Balancing a mass within 1 degree?

[XaeroR35]

Let's say I have a large structure that weights 100k pounds that is hung from a simple support at the very center (x,y = 0,0). The CG is offset by a few inches, say (6,2). I need the structure to hang within 1 degree of flat.

This seems like a simple trig problem but I am unsure how to setup the equation(s). I can add counterweight to bring it into spec, but need to know how much weight and where to put them ahead of time.

 

[rb1957]

if the structure is supported at the origin, and the CG is at some point (X,Y) then the structure will rotate under the CG is under the support ... tan(X/Y)

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

 

[drawoh]

XaeroR35,

If you are hanging the thing flexibly, the centre of mass should be directly under the hanging point. It is likely that the centre of mass is not located exactly where you think it is.

This setup shows your CofM in X and Y. You need Z as well to do your calculation

Is there any way you can mount this thing on three scales? This will tell you exactly where your CofM is. You can work out the additional masses required to move it to exactly where you want it.

--
JHG

 

[XaeroR35]

drawoh,

You made a great point. It must be too early, yes we have a Z direction component as well, and that makes the problem a lot more clear.

 

[IRstuff]

Another approach is to hang the object twice, using different points. The resultant vertical axes will intersect at the CG.

TTFN
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[hydtools]

If the x-axis is horizontal axis then the counter weight, wc, must balance the structure weight. W * x = wc * xc, sum of moments about the support attach point. Same approach applies in the z direction. The y location of the center of mass creates no rotating moment about the attachment point.

Ted

 

[drawoh]

hydtools,

I am using Z as the vertical axis, not Y.

--
JHG

 

[hydtools]

Drawoh,ok. But what is op's system ref?

Ted

 

[chicopee]

What type of a " simple support" do you have for the structure which we need to know if you want an equation to level the beam? Right now it seems that you are referring to a point load for support which I am inclined to believe that it would not be a safe procedure for a 100K load.

 

[XaeroR35]

Maybe this sketch would help make things more clear:

 

[IRstuff]

no sketch apparent

TTFN
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[hydtools]

I offer the same solution. Structure weight times 6 inches = counter balance weight times its distance from suspension vertical.
Pick counter balance weight = 1000lbs.
The counter balance weight must be located 100000*6/1000 = 600 inches from the suspension vertical to the right for perfect balance.
For 1deg out the cg is about 1.75 inches from the suspension vertical.
100000*1.75/1000 = 175 inches from suspension vertical for 1000lbs to balance offcenter cg.

Ted

 

[hydtools]

My above final solution is incorrect. At 1 deg off level the cg will be about 4.24 inches from the vertical suspension line. The counter balance of 1000lbs will need to be 424 inches from the vertical, not 175 inches. At perfectly level, the angle of a line from the suspension point through the cg is 3.43deg(arc tan 6/100).

Ted

 

[drawoh]

hydtools,

Neither the 100in or the 6in is known from the OP's test. All we have is an angle.

--
JHG

 

[hydtools]

drawoh, I was using data from op's sketch above.

Ted

 

[drawoh]

hydtools,

I don't think those are real numbers. The OP made them up to populate his sketch. If he knows X and Z as shown, I am not willing to explain his counterweights to him.

--
JHG

 

[desertfox]

Why cannot the lifting eye be repositioned if the true CofG is known?
That's surely the easiest solution.

 

[Ussuri]

Does the OP know where the COG is?

 

[desertfox]

Well reading the OP's post his problem is setting up the equations and indicates that the COG is off 6,2 so I guess he knows roughly where the COG is, that said even if he doesn't know he will have to find out roughly where it is before anyone can help him or even consider the positioning of counter weights.

 

[rb1957]

let's stop guessing ... Xaero35, what to you Know ?

do you Know the weight of the object and have an estimate for the CG location ? then you can solve for the rotation and add balance weights if required at a convenient location (away from the suspension axis). then test to confirm.

else test, and that'll show you something about the CG. then add weights to achieve your target. you might use two locations, offset parallel to the suspension axis, since you're supporting the body along an axis, and not at a point.

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