Center of Mass question

[Alex1072]

What do you guys think: If we have a foot print that fills up all 28x38 inches with wheels pretty close to the corners, is a 5 foot vertically center of mass too high? How tipsy would our robot be? The COM is actually almost dead center on the robot on the horizontal plane, but still, 5 feet is a lot isn't it?

[Cory]

That is VERY high.

Is this before you extend an arm? or after?

You want to calculate the CoM after you've got whatever appendages on your robot extended completely, as well, to make sure it isn't outside of the base of your bot.

[Alex1072]

Originally Posted by Cory

That is VERY high.

Is this before you extend an arm? or after?

You want to calculate the CoM after you've got whatever appendages on your robot extended completely, as well, to make sure it isn't outside of the base of your bot.

This is all with the arm fully extended and a vision tetra loaded.

[Cory]

As long as you're careful and don't jerk the controls back and forth, you should be ok.

If there's any way to make it lower, you might want to try it.

ie: putting motors on the base instead of at the joint on the arm.

[Alex1072]

Originally Posted by Cory

As long as you're careful and don't jerk the controls back and forth, you should be ok.

If there's any way to make it lower, you might want to try it.

ie: putting motors on the base instead of at the joint on the arm.

Mmmmh.... cheeeeeese.....

[Gary Dillard]

Consider these numbers:

Mass "m" located 60 inches high, 17 inches aft of the front wheels (assuming 4 inch wheels).

Summing the moments about the front wheels where they contact the carpet:

(60)(m)(a) = (17)(m)(g), where "a" is either the acceleration where you start driving or the decelleration where you run into something or someone. This says that you will begin to tip over at .283 g's acceleration or deceleration - that's about 9 ft/sec/sec. Most robots will get to full speed (>9 ft/sec) in less than a second, so their peak acceleration must be higher than their average acceleration of 9 ft/sec/sec. And when you run into someone (or they run into you) you will stop considerably faster. I would say you are looking for disaster here.

One interesting note (maybe a little too technical to explain in a simple post) is that this assumes rigid body motion. If your arm / mechanism is flexible it will actually accelerate or decelerate slower than your chassis because of the energy that gets stored in the spring rate. So if you have a tall rigid structure it could be more susceptible to tipping than a soft structure, assuming the decrease in acceleration of the cg more than compensates for the change in downward moment arm to the cg.

[Ianworld]

I'm going to go out on a tetra manipulator here and say that a 5 foot high COM is a bad idea. the angle of tipping that would be required to make your robot go over sideways is just a couple degrees. backwards and forwards isn't much better. Just accelerating hard willl make you go over. I remember the box ball collectors of 2002, they fell all the time because they had too much weight up top. Its a big issue. I strongly suggest trying to find a way to lower it. Even 3 or 4 feet would help a lot.

[Greg Ross]

Originally Posted by Alex1072

What do you guys think: If we have a foot print that fills up all 28x38 inches with wheels pretty close to the corners, is a 5 foot vertically center of mass too high? How tipsy would our robot be? The COM is actually almost dead center on the robot on the horizontal plane, but still, 5 feet is a lot isn't it?

Assuming your wheels are RIGHT AT the edge of the 'bot, and the center of mass is centered above your footprint, it's only 14" from the tipping point. If you were to tip any more than 13.13 degrees (tan^-1(14"/60")), you're going to go on over! Not a pretty picture.

[Projectchon]

Bacely, robots that have arms for this year, they all have a high COG (max 60in high)

PS: put thing like the 12v and the compesser lower to the ground, and on the opp. side of your arm, if your arm is at the front or back of the robot.

[Gary Dillard]

Originally Posted by Projectchon

Bacely, robots that have arms for this year, they all have a high COG (max 60in high)

Please forgive me for using numbers to back up my statements; it's one of my engineering habits.

130 pound robot, 10 pound tetra lifted ~ 8 feet in the air

Example 1: Arm weighs 40 pounds, uniformly distributed vertically; remaining chassis/drive train/battery, etc. weighs 90 pounds, cg at 3" based on center of 6" base structure.

[(40)(50)+(90)(3)+(10)(100)]/140 = 23.5 inch cg height

Example 2: Arm weighs 20 pounds, remainder weighs 110 pounds at 4" height

[(20)(50)+(110)(4)+(10)(100)]/140 = 13.5 inch cg height

Take the weight out of your tall structure and move everything down as far as possible. It's not that hard to do if you design it in to start with. Otherwise, if you're our opponent - thank you very much, and if you're on our alliance please make sure your structure is designed to be plowed across the field; we'd like to have the 10 points for putting you in the end zone with us.

[sanddrag]

If you model your robot and the tetra in Inventor and set the mass properties correctly for everything, you can get a pretty good idea of where it is. Look under the File Menu for iProperties and the View menu for Center of Gravity.

[Alex1072]

Originally Posted by Gary Dillard

Please forgive me for using numbers to back up my statements; it's one of my engineering habits.

130 pound robot, 10 pound tetra lifted ~ 8 feet in the air

Example 1: Arm weighs 40 pounds, uniformly distributed vertically; remaining chassis/drive train/battery, etc. weighs 90 pounds, cg at 3" based on center of 6" base structure.

[(40)(50)+(90)(3)+(10)(100)]/140 = 23.5 inch cg height

Example 2: Arm weighs 20 pounds, remainder weighs 110 pounds at 4" height

[(20)(50)+(110)(4)+(10)(100)]/140 = 13.5 inch cg height

Take the weight out of your tall structure and move everything down as far as possible. It's not that hard to do if you design it in to start with. Otherwise, if you're our opponent - thank you very much, and if you're on our alliance please make sure your structure is designed to be plowed across the field; we'd like to have the 10 points for putting you in the end zone with us.

Thank you everyone for the help, I already ran through the calculations and got the unhappy result after my last post. 135 fl-lb of torque needed to tip us =/
not much at all. Our arm is VERY tall at full extension, so it's not so much that its heavy, but more so that it's just big. That being said we're redesigning it to be lighter anyways. We're also considering outriggers. Our center of mass without the arm is probably obscenly low as it is, and i'm fairly sure we can't move much more stuff down.