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Center of gravity of a 3d triangle using weight

  1. Dec 17, 2014 #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?

  2. jcsd
  3. Dec 17, 2014 #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


    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.
  4. Dec 17, 2014 #3


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    Science Advisor

    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:

    But your case is presumably simpler, because it's static, right? If there are only vertical forces at the 4 sensors the math simplifies.
  5. Dec 17, 2014 #4
    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.
  6. Dec 17, 2014 #5


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    Science Advisor

    Then you can do the measurement twice and find the intersection of the lines you get from each measurement.
  7. Dec 17, 2014 #6
    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.
  8. Dec 17, 2014 #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.
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