一般形式的加速度梯形算法

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一般形式的加速度梯形算法

已知线段长度 L L L，起点速度 v 0 v_0 v0​，利用加速度梯形算法计算能达到的最大终点速度和最小终点速度。其中，最大加速度为 a m a_m am​。

计算能达到的最大终点速度 v m v_m vm​

设加加速时最大加加速度为 J m J_m Jm​，减加速时最大加加速度为 J m ′ = J m α J'_m=\dfrac{J_m}{\alpha} Jm′​=αJm​​。

加速到最大加速度时运动的距离S

首先，计算按照最大加加速度，加速到最大加速度 a m a_m am​时运动的距离 S S S，其加速度变化示意图如下

计算运动的加速度
a = { J m t , 0 ⩽ t ⩽ t 1 , a m − J m α t , 0 ⩽ t ⩽ α t 1 . \begin{aligned} a=\begin{cases} J_mt, &0\leqslant t\leqslant t_1, \\ a_m-\dfrac{J_m}{\alpha}t, &0\leqslant t\leqslant \alpha t_1. \end{cases} \end{aligned} a=⎩⎨⎧​Jm​t,am​−αJm​​t,​0⩽t⩽t1​,0⩽t⩽αt1​.​​
其中， t 1 = a m J m t_1=\dfrac{a_m}{J_m} t1​=Jm​am​​。

计算运动的速度
v = { v 0 + 1 2 J m t 2 , 0 ⩽ t ⩽ t 1 , v 1 + a m t − 1 2 J m α t 2 , 0 ⩽ t ⩽ α t 1 . \begin{aligned} v=\begin{cases} v_0+\dfrac{1}{2}J_mt^2, &0\leqslant t\leqslant t_1, \\ v_1+a_mt-\dfrac{1}{2}\dfrac{J_m}{\alpha}t^2, &0\leqslant t\leqslant \alpha t_1. \end{cases} \end{aligned} v=⎩⎪⎨⎪⎧​v0​+21​Jm​t2,v1​+am​t−21​αJm​​t2,​0⩽t⩽t1​,0⩽t⩽αt1​.​​
其中， v 1 = v 0 + 1 2 J m t 1 2 = v 0 + 1 2 a m t 1 v_1=v_0+\dfrac{1}{2}J_mt_1^2=v_0+\dfrac{1}{2}a_mt_1 v1​=v0​+21​Jm​t12​=v0​+21​am​t1​。

计算运动的距离
s = { v 0 t + 1 6 J m t 3 , 0 ⩽ t ⩽ t 1 , s 1 + v 1 t + 1 2 a m t 2 − 1 6 J m α t 3 , 0 ⩽ t ⩽ α t 1 . \begin{aligned} s=\begin{cases} v_0t+\dfrac{1}{6}J_mt^3, &0\leqslant t\leqslant t_1, \\ s_1+v_1t+\dfrac{1}{2}a_mt^2-\dfrac{1}{6}\dfrac{J_m}{\alpha}t^3, &0\leqslant t\leqslant \alpha t_1. \end{cases} \end{aligned} s=⎩⎪⎨⎪⎧​v0​t+61​Jm​t3,s1​+v1​t+21​am​t2−61​αJm​​t3,​0⩽t⩽t1​,0⩽t⩽αt1​.​​
其中， s 1 = v 0 t 1 + 1 6 J m t 1 3 = v 0 t 1 + 1 6 a m t 1 2 s_1=v_0t_1+\dfrac{1}{6}J_mt_1^3=v_0t_1+\dfrac{1}{6}a_mt_1^2 s1​=v0​t1​+61​Jm​t13​=v0​t1​+61​am​t12​。

加速到最大加速度时运动的距离 S S S为
S = s 1 + v 1 α t 1 + 1 2 a m α 2 t 1 2 − 1 6 J m α 2 t 1 3 = v 0 t 1 + 1 6 a m t 1 2 + ( v 0 + 1 2 a m t 1 ) α t 1 + 1 3 a m α 2 t 1 2 = v 0 t 1 ( 1 + α ) + 1 6 a m t 1 2 ( 1 + 3 α + 2 α 2 ) . \begin{aligned} S&=s_1+v_1\alpha t_1+\dfrac{1}{2}a_m\alpha^2t_1^2-\dfrac{1}{6}J_m\alpha^2t_1^3 \\ &=v_0t_1+\dfrac{1}{6}a_mt_1^2+(v_0+\dfrac{1}{2}a_mt_1)\alpha t_1+\dfrac{1}{3}a_m\alpha^2t_1^2 \\ &=v_0t_1(1+\alpha)+\dfrac{1}{6}a_mt_1^2(1+3\alpha+2\alpha^2). \end{aligned} S​=s1​+v1​αt1​+21​am​α2t12​−61​Jm​α2t13​=v0​t1​+61​am​t12​+(v0​+21​am​t1​)αt1​+31​am​α2t12​=v0​t1​(1+α)+61​am​t12​(1+3α+2α2).​

线段的长度 L > S L>S L>S

若 L > S L>S L>S，则整个加速过程包含加加速运动、匀加速运动和减加速运动，其加速度变化示意图如下

计算运动的加速度
a = { J m t , 0 ⩽ t ⩽ t 1 , a m , 0 ⩽ t ⩽ t 2 , a m − J m α t , 0 ⩽ t ⩽ α t 1 . \begin{aligned} a=\begin{cases} J_mt, &0\leqslant t\leqslant t_1, \\ a_m, &0\leqslant t\leqslant t_2, \\ a_m-\dfrac{J_m}{\alpha}t, &0\leqslant t\leqslant \alpha t_1. \end{cases} \end{aligned} a=⎩⎪⎪⎨⎪⎪⎧​Jm​t,am​,am​−αJm​​t,​0⩽t⩽t1​,0⩽t⩽t2​,0⩽t⩽αt1​.​​
其中， t 1 = a m J m t_1=\dfrac{a_m}{J_m} t1​=Jm​am​​。

