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#1
01-11-2012, 08:28 PM
 lemiant the Dreamer AKA: Alex FRC #4334 (Alberta Tech Alliance) Team Role: Leadership Join Date: Feb 2009 Rookie Year: 2008 Location: Calgary, AB, Canada Posts: 562
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
Attached Thumbnails

#2
01-11-2012, 09:12 PM
 Ether systems engineer (retired) no team Join Date: Nov 2009 Rookie Year: 1969 Location: US Posts: 8,722
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 counter-clockwise and tip over backwards.

Last edited by Ether : 01-14-2012 at 04:23 PM.
#3
01-11-2012, 09:17 PM
 lemiant the Dreamer AKA: Alex FRC #4334 (Alberta Tech Alliance) Team Role: Leadership Join Date: Feb 2009 Rookie Year: 2008 Location: Calgary, AB, Canada Posts: 562
Re: The physics of flipping

Quote:
 Originally Posted by Ether 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 counter-clockwise and tip over backwards.
Do you know what would be a reasonable estimate for the max acceleration of an average robot (12fps top speed, 6wd traction wheels, CIMple gearboxes, 4 CIMs)?

PS What do you use to make those pretty diagrams? (You probably realized I was just using powerpoint)
#4
01-11-2012, 09:24 PM
 Ether systems engineer (retired) no team Join Date: Nov 2009 Rookie Year: 1969 Location: US Posts: 8,722
Re: The physics of flipping

Quote:
 Originally Posted by lemiant Do you know what would be a reasonable estimate for the max acceleration of an average robot (12fps top speed, 6wd traction wheels, CIMple gearboxes, 4 CIMs)?
Use JVN's mechanical calculator, located here.

Quote:
 PS What do you use to make those pretty diagrams? (You probably realized I was just using powerpoint)
That one was also Powerpoint.
#5
01-11-2012, 09:56 PM
 lemiant the Dreamer AKA: Alex FRC #4334 (Alberta Tech Alliance) Team Role: Leadership Join Date: Feb 2009 Rookie Year: 2008 Location: Calgary, AB, Canada Posts: 562
Re: The physics of flipping

Quote:
 Originally Posted by Ether Use JVN's mechanical calculator, located here.
I haven't used JVN's calculator in a while, but I' don't think it gives acceleration.

On the other hand I plugged this into wolfram alpha (so I don't have to worry about all that unit conversion):

(21 in-lb / ((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:
 e.x. 37" long robot with a center of mass that is nearly centered on the robot has the following characteristics: Wheel base = 37 - 2*4 = 37 - 8 = 29" CoM to edge of wheelbase = 29/2 = 14.5" Max height of CoM = 1.5 * 14.5" = ~22"
I ball-parked pretty much every number in this math but it does tell us one thing:
Quote:
 If your long-orientation robot has a CoM more than 22" off the ground (or 14" for a wide-bot), you'd better think twice and prototype 10x, because you're in the danger zone!!! NOTE, this doesn't mean that less than those thresholds is safe, it's just that over them is super risky.
Ether, did I overlook anything? If this is reasonably representative for most teams then I'll move it to the top, and hope it helps some people.
#6
01-11-2012, 10:06 PM
 DonRotolo Standing on their shoulders FRC #0832 Team Role: Mentor Join Date: Jan 2005 Rookie Year: 2005 Location: Atlanta GA Posts: 7,237
Re: The physics of flipping

Quote:
 Originally Posted by lemiant Do you know what would be a reasonable estimate for the max acceleration of an average robot (12fps top speed, 6wd traction wheels, CIMple gearboxes, 4 CIMs)?
I can't hazard a guess, but I'm more worried about acceleration with a minus sign - like when I am moving at 12 fps and stop dead when I hit the bump. If F=Ma, and a is going from 12 to zero in an umptosecond, F is going to have some serious magnitude regardless of M.

I am seriously working to get our CoM below 6 inches.
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#7
01-11-2012, 10:50 PM
 Ether systems engineer (retired) no team Join Date: Nov 2009 Rookie Year: 1969 Location: US Posts: 8,722
Re: The physics of flipping

Quote:
 Originally Posted by lemiant did I overlook anything?
Yes, what Don said:

Quote:
 Originally Posted by DonRotolo I'm more worried about acceleration with a minus sign - like when I am moving at 12 fps and stop dead when I hit the bump. If F=Ma, and a is going from 12 to zero in an umptosecond, F is going to have some serious magnitude

#8
01-12-2012, 12:56 AM
 ewhitman Registered User FRC #3504 (Girls of Steel) Team Role: Mentor Join Date: Jun 2011 Rookie Year: 2004 Location: Carnegie Mellon University Posts: 12
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:
2) Keep your wheel base big and your CoM near the center of it
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#9
01-12-2012, 02:43 AM
 Aren Siekmeier on walkabout FRC #2175 (The Fighting Calculators) Team Role: Mentor Join Date: Apr 2008 Rookie Year: 2008 Location: 대한민국 Posts: 738
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
01-12-2012, 06:58 AM
 PayneTrain A Better Idiot AKA: Wil "just give me crabcake" Payne FRC #0422 (The Tex-Mex Drajones) Team Role: Mascot Join Date: Mar 2010 Rookie Year: 2009 Location: The Safety Glasses Table Posts: 2,528
Re: The physics of flipping

Quote:
 Originally Posted by DonRotolo I can't hazard a guess, but I'm more worried about acceleration with a minus sign - like when I am moving at 12 fps and stop dead when I hit the bump. If F=Ma, and a is going from 12 to zero in an umptosecond, F is going to have some serious magnitude regardless of M. I am seriously working to get our CoM below 6 inches.
I'm curious to see how much robot you're going to fit in such a small space. I think we're aiming for about 8-12 inches. Given the posts I have seen on CD, I assume your team is using a narrow intake, which probably helps a lot when trying to acheive a low COM.
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#11
01-12-2012, 07:50 AM
 IKE Not so Custom User Title AKA: Isaac Rife no team (N/A) Team Role: Mechanical Join Date: Jan 2008 Rookie Year: 2003 Location: Michigan Posts: 2,227
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 ((38-8)/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 (((28-8)/2/sqrt(3)=5.78), would be keeping the CG lower than 6" (which can be tough with 8" wheels)...
#12
01-12-2012, 05:26 PM
 SteveJanesch hopes he has enough oomph FRC #1533 (Triple Strange) Team Role: Mentor Join Date: Apr 2007 Rookie Year: 2006 Location: Greensboro, NC Posts: 103
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
01-17-2012, 12:25 PM
 Ether systems engineer (retired) no team Join Date: Nov 2009 Rookie Year: 1969 Location: US Posts: 8,722
Re: The physics of flipping

Quote:
 Originally Posted by IKE 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
I've been meaning to come back to this.

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.

Attached Files
 flip bot solution.pdf (14.5 KB, 68 views)

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