View Single Post
  #9   Spotlight this post!  
Unread 18-02-2005, 01:09
Tristan Lall's Avatar
Tristan Lall Tristan Lall is offline
Registered User
FRC #0188 (Woburn Robotics)
 
Join Date: Aug 2001
Rookie Year: 1999
Location: Toronto, ON
Posts: 2,484
Tristan Lall has a reputation beyond reputeTristan Lall has a reputation beyond reputeTristan Lall has a reputation beyond reputeTristan Lall has a reputation beyond reputeTristan Lall has a reputation beyond reputeTristan Lall has a reputation beyond reputeTristan Lall has a reputation beyond reputeTristan Lall has a reputation beyond reputeTristan Lall has a reputation beyond reputeTristan Lall has a reputation beyond reputeTristan Lall has a reputation beyond repute
Re: 500 hundreds of a inch

Quote:
Originally Posted by Elgin Clock
0.050 or 50 thousandths, is larger than you can think, and may affect form fit or function a bit with robots being 3 feet wide. (If it is an overall chassis cross member dimension).
To put this in perspective, how many teams are sure that their frame is accurate (in two dimensions) to within 1/16" (0.0625")? (Answer: many, but not a large majority, I would imagine.) How many teams can further guarantee that their frame won't distort this much under normal use? (Answer: few.) If you're cutting with a saw, good luck--the kerf isn't always easy to judge.

Quote:
Originally Posted by Elgin Clock
But, to add to that, a standard dimensional tolerance is +/- .002 of an inch on the parts I make on CAD at work. (Small cable connectors about an inch long (or less) total.) Think: the connector at the end of the cable snake that invades your home and brings you your TV and cable modem services. About that size.
I don't think that 80-20 structural framing is in the same league. Really, 2 thousandths is insignificant for the applications that 80-20 makes sense for. I'd guess that a change of 5°C would cause about that much in thermal expansion alone.

Let's see: Δd = d0αΔT; for a 28" piece of framing, α = 23 × 10-6°C-1 (we'll call it constant for small ΔT), at 20°C, in 6101 aluminum (the closest I could find to 6105, which is actually used in 80-20); Δd = (28)(23 × 10-6)(5) = 0.0032"

So if Δd = 0.0032", we're dealing with as much thermal expansion as we have machining tolerance, merely in taking it from the unheated cargo trailer to the pit area. That's why I think 0.002" is crazy, for our purposes.
Reply With Quote