Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Determining tipover point

  1. Dec 4, 2011 #1
    Hey, I'm trying to calculate the point where my robot will tip over but I don't really know where to start. Basically the robot is 24" x 50" and 40" tall. The center of mass is the center of the robot 25" from the ground. Total weight is 44lbs. The robot turns via differential steering so it can turn about its center point. I need to know the maximum rate of acceleration I can turn ( or spin ) the robot before it starts to tip.
    Any help would be much appreciated.
     
  2. jcsd
  3. Dec 4, 2011 #2
    The robot will tip over when a vertical line passing through the center of mass falls outside the base.
     
  4. Dec 4, 2011 #3
    Thank you, unfortunately that doesn't answer my question. I understand it will tip over at that point, but I am looking for the maximum speed the robot can turn at before it starts to tip over.
     
  5. Dec 4, 2011 #4
    Do you mean during acceleration or cornering or what?
     
  6. Dec 4, 2011 #5
    basically it could be either, it could be tip from acceleration, or tip from cornering, and since it can rotate about its center point, how fast it can rotate before it tips. Does that help??
     
  7. Dec 4, 2011 #6

    nvn

    User Avatar
    Science Advisor
    Homework Helper

    maxi9: Could you give us the maximum velocity of your robot, in m/s?

    The maximum forward acceleration, at which point your robot would tip over, would be at = 9.81 m/s^2.

    Regarding the centripetal acceleration, the minimum turn radius, at which point your robot would tip over, would be rho = (v^2)/(4.709 m/s^2), where v = robot velocity (m/s), and rho = turn radius (m).
     
    Last edited by a moderator: Dec 5, 2011
  8. Dec 4, 2011 #7
    Thank you nvn, The robots max velocity is 1.5m/s, fairly slow. Out of curiosity how did you determine the the rate of acceleration that would cause it to tip. Also in the second equation regarding rho, where did the acceleration 5.886 m/s^2 come from?
     
  9. Dec 4, 2011 #8

    DaveC426913

    User Avatar
    Gold Member

    A little premature to be putting numbers to it. Not enough info.

    Don't you guys care what the wheelbase is? What if his 40" robot sits on a wheelbase 2" in length and breadth?
     
  10. Dec 4, 2011 #9
    wheel base is as follows, there are 4 casters, 1 at each corner of the robot, and the two drive wheels are located on the side of the robot at its center line. Casters are 2 inches in diameter and the drive wheels are 4 inches in diameter. Has one motor per drive wheel.
     
  11. Dec 5, 2011 #10

    nvn

    User Avatar
    Science Advisor
    Homework Helper

    maxi9: In post 9, are you saying the caster wheels are almost exactly at the corner dimensions you gave in post 1? (I currently assumed they are, in post 6.) If not, then we need to know the actual wheel base (L), and track width (b), of the caster wheels. Are the tipping accelerations you need to compute a school assignment?
     
  12. Dec 5, 2011 #11
    nvn, the casters are at the corners of the robot. This is just a project im working on, I jsut want to know how fast i can get it to accelerate and turn.
     
  13. Dec 5, 2011 #12

    nvn

    User Avatar
    Science Advisor
    Homework Helper

    maxi9: The 4.709 m/s^2 in post 6 is the centripetal acceleration at the verge of tipping during cornering.

    Therefore, when the velocity of your robot is v = 1.5 m/s, the turn radius in post 6 (at which your robot would tip over during cornering) is rho = 0.478 m = 478 mm.
     
  14. Dec 5, 2011 #13
    Thank you very much, Glad I know i can take a pretty sharp turn with it. About the centripetal value from post 6, im just curious as to how you obtained that value?
     
  15. Dec 5, 2011 #14

    nvn

    User Avatar
    Science Advisor
    Homework Helper

    The centripetal acceleration when tipping occurs during cornering is, an = g*(0.5*b)/h = (9.81 m/s^2)(0.5*0.6096 m)/(0.6350 m) = 4.709 m/s^2, where b = wheel track width, and h = CG height.
     
  16. Dec 5, 2011 #15
    oh okay i see now, thank you!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Determining tipover point
  1. Representing points (Replies: 2)

Loading...