Tipping point on fence panels? I deal in temporary fence panels - and my concern is the tipping point of our product out in the field.
Panels are 6' tall x 12' long (63#)
Stands are 23" long x 6" wide
We use sand bags (30#-40#) on the bases, but for whatever reason, they don't always stay on (or people take them).
Is a 23" base an optimum base for this height - or rather the "minimum effective dose" to achieve a stable panel?   I understand that increasing it to 30" (15" on each side) would make it stable, but is it necessary for that height and still remain stable?
We are reviewing this as we are looking at a taller product (8') and I don't believe the same stands can be used on both heights, as the added height will greatly increase the chances of a tip over.
So, maybe my actual question how do I determine my center of gravity on the upright fence - so that I can accurately define the necessary base width?

 A: If the height of the fence is increased by a factor $k$, this will increase the area by the same factor, and also the wind force on it. (I assume for simplicity that windspeed does not vary with height, which is probably not true.) The average height at which the wind force is applied is also $k$ times as large. So overall the toppling moment (force x lever arm) is $k^2$ times as large. 
To counteract this, you would have to make the righting moment $k^2$ times as large. Presumably increasing the height will increase the weight of the fence by the same factor $k$. (You may have to use more material to increase strength, in which case this increase is more than $k$.) If you also increase the weight of the sandbags by $k$, then the total weight is $k$ times as much. As Lewis states, you would also have to increase the size of the base by $k$ in order to make the righting moment $k^2$ times bigger also, giving the same stability.
If an increase by factor $k$ in the size of the base or the total weight of fence plus sandbags is impractical, any combination giving a product of $k^2$ will do just as well. 
The joint keeping the base perpendicular to the panels will be under $k^2$ times as much stress, so this may need to be strengthened.
A: What you are dealing with is countering the torque induced by a wind load on the fence section by the torque that your sand bags apply to your base when the section tips. It is hard to estimate the wind load, but it should be proportional to the area of the section. If your 6' fence is stable with a 23" base and 40# bags, then for a 8' fence you should incease your base by the same fraction as the height increase (33%) as well as increase your bag weight by the same fraction to maintain the same level of stability.
A: It's worth just drawing out a diagram where you look at the fence side-on (so that the fence proper is just a straight line), tilted about this stand an angle $\phi$. To get from the edge of the stand to the middle of the fence your vector displacement would be $$\begin{bmatrix}s/2\\h/2\end{bmatrix}$$
and rotating this gives that the position of the middle of the fence is $(x, y)$ given by $$
\begin{bmatrix}x\\y\end{bmatrix} = 
\begin{bmatrix}\cos\phi & -\sin\phi\\\sin\phi & \cos \phi\end{bmatrix}\begin{bmatrix}s/2\\h/2\end{bmatrix} = \frac12
\begin{bmatrix}s~\cos\phi - h\sin\phi\\h~\cos\phi+s~\sin\phi\end{bmatrix}.$$If the majority of the mass is in the fence rather than the stand, this points to a potential energy curve that goes like $h\cos\phi + s \sin\phi$, $$U(\phi) = m_\text{fence}~g\cdot\frac12\big(h\cos\phi + s \sin\phi\big).$$(More precisely, we should replace all of those $\phi$ by $|\phi|$ in this expression as it rotates around the other side of the stand when we push it the other way, but I will just be assuming $\phi > 0.$)
By techniques which are not standard introductory physics but do get covered later (Lagrangian mechanics), a nice expression for the torque that this potential energy gives you is $$\tau(\phi) = -\frac{\partial U}{\partial \phi} = \frac12 m_\text f~g~\big(-h~\sin\phi + s\cos\phi\big).$$ It should have its maximum at $\phi=0,$ at least in this regime $s < h,$ therefore the maximum torque is $m_\text f ~ g ~ s.$ If the torque on the panel comes from steady pressures across the whole panel then the torque should go like $P~A~h/2$ while $m_\text f$ clearly also is some areal mass density $\sigma$ times the area, so we have $\sigma~A~g~s \gg~P~A~h/2$ for normal stability; the $A$ factors cancel so you would like to scale $s$ proportionately to $h$: if the stands are working for your fences at 6 feet then you want stands which are $4/3$ times the size for fences at 8 feet, assuming the mass density and the amount of force they catch is constant.
For other forces, like people running into them, you should actually see added stabilization simply because the higher fence is, hopefully, heavier -- even at a constant stand width $s$. 
There's probably some interesting stochastic analysis about how much higher you can go above that with a noisy signal before you actually topple the thing over, but any steady torque above that $mgs$ should eventually generate the necessary energy $m_\text f~g~\sqrt{h^2+s^2}$ to topple it, so this is more a concern for noisy torques that occasionally have spikes above this level but do not hit it.
There's also probably a place where my assertion that the maximum of the force is at $\phi=0$ breaks down, and it's probably when you get to $s \ge h,$ so that each leg of the stand is at least half as long as the fence is tall.