计算运动的速度
v = { v 0 + 1 2 J m t 2 , 0 ⩽ t ⩽ t 1 , v 1 + a m t , 0 ⩽ t ⩽ t 2 , v 2 + a m t − 1 2 J m α t 2 , 0 ⩽ t ⩽ α t 1 . \begin{aligned} v=\begin{cases} v_0+\dfrac{1}{2}J_mt^2, &0\leqslant t\leqslant t_1, \\ v_1+a_mt, &0\leqslant t\leqslant t_2, \\ v_2+a_mt-\dfrac{1}{2}\dfrac{J_m}{\alpha}t^2, &0\leqslant t\leqslant \alpha t_1. \end{cases} \end{aligned} v=⎩⎪⎪⎪⎨⎪⎪⎪⎧​v0​+21​Jm​t2,v1​+am​t,v2​+am​t−21​αJm​​t2,​0⩽t⩽t1​,0⩽t⩽t2​,0⩽t⩽αt1​.​​
其中
{ v 1 = v 0 + 1 2 J m t 1 2 = v 0 + 1 2 a m t 1 , v 2 = v 1 + a m t 2 = v 0 + 1 2 a m t 1 + a m t 2 , v m = v 2 + a m α t 1 − 1 2 J m α t 1 2 = v 0 + 1 2 a m t 1 + a m t 2 + 1 2 a m α t 1 = v 0 + 1 2 a m t 1 ( 1 + α ) + a m t 2 . \begin{aligned} \begin{cases} v_1&=v_0+\dfrac{1}{2}J_mt_1^2 =v_0+\dfrac{1}{2}a_mt_1, \\ v_2&=v_1+a_mt_2 =v_0+\dfrac{1}{2}a_mt_1+a_mt_2, \\ v_m&=v_2+a_m\alpha t_1-\dfrac{1}{2}J_m\alpha t_1^2 =v_0+\dfrac{1}{2}a_mt_1+a_mt_2+\dfrac{1}{2}a_m\alpha t_1 \\ &=v_0+\dfrac{1}{2}a_mt_1(1+\alpha)+a_mt_2. \end{cases} \end{aligned} ⎩⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎧​v1​v2​vm​​=v0​+21​Jm​t12​=v0​+21​am​t1​,=v1​+am​t2​=v0​+21​am​t1​+am​t2​,=v2​+am​αt1​−21​Jm​αt12​=v0​+21​am​t1​+am​t2​+21​am​αt1​=v0​+21​am​t1​(1+α)+am​t2​.​​
计算运动的距离
s = { v 0 t + 1 6 J m t 3 , 0 ⩽ t ⩽ t 1 , s 1 + v 1 t + 1 2 a m t 2 , 0 ⩽ t ⩽ t 2 , s 2 + v 2 t + 1 2 a m t 2 − 1 6 J m α t 3 , 0 ⩽ t ⩽ α t 1 . \begin{aligned} s=\begin{cases} v_0t+\dfrac{1}{6}J_mt^3, &0\leqslant t\leqslant t_1, \\ s_1+v_1t+\dfrac{1}{2}a_mt^2, &0\leqslant t\leqslant t_2, \\ s_2+v_2t+\dfrac{1}{2}a_mt^2-\dfrac{1}{6}\dfrac{J_m}{\alpha}t^3, &0\leqslant t\leqslant \alpha t_1. \end{cases} \end{aligned} s=⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧​v0​t+61​Jm​t3,s1​+v1​t+21​am​t2,s2​+v2​t+21​am​t2−61​αJm​​t3,​0⩽t⩽t1​,0⩽t⩽t2​,0⩽t⩽αt1​.​​
其中
{ s 1 = v 0 t 1 + 1 6 J m t 1 3 = v 0 t 1 + 1 6 a m t 1 2 , s 2 = s 1 + v 1 t 2 + 1 2 a m t 2 2 = v 0 t 1 + 1 6 a m t 1 2 + ( v 0 + 1 2 a m t 1 ) t 2 + 1 2 a m t 2 2 = v 0 ( t 1 + t 2 ) + 1 6 a m t 1 2 + 1 2 a m t 1 t 2 + 1 2 a m t 2 2 , s m = s 2 + v 2 α t 1 + 1 2 a m α 2 t 1 2 − 1 6 J m α 2 t 1 3 = v 0 ( t 1 + t 2 ) + 1 6 a m t 1 2 + 1 2 a m t 1 t 2 + 1 2 a m t 2 2 + ( v 0 + 1 2 a m t 1 + a m t 2 ) α t 1 + 1 3 a m α 2 t 1 2 = v 0 ( t 1 + t 2 + α t 1 ) + 1 6 a m t 1 2 ( 1 + 3 α + 2 α 2 ) + 1 2 a m t 1 t 2 ( 1 + 2 α ) + 1 2 a m t 2 2 = L . \begin{aligned} \begin{cases} s_1&=v_0t_1+\dfrac{1}{6}J_mt_1^3=v_0t_1+\dfrac{1}{6}a_mt_1^2, \\ s_2&=s_1+v_1t_2+\dfrac{1}{2}a_mt_2^2 =v_0t_1+\dfrac{1}{6}a_mt_1^2+(v_0+\dfrac{1}{2}a_mt_1)t_2+\dfrac{1}{2}a_mt_2^2 \\ &=v_0(t_1+t_2)+\dfrac{1}{6}a_mt_1^2+\dfrac{1}{2}a_mt_1t_2+\dfrac{1}{2}a_mt_2^2, \\ s_m&=s_2+v_2\alpha t_1+\dfrac{1}{2}a_m\alpha^2t_1^2-\dfrac{1}{6}J_m\alpha^2t_1^3 \\ &=v_0(t_1+t_2)+\dfrac{1}{6}a_mt_1^2+\dfrac{1}{2}a_mt_1t_2+\dfrac{1}{2}a_mt_2^2 +(v_0+\dfrac{1}{2}a_mt_1+a_mt_2)\alpha t_1+\dfrac{1}{3}a_m\alpha^2t_1^2 \\ &=v_0(t_1+t_2+\alpha t_1)+\dfrac{1}{6}a_mt_1^2(1+3\alpha+2\alpha^2) +\dfrac{1}{2}a_mt_1t_2(1+2\alpha)+\dfrac{1}{2}a_mt_2^2 \\ &=L. \end{cases} \end{aligned} ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​s1​s2​sm​​=v0​t1​+61​Jm​t13​=v0​t1​+61​am​t12​,=s1​+v1​t2​+21​am​t22​=v0​t1​+61​am​t12​+(v0​+21​am​t1​)t2​+21​am​t22​=v0​(t1​+t2​)+61​am​t12​+21​am​t1​t2​+21​am​t22​,=s2​+v2​αt1​+21​am​α2t12​−61​Jm​α2t13​=v0​(t1​+t2​)+61​am​t12​+21​am​t1​t2​+21​am​t22​+(v0​+21​am​t1​+am​t2​)αt1​+31​am​α2t12​=v0​(t1​+t2​+αt1​)+61​am​t12​(1+3α+2α2)+21​am​t1​t2​(1+2α)+21​am​t22​=L.​​
于是，可以得到两个一元方程
{ 1 2 a m t 2 2 + [ 1 2 a m t 1 ( 1 + 2 α ) + v 0 ] t 2 + v 0 t 1 ( 1 + α ) + 1 6 a m t 1 2 ( 1 + 3 α + 2 α 2 ) − L = 0 , v m = v 0 + 1 2 a m t 1 ( 1 + α ) + a m t 2 . \begin{aligned} \begin{cases} \dfrac{1}{2}a_mt_2^2+\left[\dfrac{1}{2}a_mt_1(1+2\alpha)+v_0\right]t_2 +v_0t_1(1+\alpha)+\dfrac{1}{6}a_mt_1^2(1+3\alpha+2\alpha^2)-L=0, \\ v_m=v_0+\dfrac{1}{2}a_mt_1(1+\alpha)+a_mt_2. \end{cases} \end{aligned} ⎩⎪⎨⎪⎧​21​am​t22​+[21​am​t1​(1+2α)+v0​]t2​+v0​t1​(1+α)+61​am​t12​(1+3α+2α2)−L=0,vm​=v0​+21​am​t1​(1+α)+am​t2​.​​
求解第一个一元二次方程得到 t 2 t_2 t2​，然后代入第二个方程即可得到最大终点速度 v m v_m vm​。

