Go to Post My advice is anytime your intuition tells you something, double check it with physics. - tennispro9911 [more]
Home
Go Back   Chief Delphi > Technical > Technical Discussion
CD-Media   CD-Spy  
portal register members calendar search Today's Posts Mark Forums Read FAQ rules

 
 
 
Thread Tools Rate Thread Display Modes
Prev Previous Post   Next Post Next
  #4   Spotlight this post!  
Unread 09-07-2012, 08:21
Aren Siekmeier's Avatar
Aren Siekmeier Aren Siekmeier is offline
on walkabout
FRC #2175 (The Fighting Calculators)
Team Role: Mentor
 
Join Date: Apr 2008
Rookie Year: 2008
Location: 대한민국
Posts: 735
Aren Siekmeier has a reputation beyond reputeAren Siekmeier has a reputation beyond reputeAren Siekmeier has a reputation beyond reputeAren Siekmeier has a reputation beyond reputeAren Siekmeier has a reputation beyond reputeAren Siekmeier has a reputation beyond reputeAren Siekmeier has a reputation beyond reputeAren Siekmeier has a reputation beyond reputeAren Siekmeier has a reputation beyond reputeAren Siekmeier has a reputation beyond reputeAren Siekmeier has a reputation beyond repute
Re: Turning torque

A little integral calculus helps you out here. You can add up small contributions to the total required torque from small portions of the wheel's width depending on the frictional force and the radius, take the limit, and compute the integral.

Spoiler for I guess I need a title:
We'll split the wheel in half to make it easier, starting from distance r=0 to r=W/2 from the wheel pivot point, with W the wheel width (assuming wheel pivot is in the middle, it's easy to compensate otherwise).

At each dr chunk, the normal force there is the total normal force (equal to the load, unless you're moving in and out of the ground...) on the wheel scaled to the size of dr, or L*dr/W, where L is the load on the wheel, dr is the width of the small, and W is the total wheel width.

The frictional force then is the normal force times u, the coefficient of friction, or uLdr/W. The torque is the force crossed with the radius, so we have the torque for each chunk as uL/W * rdr, which we integrate from 0 to W/2 and multiply by 2 to get our final expression, T = uLW/2.

This assumes the contact area between the wheel and the floor can be described as a line. You could integrate over a rectangular contact if you knew roughly how big that was and would find a slightly larger answer, due to the larger moment. This also assumes uniform distribution of the wheel load over the contact area.
 


Thread Tools
Display Modes Rate This Thread
Rate This Thread:

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

vB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Forum Jump


All times are GMT -5. The time now is 07:13.

The Chief Delphi Forums are sponsored by Innovation First International, Inc.


Powered by vBulletin® Version 3.6.4
Copyright ©2000 - 2017, Jelsoft Enterprises Ltd.
Copyright © Chief Delphi