# Estimate the reaction force on each leg of a 4-legged table

Assume you have a table with four legs with a weight placed somewhere inside the boundary created by the legs. The task is to determine the reaction force at each leg. Here are the problem constraints...

1. The sum of the four reaction forces must equal the total weight set on the table.
2. The table has no weight of its own.
3. You cannot directly solve for the reaction of each leg using sum of forces and sum of moments because the structure is statically indeterminate.
4. We are not allowed to introduce deflection in order to develop a 4th independent relationship between the four reactions

There are methods to estimate the reaction force on each leg. One such method uses ratios to determine the four leg reactions.

\begin{eqnarray} P_1 & = & \left(\frac{y_w-y_b}{y_f-y_b}\right)\left(\frac{x_w-x_l}{x_r-x_l}\right) \\ P_2 & = & \left(\frac{y_w-y_b}{y_f-y_b}\right)\left(\frac{x_w-x_r}{x_l-x_r}\right) \\ P_3 & = & \left(\frac{y_w-y_f}{y_b-y_f}\right)\left(\frac{x_w-x_r}{x_l-x_r}\right) \\ P_4 & = & \left(\frac{y_w-y_f}{y_b-y_f}\right)\left(\frac{x_w-x_l}{x_r-x_l}\right) \end{eqnarray} Here is my question. The above solution is only good for a square leg configuration.

Can this method be modified to work for a trapezoidal leg configuration, like this? Better yet, can this method be modified to work for any leg configuration?

Edit 1

@ja72, This is in response to your answer. Please forgive me. It's been 15 years since I've worked with summations and linear algebra. I'm having trouble solving the two torque equations for $\psi$ and $\varphi$. Please look at what I have below and let me know if I'm on the right track. First I found $F_i$ using your second equation.

$$F_i=-\frac{W+k(\psi(x_i-x_w)+\varphi(y_i-y_w)}{3k}$$ Now I substitute this into your two torque equations and attempt to solve. $$-\sum_{i=1}^n\frac{W+k(\psi(x_i-x_w)+\varphi(y_i-y_w))}{3k}(y_i-y_w)=0\tag{1}$$ $$\sum_{i=1}^n\frac{W+k(\psi(x_i-x_w)+\varphi(y_i-y_w))}{3k}(x_i-x_w)=0\tag{2}$$ Rearrange equations 1 and 2 $$\frac{nW}{3k}-\sum_{i=1}^n\frac{\psi(x_i-x_w)(y_i-y_w)+\varphi(y_i-y_w)^2}{3}=0\tag{3}$$ $$\frac{nW}{3k}+\sum_{i=1}^n\frac{\psi(x_i-x_w)^2+\varphi(y_i-y_w)(x_i-x_w)}{3}=0\tag{4}$$ $$\begin{bmatrix} \sum_{i=1}^n(x_i-x_w)(y_i-y_w)&\sum_{i=1}^n(y_i-y_w)^2\\ \sum_{i=1}^n(x_i-x_w)^2& \sum_{i=1}^n(y_i-y_w)(x_i-x_w) \end{bmatrix} \begin{bmatrix} \psi\\\varphi \end{bmatrix}= \begin{bmatrix} -\frac{nW}{k}\\ \frac{nW}{k} \end{bmatrix} \tag{5}$$ I doubt this is correct and I do not know how to solve this. Can you please help me solve for $\psi$ and $\varphi$.

Edit 2

Starting with the 2x2 given by ja72. Notice I have eliminated the $k$ terms on the right hand side since they cancel out in the end.

$$A\enspace \begin{array}{|c|} \psi \\ \varphi \end{array}= \begin{array}{|c|} W(y_w-\overline{y})\\W(x_w-\overline{x}) \end{array} \tag{1}$$ where $$A=\begin{bmatrix} n\overline{x}\overline{y}-\displaystyle{\sum_{i=1}^n(x_iy_i)} & \displaystyle{\sum_{i=1}^n(y_i^2)}-n\overline{y}^2 \\ \displaystyle{\sum_{i=1}^n(x_i^2)}-n\overline{x}^2 & n\overline{x}\overline{y}-\displaystyle{\sum_{i=1}^n(x_iy_i)} \end{bmatrix}$$ Implement variable transformation

$$a = n\overline{x}\overline{y}-\sum_i^n(x_iy_i)\quad b=\sum_i^n(y_i^2)-n\overline{y}^2$$ $$c=\sum_i^n(x_i^2)-n\overline{x}^2\quad d=n\overline{x}\overline{y}-\sum_i^n(x_iy_i)$$ now A becomes $$A=\begin{bmatrix} a&b\\c&d \end{bmatrix}$$ Solve for $\psi$ and $\varphi$ $$\begin{array}{|cc|} \psi\\\varphi \end{array}=A^{-1}\enspace \begin{array}{|cc|} W(y_w-\overline{y}\\ W(x_w-\overline{x} \end{array}$$ $$\psi=\frac{w}{ad-bc}(b(x_w-\overline{x})-d(y_w-\overline{y})) \tag{2}$$ $$\varphi=\frac{w}{ad-bc}(a(y_w-\overline{y})-c(x_w-\overline{x})) \tag{3}$$ Here is the equation for $z_w$. Again, I have removed the $k$ terms since they all cancel out. \begin{multline} z_w=\frac{1}{n}(W+ \displaystyle{\sum_{i=1}^n}\psi(x_i-x_w)- \varphi(y_i-y_w)) \end{multline} When we substitute $\psi$ and $\varphi$ we get this ugly thing. \begin{multline} z_w=\frac{W}{n}(1+\frac{1}{ad-bc} \displaystyle{\sum_{i=1}^n}(d(y_w-\overline{y})-b(x_w-\overline{x}))(x_i-x_w)- \\ (a(x_w-\overline{x})-c(y_w-\overline{y}))(y_i-y_w)) \tag{4} \end{multline} Finally we have... $$F_i=Term1-Term2+Term3$$ Where \begin{eqnarray} Term1 & = & z_m \\ Term2 & = & \psi(x_i-x_w) \\ Term1 & = & \varphi(y_i-y_w) \\ \end{eqnarray}

