Centre of Gravity - Tipping angle

[nigelshaw]

Hi,

i have, at least, im in the process of making a workwook to calculate the Centre of gravity to a given size based on its weight and footprint. I need to be more exact though and work out the angle at which the object will tip over rathe rthan tip back. I belive this is called the Moment of Inertia..

i am calculating on various things outlines as- ( based on a POS unit standing in wheels )

CWLR - Centre of wheels Left to right (600mm)
CWBF - Centre of wheels back to front(600mm)
HCW - Height of the centre of the wheels ( pivot point )
AUT - angle that the unit will be tipped to

to calculate the height of the front edge when tipped at 15 degrees, i have used this-

=SIN((15*PI()/180*B5

The return result 135.29mm ( i checked in CAD and it is correct ) this is the 90 degree vertical distance from the 600mm point to the base line when angled at 15 degrees.

There is an uneven weight to the unit on 1 side.
The left & right side weigh 32kg each
the heavy side weighs 48kg
the opposite side weighs 40kg ( a small central panel dividing left & right but spanning front to back )
The weight on the heavy side when the the opposite side is elevated to 15 degrees is 50kg ( 10kg weight transfer from back to front )

returned values
135.29mm - difference between Front raised and front flat level
10kg - Weight added
584.5mm - Distance between centre of wheels when front is elevated ( vertical distance )

CENTRE of GRAVITY

(X) = 333.33mm - centre of wheels front to back
(Y) = 196.33mm - Distance from front to back to plot a cross point
(Z) = 380.03mm - Distance from plot point in perpendicular direction u[pward

(Z) = actual centre of gravity point.

i need to work out the failure angle for this that would allow the unit to tip over rather than fall back into position. Its eithe rthis or make the unit and push it over to see if it actually falls over lol.

Would anyone know the missing calcualtion? its baffled me all day :)

i also need to get the weight displacement more accurate. If the base length is 600mm and the angled vertical length is 584.5, there is a difference of 15.5mm the opposite unit side when flat weighs 40kg so when it is angled, there is a linear displacement of 15.5mm. This must be able to be calculated as displaced weight and added on to the heany side somehow...

Hopefully, someone can help me out of direct me to a place that will assist me further


Thanks

Nigel

[Paul]

Hi Nigel,

We certainly appreciate the amount of detail you provided for your issue, but I feel this is more of a physics problem than an Excel problem. If you knew the physics formula to calculate such a design, my guess is it would be quite simple to convert to an Excel formula or macro. Would this question not be more suited to a physics or engineering site?

Hopefully there's someone around here familiar with such calculations and can help, but I can't guarantee that. If you had a function/macro that was not working properly, then more people could certainly assist.

[shg]

The object has an unstable equilibrium (i.e., balanced, but not for long) when the CG is directly above the pivot point. The pivot point would be an outside corner of the tire if it were rigid (not the center of a wheel), but of course the tire will deform.