Rollover Calcs: Find Speed to Tip Trolley

[MatsNorway]

Hi

Is this formula all i need for finding the speed i need to tip the trolley?
https://dl.dropboxusercontent.com/u/72823890/Jobb/Rollover.jpg

 

[Simon Bridge]

speed of what?

 

[SteamKing]

What do those weird lines superimposed on the trolley represent?

 

[MatsNorway]

The lines are just to find the center where i consider the CG to be. Height difference from there to tipping point is 158mm.

The speed needed to tip the trolley. (consider it a rigid unit, it is just a sketch after all)

 

[SteamKing]

To answer the OP, no, equating potential energy with kinetic energy is not sufficient IMO to determine the tipping speed of the trolley. Plus, I have my doubts that the c.g. of the trolley and load have been determined correctly.

 

[Simon Bridge]

The lines are just to find the center where i consider the CG to be. Height difference from there to tipping point is 158mm.

I have my doubts that the c.g. of the trolley and load have been determined correctly.

Me too. It looks like only the cog of the load (looks like a railway wheelset) was used. But that may have been part of the instructions ... or maybe the idea is to find the speed the trolley would have to be going for a sudden stop to spill the cargo? (In which case the pivot point is wrong.)

speed of what?

The speed needed to tip the trolley. (consider it a rigid unit, it is just a sketch after all)

Yes. Speed of what needed to tip the trolley? Speed of a hammer impacting one end? The speed of the trolley before it hits something? Hits what, where?

Note: The trolley may travel at any speed without tipping, so the literal answer to your question is "no": because there is nothing to indicate that the trolley does tip at all.

Basically we need more information in order to help you properly.
It looks like you want the maximum horizontal speed of the trolley so that, should it come to a sudden dead stop, the whole trolley will tip over. But that's a guess - what was the actual problem statement?

 

[MatsNorway]

"It looks like you want the maximum horizontal speed of the trolley so that, should it come to a sudden dead stop, the whole trolley will tip over"

This is correct.

Speed of the trolley when it gets tipped around the point being the end of the line at the wheel.

 

[sophiecentaur]

I guess your 158 figure is a attempt at the (minimum) height the CM needs to be raised for the trolley to tip over. That would give you the GPE needed, which needs to be equal to the KE of the trolley.
We have no way of knowing if the position of the CM has been calculated correctly; it could be right, if the load is a lot more massive than the trolley. But surely, a trolley that's capable of lifting any load, must have significant mass??
Your geometry make the assumption that the trolley is massless or the load has a funny internal mass distribution (but it looks like a bogey of some sort).

The only 'formulae' you will get out of us, so far, is that PE = KE. (same old same old . . .)

 

[Simon Bridge]

Using KE=PE would assume an elastic collision - all the initial kinetic energy gets used for work lifting the center of mass. So the relation given would be a reasonable back-of-envelope check giving a maximum speed where the trolley is unlikely to tip over on a sudden stop. It won't give the minimum speed to tip the trolley.

Really prefer a conservation of momentum calculation but the geometry makes that tricky.

You'd think the pivot would be the front wheel axle though wouldn't you.
Maybe it doesn't matter.

 

[sophiecentaur]

Using KE=PE would assume an elastic collision - all the initial kinetic energy gets used for work lifting the center of mass. So the relation given would be a reasonable back-of-envelope check giving a maximum speed where the trolley is unlikely to tip over on a sudden stop. It won't give the minimum speed to tip the trolley.

Really prefer a conservation of momentum calculation but the geometry makes that tricky.

You'd think the pivot would be the front wheel axle though wouldn't you.
Maybe it doesn't matter.

What more can one do, as a start? There is absolutely no way to do a 'force' based solution to this and energy is the only way into it.
The main problem with the question, as posed, is that the mass of the trolley doesn't seem to be included so the position of the CM is not known. That is a massive factor, compared with the 'efficiency' of the collision.

 

[MatsNorway]

The load on the trolley is 3.5tonnes. the trolley is about 0.2t-0.3t CoG might be off but let that be my problem.

Just found this one.
https://www.accidentreconstruction.com/research/suv/rollovers[1].pdf

 

[sophiecentaur]

Once you have got past the nasty units, that link makes a good read and seems to have most of the considerations that apply to your problem.
Quite a big beast to be dealing with here!
Once you have sorted to the CM thing and addressed the position of the pivot point (axle and not road or curb contact point) then the basic PE = KE approach will give you a pessimistic value (good for safety)

As that article points out, the sideways stability could also be relevant, the castors will give you no braking help when the trolley is going sideways and the trolley could be somewhat narrower than its length.

From the table at the end, I would know which 4X4s to avoid, if I ever wanted to buy one!

