If a baseball is dropped on a trampoline, the point under the object will move a certain distance downward before starting to travel upward again. If a bowling ball is dropped, it will deform further downwards. What is the nature of the relationship between the magnitude of this deformation and the object's mass? Linear? Square? etc.
Edit: I would like to add that the heart of what I'm asking is along the lines of this: "If a small child is jumping on a trampoline and the trampoline depresses 25% towards the ground, would an adult who weighs slighly less than four times as much be safe from depressing it all the way to the ground?"
Answer
It depends also on the shape of the object. If you assume the trampoline is circular, and the object is much smaller (like a point mass) then you can start developing the equations. You have to know the initial tension of the trampoline, and also assume the material non-elastic but supsended by perfect strings in a radial direction (with known stiffness).
After some math the static deflection (with pre-tension) obeys the following:
$$ \frac{W}{k\, R}=\tan\theta+\left(\frac{F_{0}}{k\, R}-1\right)\sin\theta $$
If $R$ is the radius of the trampoline then the dip is $H=R\,\tan\theta$ and so $\theta$ is the angle from horizontal that the cone makes. For any given angle $\theta$ the trampoline supports weight $W$ (given above) given total stiffness of $k$ and pre-tension of $F_{0}$.
So the above will give you the weight $W$ it will support given a dip $H$. It is the reciprocal of what you want, but it is solvable.
If the trampoline has $N$ linear springs each with stiffness rate of $k_i$ then the total stiffness (springs in parallel) is $k=k_i\,N$. To define the pre-tension $F_{0}$ assume that the free radius of the trampoline surface is $R$, but the springs are located at $R_0$ then the pre-tension is $F_{0}=k\,(R_0-R)$.
Example: A trampoline of 12 feet in diameter needs $F_{0}=100$ lbs total of pulling to string into a 12.5 foot ring. The stiffness is $k=\frac{100}{6}$ in pounds per inch. To dip the trampoline by 5 feet $\tan(\theta)=\frac{H}{R}=\frac{5\times12}{12\times12}$. Plug these into the above and you should get $W=115.4$ lbs.
I know I am going to confuse some people because I am treating radial quantities such as stiffness and loads as linear, but it works out (just use cylindrical coordinates).
Approximation
Small angle approximation (weight < 10% pre-tension, cone angle < 6°)
$$W = F_{0}\,\frac{H}{R}$$
Example: Using the same numbers as above a $W=10$ lbs weight will dip $H = (12\times12)\frac{W}{100} = 14.4$ inches. The full solution above gives $13.1$ inches
Theory
The deformed shape of the trampoline is a perfect cone. The distance from the center to where the springs start in the deformed state is always equal to $R$. The extension of the springs tension is then $F=F_{0}+k\,\left(\frac{R}{\cos\theta}-R\right)$ which needs to be balanced by the weight as $W = F\,\sin\theta$. The pieces come together to make the equation shown above.
Update
Solution is corrected for the fact the radial distance is constant, not the surface area of the trampoline. As the trampoline deforms its perimeter crumples and folds on itself like a napkin when lifted from the center.
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