Working with determining the corner weights on one axis isn't too bad.

1 Axis

If A is to the left of the Center of Gravity and B is to the right of the Center of Gravity. You have two equations to work with.

$$\text{Total}_\text{Moment} = \text{A}_\text{Moment} + \text{B}_\text{Moment}$$ $$\text{Total}_\text{Weight} = \text{A}_\text{Weight} + \text{B}_\text{Weight}$$

If I have a rectangle I can determine the weight on one axis for the left section and then determine the weights for the other axis. Since everything is parallel this seems to work out.


However, once I have a trapezoid, things stop coming out. How would I determine this?

Also, is there a way to determine the corner weight of a quadrilateral that doesn't have any corners that are parallel to one another.


In all the example, the Total Weight is known, the CG on the X axis and Y axis relative to a given Datum point are known. And all corner locations relative to that same Datum point are also known.

  • $\begingroup$ This is a rather complex problem (it is overconstrained) since there is an assumption that all the points of the trapezoid (or n-gon in general) always remain coplanar. This means there are only three degrees of freedom (vertical position and orientation of the plane) and typically more than three support points. To effectively answer, one must consider the table as a flexible body that can store potential energy, whose minimization yields the reaction forces. $\endgroup$ Feb 4, 2019 at 20:17
  • $\begingroup$ Would this be less complex if CD and AB were parallel and BD was perpendicular to AB? (And by less complex, I mean, solvable) $\endgroup$ Feb 4, 2019 at 20:23

1 Answer 1

  1. Problem Setup

    A flat table with $n$ legs resists the total weight $W$ and the forces and moments are balanced. Each leg is located at $(x_i,\,y_i)$ relative to the center of gravity of the table. The legs run along the z-direction.

    Find the force $F_i$ on each leg.

  2. Mathematical Model

    First the degrees of freedom are recognized and the problem is described with respect to those degrees of freedom. Since the table defines a plane, there are three possible DOF here. The total vertical translation of the table $\delta_z$, and two orientation angles $\theta_x$ and $\theta_y$ such as the cartesian coordinates of each support are $$ \vec{p}_i = \pmatrix{x_i \\ y_i \\ \delta_z + y_i \theta_x - x_i \theta_y} $$

    To make the problem solvable, we assume the legs are equally elastic (springs) with a coefficient $k$, which in the end will cancel out. This makes the leg force proportional to the deflection (taken from the z-coordinate of $\vec{p}_i$) $$ F_i = k (\delta_z + y_i \theta_x - x_i \theta_y) $$

    This means, that if we solve for the three degrees of freedeom $(\delta_z,\, \theta_x,\, \theta_y)$ we can calculate each leg force and sovle the problem.

  3. Force Balance

    The sum of the forces must equal the weight, which yields one of the three equations needed.

    $$ W = k \sum_{i=1}^n (\delta_z + y_i \theta_x - x_i \theta_y) $$

    The above is solved for the vertical translation $\delta_z = \frac{1}{n} \left( \tfrac{W}{k} + \theta_y \sum_{i=1}^n x_i - \theta_x \sum_{i=1}^n y_i \right) $ which can be used in the leg force expression as

    $$ F_i = \frac{1}{n} \left( W + k \left( \theta_x \left( n y_i - \sum_{i=1}^n y_i \right) - \theta_y \left( n x_i - \sum_{i=1}^n x_i \right) \right) \right) $$

  4. Moment Balance

    Each force results in a torque about the origin that is simply a function of the position of the leg $ (M_x)_i = y_i F_i$ and $(M_y)_i = -x_i F_i $. The sum of all the torques must be zero

    $$ \matrix{ 0 = \sum \limits_{i=1}^n y_i F_i & 0 = \sum \limits_{i=1}^n (-x_i) F_i } $$

