

If it were not about winning, it would not be called a competition  but FRC is much more than winning a competition.  JaneYoung [more] 



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#1




The physics of flipping
Today I was thinking about how acceleration/deceleration can tip a robot. I know that somehow acceleration acts on the center of mass to flip the robot and this is my theory as to how it works:
The effects of inertia act at your center of mass. They are equal in magnitude, but opposite to the direction of the acceleration of your robot (when you accelerate forwards you will flip backwards). Mathematically if you add the inertial vector (equal an opposite to acceleration), to the gravity vector, you will get the net vector of your robot. When you project this vector from your center of mass you can determine if a robot will flip. If the projection lands inside the wheelbase you will stay on the ground, but if it lands outside of your wheelbase... prepare to flip. (I have attached some diagrams illustrating this). Could one of the physics people on these forums, confirm or deny that method (in a strange twist I cannot find the right search term to take me to lessons about this). Also I'm wondering what is a ballpark value for the maximum acceleration a FRC robot sees during a match. And if someone wants to rephrase the above paragraph to use the actual terminology I would be immensely greatful. Thanks,  Alex 
#2




Re: The physics of flipping
See Figure1 attached. In this Figure, the vehicle is accelerating just fast enough that the front wheels are about to lift off the floor. In other words, there is no normal force on the front wheels.
The force causing the acceleration of the vehicle is the (horizontal) friction reaction force Ma of the floor on the bottom of the rear wheels. The weight Mg of the vehicle acts at the Center of Mass (CoM in the figure). The normal force on the rear wheels is Mg. The torque around the CoM is zero, so Mah = Mgb => a/g = b/h. If the acceleration is slightly increased, then a/g > b/h and the vehicle will rotate counterclockwise and tip over backwards. Last edited by Ether : 01142012 at 03:23 PM. 
#3




Re: The physics of flipping
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PS What do you use to make those pretty diagrams? (You probably realized I was just using powerpoint) 
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Re: The physics of flipping
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#5




Re: The physics of flipping
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On the other hand I plugged this into wolfram alpha (so I don't have to worry about all that unit conversion): (21 inlb / ((12fps / 5310 rpm) / (2*pi (inches/revolution)/inch)))/120lb to m/s^2 * 80% Getting a max acceleration for my previously stated bot of 5.3 m/s^2. This is approximately half the force of gravity. Based off that we can have a center of gravity at most ~2x as high as it is far from the front/back of the wheel base (whichever is closer), before you will flip due to accelerating on a flat field. I would probably suggest keeping it below 1.5x though, for safety. Quote:
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#6




Re: The physics of flipping
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I am seriously working to get our CoM below 6 inches. 
#7




Re: The physics of flipping
Yes, what Don said:
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#8




Re: The physics of flipping
This problem can be addressed by the concept of the Zero Moment Point. The ZMP is most often used in the context of determining tipping for humanoid robots, but it applies here.
The ZMP is the point on the ground through which a single force would act to achieve a given Center of Mass acceleration without any change in angular momentum. If the ZMP leaves your wheel base (convex hull of your points of support), your robot will begin to tip. The relevant formula is: Z = h/g*a where: Z = the location of the ZMP relative to the CoM (in m) h = the height of the CoM (in m) g = the acceleration due to gravity (9.8 m/s/s) a = the acceleration of the CoM (in m/s/s) This means that while sitting still or traveling at a constant speed (a = 0), the ZMP is directly under the CoM (Z=0), which is hopefully inside your wheel base. However, if your CoM is 50 cm high, and you accelerate forward at 5 m/s/s, your ZMP will move backwards 25.5 cm. (0.255 = 0.5/9.8*5) If your CoM is less than 25.5 cm in front of your rear wheels, you will begin to tip because the ZMP will have left your wheel base. If I haven't explained the math well enough, the moral is: 1) Keep your CoM low 2) Keep your wheel base big and your CoM near the center of it 
#9




Re: The physics of flipping
The negative acceleration pointed out by Don is definitely the one we're worried about too. We're trying to keep the angle between the floor and the line between our rear contact and the CG less than 45 degrees so that we can experience as much as 1g laterally and be just fine. And even lower is even better. This includes when we are crossing the bump/bridge, so depending on wheel profile, the angle on flat ground needs to be potentially as small as 22 degrees, making a short wheel base harder and harder.

#10




Re: The physics of flipping
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#11




Re: The physics of flipping
Some other factors to take into consideration:
Torque at the wheel. Larger torque on the rear wheels also adds to the moment of tipping. For the same traction force and acceleration, the larger the rear wheel will require larger torque and thus be more prone to tipping (works both ways including dynamic braking). Another big factor is if you plan to go on the ramp. As you climb the incline, the normal force on the rear wheels will increase and thus have a higher traction potential. If you then apply a large torque to accelerate up the ramp, you can flip over as well. Tilt your Powerpoint model 15 degrees, and you will see how the gravity vector will now angle more rearward. http://www.chiefdelphi.com/media/papers/2321 Last but certainly not least, wheel and carpet compliance can be very big in flipping. The compliance in the carpet and your wheels/tires can act as a spring. When timed wrong, shifting back and forth can exercise those springs. The more compliant the wheel/tire (pneumatic), the more effect you can get. This can start to get complicated pretty quick. A good "rule of thumb" for a drivetrain that you plan on having traction limited (gearing and motors powerful enough to spin the tires) is... Assuming a Coefficient of friction of 1, then the acceleration and traction vector can be equal, and thus your CG needs to fall inside of 45 degree lines from your tire contact patches. Higher COF means a lower angle (though not much lower). Wanting to traverse inclines means reducing angle as well. Assuming the 15 degree ramp is a goal, keeping your CG below 30 degree lines should be fairly conservative. For a long robot with 8" wheels ((388)/2/sqrt(3)=8.67), this would be keeping the CG in the center below 9" and lower the more offset from center it becomes. For a wide bot with 8" wheels (((288)/2/sqrt(3)=5.78), would be keeping the CG lower than 6" (which can be tough with 8" wheels)... 
#12




Re: The physics of flipping
At the risk of telling you something you already know...
It's not just acceleration that causes tipping  it's the fact that the accelerating force is applied away from the center of gravity (by the wheels, on the floor) that imparts the torque around the COG that can tip the robot. If your accelerating force were applied parallel to the floor and at the same height as the COG there'd be no torque to cause the tipping (ignoring friction, of course). Great question, Alex, thanks for posting it. And great responses, too.  Steve 
#13




Re: The physics of flipping
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If the ratio of the vehicle mass to the mass of the rear wheels is sufficiently greater than (1+2*r/h), then the radius drops out of the formula for tipping acceleration, and the result is the formula posted earlier in this thread. See attachment. 
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