Will this cart tip over?

[oldmekkie]

Hi All!

I've run into a stability problem that's stumped quite a few so far. I've searched this forum and the internet,as well. The problem seems simple: a 4 wheel cart with a mass of 73 lbs hits it's front wheels against a barrier. The barrier does not obstruct rotation of the cart should it tip. The CG of the system (cart + load) is 5.64" in the "X" direction ("X" direction is parallel to the ground) and 30.58" in "Y" (vertical). These CG measurements are taken relative to the front wheel axis. The CG is also between the front and rear wheel axis so the cart is not tipping at zero acceleration on the system. The initial velocity of the cart is 31.5 inches per second. What is the formula to calculate whether the cart tips over?

The goal is to determine if tipping will occur for a variety of X, Y, and mass values.

If you're a wiz at dynamics, assume the cart's wheels are 3" diameter and the height of the obstruction is 0.4". So in some (likely most) instances the cart will roll over the barrier. What are the formulas to determine if it tips over or stabilizes after impact?

Your help is kindly appreciated!

Thanks in advance!

 

[KENAT]

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[gruntguru]

Sounds ridiculously simple to me so I'm probably wrong.

The toppling limt occurs when (in side view) a line drawn through the wheel-contact-point on the barrier and the wheel axis, passes through the CG. If the CG is above this line the cart will topple. If the CG is below the line, the wheel will roll over the barrier.

This is since all force vectors act along that line at the toppling limit. The wheel is on the verge of lifting off the ground so the wheel force acts from the remaining contact point through the axle. The rear wheel is also on the verge of lifting off the ground so there is no force there. The inertial forces (gravity + acceleration) oppose the only external force (front wheel force already mentioned).

The formula then comes down to basic trig.

je suis charlie

 

[Jughandle]

Similar to post #2 by Kenat, I would set up an energy balance equation using x, y, and m (as the parameters in which you are interested). Assume no dissipative effects for starters (i.e. no rubber tires that absorb energy), and wheels that are a relatively small percentage of the total mass.
Chronology will be:
-Initial KE = ½ mv^2 + [rotational]
-The cart and wheel assembly hits the barrier, cart comes to an abrupt stop, fudge in a wheel factor, or experimentally determine how much wheels continue to spin.
-Now the cart and wheel assembly start to flip.
-Tipping point occurs when the CG rises to equal the initial KE.

 

[rb1957]

I don't think it's that simple, though certainly it won't tip if the contact point is above the CG.

the forward KE of the cart is turned into rotation energy around the contact point and this rotational energy is working against gravity. there's also a potential energy term (but this my be the same as my "working against gravity").

more complexity ... the initial KE is easy to figure out. as the cart impacts the stop, how much force is created ? (the typical "impact force" problem; adding a soft pad to the contact point will reduce the tipping of the cart). This force creates a moment about the contact point, which creates the rotational energy as the cart pivots about the contact point. The work done by this rotational energy is reduced by gravity (potential energy) so there's some velocity which is the critical limit for tipping (less than this it doesn't tip, more than it "over she goes !").

not a simple calc.

another day in paradise, or is paradise one day closer ?

 

[LittleInch]

I think you can this on an energy basis.

The CG of the cart needs to rise approx. 0.515" ( hypotenuse of the dimensions given minus Y) before it reaches the point where it is in front of the front axle and hence falls forward. The velocity at this critical point can be close to zero for the ultimate case and hence no rotational energy is present.

Hence it needs to gain that potential energy from somewhere.

this is the transfer of kinetic energy from the horizontal movement of the cart to the potential energy of the mass.

Whether your cart tips over will depend on how that kinetic energy is otherwise absorbed by say some foam or other spring like device or perhaps some brakes sufficient to reduce the kinetic energy at the final point to less than what is needed to raise the CG a fairly minute amount of 0.51"

I can't work imperial units so easily to work this one out, but I think that's one way to approach it.

I think the answer in terms of velocity will be very low - this seems a high Cof G for such small wheels and very close to the front axle

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Denial]

I was going to make a similar suggestion to LittleInch's.[ ] It is certainly true that the cart will not somersault unless its initial kinetic energy exceeds its gain in gravitational potential energy when its centre of gravity rises to its highest point.[ ] But as a predictor for whether the cart will somersault this approach is conservative because it assumes perfect KE–PE conversion.

It also assumes, as is assumed by the original question, that somersaulting is what will tend to happen.[ ] Given that the wheel diameter is ~3" and the obstruction height is ~0.4", isn't it possible that the cart's front wheel will run over the obstruction?[ ] (This situation can also be investigated by the same sort of energy approach that is used above, but with all the same caveats.)

 

[rb1957]

actually, since the contact is so low (within the tire) I think the problem is more complicated ...

when it hits the obstruction, will the cart pivot about the contact point, only 0.4" above the ground with tire 1.5" radius, or will it simply roll over it ? now i think the problem is that there's a limiting slow speed above which the cart will roll over the obstacle ?

another day in paradise, or is paradise one day closer ?

 

[oldmekkie]

Good points all. I set it as 2 problems given the notably higher complexity in solving for the wheel rolling over the obstruction. The simpler "hit, stop, rotate, and maybe crash" was giving me fits at first. Even still, there are a lot of assumptions with the simpler version. I've had the benefit of conducting physical testing but even with that there are a lot of assumptions.

The latter problem of accounting for the wheel rolling over the obstruction is notably more challenging. Assumptions which can be made (though not necessarily true in reality) are: system is rigid to itself, wheels and obstacle do not compress, there is no slippage or friction in the system, and secondary impact of rear wheels is ignored, both front wheels hit the obstruction at the same time (e.g. no "spin" of the system is introduced.) My initial pass at setting up the problem indicated that the calculations centered around determining if the CGx passed the center axis of the wheel at various time intervals. I couldn't quite derive formulas that worked successfully so I abandoned that approach.

I rarely encounter this type of problem in my work so there's a bit of rust on the gray matter.

Thanks again to those providing comment.

 

[GregLocock]

For the very closely related problem of working out the forces in a compliant suspension when a compliant tire hits a square step I had some success with an excel based cam following model. However, it was indicative rather than absolute. Further work on the same subject needed a full blown multi body dynamics model, and all the real world data to correlate it. In essence the model was fine tuned by varying the mass of the little bit of tire where it struck the step, and the stiffness and damping properties of the sidewall at that point.




Cheers

Greg Locock


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[rb1957]

a thought ... for it to roll over the force on the rear tires is down (and deflection away from the surface is zero, of course, although it'd still have velocity along the surface), for it to tip (or at least start to tip) the force is zero (as it deflects upwards, rotating about the contact point).

another day in paradise, or is paradise one day closer ?

 

[gruntguru]

If it tips it actually rotates about the front axle.

The simple model I proposed above is correct if you assume rigid components, zero friction and zero rotational inertia. ie it works for the case of pulling the cart forward slowly with a horizontal string attached to the CG.

je suis charlie