线段的长度 L ⩽ S L\leqslant S L⩽S

若 L ⩽ S L\leqslant S L⩽S，则整个加速过程只包含加加速运动和减加速运动，其加速度变化示意图如下

计算运动的加速度
a = { J m t , 0 ⩽ t ⩽ t 1 , J m t 1 − J m α t , 0 ⩽ t ⩽ α t 1 . \begin{aligned} a=\begin{cases} J_mt, &0\leqslant t\leqslant t_1, \\ J_mt_1-\dfrac{J_m}{\alpha}t, &0\leqslant t\leqslant \alpha t_1. \end{cases} \end{aligned} a=⎩⎨⎧​Jm​t,Jm​t1​−αJm​​t,​0⩽t⩽t1​,0⩽t⩽αt1​.​​
计算运动的速度
v = { v 0 + 1 2 J m t 2 , 0 ⩽ t ⩽ t 1 , v 1 + J m t 1 t − 1 2 J m α t 2 , 0 ⩽ t ⩽ α t 1 . \begin{aligned} v=\begin{cases} v_0+\dfrac{1}{2}J_mt^2, &0\leqslant t\leqslant t_1, \\ v_1+J_mt_1t-\dfrac{1}{2}\dfrac{J_m}{\alpha}t^2, &0\leqslant t\leqslant \alpha t_1. \end{cases} \end{aligned} v=⎩⎪⎨⎪⎧​v0​+21​Jm​t2,v1​+Jm​t1​t−21​αJm​​t2,​0⩽t⩽t1​,0⩽t⩽αt1​.​​
其中
{ v 1 = v 0 + 1 2 J m t 1 2 , v m = v 1 + J m t 1 α t 1 − 1 2 J m α t 1 2 = v 0 + 1 2 J m t 1 2 + 1 2 J m α t 1 2 = v 0 + 1 2 J m t 1 2 ( 1 + α ) . \begin{aligned} \begin{cases} v_1&=v_0+\dfrac{1}{2}J_mt_1^2, \\ v_m&=v_1+J_mt_1\alpha t_1-\dfrac{1}{2}J_m\alpha t_1^2 =v_0+\dfrac{1}{2}J_mt_1^2+\dfrac{1}{2}J_m\alpha t_1^2 \\ &=v_0+\dfrac{1}{2}J_mt_1^2(1+\alpha). \end{cases} \end{aligned} ⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧​v1​vm​​=v0​+21​Jm​t12​,=v1​+Jm​t1​αt1​−21​Jm​αt12​=v0​+21​Jm​t12​+21​Jm​αt12​=v0​+21​Jm​t12​(1+α).​​
计算运动的距离
s = { v 0 t + 1 6 J m t 3 , 0 ⩽ t ⩽ t 1 , s 1 + v 1 t + 1 2 J m t 1 t 2 − 1 6 J m α t 3 , 0 ⩽ t ⩽ α t 1 . \begin{aligned} s=\begin{cases} v_0t+\dfrac{1}{6}J_mt^3, &0\leqslant t\leqslant t_1, \\ s_1+v_1t+\dfrac{1}{2}J_mt_1t^2-\dfrac{1}{6}\dfrac{J_m}{\alpha}t^3, &0\leqslant t\leqslant \alpha t_1. \end{cases} \end{aligned} s=⎩⎪⎨⎪⎧​v0​t+61​Jm​t3,s1​+v1​t+21​Jm​t1​t2−61​αJm​​t3,​0⩽t⩽t1​,0⩽t⩽αt1​.​​
其中
{ s 1 = v 0 t 1 + 1 6 J m t 1 3 , s m = s 1 + v 1 α t 1 + 1 2 J m t 1 α 2 t 1 2 − 1 6 J m α 2 t 1 3 = v 0 t 1 + 1 6 J m t 1 3 + ( v 0 + 1 2 J m t 1 2 ) α t 1 + 1 3 J m α 2 t 1 3 = v 0 t 1 ( 1 + α ) + 1 6 J m t 1 3 ( 1 + 3 α + 2 α 2 ) = L . \begin{aligned} \begin{cases} s_1&=v_0t_1+\dfrac{1}{6}J_mt_1^3, \\ s_m&=s_1+v_1\alpha t_1+\dfrac{1}{2}J_mt_1\alpha^2t_1^2-\dfrac{1}{6}J_m\alpha^2t_1^3 \\ &=v_0t_1+\dfrac{1}{6}J_mt_1^3 +(v_0+\dfrac{1}{2}J_mt_1^2)\alpha t_1 +\dfrac{1}{3}J_m\alpha^2t_1^3 \\ &=v_0t_1(1+\alpha)+\dfrac{1}{6}J_mt_1^3(1+3\alpha+2\alpha^2) \\ &=L. \end{cases} \end{aligned} ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​s1​sm​​=v0​t1​+61​Jm​t13​,=s1​+v1​αt1​+21​Jm​t1​α2t12​−61​Jm​α2t13​=v0​t1​+61​Jm​t13​+(v0​+21​Jm​t12​)αt1​+31​Jm​α2t13​=v0​t1​(1+α)+61​Jm​t13​(1+3α+2α2)=L.​​
于是，可以得到两个一元方程
{ 1 6 J m ( 1 + 3 α + 2 α 2 ) t 1 3 + v 0 ( 1 + α ) t 1 − L = 0 , v m = v 0 + 1 2 J m t 1 2 ( 1 + α ) . \begin{aligned} \begin{cases} \dfrac{1}{6}J_m(1+3\alpha+2\alpha^2)t_1^3+v_0(1+\alpha)t_1-L=0, \\ v_m=v_0+\dfrac{1}{2}J_mt_1^2(1+\alpha). \end{cases} \end{aligned} ⎩⎪⎨⎪⎧​61​Jm​(1+3α+2α2)t13​+v0​(1+α)t1​−L=0,vm​=v0​+21​Jm​t12​(1+α).​​
求解第一个一元三次方程得到 t 1 t_1 t1​，然后代入第二个方程即可得到最大终点速度 v m v_m vm​。

计算能达到的最小终点速度 v m v_m vm​

设减减速时最大加加速度为 J m J_m Jm​，加减速时最大加加速度为 J m ′ = J m β J'_m=\dfrac{J_m}{\beta} Jm′​=βJm​​。

对于减速运动，由于速度值不小于零，因此先判断是否能够减速到零。

判断能否减速到零

按照给定的条件，从零反向加速，计算能达到的最大终点速度 v m a x v_{max} vmax​。如果 v m a x ⩾ v 0 v_{max}\geqslant v_0 vmax​⩾v0​，则能达到的最小终点速度 v m = 0 v_m=0 vm​=0。以下均假设 v m a x < v 0 v_{max}<v_0 vmax​<v0​，即 v m > 0 v_m>0 vm​>0成立。

减速到最大加速度时运动的速度 V V V和距离 S S S

计算按照最大加加速度，减速到最大加速度 a m a_m am​时运动的速度 V V V和距离 S S S，其加速度变化示意图如下

计算运动的加速度
a = { − J m β t , 0 ⩽ t ⩽ β t 1 , − a m + J m t , 0 ⩽ t ⩽ t 1 . \begin{aligned} a=\begin{cases} -\dfrac{J_m}{\beta}t, &0\leqslant t\leqslant\beta t_1, \\ -a_m+J_mt, &0\leqslant t\leqslant t_1. \end{cases} \end{aligned} a=⎩⎨⎧​−βJm​​t,−am​+Jm​t,​0⩽t⩽βt1​,0⩽t⩽t1​.​​
其中， t 1 = a m J m t_1=\dfrac{a_m}{J_m} t1​=Jm​am​​。

