# Motion profile, the maximum velocity allowed from stop

I´m trying to calculate what the maximum allowed velocity for a mechanical axis traveling towards a stop. The maximum velocity must be a secure speed so the axis have is able to stop in time.

I have:

• current position
• current distance to stop
• current velocity
• acceleration/deceleration constant
• jerk constant

I did found this formula

$$\frac{v^2}{2a}+\frac{va}{2j}=d$$

where

• $a$ = acceleration,
• $v$ = velocity
• $j$ = jerk
• $d$ = stop distance

but I don't know how to calculate the $v$ instead.

Is this the correct formula to use and in that case, how should I use it?

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In case the formula is correct (and I don't know), it is a quadratic equation with respect to $v$, so apply the corresponding formula. – enzotib Aug 13 '12 at 10:33
Ah alright, have I got it right like this? v = -(a/2j)+sqrt((a/2j)^2-2a*d) ) / a – stamp Aug 13 '12 at 11:41
I find $a\left(-\frac{a}{2j}\pm\sqrt{\left(\frac{a}{2j}\right)^2+\frac{2d}{a}}\right)$. – enzotib Aug 13 '12 at 12:25
@stamp: I don't think your question has anything to do with algebraic geometry. Maybe you should retag your question as "physics", or something similar. – Nils Matthes Aug 13 '12 at 13:32
This seems like a question along the lines of find and solve eqns of motion of a rigid cylinder rolling toward a stop, or a rigid cylinder sliding toward a stop. Is that right? – Neal Aug 13 '12 at 14:41

I presume $a$ is the maximum allowed acceleration, $j$ is the maximum allowed rate of change in acceleration, and you want to arrive at the stop with zero velocity and zero acceleration. You can run it backwards to find the velocity attained when you start at the stop. Then you ramp the acceleration up to maximum from zero. This phase lasts until $t_1=\frac aj.$ During this phase, $$v_1(t)=\int_0^t a(t')dt'=\frac{jt^2}2$$
$$d_1(t)=\int_0^t v(t')dt'=\frac {jt^3}6$$ so the allowable velocity is $$v_{1a}=\sqrt[\uproot {4}\scriptstyle 3]{\frac {9d_1^2j}2}$$covering a distance $$D=\int_0^{\frac aj}v(t)dt=\left. \frac {jt^3}6 \right|_0^{\frac aj}=\frac{a^3}{6j^2}$$
at a final velocity of $$V=\frac {a^2}{2j}$$ When you are farther away, you accelerate at $a$ and the $j$ doesn't enter. You then have $$v_2(t)=V+a(t-t_1)$$ and reach $$d_2(t)=D+\int_{t_1}^t v_2(t')dt'=D+\frac 12 a(t-t_1)^2$$ so the allowable velocity is now $$v_{2a}=V+\sqrt{2a(d_2-D)}$$