缩放贝塞尔曲线中的切线

Question

I am currently working on a project about Bezier curves and their properties. I am struggling to understand a concept. I can't seem to understand why you need a scale for the tangent line in a Bezier curve.

This is my code that does the animation.

from tkinter import Tk, Canvas
from graphics_template import *
import math, time

vp_width, vp_height = 1024, 768
w_xmin, w_ymin, w_xmax = -3, -3, 10
w_ymax = w_ymin + (w_xmax - w_xmin)/vp_width * vp_height

B2 = [[0.0, 0.0], # point 0
      [7.0, 0.0], # point 1
      [1.0, 4.0]] # point 2
V_pos = [] # position
V_vec = [] # velocity vector
animation_done = False
DELTA_TDRAW = 0.02  # 50 fps

def eval_Bezier2(P, t):
    # P(t) = (1-t)^2P[0] + 2t(1-t) P[1] + t^2P[2]
    res = [0.0, 0.0]
    for xy in range(2):
        res[xy] = (1-t)*(1-t)*P[0][xy] + 2*t*(1-t)*P[1][xy] + t*t*P[2][xy]
    return res

def eval_dBezier2(P, t):
    # P'(t) = -2(1-t)P[0] + 2(1-t)P[1]-2tP[1] + 2tP[2]
    res = [0.0, 0.0]
    for xy in range(2):
        res[xy] = -2*(1-t)*P[0][xy] + 2*(1-t)*P[1][xy]-2*t*P[1][xy] + 2*t*P[2][xy]
    return res

def draw_Bezier (P, nsteps):
    xi = P[0][0]
    yi = P[0][1]
    t_delta = 1/nsteps
    t = t_delta
    for ti in range(nsteps):
        p = eval_Bezier2(P, t)
        draw_line(canvas, xi, yi, p[0], p[1], rgb_col(255, 0 ,0))
        draw_small_square(canvas, xi, yi, rgb_col(255, 255, 0))
        xi = p[0]
        yi = p[1]
        t += t_delta
    for i in range(len(P)):
        draw_small_square(canvas, P[i][0], P[i][1], rgb_col(0, 255, 0))

def do_animation (t):
    global animation_done
    v_factor = 5 # reparameterization
    u = t/v_factor
    if (t > v_factor):  #animation stops at t = v_factor    
        animation_done = True
    else:
        current_pos = eval_Bezier2(B2, u);
        V_pos[0] = current_pos[0]
        V_pos[1] = current_pos[1]
        current_vel = eval_dBezier2(B2, u);
        V_vec[0] = current_vel[0]
        V_vec[1] = current_vel[1]

def draw_scene ():
    draw_grid(canvas)
    draw_axis(canvas)
    draw_Bezier(B2, 20)
    draw_dot(canvas, V_pos[0], V_pos[1], rgb_col(0,255,0))
    draw_line(canvas, V_pos[0], V_pos[1], (V_pos[0] + V_vec[0]), (V_pos[1] + V_vec[1]), rgb_col(0,255,0))

def init_scene ():
    #no data inits needed
    V_pos.append(0.0)
    V_pos.append(0.0)
    V_vec.append(0.0)
    V_vec.append(0.0)
    do_animation(0.0)    
    draw_scene()

window = Tk()
canvas = Canvas(window, width=vp_width, height=vp_height, bg=rgb_col(0,0,0))
canvas.pack()

init_graphics (vp_width, vp_height, w_xmin, w_ymin, w_xmax)

# time.perf_counter() -> float. Return the value (in fractional seconds)
# of a performance counter, i.e. a clock with the highest available resolution
# to measure a short duration. It does include time elapsed during sleep and
# is system-wide. The reference point of the returned value is undefined,
# so that only the difference between the results of consecutive calls is valid.

init_time = time.perf_counter()
prev_draw_time = 0
init_scene ()

while (not animation_done):
    draw_dt = time.perf_counter() - init_time - prev_draw_time
    if (draw_dt > DELTA_TDRAW): # 50 fps
        prev_draw_time += DELTA_TDRAW
        do_animation(prev_draw_time)
        canvas.delete("all")
        draw_scene()
        canvas.update()

As you can see, the formulas work and are correct. As you also can see in the do_animation() function, there is a v_factor that makes the t smaller. This is because the animation goes at 50 frames per second. Every t that comes in is 0.02, this is divided by 5 to make the point move accross the curve for 5 seconds before reaching u = 1.

As you may notice, I divided the velocity vector by the v_factor, I don't understand this concept. I know my v_factor is a scale, but I don't understand why I need it. Doesn't the derivative of the bezier curve just output the correct velocity at every point in the curve? I only know that when I remove the v_factor from there, my velocity vector would become too big to fit my screen. As I said, I know it's a scale, but why do I need it? Why don't I need it for the V_pos vector? I tried to understand this concept from this stackoverflow post: How to calculate tangent for cubic bezier curve? , but unfortunately to no succes.

Answer 1

The velocity at any point B(t) is the derivative vector B'(t) so you got that part right, but the derivative (remember not to call it the tangent: tangents are true lines without a start or end point) at a point tells you the velocity "over one unit of time". And inconveniently, an entire Bezier curve only covers a single unit of time (with the t parameter running from 0 to 1)...

So, if we space our t values using some smaller-than-one value, like your curve does by using 20 points (so a t interval of 1/20 ), we need to scale all those derivatives by 1/(point count) too.

If we look at the unscaled derivatives, they are unwieldy and huge:

But if we scale them by 1/20 so that they represent the velocity vector for the time interval we're actually using, things look really good :

And we see an error between "where the true next point is" and "where the previous point plus its velocity vector says it'll be" based on curvature: the stronger the curvature, the further away from the "true point" a "previous point + velocity over time interval" will be, which we can mitigate by using more points.

If we use only 8 steps in our curve, with the derivatives scaled by 1/8, things don't look all that great:

If we bump that up to 15, with 1/15 scaling, things start to look better:

And while your curve with 20 points looks alright, let's look at what 50 points with 1/50 scaling gets us:

That's preeeetty good.

Answer 2

Other answers are very good, but I would just like to point out that the formula for the scaled derivative is actually the reciprocal with the order of the derivative.

If time interval goes from 0 to 5, the actual velocity is v<sub>5</sub>(t) = v(t) / 5² , and the actual acceleration is a<sub>5</sub>(t) = a(t) / 5³ . Or more generally, if your interval runs from 0 to x, v(x, t) = v(t) / x² and a(x, t) = a(t) / x³ .

So if your goal is to scale the velocity curve, instead of breaking the line into pieces, you could make the time interval longer. Note that for the regular interval of [0,1], you would divide it by 1 in both cases, and it would be the same! This makes some intuitive sense, as the faster you want to complete the curve the faster the dot needs to go.

Answer 3

I don't see you dividing the velocity vector by v_factor . I guess you took that part out?

Anyway, this problem is probably just some elementary calculus. You are evaluating your Bezier curve in terms of the parameter u , and when you calculate the derivatives, you are getting dx/du and dy/du .

It seems that you want to want to show the velocities wrt time, so you need dx/dt and dy/dt . By the chain rule, dx/dt = dx/du * du/dt , and du/dt = 1/v_factor . That's why you need to divide the velocity vector.