Created by Created August 05, 2020 Last modified August 08, 2020 Tags No tags.

### Summary

Exploring the capabilities of an analog computer, simulating a positional control system.

### Description

UPDATE 8th Aug 2020 i) correct spring constant definition; ii) add details of scaling factors for equation constants.

Linear positional control system. For example, moving a weight with a spring.

The weight is attached to one end of the spring. When the other end of the spring is moved (step) the weight moves to catch up (displacement).

The negative of the speed of the weight's movement is negspeed.

Equation scaling (reality) = (simulation):-

1 unit distance (arbitrarily, 1 meter) = 1volt (at displacement output)

1 unit of time (Second) = 10mS (set by R1C1 = R4C2)

Speed, 1meter/sec = 1volt/10mS (at output of acceleration integrator, OA1)

Acceleration, 1 meter/sec/sec = 10uA= 1volt/10mS across 100nF

In relation to the weight-with-spring example. Xdouble-dot - Xdot + X = 0 (then constant)

The differential equation has unity scaling: all multiplying constants are "1". This works well with the SI unit definitions, for example:

Force: 1 Newton = 1 meter/sec/sec X 1 kg

For other parts of the simulation:

Mass 1kg = 100nF at acceleration integrator, OA1 AND pot R9, 1/K = 1 (NB reciprocal)

Spring constant, unit force for unit extension (1 Newton per meter) = 1 = 10uA/volt in R1

Loss, unit force for unit speed (1 Newton per meter/sec) = 1 when pot R8 K=1

Leverage - displacement multiplier (not shown) related to R10's K value.