重力火車 - 推理遊戲
![Kelly avatar](/img/cat3.jpg)
By Kelly
at 2009-08-15T06:41
at 2009-08-15T06:41
Table of Contents
加速度 會隨著 離地球中心的距離r 變化,如下:
GM
r"﹦── → (r^2)r"﹦GM《一撇是對時間t微分:速度、兩撇是微兩次:加速度》
r^2
設 r ﹦ α cos(at)+ β sin(bt)《用 三角函數 比用 exp 好算》
則 r'﹦- a α sin(at)+ b β cos(bt)
r"﹦-(a^2)α cos(at)-(b^2)β sin(bt)
用一些條件來決定參數~寫出完整的r:
│r ﹦R《從地表出發》→ R ﹦α → r ﹦R cos(at)+β sin(bt)
當t為零時│r'﹦0《一開始速度為零》→ 0 ﹦bβ →β sin(bt) ﹦0 → r ﹦R cos(at)
│r"﹦-g《地表加速度約為-g》→ -g ﹦-(a^2)R → a ﹦√(g/R)
得到 r ﹦R cos(√(g/R)t)
0 T/2 T/2
最後:∫dr ﹦-R√(g/R)∫sin(√(g/R)t) dt → 1 ﹦[-cos(√(g/R)t)]∕
R 0 0
→ 1 ﹦1-cos(√(g/R)(T/2)) → cos(√(g/R)(T/2)) ﹦0 →√(g/R)(T/2) ﹦π/2
→ T ﹦π√(R/g)《其實好像根本就不需要微分在積分唄,只是習慣而已,好傻的感覺》
算出來跟 euleramon 一樣(點頭)
P.S.我在想,如果加上 離心力 與 科氏力 的話要怎麼算?(應該跟所在緯度有關吧?)
或許就不會掉到"對面"了,那進去的路徑會是怎樣?螺旋嗎?科科
--
GM
r"﹦── → (r^2)r"﹦GM《一撇是對時間t微分:速度、兩撇是微兩次:加速度》
r^2
設 r ﹦ α cos(at)+ β sin(bt)《用 三角函數 比用 exp 好算》
則 r'﹦- a α sin(at)+ b β cos(bt)
r"﹦-(a^2)α cos(at)-(b^2)β sin(bt)
用一些條件來決定參數~寫出完整的r:
│r ﹦R《從地表出發》→ R ﹦α → r ﹦R cos(at)+β sin(bt)
當t為零時│r'﹦0《一開始速度為零》→ 0 ﹦bβ →β sin(bt) ﹦0 → r ﹦R cos(at)
│r"﹦-g《地表加速度約為-g》→ -g ﹦-(a^2)R → a ﹦√(g/R)
得到 r ﹦R cos(√(g/R)t)
0 T/2 T/2
最後:∫dr ﹦-R√(g/R)∫sin(√(g/R)t) dt → 1 ﹦[-cos(√(g/R)t)]∕
R 0 0
→ 1 ﹦1-cos(√(g/R)(T/2)) → cos(√(g/R)(T/2)) ﹦0 →√(g/R)(T/2) ﹦π/2
→ T ﹦π√(R/g)《其實好像根本就不需要微分在積分唄,只是習慣而已,好傻的感覺》
算出來跟 euleramon 一樣(點頭)
P.S.我在想,如果加上 離心力 與 科氏力 的話要怎麼算?(應該跟所在緯度有關吧?)
或許就不會掉到"對面"了,那進去的路徑會是怎樣?螺旋嗎?科科
--
Tags:
推理遊戲
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By Doris
at 2009-08-16T14:07
at 2009-08-16T14:07
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at 2009-08-16T18:30
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