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Balancing a mass within 1 degree?

Balancing a mass within 1 degree?

Balancing a mass within 1 degree?

(OP)
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.

RE: Balancing a mass within 1 degree?

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 ?

RE: Balancing a mass within 1 degree?

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

RE: Balancing a mass within 1 degree?

(OP)
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.

RE: Balancing a mass within 1 degree?

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

RE: Balancing a mass within 1 degree?

hydtools,

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

--
JHG

RE: Balancing a mass within 1 degree?

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

Ted

RE: Balancing a mass within 1 degree?

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.

RE: Balancing a mass within 1 degree?

(OP)
Maybe this sketch would help make things more clear:

RE: Balancing a mass within 1 degree?

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

RE: Balancing a mass within 1 degree?

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

RE: Balancing a mass within 1 degree?

hydtools,

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

--
JHG

RE: Balancing a mass within 1 degree?

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

Ted

RE: Balancing a mass within 1 degree?

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

RE: Balancing a mass within 1 degree?

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

RE: Balancing a mass within 1 degree?

Does the OP know where the COG is?

RE: Balancing a mass within 1 degree?

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.

RE: Balancing a mass within 1 degree?

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 ?

RE: Balancing a mass within 1 degree?

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

Ted

RE: Balancing a mass within 1 degree?

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

RE: Balancing a mass within 1 degree?

(OP)

Quote (chicopee)

It is done everday while installing subsea equipment.

It is done everyday while installing subsea equipment.

RE: Balancing a mass within 1 degree?

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 ?

RE: Balancing a mass within 1 degree?

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.

RE: Balancing a mass within 1 degree?

i'd expect it to be less underwater

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

RE: Balancing a mass within 1 degree?

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.

RE: Balancing a mass within 1 degree?

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 ?

RE: Balancing a mass within 1 degree?

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

RE: Balancing a mass within 1 degree?

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 ?

RE: Balancing a mass within 1 degree?

rb1957,

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

--
JHG

RE: Balancing a mass within 1 degree?

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 ?

RE: Balancing a mass within 1 degree?

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.

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