    These are the two other equations needed to solve for $\theta_x$ and $\theta_y$. Use the force expression to get

$$\begin{aligned} 0&=\tfrac{W\left(\sum_{i=1}^{n}y_{i}\right)+k\left(\left(n\sum_{i=1}^{n}y_{i}^{2}-\left(\sum_{i=1}^{n}y_{i}\right)^{2}\right)\theta_{x}-\left(n\left(\sum_{i=1}^{n}x_{i}y_{i}\right)-\left(\sum_{i=1}^{n}x_{i}\right)\left(\sum_{i=1}^{n}y_{i}\right)\right)\theta_{y}\right)}{n}\\ 0&=\tfrac{-W\left(\sum_{i=1}^{n}x_{i}\right)+k\left(-\left(n\sum_{i=1}^{n}x_{i}y_{i}-\left(\sum_{i=1}^{n}x_{i}\right)\left(\sum_{i=1}^{n}y_{i}\right)\right)\theta_{x}+\left(n\sum_{i=1}^{n}x_{i}^{2}-\left(\sum_{i=1}^{n}x_{i}\right)^{2}\right)\theta_{y}\right)}{n} \end{aligned}$$

  1. Deflection Solution

    The above 2×2 system is solved as $ \theta_x = \frac{W\,n}{k} C_x$ and $ \theta_y = \frac{W\,n}{k} C_y$ where

    $$ \begin{aligned} C_{x} & = \tfrac{\left(\sum_{i=1}^{n}x_{i}\right)\left(\sum_{i=1}^{n}x_{i}y_{i}\right)-\left(\sum_{i=1}^{n}x_{i}^{2}\right)\left(\sum_{i=1}^{n}y_{i}\right)}{\left(n\left(\sum_{i=1}^{n}y_{i}^{2}\right)-\left(\sum_{i=1}^{n}y_{i}\right)^{2}\right)\left(n\left(\sum_{i=1}^{n}x_{i}^{2}\right)-\left(\sum_{i=1}^{n}x_{i}\right)^{2}\right)-\left(n\left(\sum_{i=1}^{n}x_{i}y_{i}\right)-\left(\sum_{i=1}^{n}x_{i}\right)\left(\sum_{i=1}^{n}y_{i}\right)\right)} \\ C_{y} & = \tfrac{\left(\sum_{i=1}^{n}x_{i}\right)\left(\sum_{i=1}^{n}y_{i}^{2}\right)-\left(\sum_{i=1}^{n}y_{i}\right)\left(\sum_{i=1}^{n}x_{i}y_{i}\right)}{\left(n\left(\sum_{i=1}^{n}y_{i}^{2}\right)-\left(\sum_{i=1}^{n}y_{i}\right)^{2}\right)\left(n\left(\sum_{i=1}^{n}x_{i}^{2}\right)-\left(\sum_{i=1}^{n}x_{i}\right)^{2}\right)-\left(n\left(\sum_{i=1}^{n}x_{i}y_{i}\right)-\left(\sum_{i=1}^{n}x_{i}\right)\left(\sum_{i=1}^{n}y_{i}\right)\right)} \end{aligned} $$

  2. Force Solution

    Using the above angles $\theta_x$ and $\theta_y$ in the force equation from step 3) yields the following solution (where $k$ cancels out).

$$ \boxed{ F_i = W \left( \frac{1}{n} + \left( n y_i - \sum_{i=1}^n y_i \right) C_x - \left( n x_i - \sum_{i=1}^n x_i \right) C_y \right) } $$

  1. Example

    I wrote a quick (Fortran) computer program to test the above. I used 4 legs at $A=(-1,-1)$, $B=(1,-1)$, $C=(-2,2)$ and $D=(2,5)$.


The source code is avaiable on GitHub as a gist.

  • $\begingroup$ @Questionable - As I said, this problem is far more complex than one imagines. Look over the solution above (if you can follow) where I simplify the general problem down to a 2×2 system of equations. $\endgroup$ Feb 7, 2019 at 19:29

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