计算运动的速度
v = { v 0 − 1 2 J m β t 2 , 0 ⩽ t ⩽ β t 1 , v 1 − a m t + 1 2 J m t 2 , 0 ⩽ t ⩽ t 1 . \begin{aligned} v=\begin{cases} v_0-\dfrac{1}{2}\dfrac{J_m}{\beta}t^2, &0\leqslant t\leqslant\beta t_1, \\ v_1-a_mt+\dfrac{1}{2}J_mt^2, &0\leqslant t\leqslant t_1. \end{cases} \end{aligned} v=⎩⎪⎨⎪⎧​v0​−21​βJm​​t2,v1​−am​t+21​Jm​t2,​0⩽t⩽βt1​,0⩽t⩽t1​.​​
其中
{ v 1 = v 0 − 1 2 J m β t 1 2 = v 0 − 1 2 a m β t 1 , v m = v 1 − a m t 1 + 1 2 J m t 1 2 = v 0 − 1 2 a m β t 1 − 1 2 a m t 1 = v 0 − 1 2 a m t 1 ( 1 + β ) . \begin{aligned} \begin{cases} v_1&=v_0-\dfrac{1}{2}J_m\beta t_1^2=v_0-\dfrac{1}{2}a_m\beta t_1, \\ v_m&=v_1-a_mt_1+\dfrac{1}{2}J_mt_1^2=v_0-\dfrac{1}{2}a_m\beta t_1-\dfrac{1}{2}a_mt_1 \\ &=v_0-\dfrac{1}{2}a_mt_1(1+\beta). \end{cases} \end{aligned} ⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧​v1​vm​​=v0​−21​Jm​βt12​=v0​−21​am​βt1​,=v1​−am​t1​+21​Jm​t12​=v0​−21​am​βt1​−21​am​t1​=v0​−21​am​t1​(1+β).​​
计算运动的距离
s = { v 0 t − 1 6 J m β t 3 , 0 ⩽ t ⩽ β t 1 , s 1 + v 1 t − 1 2 a m t 2 + 1 6 J m t 3 , 0 ⩽ t ⩽ t 1 . \begin{aligned} s=\begin{cases} v_0t-\dfrac{1}{6}\dfrac{J_m}{\beta}t^3, &0\leqslant t\leqslant\beta t_1, \\ s_1+v_1t-\dfrac{1}{2}a_mt^2+\dfrac{1}{6}J_mt^3, &0\leqslant t\leqslant t_1. \end{cases} \end{aligned} s=⎩⎪⎨⎪⎧​v0​t−61​βJm​​t3,s1​+v1​t−21​am​t2+61​Jm​t3,​0⩽t⩽βt1​,0⩽t⩽t1​.​​
其中
{ s 1 = v 0 β t 1 − 1 6 J m β 2 t 1 3 = v 0 β t 1 − 1 6 a m β 2 t 1 2 , s m = s 1 + v 1 t 1 − 1 2 a m t 1 2 + 1 6 J m t 1 3 = v 0 β t 1 − 1 6 a m β 2 t 1 2 + ( v 0 − 1 2 a m β t 1 ) t 1 − 1 3 a m t 1 2 = v 0 t 1 ( 1 + β ) − 1 6 a m t 1 2 ( β 2 + 3 β + 2 ) . \begin{aligned} \begin{cases} s_1&=v_0\beta t_1-\dfrac{1}{6}J_m\beta^2t_1^3=v_0\beta t_1-\dfrac{1}{6}a_m\beta^2t_1^2, \\ s_m&=s_1+v_1t_1-\dfrac{1}{2}a_mt_1^2+\dfrac{1}{6}J_mt_1^3 =v_0\beta t_1-\dfrac{1}{6}a_m\beta^2t_1^2+(v_0-\dfrac{1}{2}a_m\beta t_1)t_1-\dfrac{1}{3}a_mt_1^2 \\ &=v_0t_1(1+\beta)-\dfrac{1}{6}a_mt_1^2(\beta^2+3\beta+2). \end{cases} \end{aligned} ⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧​s1​sm​​=v0​βt1​−61​Jm​β2t13​=v0​βt1​−61​am​β2t12​,=s1​+v1​t1​−21​am​t12​+61​Jm​t13​=v0​βt1​−61​am​β2t12​+(v0​−21​am​βt1​)t1​−31​am​t12​=v0​t1​(1+β)−61​am​t12​(β2+3β+2).​​
减速到最大加速度时运动的速度 V V V和 S S S为
{ V = v m = v 0 − 1 2 a m t 1 ( 1 + β ) , S = s m = v 0 t 1 ( 1 + β ) − 1 6 a m t 1 2 ( β 2 + 3 β + 2 ) . \begin{aligned} \begin{cases} V=v_m=v_0-\dfrac{1}{2}a_mt_1(1+\beta), \\ S=s_m=v_0t_1(1+\beta)-\dfrac{1}{6}a_mt_1^2(\beta^2+3\beta+2). \end{cases} \end{aligned} ⎩⎪⎨⎪⎧​V=vm​=v0​−21​am​t1​(1+β),S=sm​=v0​t1​(1+β)−61​am​t12​(β2+3β+2).​​

V ⩽ 0 V\leqslant0 V⩽0或者线段的长度 L ⩽ S L\leqslant S L⩽S

若 V ⩽ 0 V\leqslant0 V⩽0或者 L ⩽ S L\leqslant S L⩽S，则整个减速过程只包含加减速运动和减减速运动，其加速度变化示意图如下

