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.

 

[robyengIT]

My math memory has grown a bit rusty : am I wrong somewhere ?

 

[hydtools]

robyengIT, no your memory is good. The op does not give explicitly your Lg.

Ted

 

[chicopee]

My two cents again, lifting 100K lbs (=100,000 lbs) which is 50 tons with a single line is shear madness.

 

[XaeroR35]

It is done everday while installing subsea equipment.

It is done everyday while installing subsea equipment.

 

[rb1957]

1st I've heard of "subsea" ...

is bouyancy an issue ?

is this only assembled underwater ? if so, can you test (with currents and such) ?

or is this "by analysis only" ?



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

 

[Ussuri]

You said you need it to hang by 1 degree. You should be aware that the COG location will change once you lower it subsea, so if you get hanging at 1 degree in air, it will not be at 1 degree subsea, and vice versa.

If the angle is critical (i.e for mating up subsea) then I would suggest using a multi point lift and setting your sling lengths to the give you the angle required.

 

[rb1957]

i'd expect it to be less underwater

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

 

[Dougt115]

By hanging it from a single point there is no way to perform your calculation. Even with the angle given. The CG can be located anywhere along the vertical axis.

Easiest solution is hang it from a second then third points and find the intersecting axis point or use a variable counter weight until the angle meets requirements.

Also weigh it at three points equidistant from the lift point and then counter balance until all three scales are equal.

 

[rb1957]

is he trying to determine where the CG is ? i read it as he wanted the object to rotate less than 1deg from vertical, and we were suggesting ways to achieve this.

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

 

[drawoh]

rb1957,

My interpretation is that he is trying to hang the object less than one degree from vertical, and he does not know where the centre of mass is.

--
JHG

 

[rb1957]

but i don't think he cares (ie he's not trying to find the CG), but if he knows the weight of the body he can adjust the rotation by adding a balance weight.

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

 

[drawoh]

rb1957,

In the OP, he is trying to set up the calculation, which does require the centre of mass.

--
JHG

 

[rb1957]

in the OP he assumed a CG position, and i think later posts suggests he knows where it is. I think we're all trying to tell him it's easy to figure out the rotation, tan(X/Y), like he shows in his sketch, 13 Nov 15:13, and from that is should be easy to figure out much much weight to put where to bring the CG to the 1deg line.

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

 

[Dougt115]

M2 * X = M1 * l* sin(theta)

M1 is the block weight
l is the cg distance from the horizontal plan containing the attachment point

M2 is the mass to add
X is the distance from the hook point

vector is opposite to the slope of the block (uphill direction)

Rotation created by M2 * x must equal M1 * (CG offset from vertical) before lifting to keep the block balanced. After lifting the horizontal distance can be found from L * sin (theta). M2 * x is a constant. The vector is reverse of the slope, I believe it was 6,2 for an offset, so the vector would be on the line of -6, -2.

Also if you balance the unit by trial and error you can used these same equations to find the CG of the original block.