Math/Physics Question: Stability of an abruptly stopped object
March 5, 2013 9:16 PM   Subscribe

If you have an object with a given base length and a given CG height, how fast can it travel before you have to worry about it running over an obstacle and falling over?

This is for a work problem. I have a cart of given length and CG height. I'm trying to specify a "safe speed" which will guarantee that, if the cart were to run over an obstacle, it would not topple.

My first pass at this problem went as follows: The potential energy required to rotate the CG over the front wheels is:
delta_height = sqrt(cg_height^2 + (base_length/2)^2) - cg_height)
that is, the radius from the front wheels to the CG, minus the original Cg height
Ep = delta_height * mass * g

The kinetic energy at a given speed is:
Ek = 1/2 * mass * speed^2

Assuming all of the kinetic energy is transferred into potential energy when the front wheels hit an obstacle I equate the two E values and solve for speed:

speed_max = sqrt( 2 * delta_height * g)

This gives me a very pretty low value for speed_max (slower than a slow walk, for a base_length/cg_height ratio of about 1.5) which seems unreasonable. Have I made a mistake? Are my assumptions too stringent?
posted by Popular Ethics to Science & Nature (16 answers total)
you can go any speed and tip over an obstacle, and this is easy to verify. Just rest it against an obstacle and lean on it.
posted by empath at 9:19 PM on March 5, 2013 [1 favorite]

I think you just need to use the torque equations to figure out how much force would be required to tip it at rest, and then work out the velocity from F = m*v^2, no?
posted by empath at 9:29 PM on March 5, 2013

Your Ek=Ep would have been the first thing to pop into my mind, but ponder this... if the CG were at 0 height the cart wouldn't rotate at all. Sort of like an arrow shot straight into a wall would pierce, shatter, or bounce back. I would guess that you have to take into account the proportion of the forward velocity that goes into rotation, which at CGh=0 would be none, and at CGh=1/2 base would be 1/2. You would also need to take into account the amount of energy that was converted into rotational energy as in addition to raising the CG a height you're also rotating the cart.
posted by zengargoyle at 9:35 PM on March 5, 2013 [1 favorite]

Right, you'd have to treat the cart as a lever with the point that it hits the obstacle as the fulcrum.
posted by empath at 9:43 PM on March 5, 2013

I think zengargoyle is right -- you have to decompose the horizontal velocity vector into a component directed from the CG to the stopped front corner of the cart, and a perpendicular component that gives the force causing the CG to rotate up and over the corner. That's sinθ, where θ is the angle between the floor and the line from the stopped corner to the CG. So I think you can raise your tipover speed by that factor (i.e. divide by sinθ)
posted by spacewrench at 9:44 PM on March 5, 2013

Response by poster: I was thinking I could ignore the rotational energy by examining the case where the cart rotates and comes to a halt with the CG exactly over the top of the front wheels (i.e. kinetic energy, linear and rotational, is completely converted to potential energy). Any more energy and the cart keeps rotating and topples.

empath - I think you've got your force equations confused. F = ma, E = 1/2 mv^2. I chose an energy analysis over a force analysis because I thought the assumptions would be simpler.
posted by Popular Ethics at 10:12 PM on March 5, 2013

your approach was the first that sprung to mind for me as well, but that's actually modelling the cart hooking onto a hanging cable and swinging upwards by your delta_height - this will obviously require far less Ek.

I think you can ignore rotational energy (for the reason you described), and just assume that the Ek component on the vector from CG to front wheel is dissipated as heat/noise, and the component tangent to that is all that's available for the Ep increase.

So, if the line from front wheel to CG is at a 45º line to the ground, you need 1.414 times the Ek to topple (so ~1.2 times the speed you originally calculated). If the cart is lower to the ground, say the line from wheel to CG is inclined 20º, you'd almost 3 times the Ek and ~1.7 times the velocity.
posted by russm at 11:47 PM on March 5, 2013

I remember something said to me once about skyscrapers, that the reason they don't fall over in high wind is their basements. The structure needs to extend down into the earth in order to have the ground to push back against the wind.

