B.5.MPC.R:5.1 - A robot receives a new deadline, then keeps moving during a pause

A robot must travel d=1 m along a straight guide. In the stipulated regime, constant speed obeys v=k*u, where u is a dimensionless command and k=0.2 m/s. The model initially idealizes starting and stopping as immediate. The available command range is 0<u<=0.8.

With u=0.5, the relation d=vT gives T=d/(ku)=10 s. The question then changes: find the command for T=5 s. Keep the relations but choose u as the unknown:

u=d/(kT)=1/(0.25)=1.

The algebra supplies a value outside the available range. The corresponding lower bound on duration is T>=1/(0.2*0.8)=6.25 s. This answers the feasibility question before another controller is written. Changing the deadline or the available motion capability is now a meaningful choice.

Suppose the requester chooses T=8 s. The required command is u=0.625, giving v=0.125 m/s. A controller applies that command, counts eight seconds of active program execution and then issues stop.

During a diagnostic run, a debugger freezes the program and its timer for three seconds. In this example the output retains u=0.625 and the motor continues to run. The stop is therefore issued after eleven seconds of physical operation, giving d=0.125*11=1.375 m under the model.

The disagreement concerns execution timing. The equation and chosen command remain appropriate to eight seconds of motion. If the question concerns an uninterrupted run, observe that run under its intended timing. If such pauses can occur in the required operation, provide a stopping mechanism whose elapsed-time behavior survives the pause, or revise the control arrangement accordingly. Test its physical completion at the condition the use needs.

For example, suppose an independent timer can remove the motor command and continues to run while the controller is paused. Start it with the command and set it to remove the command after eight physical seconds. Under the example’s immediate-start/stop model, motion then lasts eight seconds and covers 0.125*8=1 m, including a run with the three-second program pause. Compare the timer’s actual behavior with those conditions before relying on that performance. If no such mechanism is available, the next contribution is a way to stop motion with that timing or a revised operating requirement.

A real stopping delay or speed transient would be another changed premise. Incorporate it when predicting final displacement; acknowledging stop does not measure that displacement. The example’s next useful result is a duration bound or a repaired timing design with an identified performance question.