How Does a Calculator-Based Ranger (CBR) Detect Motion?

  • Thread starter Thread starter p.ella
  • Start date Start date
  • Tags Tags
    Calculator
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 3K views
p.ella
Messages
41
Reaction score
0
How does a calculator based ranger, or cbr, work? like the motion detector, with the TI- 83 calculator and computer. Like, explain the ultrasonic waves and sound stuff DUMBED DOWN please (:
Pleaseee reply ASAP, need it for tmrw! PLEAASSEE!
 
Physics news on Phys.org
Where were you supposed to get this information?
 
SammyS said:
Where were you supposed to get this information?

we used cbrs for a lab, the teacher just wants us to study the whole idea of how they work, really brief like a couple of sentences.
 
I Googled it. Not much on how it works. In my view, that's bad. They (suppliers) supply lots of labs for these. One should include a discussion of how they work !

Here's a short description.

They produce a brief burst of ultrasonic waves (high frequency -- short wavelength). Then the sensor "listens" for the reflection of similar frequency ultrasonic waves. It uses the time between when the burst was sent and when the return sound was detected to determine how far the wave travels to the object and back. It does this by using a reasonable value for the speed of sound. The distance to object that reflects the wave is 1/2 of the distance traveled by the sound wave. (There & back)
 
SammyS said:
I Googled it. Not much on how it works. In my view, that's bad. They (suppliers) supply lots of labs for these. One should include a discussion of how they work !

Here's a short description.

They produce a brief burst of ultrasonic waves (high frequency -- short wavelength). Then the sensor "listens" for the reflection of similar frequency ultrasonic waves. It uses the time between when the burst was sent and when the return sound was detected to determine how far the wave travels to the object and back. It does this by using a reasonable value for the speed of sound. The distance to object that reflects the wave is 1/2 of the distance traveled by the sound wave. (There & back)

Ahhhh thank you sooo much! I really appreciate your time and effort, that's EXACTLY what I needed (: And yea, you're right; the suppliers should give more information :/ Nevertheless, you really helped me (:
 
You're welcome.