It's really a similar situation here, but even more of an idealized example. It's like we're sliding a frictionless block across the ground and then it doesn't even hit anything physical, it's just that the two front bottom corners instantly stop and are locked in place. Other than gravity, there's absolutely nothing to keep it from tipping forward.

Maybe that's why the calculation is returning a seemingly low number, we can't help but put it in real life terms where there's going to be friction and an impact, and not all of the energy being converted to potential.

One thing I want to call attention to is that if you're saying you're using a base_length/cg_height ratio of about 1.5, is that it means the base_height is 2*cg_height, and so it's taller than it is wide. I don't know if that helps conceptualize it easily tipping over. It seems like if you change the base height to 1, it does require 25% more speed to get it to tip.
posted by cali59 at 11:55 PM on March 5, 2013

Are you designing a cart, or a safety rule for an existing cart? If it's the latter, a physical test would probably shed a lot of light on this fairly quickly.

IANAEngineer, but other complicating factors (which you may already be considering but left out of the question for simplicity):

*The weight and position of whatever the cart is carrying
*Whether the load itself might shift as the cart suddenly decelerates
*The height of the obstacle
*The diameter of the wheels
posted by jon1270 at 1:53 AM on March 6, 2013 [1 favorite]

I think you may have lost a bit of information:
delta_height = sqrt(cg_height^2 + (base_length/2)^2) - cg_height [removed superfluous closing bracket]
that is, the radius from the front wheels to the CG, minus the original Cg height
Ep = delta_height * mass * g

The kinetic energy at a given speed is:
Ek = 1/2 * mass * speed^2

Assuming all of the kinetic energy is transferred into potential energy when the front wheels hit an obstacle I equate the two E values and solve for speed:

speed_max = sqrt( 2 * delta_height * g)
so writing in full:
where r is the ratio base/height. This still has an extra factor of h that you may have missed, given at the end you say
This gives me a very pretty low value for speed_max (slower than a slow walk, for a base_length/cg_height ratio of about 1.5) which seems unreasonable. Have I made a mistake? Are my assumptions too stringent?
and don't mention what height you used, just what ratio.
posted by edd at 3:21 AM on March 6, 2013

Response by poster: Thanks for your help all - closer!

russm: I'm not sure you can treat energy as a vector like that (though I could be wrong). I was always taught that energy was a scalar. With regard to the Zengargoyle's arrow example, you can't guarantee that even at cg_height = 0 that the cart won't rotate. A little unstable bounce during collision would be all that's necessary to get the cart rotating, at which point it will keep rotating until all that energy is spent lifting the CG (and making noise, and other leftovers I'm ignoring).

jon1270: I'm setting safety rules for an existing cart - so it wouldn't be wise to perform a test. For the same reason, I can't change the cart dimensions.

cali59: I know I'm simplifying the problem significantly. Friction, noise and the height of the obstacle will change the result. I am trying to bound the problem by taking the worst possibility - the front wheels hit a thin wedge which stops them instantly, or the wheel bearings seize instantly and the wheels have such good grip on the floor that the cart rotates.

edd:Your expansion gives the same result as I was getting before. To put numbers to the problem (sorry about the superfluous bracket):

base_length = 24 in
cg_height = 16 in (assumed to be centered in base_length)

which gives:
delta_height = 4 in
speed_max = 55.6 in/s, or 1.4 m/s or about walking speed.

The units check out, so I'm tempted to conclude the formula is correct in so far as my assumptions are. I'm surprised that the result is so slow - my instincts were that a base_length/cg_height ratio of 1.5 would have been plenty stable.
posted by Popular Ethics at 5:20 AM on March 6, 2013

You know, I think your energy analysis is treating the problem like a pole-vaulter: you're assuming all the energy can be converted into lifting/rotating the CG around the stuck wheel. But since you don't have an efficient place to store the energy (no vaulting pole) you can only get good conversion if there's a smooth transition from horizontal to radial motion.

Consider the case where the CG is running along the floor: when the cart stops, the CG has to start moving directly upwards. If you don't have the "little unstable bounce during collision" there's no force to get the CG moving up. And you're not accounting for the energy dissipated in the bounce.