计算运动的加速度
a = { − J m β t , 0 ⩽ t ⩽ β t 1 , − J m t 1 + J m t , 0 ⩽ t ⩽ t 1 . \begin{aligned} a=\begin{cases} -\dfrac{J_m}{\beta}t, &0\leqslant t\leqslant\beta t_1, \\ -J_mt_1+J_mt, &0\leqslant t\leqslant t_1. \end{cases} \end{aligned} a=⎩⎨⎧​−βJm​​t,−Jm​t1​+Jm​t,​0⩽t⩽βt1​,0⩽t⩽t1​.​​
计算运动的速度
v = { v 0 − 1 2 J m β t 2 , 0 ⩽ t ⩽ β t 1 , v 1 − J m t 1 t + 1 2 J m t 2 , 0 ⩽ t ⩽ t 1 . \begin{aligned} v=\begin{cases} v_0-\dfrac{1}{2}\dfrac{J_m}{\beta}t^2, &0\leqslant t\leqslant\beta t_1, \\ v_1-J_mt_1t+\dfrac{1}{2}J_mt^2, &0\leqslant t\leqslant t_1. \end{cases} \end{aligned} v=⎩⎪⎨⎪⎧​v0​−21​βJm​​t2,v1​−Jm​t1​t+21​Jm​t2,​0⩽t⩽βt1​,0⩽t⩽t1​.​​
其中
{ v 1 = v 0 − 1 2 J m β t 1 2 , v m = v 1 − J m t 1 t 1 + 1 2 J m t 1 2 = v 0 − 1 2 J m β t 1 2 − 1 2 J m t 1 2 = v 0 − 1 2 J m t 1 2 ( 1 + β ) . \begin{aligned} \begin{cases} v_1&=v_0-\dfrac{1}{2}J_m\beta t_1^2, \\ v_m&=v_1-J_mt_1t_1+\dfrac{1}{2}J_mt_1^2=v_0-\dfrac{1}{2}J_m\beta t_1^2-\dfrac{1}{2}J_mt_1^2 \\ &=v_0-\dfrac{1}{2}J_mt_1^2(1+\beta). \end{cases} \end{aligned} ⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧​v1​vm​​=v0​−21​Jm​βt12​,=v1​−Jm​t1​t1​+21​Jm​t12​=v0​−21​Jm​βt12​−21​Jm​t12​=v0​−21​Jm​t12​(1+β).​​
计算运动的距离
s = { v 0 t − 1 6 J m β t 3 , 0 ⩽ t ⩽ β t 1 , s 1 + v 1 t − 1 2 J m t 1 t 2 + 1 6 J m t 3 , 0 ⩽ t ⩽ t 1 . \begin{aligned} s=\begin{cases} v_0t-\dfrac{1}{6}\dfrac{J_m}{\beta}t^3, &0\leqslant t\leqslant\beta t_1, \\ s_1+v_1t-\dfrac{1}{2}J_mt_1t^2+\dfrac{1}{6}J_mt^3, &0\leqslant t\leqslant t_1. \end{cases} \end{aligned} s=⎩⎪⎨⎪⎧​v0​t−61​βJm​​t3,s1​+v1​t−21​Jm​t1​t2+61​Jm​t3,​0⩽t⩽βt1​,0⩽t⩽t1​.​​
其中
{ s 1 = v 0 β t 1 − 1 6 J m β 2 t 1 3 , s m = s 1 + v 1 t 1 − 1 2 J m t 1 t 1 2 + 1 6 J m t 1 3 = v 0 β t 1 − 1 6 J m β 2 t 1 3 + ( v 0 − 1 2 J m β t 1 2 ) t 1 − 1 3 J m t 1 3 = v 0 t 1 ( 1 + β ) − 1 6 J m t 1 3 ( β 2 + 3 β + 2 ) = L . \begin{aligned} \begin{cases} s_1&=v_0\beta t_1-\dfrac{1}{6}J_m\beta^2t_1^3, \\ s_m&=s_1+v_1t_1-\dfrac{1}{2}J_mt_1t_1^2+\dfrac{1}{6}J_mt_1^3 =v_0\beta t_1-\dfrac{1}{6}J_m\beta^2t_1^3+(v_0-\dfrac{1}{2}J_m\beta t_1^2)t_1-\dfrac{1}{3}J_mt_1^3 \\ &=v_0t_1(1+\beta)-\dfrac{1}{6}J_mt_1^3(\beta^2+3\beta+2) \\ &=L. \end{cases} \end{aligned} ⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧​s1​sm​​=v0​βt1​−61​Jm​β2t13​,=s1​+v1​t1​−21​Jm​t1​t12​+61​Jm​t13​=v0​βt1​−61​Jm​β2t13​+(v0​−21​Jm​βt12​)t1​−31​Jm​t13​=v0​t1​(1+β)−61​Jm​t13​(β2+3β+2)=L.​​
于是，可以得到两个一元方程
{ 1 6 J m ( β 2 + 3 β + 2 ) t 1 3 − v 0 ( 1 + β ) t 1 + L = 0 , v m = v 0 − 1 2 J m t 1 2 ( 1 + β ) . \begin{aligned} \begin{cases} \dfrac{1}{6}J_m(\beta^2+3\beta+2)t_1^3-v_0(1+\beta)t_1+L=0, \\ v_m=v_0-\dfrac{1}{2}J_mt_1^2(1+\beta). \end{cases} \end{aligned} ⎩⎪⎨⎪⎧​61​Jm​(β2+3β+2)t13​−v0​(1+β)t1​+L=0,vm​=v0​−21​Jm​t12​(1+β).​​
求解第一个一元三次方程得到 t 1 t_1 t1​，然后代入第二个方程即可得到最小终点速度 v m v_m vm​。

线段长度 L > S L>S L>S

若 L > S L>S L>S，则整个减速过程包含加减速运动、匀减速运动和减减速运动，其加速度变化示意图如下

计算运动的加速度
a = { − J m β t , 0 ⩽ t ⩽ β t 1 − a m , 0 ⩽ t ⩽ t 2 , − a m + J m t , 0 ⩽ t ⩽ t 1 . \begin{aligned} a=\begin{cases} -\dfrac{J_m}{\beta}t, &0\leqslant t\leqslant\beta t_1 \\ -a_m, &0\leqslant t\leqslant t_2, \\ -a_m+J_mt, &0\leqslant t\leqslant t_1. \end{cases} \end{aligned} a=⎩⎪⎪⎨⎪⎪⎧​−βJm​​t,−am​,−am​+Jm​t,​0⩽t⩽βt1​0⩽t⩽t2​,0⩽t⩽t1​.​​
其中， t 1 = a m J m t_1=\dfrac{a_m}{J_m} t1​=Jm​am​​。

