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Subject: 
Re: Brickworld 2007 - Indy 5.00
Newsgroups: 
lugnet.events.brickworld, lugnet.robotics
Date: 
Sun, 17 Jun 2007 17:01:53 GMT
Viewed: 
10172 times
  
In lugnet.events.brickworld, Brian Davis wrote:
Brickworld 2007, June 21-24, Chicago

Indy 5.00:

Autonomous robots will run head-to-head races of five laps around a
8'x10' oval
gradient track.

Arena:

The oval will be centered on a 4' by 8' surface, with no rails or other
physical limits.

Okay, I'm confused.  How can an 8' x 10' track be centered on a 4' by 8'
surface?  What is the actual width of the track?  If the whole track fits
on a
4' x 8' surface, then I would assume the width must be less then 2'.

That appears to be a typo.  The track fits on a FLL table, so it's a 44"x92"
inch oval.  And the lane is about 2ft wide.

It's pretty easy to build and program a robot for this event.  It just needs
to follow a specific shade (actual value will vary).  Of course, making it
go fast is another issue...  :)

Steve



Message has 1 Reply:
  Re: Brickworld 2007 - Indy 5.00
 
(...) Thank you Steve, Brian, and Bryan for the clarification on the track size. One more question: Would it be possible for whoever currently has the track to post some approximate NXT light sensor readings for the various positions across the (...) (17 years ago, 20-Jun-07, to lugnet.events.brickworld, lugnet.robotics)

Message is in Reply To:
  Re: Brickworld 2007 - Indy 5.00
 
(...) Okay, I'm confused. How can an 8' x 10' track be centered on a 4' by 8' surface? What is the actual width of the track? If the whole track fits on a 4' x 8' surface, then I would assume the width must be less then 2'. Gus (17 years ago, 17-Jun-07, to lugnet.events.brickworld, lugnet.robotics)

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