這次換我有問題了XD

基本上是這樣的,我在寫一段程式碼,要模擬一顆球在碗裡移動.假設沒有滾動,但有摩擦力.那理論上來說過一陣子,球就會靜止停在碗底.

我這樣推公式的:

Assuming that the "bowl" look like y=x^2

a=g(-sin(tan^-1 (dy/dx))-\mu_k  cos(tan^-1(dy/dx))*v/|v|)

This First part is the gravity "drive", and the second part is the frictional force part. the v/|v| is to make sure that the frictional force is acting against the direction of velocity.

With a in mind, we could - to first approximation,

v(t)=a(t-1)*time(t)+v(t-1)

x(t)=0.5*v(t)*time(t)+0.5*v(t-1)*t+x(t-1)

但這是我跑出來的結果:

t   x   v    a

  1    0.100000    0.500000   -6.864065

  2    0.512719   -0.872813   -3.668453

  3   -1.092503   -1.973349   11.173711

  4   -4.539974    2.496136    9.392545

  5    3.498467    7.192408  -10.606819

  6   25.324187    0.828317  -10.096752

  7   26.039503   -6.239409   -9.902167

  8   -4.582591  -14.161143   10.483327

  9  -70.434501   -4.726148   10.035241

 10  -91.410695    5.309093    9.972502

 11  -53.257318   16.278845    9.952619

 12   61.348947   28.221989  -10.040417

 13  254.652016   15.169446  -10.009798

 14  361.647149    1.155729  -10.006903

 15  359.924149  -13.854626   -9.993044

 16  225.212345  -29.843497   -9.988875

 17  -68.258279  -46.824585   10.036356

 18 -515.562769  -28.759143   10.004844

 19 -798.037065   -9.749938   10.003131

 20 -885.280126   10.256323    9.997174

大家可以看的出來,這是越甩越遠...

我現在不是很確定哪邊出問題了...

這是cos(atan(dy/dx), -sin(atan(dy/dx)) 看起來數字正負號很正確

    0.049938    0.998752

    0.055470    0.998460

    0.062378    0.998053

    0.071247    0.997459

    0.083046    0.996546

    0.099504    0.995037

    0.124035    0.992278

    0.164399    0.986394

    0.242536    0.970142

    0.447214    0.894427

    1.000000   -0.000000

    0.447214   -0.894427

    0.242536   -0.970142

    0.164399   -0.986394

    0.124035   -0.992278

    0.099504   -0.995037

    0.083046   -0.996546

    0.071247   -0.997459

    0.062378   -0.998053

    0.055470   -0.998460

    0.049938   -0.998752

難道問題出在滾動?