Trapezoidal Motion Profile Using Discrete Method

[martinrand]

Thanks, I'll certainly take another look.
Potentially, I could generate an sigmoid function to represent position, using parameters from a trapezoidal profile.

I suppose calculating position at a given time would then be as simple as plugging t into the equation. This would then do away with the need to have 3 section for accel, max velocity and decel.

Does that sound feasible?
This was how I originally assumed things were done.

[Ether]

The a(t), v(t), and x(t) functions in the paper are stand-alone.

If you are not going to use velocity or acceleration in your control algorithm, you don't need to compute a(t) or v(t). Just use x(t).

[martinrand]

Any idea how I would take 3 variables for a trapezoidal profile (Accel, Max Velocity and Target Position, *decel=-accel*) and convert that into x(t)?

[Ether]

Originally Posted by martinrand

Any idea how I would take 3 variables for a trapezoidal profile (Accel, Max Velocity and Target Position, *decel=-accel*) and convert that into x(t)?

http://www.chiefdelphi.com/media/papers/download/4284

http://www.chiefdelphi.com/media/papers/download/4287

http://www.chiefdelphi.com/media/papers/download/4288

Knock yourself out.

[martinrand]

Thanks for that!

Afraid I can't open the .exe as I'm using a mac. Any chance you have a mac version or source code available?
Thanks

[Ether]

Originally Posted by martinrand

Afraid I can't open the .exe as I'm using a mac. Any chance you have a mac version or source code available?

You have the equations. Just put them into a spreadsheet or LabVIEW. Or install Maxima on your Mac and run the script I gave you.

[Ether]

Originally Posted by martinrand

Potentially, I could generate an sigmoid function to represent position

Which particular sigmoid function did you have in mind?

Assuming you meant the Logistic Function:

x(t) = L/(1+exp(-k*(t-t_m))),

where:

t_m = the t-value of the sigmoid's midpoint

L = the curve's maximum value

k = the steepness of the curve

...that would be a poor choice.

To see why, suppose you want to go from x=0 and v=0 and a=0 at t=0 to x=5ft and v=0 and a=0 at t=3seconds.

Then L and t_m would be L=5 t_m=3/2.

The only free parameter left is k; try k=2, k=3, k=4, k=5.

As can be seen from the graphs, there is no acceptable value for k.

k=2: a, v, and x are non-zero at beginning and end points, and x has not reached 5 at t=3.

k=3: better, but a and v are still nonzero at endpoints

k=4: a is still not quite zero at endpoints. max accel has grown to 7.7

k=5: endpoints look good, but max accel is now 12, and max jerk is 77

Compare those graphs to the Figure on page 1 of this paper, which uses a sinusoidal function for x(t), resulting in

x=0, v=0, and a=0 at t=0, and

x=5, v=0, a=0 at t=3, and

max accel=3.5, max speed=3.3, max jerk=7.3