计算运动的速度
v = { v 0 − 1 2 J m β t 2 , 0 ⩽ t ⩽ β t 1 v 1 − a m t , 0 ⩽ t ⩽ t 2 , v 2 − a m t + 1 2 J m t 2 , 0 ⩽ t ⩽ t 1 . \begin{aligned} v=\begin{cases} v_0-\dfrac{1}{2}\dfrac{J_m}{\beta}t^2, &0\leqslant t\leqslant\beta t_1 \\ v_1-a_mt, &0\leqslant t\leqslant t_2, \\ v_2-a_mt+\dfrac{1}{2}J_mt^2, &0\leqslant t\leqslant t_1. \end{cases} \end{aligned} v=⎩⎪⎪⎪⎨⎪⎪⎪⎧​v0​−21​βJm​​t2,v1​−am​t,v2​−am​t+21​Jm​t2,​0⩽t⩽βt1​0⩽t⩽t2​,0⩽t⩽t1​.​​
其中
{ v 1 = v 0 − 1 2 J m β t 1 2 = v 0 − 1 2 a m β t 1 , v 2 = v 1 − a m t 2 = v 0 − 1 2 a m β t 1 − a m t 2 , v m = v 2 − a m t 1 + 1 2 J m t 1 2 = v 0 − 1 2 a m β t 1 − a m t 2 − 1 2 a m t 1 = v 0 − 1 2 a m t 1 ( 1 + β ) − a m t 2 . \begin{aligned} \begin{cases} v_1&=v_0-\dfrac{1}{2}J_m\beta t_1^2=v_0-\dfrac{1}{2}a_m\beta t_1, \\ v_2&=v_1-a_mt_2=v_0-\dfrac{1}{2}a_m\beta t_1-a_mt_2, \\ v_m&=v_2-a_mt_1+\dfrac{1}{2}J_mt_1^2=v_0-\dfrac{1}{2}a_m\beta t_1-a_mt_2-\dfrac{1}{2}a_mt_1 \\ &=v_0-\dfrac{1}{2}a_mt_1(1+\beta)-a_mt_2. \end{cases} \end{aligned} ⎩⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎧​v1​v2​vm​​=v0​−21​Jm​βt12​=v0​−21​am​βt1​,=v1​−am​t2​=v0​−21​am​βt1​−am​t2​,=v2​−am​t1​+21​Jm​t12​=v0​−21​am​βt1​−am​t2​−21​am​t1​=v0​−21​am​t1​(1+β)−am​t2​.​​
计算运动的距离
s = { v 0 t − 1 6 J m β t 3 , 0 ⩽ t ⩽ β t 1 s 1 + v 1 t − 1 2 a m t 2 , 0 ⩽ t ⩽ t 2 , s 2 + v 2 t − 1 2 a m t 2 + 1 6 J m t 3 , 0 ⩽ t ⩽ t 1 . \begin{aligned} s=\begin{cases} v_0t-\dfrac{1}{6}\dfrac{J_m}{\beta}t^3, &0\leqslant t\leqslant\beta t_1 \\ s_1+v_1t-\dfrac{1}{2}a_mt^2, &0\leqslant t\leqslant t_2, \\ s_2+v_2t-\dfrac{1}{2}a_mt^2+\dfrac{1}{6}J_mt^3, &0\leqslant t\leqslant t_1. \end{cases} \end{aligned} s=⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧​v0​t−61​βJm​​t3,s1​+v1​t−21​am​t2,s2​+v2​t−21​am​t2+61​Jm​t3,​0⩽t⩽βt1​0⩽t⩽t2​,0⩽t⩽t1​.​​
其中
{ s 1 = v 0 β t 1 − 1 6 J m β 2 t 1 3 = v 0 β t 1 − 1 6 a m β 2 t 1 2 , s 2 = s 1 + v 1 t 2 − 1 2 a m t 2 2 = v 0 β t 1 − 1 6 a m β 2 t 1 2 + ( v 0 − 1 2 a m β t 1 ) t 2 − 1 2 a m t 2 2 = v 0 ( β t 1 + t 2 ) − 1 6 a m β 2 t 1 2 − 1 2 a m β t 1 t 2 − 1 2 a m t 2 2 , s m = s 2 + v 2 t 1 − 1 2 a m t 1 2 + 1 6 J m t 1 3 = v 0 ( β t 1 + t 2 ) − 1 6 a m β 2 t 1 2 − 1 2 a m β t 1 t 2 − 1 2 a m t 2 2 + ( v 0 − 1 2 a m β t 1 − a m t 2 ) t 1 − 1 3 a m t 1 2 = v 0 ( β t 1 + t 2 + t 1 ) − 1 6 a m t 1 2 ( β 2 + 3 β + 2 ) − 1 2 a m t 1 t 2 ( β + 2 ) − 1 2 a m t 2 2 = L . \begin{aligned} \begin{cases} s_1&=v_0\beta t_1-\dfrac{1}{6}J_m\beta^2t_1^3=v_0\beta t_1-\dfrac{1}{6}a_m\beta^2t_1^2, \\ s_2&=s_1+v_1t_2-\dfrac{1}{2}a_mt_2^2=v_0\beta t_1-\dfrac{1}{6}a_m\beta^2t_1^2 +(v_0-\dfrac{1}{2}a_m\beta t_1)t_2-\dfrac{1}{2}a_mt_2^2 \\ &=v_0(\beta t_1+t_2)-\dfrac{1}{6}a_m\beta^2t_1^2-\dfrac{1}{2}a_m\beta t_1t_2-\dfrac{1}{2}a_mt_2^2, \\ s_m&=s_2+v_2t_1-\dfrac{1}{2}a_mt_1^2+\dfrac{1}{6}J_mt_1^3 \\ &=v_0(\beta t_1+t_2)-\dfrac{1}{6}a_m\beta^2t_1^2-\dfrac{1}{2}a_m\beta t_1t_2-\dfrac{1}{2}a_mt_2^2 +(v_0-\dfrac{1}{2}a_m\beta t_1-a_mt_2)t_1-\dfrac{1}{3}a_mt_1^2 \\ &=v_0(\beta t_1+t_2+t_1)-\dfrac{1}{6}a_mt_1^2(\beta^2+3\beta+2) -\dfrac{1}{2}a_mt_1t_2(\beta+2)-\dfrac{1}{2}a_mt_2^2 \\ &=L. \end{cases} \end{aligned} ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​s1​s2​sm​​=v0​βt1​−61​Jm​β2t13​=v0​βt1​−61​am​β2t12​,=s1​+v1​t2​−21​am​t22​=v0​βt1​−61​am​β2t12​+(v0​−21​am​βt1​)t2​−21​am​t22​=v0​(βt1​+t2​)−61​am​β2t12​−21​am​βt1​t2​−21​am​t22​,=s2​+v2​t1​−21​am​t12​+61​Jm​t13​=v0​(βt1​+t2​)−61​am​β2t12​−21​am​βt1​t2​−21​am​t22​+(v0​−21​am​βt1​−am​t2​)t1​−31​am​t12​=v0​(βt1​+t2​+t1​)−61​am​t12​(β2+3β+2)−21​am​t1​t2​(β+2)−21​am​t22​=L.​​
于是，可以得到两个一元方程
{ 1 2 a m t 2 2 + [ 1 2 a m t 1 ( β + 2 ) − v 0 ] t 2 + 1 6 a m t 1 2 ( β 2 + 3 β + 2 ) − v 0 t 1 ( 1 + β ) + L = 0 , v m = v 0 − 1 2 a m t 1 ( 1 + β ) − a m t 2 . \begin{aligned} \begin{cases} \dfrac{1}{2}a_mt_2^2+\left[\dfrac{1}{2}a_mt_1(\beta+2)-v_0\right]t_2 +\dfrac{1}{6}a_mt_1^2(\beta^2+3\beta+2)-v_0t_1(1+\beta)+L=0, \\ v_m=v_0-\dfrac{1}{2}a_mt_1(1+\beta)-a_mt_2. \end{cases} \end{aligned} ⎩⎪⎨⎪⎧​21​am​t22​+[21​am​t1​(β+2)−v0​]t2​+61​am​t12​(β2+3β+2)−v0​t1​(1+β)+L=0,vm​=v0​−21​am​t1​(1+β)−am​t2​.​​
求解第一个一元二次方程得到 t 2 t_2 t2​，然后代入第二个方程即可得到最小终点速度 v m v_m vm​。

在最短时间内运动的距离 Δ S \Delta S ΔS

已知起点速度 v 0 v_0 v0​，最大速度 v m ( ⩾ v 0 ) v_m(\geqslant v_0) vm​(⩾v0​)，最大加速度为 a m a_m am​，加加速时最大加加速度为 J m J_m Jm​，减加速时最大加加速度为 J m ′ = J m α J'_m=\dfrac{J_m}{\alpha} Jm′​=αJm​​，按照加速度梯形算法计算在最短时间内运动的距离 Δ S \Delta S ΔS。

