Thread: Team Update #18
View Single Post
  #13   Spotlight this post!  
Unread 17-03-2011, 11:49
jspatz1's Avatar
jspatz1 jspatz1 is offline
Registered User
AKA: Jeff
FRC #1986 (Team Titanium)
Team Role: Mentor
 
Join Date: Jan 2008
Rookie Year: 2007
Location: Lee's Summit, MO
Posts: 836
jspatz1 has a reputation beyond reputejspatz1 has a reputation beyond reputejspatz1 has a reputation beyond reputejspatz1 has a reputation beyond reputejspatz1 has a reputation beyond reputejspatz1 has a reputation beyond reputejspatz1 has a reputation beyond reputejspatz1 has a reputation beyond reputejspatz1 has a reputation beyond reputejspatz1 has a reputation beyond reputejspatz1 has a reputation beyond repute
Send a message via AIM to jspatz1
Re: Team Update #18

Quote:
Originally Posted by JesseK View Post
Does a standard scale, such as the ones used to inspect the robots at competition, give us the weight or mass of the robot? In other words, is the number we see equal to mass*[accel due to gravity] or equal to only mass?

Using a direct weight-equals-mass calculation, I get 0.586 feet stopping distance and 488N for stopping @ 1/4".

Dividing by the acceleration of gravity, I get a stopping distance of 0.060 feet (~3/4") and ~50N to stopping @ 1/4".
In imperial units, a scale measuring pounds gives the weight of an object, not the mass. The imperial unit of mass is slugs, which is equivelent to approx. 32 lb-mass. When doing Newtonian calculations in imperial units, you must use the units slugs, lb-force, feet, and seconds. This yields the results you mentioned first, .586 ft and ~480 N (107 lbs).
__________________
Reply With Quote