Thread: Arc Driving
View Single Post
  #13   Spotlight this post!  
Unread 22-01-2007, 20:13
maniac_2040's Avatar
maniac_2040 maniac_2040 is offline
Registered User
AKA: Matt
FRC #3302 (Turbo Trojans)
Team Role: Programmer
 
Join Date: Jan 2006
Rookie Year: 2005
Location: Clawson, Michigan
Posts: 34
maniac_2040 is infamous around these partsmaniac_2040 is infamous around these partsmaniac_2040 is infamous around these partsmaniac_2040 is infamous around these parts
Send a message via MSN to maniac_2040
Re: Arc Driving

Quote:
Originally Posted by worldbringer View Post
I'm pretty sure that f(x) = tan-1(d/x) would work.

If you draw lines through the front and back wheels (basically, through their axles) as TimCraig said, I am assuming that the intersection of those lines would be the center point of the arc. The line from the point between the two back wheels to the point between the front wheels gives you d (I'll call it line D). Line D is perpendicular to the line drawn from the back wheels to the center of the circle/arc (line B). Thus lines D and B make a right triangle with hypotenuse drawn from the front wheels to the center of the circle.

The tangent function is described as the ratio of length d to length b. b is the arc radius. Angle A is the angle between b and the hypotenuse.
tan(A) = d/x
A = tan-1(d/x)

I don't think it matters if the front wheels don't share an axle, because the lines are drawn from the midpoint between the two wheels.
Thanks, I will test this to see if it works out. Sounds right, but could you perhaps maybe post a picture? It's hard to visualize what you are saying.