加速到最大加速度时速度的变化量 Δ V \Delta V ΔV

首先，计算按照最大加加速度，加速到最大加速度 a m a_m am​时速度的变化量 Δ V \Delta V ΔV，其加速度变化示意图如下

t 1 = a m J m , Δ V = 1 2 a m ( t 1 + α t 1 ) = 1 2 a m t 1 ( 1 + α ) . \begin{aligned} &t_1=\dfrac{a_m}{J_m}, \\ &\Delta V=\dfrac{1}{2}a_m(t_1+\alpha t_1)=\dfrac{1}{2}a_mt_1(1+\alpha). \end{aligned} ​t1​=Jm​am​​,ΔV=21​am​(t1​+αt1​)=21​am​t1​(1+α).​

速度变化量 ( v m − v 0 ) > Δ V (v_m-v_0)>\Delta V (vm​−v0​)>ΔV

若 ( v m − v 0 ) > Δ V (v_m-v_0)>\Delta V (vm​−v0​)>ΔV，则加速阶段包含匀加速过程，其加速度变化示意图如下

t 1 = a m J m , v m − v 0 = 1 2 a m t 1 ( 1 + α ) + a m t 2 ⟹ t 2 = v m − v 0 a m − 1 2 t 1 ( 1 + α ) , Δ S = v 0 ( t 1 + t 2 + α t 1 ) + 1 6 a m t 1 2 ( 1 + 3 α + 2 α 2 ) + 1 2 a m t 1 t 2 ( 1 + 2 α ) + 1 2 a m t 2 2 . \begin{aligned} &t_1=\dfrac{a_m}{J_m}, \\ &v_m-v_0=\dfrac{1}{2}a_mt_1(1+\alpha)+a_mt_2 \Longrightarrow t_2=\dfrac{v_m-v_0}{a_m} -\dfrac{1}{2}t_1(1+\alpha), \\ &\Delta S=v_0(t_1+t_2+\alpha t_1)+\dfrac{1}{6}a_mt_1^2(1+3\alpha+2\alpha^2) +\dfrac{1}{2}a_mt_1t_2(1+2\alpha) +\dfrac{1}{2}a_mt_2^2. \end{aligned} ​t1​=Jm​am​​,vm​−v0​=21​am​t1​(1+α)+am​t2​⟹t2​=am​vm​−v0​​−21​t1​(1+α),ΔS=v0​(t1​+t2​+αt1​)+61​am​t12​(1+3α+2α2)+21​am​t1​t2​(1+2α)+21​am​t22​.​

速度变化量 ( v m − v 0 ) ⩽ Δ V (v_m-v_0)\leqslant\Delta V (vm​−v0​)⩽ΔV

若 ( v m − v 0 ) ⩽ Δ V (v_m-v_0)\leqslant\Delta V (vm​−v0​)⩽ΔV，则加速阶段不包含匀加速过程，其加速度变化示意图如下

v m − v 0 = 1 2 J m t 1 2 ( 1 + α ) ⟹ t 1 = 2 ( v m − v 0 ) J m ( 1 + α ) , Δ S = v 0 t 1 ( 1 + α ) + 1 6 J m t 1 3 ( 1 + 3 α + 2 α 2 ) . \begin{aligned} &v_m-v_0=\dfrac{1}{2}J_mt_1^2(1+\alpha) \Longrightarrow t_1=\sqrt{\dfrac{2(v_m-v_0)}{J_m(1+\alpha)}}, \\ &\Delta S=v_0t_1(1+\alpha)+\dfrac{1}{6}J_mt_1^3(1+3\alpha+2\alpha^2). \end{aligned} ​vm​−v0​=21​Jm​t12​(1+α)⟹t1​=Jm​(1+α)2(vm​−v0​)​ ​,ΔS=v0​t1​(1+α)+61​Jm​t13​(1+3α+2α2).​

加速度梯形算法的完整运动过程

假设已知各运动阶段的时间，如下示意图

完整的运动过程包含7个阶段，依次为加加速阶段、匀加速阶段、减加速阶段、匀速阶段、加减速阶段、匀减速阶段、减减速阶段，设起始速度为 v 0 v_0 v0​，加加速阶段最大加加速度为 J m J_m Jm​，减减速阶段最大加加速度为 J n J_n Jn​，则

