com/questions/13462088/cropping-out-whitespace-from-a-user-drawn-image). This worked for a while, but now I want to give the user the ability to erase. This is a problem because if for example the user erases the topmost point the bounding box would change, but I wouldn't the new dimensions of the box.
Right now I'm going through the entire image, pixel by pixel, to determine the bounding box. This isn't bad for one image, but I'm going to have ~70, it's way too slow for 70. I also thought about keeping every pixel in an arraylist, but I don't feel like that would work well at all.
Is there an algorithm that would help me solve this? Perhaps something already built in? Speed is more important to me than accuracy. If there is some whitespace left on each side it won't be a tragedy.
Thank you so much.
You mentioned that you are keeping track of the min and max values for X and Y co-ordinates (that also seems the solution you have chosen in the earlier question).
In similar way to this, you should be able to find the min and max X & Y co-ordinates for the erased area, from the erase event...
When the user erases part of the image, you can simply compare the co-ordinates of the erased part with the actual image to find the final co-ordinates.
There is a related problem of trying to see if 2 rectangles overlap:
Determine if two rectangles overlap each other?
You can use similar logic (though slightly different) and figure out the final min/max X & Y values.