Center of gravity of a 3d triangle using weight

In summary: D space). Draw a line from the fulcrum point to the center of mass. This is the line of intersection of the load indicators.The fulcrum point is the point where the two lines intersect. From there, you can use geometry to determine the center of mass.
  • #1
cps.13
40
1
Hi all,

I am after some help to find out how to do the following:

I have a triangular object (weighing around 40,000kg) which I can measure and weigh. I have 4 load indicators I can position anywhere beneath the object, these will give me a very accurate reading of the weight of the object at point were the load indicator is positioned.

What I need to be able to do is calculate the center of gravity using only the info below:
1) The individual weights of each load indicator
2) The distance of each load indicator from a specified point, either a side, corner, center etc.
3) Size of the object

The weight of the object is not uniformly distributed so I cannot simply calculate the center of mass.

If it is not possible with the 3 bits of information I have, can you advise what information I would need to do it?

Can anybody help?

Thanks,
 
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  • #2
If it helps I have been reading this to start (source: http://www.grc.nasa.gov/WWW/K-12/airplane/cg.html)

If the mass of the object is not uniformly distributed, we must use calculus to determine center of gravity. We will use the symbol S dw to denote the integration of a continuous function with respect to weight. Then the center of gravity can be determined from:

cg * W = S x dw

where x is the distance from a reference line, dw is an increment of weight, and W is the total weight of the object. To evaluate the right side, we have to determine how the weight varies geometrically. From the weight equation, we know that:

w = m * g

where m is the mass of the object, and g is the gravitational constant. In turn, the mass m of any object is equal to thedensity, rho, of the object times the volume, V:

m = rho * V

We can combine the last two equations:

w = g * rho * V

then

dw = g * rho * dV

dw = g * rho(x,y,z) * dx dy dz

If we have a functional form for the mass distribution, we can solve the equation for the center of gravity:

cg * W = g * SSS x * rho(x,y,z) dx dy dz

where SSS indicates a triple integral over dx. dy. and dz. If we don't know the functional form of the mass distribution, we can numerically integrate the equation using a spreadsheet. Divide the distance into a number of small volume segments and determining the average value of the weight/volume (density times gravity) over that small segment. Taking the sum of the average value of the weight/volume times the distance times the volume segment divided by the weight will produce the center of gravity.
 
  • #3
cps.13 said:
I have 4 load indicators I can position anywhere beneath the object, these will give me a very accurate reading of the weight of the object at point were the load indicator is positioned.
I don't think you can get a 3D position of the CoM from just one measurement (one orientation of the object). But you can the 2D position in the horizontal plane. It'S similar to computing the center of pressure on a force platform with 4 sensors:
http://www.kwon3d.com/theory/grf/cop.html

But your case is presumably simpler, because it's static, right? If there are only vertical forces at the 4 sensors the math simplifies.
 
  • #4
A.T. said:
I don't think you can get a 3D position of the CoM from just one measurement (one orientation of the object). But you can the 2D position in the horizontal plane. It'S similar to computing the center of pressure on a force platform with 4 sensors:
http://www.kwon3d.com/theory/grf/cop.html

But your case is presumably simpler, because it's static, right? If there are only vertical forces at the 4 sensors the math simplifies.

Thanks I will look at that center of pressure.

You are correct the object is static.

It is possible to rotate the object.

i.e. lower onto the load indicators with them in one position. Rotate, lower onto the load cells in a second position. If this makes things any easier.
 
  • #5
cps.13 said:
It is possible to rotate the object.

i.e. lower onto the load indicators with them in one position. Rotate, lower onto the load cells in a second position. If this makes things any easier.
Then you can do the measurement twice and find the intersection of the lines you get from each measurement.
 
  • #6
A.T. said:
Then you can do the measurement twice and find the intersection of the lines you get from each measurement.

Sorry to be a paint but could you explain that in a bit more depth, this is the first time I have looked into this so all a bit new to me.
 
  • #7
In plane geometry there is a theorem that the COM of a triangle is where the geometric medians meet: http://en.wikipedia.org/wiki/Median_(geometry)
This can be extended to non-uniform mass distribution.
Calculate the fulcrum point between each pair of points where load indicators are located.
Connect each fulcrum point with its opposite load indicator point.
Ideally the line segments so constructed will converge at the COM.
If there is a small error in one or more of the lines, they will form a much smaller triangle, within which the COM lies.
 

1. What is the center of gravity of a 3D triangle?

The center of gravity of a 3D triangle is the point where the entire weight of the triangle can be considered to act. It is the average position of all the individual particles that make up the triangle, taking into account their respective masses and positions.

2. How is the center of gravity of a 3D triangle calculated?

The center of gravity of a 3D triangle is calculated by dividing the sum of the moments of all the individual particles by the total weight of the triangle. The moment of a particle is the product of its mass and its distance from a chosen reference point.

3. Does the center of gravity of a 3D triangle change with its orientation?

Yes, the center of gravity of a 3D triangle changes with its orientation. As the triangle is rotated, the positions of the individual particles change, thus altering the calculation of the center of gravity.

4. What is the significance of knowing the center of gravity of a 3D triangle?

Knowing the center of gravity of a 3D triangle is important in understanding the stability and balance of the triangle. It can also be used in designing structures or objects that need to be balanced or maneuvered in a specific way.

5. Can the center of gravity of a 3D triangle be outside of the triangle?

Yes, the center of gravity of a 3D triangle can be outside of the triangle. This can happen if the triangle is not symmetrical or if the distribution of mass is uneven. In this case, the center of gravity will be located at a point outside of the triangle's boundaries.

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