How far does a trampoline vertically deform based on the mass of the object?

Question

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 1

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.

Answer 2

Since @jalaxiou provided a solution for the static case looking at the balance of forces, I will try to provide a solution for the dynamic case from energetic considerations.

Ǹotation and energies

I will first restrict my self to a 1D trampoline, i.e. a rope with 2 springs of stiffness $k$. The length of the string + the springs without pretention is $2R_0$, ant the diameter of the trampoline is $2R>2R_0$. When a point-mass in the middle of the trampoline is at a depth $H$ below the initial level, each spring is elongated by $\sqrt{H^2+R^2}-R_0$. The elastic potential energy of the 2 springs is therefore :

$$E_{\text{elastic}}=k(\sqrt{H^2+R^2}-R_0)^2 $$

The gravitational potential energy of the mass is $$E_{\text{gravitational}}=-mgH.$$

Static solution The static solution corresponds to the minimum value of $E_{\text{gravitational}}+E_{\text{elastic}}$ obtained by derivation i*H*. However I get a fourth order equation which is not trivial to solve analytically, except for small $\frac{H}{R}$, where $H\simeq\frac{mg}{2k(1-R_0/R}$.

Dynamic solution For the dynamic solution, you add the kinetic energy $E_{\text{kinetic}}=\tfrac12mv^2$ of the mass into the game. If you suppose that there is no friction (yo have an ideal trampoline), the total energy $E=E_{\text{elastic}}+E_{\text{gravitational}}+E_{\text{kinetic}}$ is conserved during the motion. If the motion is vertical, the kinetic energy is 0 at the top and the bottom of the movement, since it is by definition the point where the vertical speed is 0. If you make jump of height $H_J$ above the trampoline, the energy at the top of the jump is $mgH_J+k(R-R_0)^2$. comparing thin energy with the one at the bottom of the movement, we have

\begin{gather} mgH_J+k(R-R_0)^2=-mgH+k(\sqrt{H^2+R^2}-R_0)^2 \\ mg(H+H_J)=k(H^2-2R_0(\sqrt{R^2+H^2}-R)) \end{gather} whch can also me converted to a fourth order polynomial in $H$, with probably ugly but analytical solutions. In the limit of small deformation $H\ll R$, we have \begin{gather} mg(H+H_J)\simeq k(H^2-2R_0\frac{H^2}{2R})=k(1-\frac{R_0}R)H^2 \end{gather} which can be solved for H and gives: $$ H=\frac{mg}{2k(1-\frac{R_0}{R})}\left(1+\sqrt{1+\frac{4k}{mg}(1-\frac{R_0}{R})H_J}\right). $$ This ugly expression simplifies for small jumps, when $H_J\ll\frac{mg}{4k(1-R_0/R)}$ into $$ H\simeq\frac{mg}{k(1-\frac{R_0}{R})}+H_J $$ If you look at the total amplitude of the movement, between $z=+H_J$ and $z=-H$, you see oscillations centered around the static equilibrium postion which is expected.

Then, if you want a circular trampoline, with $n$ spring instead of a rope with 2 springs, you just have to replace $k$ by $\frac n2 k$ in the equations above by symmetry.

Answer 3

It is not a simple relationship. A trampoline will have various natural modes, which are shapes of deformation that correspond to different frequencies. Any dynamic collision, will excite many of these modes, which will continue until they gradulay damp out. For a very low frequency applied force,for example how far will the baseball depress the trampoline if it just sits on it for a long time, (i.e. a static situation), the answer should be a linear function of the ball's weight (or nearly so), but for dynamic processes all sorts of transient dynamics come into play to complicate things.

Answer 4

Interesting! I don't have time to investigate this in detail, and it's been a while since Itouched mechanical physics, but it sounds to me that a half-decent first approximation to the static case would be to consider the trampoline as a mesh of perfects springs. Somebody please correct my hunch, but: In the limit of infinitely finely woven springs, a point mass interacts with the mesh as though it were instead suspended by a string of 2*2=4 (2 springs per dimension) times the spring constant, right?

Now, for a suitably regular non-point object, let's say a cube, neighboring springs in the interior of the contact surface (here the base square) between the object and the trampoline won't matter (right?). Thus only the circumference of the object should matter, and depress the trampoline like a continuum of point masses... so we should get a depression increasing linearly with the object's mass and decreasing linearly with the side lengths of its base.

Of course, as has been mentioned, what's really needed here is a good/realistic model for an elastic membrane.

Answer 5

To answer your edit:

A trampoline is HIGHLY non linear. The force to push the next inch down goes up at least quadratically with displacement.

Edit: Some playing with Desmos says that a cubic fits really closely up to 45 degrees

So if kidlet deforms the mat by 25% of the vertical distance, a person massing 4 Kidlets will deform it by something under 50%

First approximation: the trampoline is in the shape of a cone with the weight at the centre. If you have an angle of depression angle of $\theta$ degrees, then the upward force is proportional to spring tension times the sine of the depression angle. $$F=T \sin(\theta)$$

Spring tension = some constant * it's extension.

$$T=kx$$

$$x=\frac{R}{\cos(\theta)}-R$$ or $$x=R(\frac{1}{\cos(\theta)} -1)$$

So now

$$F=kR(\frac{1}{\cos(\theta)} -1)( \sin(\theta))$$