运动各阶段加速度为
{ 0 ⩽ t ⩽ t 1 , a = J m t , 0 ⩽ t ⩽ t 2 , a = J m t 1 , 0 ⩽ t ⩽ α t 1 , a = J m t 1 − J m α t , 0 ⩽ t ⩽ t 3 , a = 0 , 0 ⩽ t ⩽ β t 4 , a = − J n β t , 0 ⩽ t ⩽ t 5 , a = − J n t 4 , 0 ⩽ t ⩽ t 4 , a = − J n t 4 + J n t . \begin{aligned} \begin{cases} 0\leqslant t\leqslant t_1, &a=J_mt, \\ 0\leqslant t\leqslant t_2, &a=J_mt_1, \\ 0\leqslant t\leqslant \alpha t_1, &a=J_mt_1-\dfrac{J_m}{\alpha}t, \\ 0\leqslant t\leqslant t_3, &a=0, \\ 0\leqslant t\leqslant \beta t_4, &a=-\dfrac{J_n}{\beta}t, \\ 0\leqslant t\leqslant t_5, &a=-J_nt_4, \\ 0\leqslant t\leqslant t_4, &a=-J_nt_4+J_nt. \end{cases} \end{aligned} ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​0⩽t⩽t1​,0⩽t⩽t2​,0⩽t⩽αt1​,0⩽t⩽t3​,0⩽t⩽βt4​,0⩽t⩽t5​,0⩽t⩽t4​,​a=Jm​t,a=Jm​t1​,a=Jm​t1​−αJm​​t,a=0,a=−βJn​​t,a=−Jn​t4​,a=−Jn​t4​+Jn​t.​​
运动各阶段速度为
{ 0 ⩽ t ⩽ t 1 , v = v 0 + 1 2 J m t 2 , 0 ⩽ t ⩽ t 2 , v = v 1 + J m t 1 t , 0 ⩽ t ⩽ α t 1 , v = v 2 + J m t 1 t − 1 2 J m α t 2 , 0 ⩽ t ⩽ t 3 , v = v 3 , 0 ⩽ t ⩽ β t 4 , v = v 4 − 1 2 J n β t 2 , 0 ⩽ t ⩽ t 5 , v = v 5 − J n t 4 t , 0 ⩽ t ⩽ t 4 , v = v 6 − J n t 4 t + 1 2 J n t 2 . ⟸ { v 1 = v 0 + 1 2 J m t 1 2 , v 2 = v 1 + J m t 1 t 2 , v 3 = v 2 + J m t 1 α t 1 − 1 2 J m α t 1 2 = v 2 + 1 2 J m α t 1 2 , v 4 = v 3 , v 5 = v 4 − 1 2 J n β t 4 2 , v 6 = v 5 − J n t 4 t 5 , v e = v 6 − J n t 4 2 + 1 2 J n t 4 2 = v 6 − 1 2 J n t 4 2 . \begin{aligned} \begin{cases} 0\leqslant t\leqslant t_1, &v=v_0+\dfrac{1}{2}J_mt^2, \\ 0\leqslant t\leqslant t_2, &v=v_1+J_mt_1t, \\ 0\leqslant t\leqslant \alpha t_1, &v=v_2+J_mt_1t-\dfrac{1}{2}\dfrac{J_m}{\alpha}t^2, \\ 0\leqslant t\leqslant t_3, &v=v_3, \\ 0\leqslant t\leqslant \beta t_4, &v=v_4-\dfrac{1}{2}\dfrac{J_n}{\beta}t^2, \\ 0\leqslant t\leqslant t_5, &v=v_5-J_nt_4t, \\ 0\leqslant t\leqslant t_4, &v=v_6-J_nt_4t+\dfrac{1}{2}J_nt^2. \end{cases} \Longleftarrow \begin{cases} v_1&=v_0+\dfrac{1}{2}J_mt_1^2, \\ v_2&=v_1+J_mt_1t_2, \\ v_3&=v_2+J_mt_1\alpha t_1-\dfrac{1}{2}J_m\alpha t_1^2 \\ &=v_2+\dfrac{1}{2}J_m\alpha t_1^2, \\ v_4&=v_3, \\ v_5&=v_4-\dfrac{1}{2}J_n\beta t_4^2, \\ v_6&=v_5-J_nt_4t_5, \\ v_e&=v_6-J_nt_4^2+\dfrac{1}{2}J_nt_4^2 \\ &=v_6-\dfrac{1}{2}J_nt_4^2. \end{cases} \end{aligned} ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​0⩽t⩽t1​,0⩽t⩽t2​,0⩽t⩽αt1​,0⩽t⩽t3​,0⩽t⩽βt4​,0⩽t⩽t5​,0⩽t⩽t4​,​v=v0​+21​Jm​t2,v=v1​+Jm​t1​t,v=v2​+Jm​t1​t−21​αJm​​t2,v=v3​,v=v4​−21​βJn​​t2,v=v5​−Jn​t4​t,v=v6​−Jn​t4​t+21​Jn​t2.​⟸⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​v1​v2​v3​v4​v5​v6​ve​​=v0​+21​Jm​t12​,=v1​+Jm​t1​t2​,=v2​+Jm​t1​αt1​−21​Jm​αt12​=v2​+21​Jm​αt12​,=v3​,=v4​−21​Jn​βt42​,=v5​−Jn​t4​t5​,=v6​−Jn​t42​+21​Jn​t42​=v6​−21​Jn​t42​.​​
运动各阶段距离为
{ 0 ⩽ t ⩽ t 1 , s = v 0 t + 1 6 J m t 3 , 0 ⩽ t ⩽ t 2 , s = s 1 + v 1 t + 1 2 J m t 1 t 2 , 0 ⩽ t ⩽ α t 1 , s = s 2 + v 2 t + 1 2 J m t 1 t 2 − 1 6 J m α t 3 , 0 ⩽ t ⩽ t 3 , s = s 3 + v 3 t , 0 ⩽ t ⩽ β t 4 , s = s 4 + v 4 t − 1 6 J n β t 3 , 0 ⩽ t ⩽ t 5 , s = s 5 + v 5 t − 1 2 J n t 4 t 2 , 0 ⩽ t ⩽ t 4 , s = s 6 + v 6 t − 1 2 J n t 4 t 2 + 1 6 J n t 3 . ⟸ { s 1 = v 0 t 1 + 1 6 J m t 1 3 , s 2 = s 1 + v 1 t 2 + 1 2 J m t 1 t 2 2 , s 3 = s 2 + v 2 α t 1 + 1 2 J m t 1 α 2 t 1 2 − 1 6 J m α 2 t 1 3 = s 2 + v 2 α t 1 + 1 3 J m α 2 t 1 3 , s 4 = s 3 + v 3 t 3 , s 5 = s 4 + v 4 β t 4 − 1 6 J n β 2 t 4 3 , s 6 = s 5 + v 5 t 5 − 1 2 J n t 4 t 5 2 , s e = s 6 + v 6 t 4 − 1 2 J n t 4 3 + 1 6 J n t 4 3 = s 6 + v 6 t 4 − 1 3 J n t 4 3 . \begin{aligned} \begin{cases} 0\leqslant t\leqslant t_1, &s=v_0t+\dfrac{1}{6}J_mt^3, \\ 0\leqslant t\leqslant t_2, &s=s_1+v_1t+\dfrac{1}{2}J_mt_1t^2, \\ 0\leqslant t\leqslant \alpha t_1, &s=s_2+v_2t+\dfrac{1}{2}J_mt_1t^2-\dfrac{1}{6}\dfrac{J_m}{\alpha}t^3, \\ 0\leqslant t\leqslant t_3, &s=s_3+v_3t, \\ 0\leqslant t\leqslant \beta t_4, &s=s_4+v_4t-\dfrac{1}{6}\dfrac{J_n}{\beta}t^3, \\ 0\leqslant t\leqslant t_5, &s=s_5+v_5t-\dfrac{1}{2}J_nt_4t^2, \\ 0\leqslant t\leqslant t_4, &s=s_6+v_6t-\dfrac{1}{2}J_nt_4t^2+\dfrac{1}{6}J_nt^3. \end{cases} \Longleftarrow \begin{cases} s_1&=v_0t_1+\dfrac{1}{6}J_mt_1^3, \\ s_2&=s_1+v_1t_2+\dfrac{1}{2}J_mt_1t_2^2, \\ s_3&=s_2+v_2\alpha t_1+\dfrac{1}{2}J_mt_1\alpha^2t_1^2-\dfrac{1}{6}J_m\alpha^2t_1^3 \\ &=s_2+v_2\alpha t_1+\dfrac{1}{3}J_m\alpha^2t_1^3, \\ s_4&=s_3+v_3t_3, \\ s_5&=s_4+v_4\beta t_4-\dfrac{1}{6}J_n\beta^2t_4^3, \\ s_6&=s_5+v_5t_5-\dfrac{1}{2}J_nt_4t_5^2, \\ s_e&=s_6+v_6t_4-\dfrac{1}{2}J_nt_4^3+\dfrac{1}{6}J_nt_4^3 \\ &=s_6+v_6t_4-\dfrac{1}{3}J_nt_4^3. \end{cases} \end{aligned} ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​0⩽t⩽t1​,0⩽t⩽t2​,0⩽t⩽αt1​,0⩽t⩽t3​,0⩽t⩽βt4​,0⩽t⩽t5​,0⩽t⩽t4​,​s=v0​t+61​Jm​t3,s=s1​+v1​t+21​Jm​t1​t2,s=s2​+v2​t+21​Jm​t1​t2−61​αJm​​t3,s=s3​+v3​t,s=s4​+v4​t−61​βJn​​t3,s=s5​+v5​t−21​Jn​t4​t2,s=s6​+v6​t−21​Jn​t4​t2+61​Jn​t3.​⟸⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​s1​s2​s3​s4​s5​s6​se​​=v0​t1​+61​Jm​t13​,=s1​+v1​t2​+21​Jm​t1​t22​,=s2​+v2​αt1​+21​Jm​t1​α2t12​−61​Jm​α2t13​=s2​+v2​αt1​+31​Jm​α2t13​,=s3​+v3​t3​,=s4​+v4​βt4​−61​Jn​β2t43​,=s5​+v5​t5​−21​Jn​t4​t52​,=s6​+v6​t4​−21​Jn​t43​+61​Jn​t43​=s6​+v6​t4​−31​Jn​t43​.​​