View Single Post
  #9   Spotlight this post!  
Unread 23-05-2012, 15:12
Ken Streeter's Avatar
Ken Streeter Ken Streeter is online now
Let the MAYHEM begin!
FRC #1519 (Mechanical Mayhem)
Team Role: Engineer
 
Join Date: Feb 2005
Rookie Year: 2005
Location: Team: Milford, NH; Me: Bedford, NH
Posts: 470
Ken Streeter has a reputation beyond reputeKen Streeter has a reputation beyond reputeKen Streeter has a reputation beyond reputeKen Streeter has a reputation beyond reputeKen Streeter has a reputation beyond reputeKen Streeter has a reputation beyond reputeKen Streeter has a reputation beyond reputeKen Streeter has a reputation beyond reputeKen Streeter has a reputation beyond reputeKen Streeter has a reputation beyond reputeKen Streeter has a reputation beyond repute
Re: The physics of pushing

Quote:
Originally Posted by Ether View Post
assuming you're pushing against an identical robot, this limits pushing force F to approximately (W*k-tau)/h, where h is the effective contact point of the pushing force and tau is the torque on the rear wheels
Awesome diagram! I hadn't ever really tried to figure out the solution before, but with your diagram, it's pretty easy! Thanks, Ether!

However, I think there's an extra variable (tau) in your solution. From looking at your diagram, I get the maximum pushing force as being (W*k)/h, with F=tau. However, I think I must be missing the reason that you have included "tau" in the calculation. Can you explain? I set the rotational torques about the ground contact point equal to one another (F*h = W*k) and then solved for F. However, either I'm missing something (the reason for tau being in the answer) or you made a minor mistake.

When coming up with physical solutions like this, I like considering each of the independent variables and how making them bigger / smaller effects the result, particularly at the extremes. (W*k)/h seems to make sense from a number of practical ways:
  • The lower you can get the pushing point ("h"), the harder the robot can push without tipping over. If the height can be made to be zero, then this constraint (tipping over) is never a limiting factor. Very tall robots can't push well at all, as they'll simply tip over instead.
  • The heavier you can make the robot ("W"), the harder the robot can push. A weightless robot can't push at all, and an infinitely heavy robot won't have it's pushing power limited by tipping over.
  • The further you can move the CG forward of the rear wheels ("k"), the harder the robot can push. If the CG is directly over the rear wheels (k=0), then the slightest push will tip the robot over. If the CG is infinitely far in front of the rear wheels, then the robot isn't limited in pushing by tipping over, either.
In any case, thanks, Ether! I'll have to encourage our team to consider this factor when designing gear ratios for pushing next year -- we've definitely had more pushing power than we could use during these past few years! Accordingly, we probably could have geared higher to get a little more speed without compromising our pushing ability (barring electrical energy considerations.)

PS: "You must spread some Reputation around before giving it to Ether again."
__________________
Ken Streeter - Team 1519 - Mechanical Mayhem (Milford Area Youth Homeschoolers Enriching Minds)
2015 NE District Winners with 195 & 2067, 125 & 1786, 230 & 4908, and 95 & 1307
2013 World Finalists & Archimedes Division Winners with 33 & 469
2013 & 2012 North Carolina Regional Winners with teams 435 & 4828 and 1311 & 2642
2011, 2010, 2006 Granite State Regional Winners with teams 175 & 176, 1073 & 1058, and 1276 & 133
Team 1519 Video Gallery - including Chairman's Video, and the infamous "Speed Racer!"

Last edited by Ken Streeter : 23-05-2012 at 15:24. Reason: re-considering my initial conclusion -- I think the "tau" must be in there for a reason, but I can't think of what it is...