数学：缓和，使用具有时间约束的Hermite曲线缓和位移

Question

I'm trying to write a method that interpolates from 0 to x (position of an object in one dimension) over time using acceleration at the beginning and deceleration at the end (ease out / ease in) with the only constraints that the total time is provided , as well as the duration of the acceleration and deceleration . the motion should replicate the inertia effect and I'm considering a Hermite curve for the non-linear portions.

double Interpolate(
    double timeToAccel, double timeCruising, double timeToDecel,
    double finalPosition,
    double currentTime)
{
    //...
}

Can someone point me out to a portion of code that does that? I don't know how to integrate the Hermite curve, hence don't know how much I'll move in the accelerating portion or in the decelerating portion, and in turn I can't figure out what will be the speed in the linear portion.

Thanks.

Some reference to illustrate my question.

Edit :

• start and end speeds are null, and the current time is also part of the parameters in the method, I've updated the signature.
• basically the idea is to imagine a move at constant speed on a distance d, this gives a total duration. Then we add the acceleration and deceleration phases, while maintaining the same duration, hence we have an unknown new cruise speed to determinate (because we move less in the Hermite phases than in the linear phases they have replaced). Maybe the amount of move lost in the Hermite phases, compared to a linear move of the same duration is the ratio between the top and bottom area in the curves, just an idea from a non expert.

Edit : Roman and Bob10 have provided full working solutions. I implemented the code from Roman. Thanks to you both, guys! I appreciate your perfect support and your detailed solutions, you saved me long searches and trials.

Answer 1

First, let's make a cubic hermite spline function:

/*
  t  - in interval <0..1>
  p0 - Start position
  p1 - End position
  m0 - Start tangent
  m1 - End tangent
*/
double CubicHermite(double t, double p0, double p1, double m0, double m1) {
   t2 = t*t;
   t3 = t2*t;
   return (2*t3 - 3*t2 + 1)*p0 + (t3-2*t2+t)*m0 + (-2*t3+3*t2)*p1 + (t3-t2)*m1;
}

Now your task is to calculate the p0, p1, m0 and m1 for both ease-in and ease-out portions. Let us add a few variables to make the math a bit easier to write:

double Interpolate(
    double timeToAccel, double timeCruising, double timeToDecel,
    double finalPosition,
    double currentTime) {

    double t1 = timeToAccel;
    double t2 = timeCruising;
    double t3 = timeToDecel;
    double x = finalPosition;
    double t = currentTime;

We need to specify where should the object be when it stops accelerating and starts decelerating. You can specify these however you please and still produce a smooth movement, however, we would like a somewhat "natural" solution.

Let's assume that the cruising speed is v . During crusing, the object travels distance x2 = v * t2 . Now, when the object accelerates from 0 to speed v, it travels distance x1 = v * t1 / 2 . Same for deceleration x3 = v * t3 / 2 . Put all together:

x1 + x2 + x3 = x

v * t1 / 2 + v * t2 + v * t3 / 2 = x

From that we can calculate our speed and the distances:

    double v = x / (t1/2 + t2 + t3/2);
    double x1 = v * t1 / 2;
    double x2 = v * t2;
    double x3 = v * t3 / 2;

And now that we know everything, we just feed it into our cubic hermite spline interpolator

    if(t <= t1) {
       // Acceleration
       return CubicHermite(t/t1, 0, x1, 0, v*t1);
    } else if(t <= t1+t2) {
       // Cruising
       return x1 + x2 * (t-t1) / t2;
    } else {
       // Deceleration
       return CubicHermite((t-t1-t2)/t3, x1+x2, x, v*t3, 0);
    }
}

I tested this in Excel, here's the equivalent VBA code to play with. There are some divisions by zero for boundary conditions, I leave fix to this as an excercise to the reader

Public Function CubicHermite(t As Double, p0 As Double, p1 As Double, _
m0 As Double, m1 As Double) As Double
   t2 = t * t
   t3 = t2 * t
   CubicHermite = (2 * t3 - 3 * t2 + 1) * p0 + _
(t3 - 2 * t2 + t) * m0 + (-2 * t3 + 3 * t2) * p1 + (t3 - t2) * m1
End Function

Public Function Interpolate(t1 As Double, t2 As Double, t3 As Double, _
x As Double, t As Double) As Double
    Dim x1 As Double, x2 As Double, x3 As Double

    v = x / (t1 / 2 + t2 + t3 / 2)
    x1 = v * t1 / 2
    x2 = v * t2
    x3 = v * t3 / 2

    If (t <= t1) Then
       Interpolate = CubicHermite(t / t1, 0, x1, 0, v*t1)
    ElseIf t <= t1 + t2 Then
       Interpolate = x1 + x2 * (t - t1) / t2
    Else
       Interpolate = CubicHermite((t-t1-t2)/t3, x1+x2, x, v*t3, 0)
    End If
End Function

Answer 2

This is straightforward using normal constant acceleration. Then the question becomes what velocity (v) do you need to accelerate to in order to complete the trip in the right amount of time, and this will tell you the acceleration you need to get to that velocity.

If the total time is t_t and the time of acceleration is t_a, then you have the distance traveled as the two accelerating and decelerating parts, and the constant velocity part:

x = 2*(a*t_a*t_a/2) + v*(t_t-2*t_a)

This can be solved for the acceleration by subbing in v=a*t_a, to find

a = x/(t_a*(t_t - t_a))

Here's some Python code that uses and plots the result of these equations, that shows both how to use the equations and what the result looks like:

from pylab import *

t_a, t_t, D = 3., 10., 1.  # input values

a = D/(t_a*(t_t - t_a))
segments = (t_a, a), (t_t-2*t_a, 0.), (t_a, -a)  # durations and accelerations for each segment

t0, x0, v0 = 0.0, 0.0, 0.0  #initial values for the segment
tdata, xdata = [], []
for t_segment, a in segments: # loop over the three segments
    times = arange(0, t_segment, .01)
    x = x0 + v0*times + .5*a*times*times
    xdata.append(x)
    tdata.append(times+t0)
    x0 = x[-1] # the last x calculated in the segment above
    v0 += a*t_segment
    t0 += t_segment

plot(tdata[0], xdata[0], 'r', tdata[1], xdata[1], 'r', tdata[2], xdata[2], 'r')
xlabel("time")
ylabel("position")
show()