Position = FUNCTION(VELOCITY, ACCELERATION, JERK)

[Boothby171]

Does anyone have a reference for equations of motion for an S-Curve?

I'm looking for triangular acceleration ramps (step jerks), and eventually want to be able to calculate the force (acceleration) and therefore the power required to get a piece from point (a) to point (b) in a known time, with a known peak jerk, acceleration, and velocity.

You know--the usual.

I had done the math for a sinusoidal acceleration (offset half-cosine jerk) years ago, and, um, I lost it...really!

 

[nbucska]

a=dv/dt;
v=dx/dt;
to do it backwards:
v=v0 + integral(a*dt); and
x=x0 + integral(v*dt) ; where "*" marks multiplication

Both intergals are from beginning of the motion
at t=0 to "t" time. Both v0 and x0 are measured at t=0.







<nbucska@pcperipherals DOT com> subj: eng-tips
read FAQ240-1032

 

[Boothby171]

Thank you, nbucska, but I was looking for a more specific solution, given the following acceleration profile:

[tt]___/\ _________ /\__
\/ \/[/tt]

 

[nbucska]

Even if you have forgotten integration, a simple program
-- even in GWBASIC -- can solve your problem.



<nbucska@pcperipherals DOT com> subj: eng-tips
read FAQ240-1032

 

[Boothby171]

nbucs,

and not to forget that

a=a0 + integral(j*dt)

I need the overall equation of motion (position as a function of...)

given:

@ t=0
x=0
v=0
a=0
j=0

@ t=t1
x1=integral(integral(integral(j)))
v1=integral(integral(j))
a1=integral(j)
j1=jmax (j has gone from j=0 at t=0, to j=jmax at t=t1)

@ t=t2
x2=x1 + integral(integral(integral(j)))
v2=v1 + integral(integral(j))
a2=a1 + integral(j)=0
j2=0 (j has gone from j=jmax at t=t1, to j=0 at t=t2)

from t2 to t3, j=0, a=0, v=constant, x=x2+v*(t-t2), etc.

Once I have that equation, I plan on turning it inside out a number of different ways, so that I can isolate (and thereby solve for), v, a, j, x(total), etc.

I'm not asking anyone to do it for me. But I am hoping there's another geek out there just like me, who's already DONE it, and can forward that info on to me.

Thanks.

 

[nbucska]

Assuming the "a" is really triangle, do this for each of
the three section:

The acceleration is linear: a(t)= a0*t where a0 =slope
v(t)= integral a(t)*dt
x(t)=integral(v(t)*dt

the integral of F(x)=x^n is x^(n+1)/(n+1) +C
where ^ indicates exponent.
i.e. the successive integrations of Q = Q*x^0 will give
Q*x ; Q*x^2/2; Q*x^3/6; etc.

What is the difficulty?




<nbucska@pcperipherals DOT com> subj: eng-tips
read FAQ240-1032

 

[Boothby171]

Nothing, really.

If I sat down and did it, it would have taken me less than an hour.

In trying to find the solution already "out there," I've probably spent twice that amount!

The problem is keeping track of "time" during the various time-spans (step functions based on threshholds (t-t3)), keeping track of subsequent integrations, and then summing up all the equations, simplifying, etc.

The end result should be "enterable" as an equation or series of equations in EXCEL (my programme de choix).

 

[nbucska]

Do it on computer. Ramp up for 1000 steps, down for 2000
and up again for 1000. in each step calculate a-->v-->x.

That is all. A hard 30 minutes work and you can play with it all you want.


<nbucska@pcperipherals DOT com> subj: eng-tips
read FAQ240-1032