 

[MatsNorway]

What you mean PE = KE? same formula i have? mgh = 1/2 x m x v^2

The stability comes from the contact point. It might rotate around axle center once you crash but the stability comes from the contact points. New versjon has no wheels so its not vital mind you.
https://dl.dropboxusercontent.com/u/72823890/Jobb/New%20config.jpg

 

[sophiecentaur]

What you mean PE = KE? same formula i have? mgh = 1/2 x m x v^2

The stability comes from the contact point. It might rotate around axle center once you crash but the stability comes from the contact points. New versjon has no wheels so its not vital mind you.

Yes. I just put it in its basic form.
I'm not sure I agree about the stability relating to the impact point. Why would the trolley not rotate about the wheel spindles? Why would the front wheels leave the ground unless the impact were well below the level of the spindles?

Bearing in mind the relative masses of trolly and cargo, I would suggest that any impulse on the base of the trolley is more likely to bust the trolley than tip it over. But I can't imagine this is a scenario you would really want to contemplate.

 

[MatsNorway]

It will probably go around wheel centre unless its a solid round object. its kinda hypotetical. Only relevant with great grip and good brakes. A bigger wheel on the same obstacle would also easier skip over rather than hook it and flip the trolley. It would also add a factor where the roll centre would follow the path of the wheel. I have no idea how that would work. But stability is from either the impact point or the furthermost outside of the device assuming sufficient grip.

My criterium was that it should not flip over in walking pace or about 6-7km`t close to powerwalk who I've been told is at 8km`t

 

[sophiecentaur]

How could you get it to fast walking pace? You been working out?

 

[MatsNorway]

Walking pace with the wheels would have been difficult you say? I am not sure on the resistance offered but it was discussed and considered likely to be a pain when static and one want to turn it a little bit.

 

[Simon Bridge]

What more can one do, as a start? There is absolutely no way to do a 'force' based solution to this and energy is the only way into it.

Conservation of momentum.
That would require a more detailed modelling of the cart and load since we need the moment of inertia about the front axle.

The main problem with the question, as posed, is that the mass of the trolley doesn't seem to be included so the position of the CM is not known. That is a massive factor, compared with the 'efficiency' of the collision.

i.e. post #1 is modelling the situation as if the trolley com is not important ... which we will agree is a poor model.

The question put to us is if the conservation of energy relation is appropriate to finding the tipping speed.

The answer is: "not really but together with some specific assumptions it's OK."

It's a situation where we really need to know what the calculation is for. i.e. if the scenario has been given as part of a course in engineering, then that would get a different sort of calculation to, say, a crash-scene investigation or determining safety protocols in a workplace.

 

[MatsNorway]

It is going to be made and used.

 

[sophiecentaur]

It is going to be made and used.

That should be impressive to see. Be sure to post some pictures of a working system.

I imagine that a thing this big would require some insurance so your insurer would give some advice.
I would suggest a speed limiting brake would be worth while (essential) and a dead man's handle would avoid the ghastly result of this thing getting on a slope.

 

[MatsNorway]

You did not read it all. I am ditching the wheels and there is no need for those things mentioned. Indoors use only and only in a workshop.

 

[sophiecentaur]

How about Friction? With three tonnes, it could be a problem.

 

[MatsNorway]

Moved by truck. The new solution allows the wheelset to be rotated on top. Rather than rotate the entire device. less friction now as i can use steel wheels.

 

[Simon Bridge]

It is going to be made and used.

In which case, I think that the best advise is "hire an engineer".

The specs, such as there are, have been a moving target so there is a limit to what we can do for you here.
The best we can say here is that a top speed fro that calculation will not be expected to tip the cart should it come to a sudden stop. You don't want any part of your body between the cart and any obstacle capable of making it stop so quickly.

The components will certainly rattle about - there is some way to secure the load right?
Do not neglect safety considerations - depending on jurisdiction there will likely also be industrial regulations about this sort of thing.

Good luck.

 

[MatsNorway]

You are just confused the new design it is not going to be moved without a truck. consern then is the loads during transport.

 

[Simon Bridge]

You are just confused the new design it is not going to be moved without a truck. consern then is the loads during transport.

I certainly am - like I said: the design is a moving target.
Don't know what "loads during transport" means or what the concern about that would be.
It would certainly change the nature of the calculation.

Advise still stands:
Hire and engineer.
Do not neglect safety.

 

[MatsNorway]

Well then i have no idea what you mean with moving target.

 

[MatsNorway]

That should be impressive to see. Be sure to post some pictures of a working system.

Works like a charm. Only difference for the future is that we will cut the feet of it to get it lower.