Even where the CG is up 45 degrees from the stopped wheel, it has to change from horizontal to radial starting at 45 degrees. You'll certainly lose a lot of energy in that transition.
posted by spacewrench at 7:30 AM on March 6, 2013 [1 favorite]

Yes, I don't think you should be framing it in terms of energy at first - if energy is disappearing into relatively ill-defined stuff then it's not going to be a useful quantity to work with, and there are indeed good reasons Popular Ethics pointed towards for not thinking about considering components of kinetic energy in different directions - even if you can get away with it sometimes you're at best making it easy for yourself to make an error.

So, treat it as an inelastic collision where basically KE+PE isn't conserved but momentum is, and then you want to consider the moment the collision happens as effectively eliminating the component of the momentum along the line between the centre of mass and the wheel (it's been sunk into the combined wheel-Earth system with no noticeable change in velocity of the Earth), and then consider the tangential momentum as providing angular momentum around the wheel (assuming the wheel is now basically anchored to the ground by the obstacle).

Only after that do you consider the energy, as when the system starts rotating after the inelastic bit has happened is it now a conserved quantity (with potential energy) with no losses to heat, sound and other annoyances.

You don't want to be decomposing energy into different directions as you'll probably start getting things wrong (even if in principle you could do it right).
posted by edd at 8:47 AM on March 6, 2013 [2 favorites]

I will agree with edd that this is definitely an interesting angular mechanics problem. I like his approach, although determining the rotational moment of inertia of your cart may be thoroughly nontrivial, especially as you begin loading it down with cargo.

Furthermore, is the toppling of the cart itself the only hazard? What about the stuff on or in the cart? Is there not a danger of cargo sliding around or falling to the floor, even if the cart manages to remain upright at the end of the dramatic halt?

I know you've specifically said you don't like the idea, but I would still strongly recommend an empirical solution: Decide how much wheelie-oriented jostling you can tolerate, reduce that figure significantly to provide you a comfortable margin of error, and then, in carefully controlled conditions (including, but not limited to, assistants to help control a bucking cart), determine what speed provides that amount of jostling upon a sudden halt.

Finally, even if you arrive at a quantitative answer, how helpful will that be? Do these carts have speedometers on them? Even if, at your next staff meeting, you announce that the maximum safe cart speed is 2.4 m/s, will that even be meaningful for the people pushing them around?
posted by unnatural logarithm at 3:42 PM on March 6, 2013 [1 favorite]

I'm setting safety rules for an existing cart - so it wouldn't be wise to perform a test.

This strikes me as a bizzare thing to say. The difficulty of establishing a mathematical method here only illustrates how likely you are to be wrong if you don't do real-world testing. Which is to say, extremely likely.
posted by jon1270 at 4:52 PM on March 6, 2013

I think that the second this became a real-world problem (and not a fictional ideal base with some arbitrary CG, eg) the chances of calculating the answer dropped to zero. For example, take "run over an obstacle" -- only one case is that the cart stops. Another case is a speed bump: the front wheels make it over the obstacle, the cart is now bouncing about, and then the back wheels hit. Now the back wheels are atop the obstacle, the entire cart is pitched forward by some amount, the CG is therefore leaning forward -- how much bounce can you tolerate now? This is all going to depend on suspension, whether you have pneumatic tires or hard rubber, how well inflated the tires are, etc.

Or take the case of a curb cut (going down from a sidewalk down to a road). Taken straight on, front axle perpendicular to the slope, it may be possible to figure out a safe speed to hit the transition between the slope of the curb cut and the slope of the ground. But even a few degrees out of true, the dynamics are totally different -- one of the front wheels hits that transition first, and the entire apparatus will flex, torque, and basically trip itself in a very complicated way. (This, by the way, is the failure mode that brought me about 5° of tipping away from depositing a 2500lb Bridgeport mill on its side in the middle of Massachusetts Ave. in Cambridge.)
posted by range at 7:27 PM on March 6, 2013

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