Question

Let me further explain my question. Let's say my Velocity is at -5.4028000 I would love it to return to zero like so.

-5.4,
-5.3,
-5.2
... 
-0.2,
-0.1,
0

Same with positive numbers. For them to return to 0, In groups of 0.1's. It would return to 0, every time it was run. Because this will be placed in a update loop.

Things I have tried:

if(vx > 0)
{
    vx-=0.1;
}
if(vx < 0)
{
    vx+=0.1;
}

But this just locks my VX into:

VX:0.04999999

OR

VX:0.09999999

Answer 1

The reason your number does not return back to zero is that the last subtraction may be subtracting a wrong number due to rounding errors. Since neither 5.4 nor 0.1 can be represented precisely in the format of double, your last subtraction (or your last addition in case of negative numbers) would "overshoot".

This is a problem for Math.min(...) and Math.max(...). Basically, we're going to only add or subtract the exact amount we need to reach zero.

if (vx > 0) {
    vx -= Math.min(0.1, vx);
}
if (vx < 0) {
    vx += Math.min(0.1, -vx);
}

Answer 2

您需要确保最终的增量/减量不大于数字的大小：

if(vx > 0) {
    vx -= Math.min(0.1, vx);
}
if(vx < 0) {
    vx += Math.min(0.1, Math.abs(vx));
}

Answer 3

假设你想要每秒0.1的变化率并且不使用任何花哨的库，你可以这样做：

while (Math.abs(vx) - 0.1 > 0.0) {
    vx += -Math.signum(vx) * 0.1
    Thread.sleep(1000)
}
vx = 0.0;

How do I make a number slowly return back to zero

3 个答案: