Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.




HELLO:I have a bit of a math curiosity. I need to understand how the formula works. This is from a flow sensor,going to an ADC,on a 5 volt ref.. Here it is...

(240 - 238) = 2 (* 2) = 4 (\ 5) = .8 (+ 30) = 30.8 original formula...

(240 - 238) = 2 (\ 2.5) = .8 (+ 30) = 30.8 my reduced formula...

240 = ADC_count from sensor
238 = ADC_count at 0 flow
\2.5 or * 2 \ 5 = ???? mystery number
+ 30 = lowest value displayed on screen (50 is the highest)
30.8 = flow rate

OK so ADC value is from 238 to 288,translates to 30 to 50 on screen...
The formula works,but how???
What is the \2.5???
Is there an easier way to work this without the 3 constants (-238 offset),(\2.5 mystery number),(+30 lowest reading on screen)???
If any one has a good explanation and/or a better way to rewrite this formula that's great....THANK YOU FOR ANY HELP.....


Mystery number is convertion range = range 1 / range 2 = (288-238) / (50-30) = 50/20=2.5


WOW!!!! That was easy...Sometimes I just need to see it...Thank you very

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Close Box

Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

Register now while it's still free!

Already a member? Close this window and log in.

Join Us             Close