Re: MOTION DETECTORS

[LUDWIK KOWALSKI <KOWALSKIL@alpha.montclair.edu>]

Jim Riley provided a reasonable tentative explanation of the behaviour of 
ultrasonic rangers, as numerically illustrated by Brian. He wrote:

> ...  If you point the motion detector at the floor the pulsed sound will
> bounce back and forth between floor and ceiling. The detector determines
> distance by measuring the time between the issuance of a pulse and the 
> first echo that comes back to it, ignoring all other echos. Under the 
> right sample rate and ceiling to floor height you can get a multiply 
> reflected echo from a previous pulse returning to the detector before 
> the echo from the current pulse. This is exacerbated by highly reflecting 
> surfaces such as a linoleum floor and acoustical tile ceiling.  Changing 
> the sample rate allows the reflections to die off between samples so that 
> the first echo to return is the desired one.  Putting carpet on the floor 
> dampens the first reflection so that subsequent reflections are not large 
> enough to be detected. 

Can it be as simple as this? The ultrasonic frequency is 40 kHz. Assuming
v=340 m/s we can calculate lambda=0.0085 m. How long are the wave packets?
The minimum measurable distance of 0.4 m (specified in the manual) implies 
that packets are about 0.3 m long (nearly 50 wavelengths). If the carpet 
were a perfect absorber then the distance to the floor would not be 
measurable (nothing is reflected). A good reflection is a help, not a
hindrance. Jim is saying that too good refections are responsible for
the problem we encountered. 

Let us think in terms of "sound particles" emitted at the rate of 20 per 
second (every 50 ms). Why? Because our distance of 3 meters was measured 
correctly at the rate of 10 and wrongly at the rate of 20. At the speed
of 340 m/s the round trip (6 m) time for each bullet is 17.6 ms. This is 
shorter than 50 ms. The echo of bullet i should arrive to the detector 
long before the i+1 bullet is fired. Now let us think about bullet i-1.
Its second echo arrives at 2*17.6 +dt after its firing, where dt is the
time to cover the distance between the detector and the ceiling twice.
Our ceiling is not flat, it is a cavity with several thin pipes below
it, at different distances. It is quite possible that the estimated dt
is about what is needed to confuse the detector. I think that Jim's 
explanation is correct, as already illustrated by Brian.

We tried two detectors and two ULI boxes; both behaved in the same way.
It is clearly not the case of "a bad piece of equipment". Everybody can 
do what Jim Riley did to demonstrate "ways in which scientific ideas 
are tested".
                                               Ludwik Kowalski