From center of mass to individual points with mass

Grasshopper
1

Hi all,

calculating from a bunsh of point with different mass the center of mass can be done like this
grafik
is there a way to calculate m1,m2,m3 etc. if Xcom,Ycom,Mass_com and all individual point positions are known ?.

2

Not if you have more than about 2 masses, I think, because you only have 2 equations…

1 Like
3

Your total mass equation gives you a third equation so you could get an answer with 3 points.
You might be able to find more, if you have some rotational information like moment of inertia.

I think if you have more points there is probably not one unique solution. I’m not 100% sure it would work, but it might be fun to try seeing if Galapagos can help you find valid solutions.

4

True,
i tried already your solution with 3 equations between 3 points.
one with geometry solution.

import Rhino as rh

a = []
for i in x:
    L = []
    for j in x:
        if i != j:
            L.append(j)
    Li1 = rh.Geometry.Line(i,y)
    Li2 = rh.Geometry.Line(L[0],L[1])
    t = rh.Geometry.Intersect.Intersection.LineLine(Li1,Li2)[2]
    IntPt = Li2.PointAt(t)
    IntDis = rh.Geometry.Point3d.DistanceTo(IntPt,y)
    Dis = rh.Geometry.Point3d.DistanceTo(i,IntPt)
    weight = IntDis/Dis
    a.append(weight)

grafik
one with a matrix solution

import os
from clr import AddReferenceToFileAndPath as addref
location = os.path.join(os.path.expanduser("~"),
'C:/Users/Florian/Desktop/MathNet.Numerics/lib/net40/MathNet.Numerics.dll')
addref(location)

import MathNet.Numerics.LinearAlgebra as la
from System import Array as sys_array
def array(*x): return sys_array[float](x) 

A = la.Double.Matrix.Build.DenseOfRowArrays(
array(*[i.X for i in x]),
array(*[i.Y for i in x]),
array(*(1,1,1))
)

b = la.Double.Vector.Build.DenseOfArray(array(y[0],y[1],1))

a = A.Solve(b)

Both solutions gives the same results

I ask myself if triangles between 4 points could be made if the mass point is in between.
So 4 points with 1 mass point gives 2 triangles.
Capture
and in suspension points with two results i would divide the sum by the amount of triangles to get the average.
A exact solution is not required only a non optimum solution is good enougth.
My onkel gives me already a solution to deal with 4 points but i was not able to build the equations with his explenation…maybe someone give it a try.

The translated text is this.

Assuming that the individual masses in the quadrants are equal and add up to a sum (M1, M2, M3, M4).
M1, M2, M3, M4 = Sum of the (same) masses in the respective quadrant
X1, X2, X3, X4 = X coordinate of the centers of mass of M1, M2, M3 …
ie. the mean distances of the X - axis in the respective quadrant
Y1, Y2, Y3, Y4 = Y coordinates of the center of gravity of M1, M2 …
ie. the mean distances from the Y axis in the respective quadrant
(all distance values as amount positive!)
G = total weight in the coordinate origin = M1 + M2 + M3 + M4

If you let the masses (weights) turn around the X- and Y-axis follows
Equation system with 4 unknowns: