Smooth fly movement like flappy bird or jet pack joy ride with gravity and Accelaration

问题

I am developing a simple game in which a character fly when you tap/click the screen. keep tapping the character will fly (some what similar to flappy bird and jet pack). However the movement is not smooth at all, as of jet pack.

Here is sample of my code.

Varaible Initilization

maxSpeedLimit  = spriteHeight/10;
speed = maxSpeedLimit/2; //half of the max speed

touch event

public void onTapOrClick(int action) {
    if (action == UP) {
        sprite.up= true;
    }
    else {
        sprite.up = false;
    }
}

Sprite update called from game loop

public void update() {
    if (up) {
        y -= speed; //fly up
    }
    else {
        y += speed;// fly down
    }
    if (speed < maxSpeedLimit) {
        speed++; // May be cheep way to add **velocity/acceleration**
    }   
}

I think speed++ is not a smooth way to increase the speed, I am not sure but may be adding some time related variable to increment may improve it, moreover adding a gravity will make it more realistic but i have no idea how to do it, I have read few blogs, first thing I am not able to search with the correct keyword, and second thing is they are so hard understand because they contains platform related codes. Please help.

I am making this game in android, but code in any language is accepted (HTML5, javascript, Android, Flash or any).

Q: How to add acceleration and gravity to an object (sprite) which fly when user tap or click and fall on release? Something similar to jet pack joyride (only up and down movement)

UPDATED

After @scottt advise I have implemented the dY += gravity + flapp I can feel now gravity, however there are 2 issues.

1. My screen height is 480, and sprite immediately touch upper(y=0) and lower(y=480) boundary, becuase i think it keep increasing the speed of the sprite.
2. when it touches the ground, it seems very havey and take much time to lift the sprite up in the air.

Some how there should be some limit to dY which is constantly being added to y location.

Here is update code.

int downSpeed = 1;
int upSpeed = -2;
int dy = 0;
private void update() {
    if (flapping) {
        upSpeed = -2; //if flying speed
    }
    else {
        upSpeed = 0;
    }

    dy  +=  downSpeed + upSpeed;
    if (dy < -10) {
        dy = -10; //limit for rise speed
    }
    else if (dy > 8) {
        dy = 8; //limit for gravity
    }
    y += dy; // add value in y location
    if (y > GAME_HEIGHT - sprite.getHeight()) {
        y = GAME_HEIGHT  - sprite.getHeight(); // reset y, if touch ground
        dy = 0; //reset speed, otherwise it make it very heavy to rise
    }
    else if ( y < 0) {
        y = 0; //reset y if touch upper limit
        dy = 0; //reset speed, otherwise take much time to fall (as it would be in negative)
    }
}

回答1:

Gravity is just a downward acceleration. An acceleration is in turn just a change in velocity (directed speed). For the following discussion, I make the following assumptions to keep things simpler:

• your horizontal speed is constant
• input tapping is called 'flapping'
• gravity behaves normally
• simplistic, rather than a scientific approach/verbiage is OK for our purposes

For each pass through the update loop, all of the various accelerations must be summed and the total added to the current speed. In your situation, there are 2 possible accelerations, gravity and possibly flapping. Gravity is constant and is a negative value (is works downward), while flapping only occurs during tapping as is positive (upwards).

Let's set gravity to -10 pixels per loop, tapping to be +25 pixels per loop, and initial height to 500. Some initial definitions are:

static final int gravity = -10;   // constant downward acceleration
static final int flapping = 25;   // upward acceleration whenever isFlapping is true

Boolean isFlapping = false;   // Is the bird flapping

int dY = 0; // current vertical speed
y = 500;    // current vertical position

Each time through the loop, without flapping, the speed calculation would be:

dY += gravity + flapping;

So the first time through, the speed calculation would be dY = 0 + (-10) + 0 = -10. The second time, dy = -10 + (-10) + 0 = -20. The 5th time, dy = -40 + (-10) = -50. Each time through, the downward speed is 10 more than the time before.

The height is simple. Each time through, the height changes by the amount of vertical acceleration. So:

y += dY;

So the first time through, the height would be y = 500 + (-10) = 490. The second time, y = 490 + (-20) = 470. And the 5th time, y = 400 + (-50) = 350. Because the rate of falling increases each time through, the bird will plummet faster and faster until splat!

That's where flapping comes in. Each time through the loop where flapping is occurring, a +25 will be applied to the dY calculation. So lets assume the bird starts flapping in 6th iteration. The dy calculation would be dy = -50 + (-10) + 25 = -35 and the height would be y = 350 + (-35) = 315. The next time through would give dy = -35 + (-10) + 25 = -20 and the height would be y = 350 + (-20) = 295. Still falling, but more slowly. The time after: that dy = -20 + (-10) + 25 = -5 and y = 295 + (-5) = 290. The time after that finally shows a gain in height: dy = -5 + (-10) + 25 = 10 and y = 290 + 10 = 300.

All that said, you'll definitely need to play with the numbers until you get a satisfying result.

TLDR: You don't want to change the height directly using gravity and flapping. Instead you want to use gravity and flapping to calculate the speed for each iteration and then use that to adjust the height.

来源：https://stackoverflow.com/questions/22442495/smooth-fly-movement-like-flappy-bird-or-jet-pack-joy-ride-with-gravity-and-accel