• As you note, the system is statically indeterminate. You have 4 unknowns but only 3 equations. You need to introduce other equations. This is usually done by allowing the legs to have some elasticity. This approach is realistic. What is the physical rationale behind your approach of using ratios? Do the ratios add up to 1? – sammy gerbil Aug 23 '16 at 18:40
• @sammygerbil, thanks for your comment. Yes you are correct, introducing elasticity would make the system determinate. However, this is easier said than done. For example, let's say instead of table legs the reactions are actually outriggers for a crane. Determining the flexibility of the structure of the undercarriage of the machine is very time consuming and complicated. Using ratios is a simple approach to allow the reactions to be found without having to analyze the entire structure. The problem arises when the outrigger pattern is not square, thus my question. – Kyle Aug 23 '16 at 18:49
• Your approach seems to be the same as the simple linear blending used by Olin Lathrop in the possible duplicate. It gives a definite & consistent answer, but I don't know what the physical justification is - possibly related to the Moment Distribution Method.... Anyway, I suspect the method relies on the supports being arranged in orthogonal pairs. When applied to the trapezium, I think the equations will be more complex. Have you tried deriving them? – sammy gerbil Aug 23 '16 at 19:17
• – sammy gerbil Aug 23 '16 at 21:38
• I have edited my answer per your comments. – John Alexiou Aug 24 '16 at 19:00

To solve this problem make the legs elastic, but nearly rigid. The reaction force on each leg is going to be a spring force. There are three degrees of freedom: a) two tilt angles and b) Overall height.

Let's assume each leg attachment is at $(x_i,y_i,z_i)$ where $z_i$ is the unknown "lift" of the table leg from the horizontal plane. This lift is a function of the center of mass height $z_w$ and two angles

$$z_i = z_w + \psi (x_i - x_w) - \varphi (y_i - y_w)$$

The force in each leg is $F_i =-k z_i$ so the total weight must be

$$\left. W =- k \sum_i^n z_i \;\right\} z_w =- \frac{W+k (\sum \limits_i^n \psi (x_i - x_w) - \varphi (y_i - y_w) )}{3 k}$$

Now the sum of the two torque components along x and y for each leg force is zero

$$\sum_i^n {\tau_x}_i = \sum_i^n F_i (y_i-y_w) = 0$$ $$\sum_i^n {\tau_y}_i = -\sum F_i (x_i-x_w) = 0$$

These two equations are solved for $\varphi$ and $\psi$ and when used back in the force equation $F_i = -k (z_w + \psi (x_i - x_w) - \varphi (y_i - y_w))$ miraculously the stiffness $k$ cancels out.

Edit 1

To solve for the tilt angles you make the following 2×2 system of equations

$$\begin{bmatrix} n \bar{x} \bar{y} - \sum \limits_i^n \left(x_i y_i\right) & \sum \limits_i^n \left( y_i^2 \right) - n \bar{y}^2 \\ \sum \limits_i^n \left( x_i^2 \right) - n \bar{x}^2 & n \bar{x} \bar{y} - \sum \limits_i^n \left(x_i y_i\right) \end{bmatrix} \begin{vmatrix} \psi \\ \varphi \end{vmatrix} = \begin{vmatrix} \frac{W}{k} \left(y_w - \bar{y}\right) \\ -\frac{W}{k} \left(x_w - \bar{x}\right) \end{vmatrix}$$

where $\bar{x} = \frac{1}{n} \sum \limits_i^n x_i$ and $\bar{y} = \frac{1}{n} \sum \limits_i^n y_i$

• please see below. My comment was too long to post in the comment section. Not sure what to do except post as an answer? – Kyle Aug 24 '16 at 18:00
• I have deleted my answer and added an appendix to my original question. Thanks for the direction. – Kyle Aug 24 '16 at 18:54
• I have posted an edit in my original question above. It seems to be working properly. One question though. When solving for $\psi$ the correct matrix multiplication gives... $$\psi=\frac{w}{ad-bc}(d(y_w-\overline{y})-b(x_w-\overline{x}))$$ However, this produces results where correct values are assigned to incorrect legs. To correct this behavior I needed to switch the $b$ and $d$ terms like this. $$\psi=\frac{w}{ad-bc}(b(x_w-\overline{x})-d(y_w-\overline{y}))$$ Any idea why? – Kyle Aug 25 '16 at 14:40
• Also, can you post your final solution so I can use it to correlate results. – Kyle Aug 25 '16 at 14:42
• I have a question. Here is a scenario with four table legs. Notice $P_4$ is zero. Theoretically running the same calculation without $P_4$ should give the same values for $P_1, P_2, P_3$ right? Here is the scenario with 4 legs $$\begin{array}{ccc} &x&y\\ P_1 & 100 & 100\\ P_2 & -100 & 100\\ P_3 & -100 & -100\\ P_4 & 100 & -100\\ W & -50 & 50 \end{array} \qquad \begin{array}{ccc} n & = & 4\\ W & = & 1000 \end{array} \qquad \begin{array}{ccc} P_1 & = & 250 \\ P_2 & = & 500\\ P_3 & = & 250 \\ P_4 & = & 0\\ \end{array}$$ – Kyle Aug 25 '16 at 21:11