Trampoline physics - what is the right time to crouch and to jump to maximize your height?

Please bear in mind that I'm neither a physics guy nor particularly a math guy. So I will probably need a bit more hand-holding than just a complex equation.

For a game, I'm trying to calculate a reasonable simulation of a human body on a trampoline. It seems to me that there are two phases to consider once the body has contacted the trampoline: the deceleration of the body once it has contacted the surface, which gets stored as potential energy in the trampoline, and then the combination of the release of that potential energy once the body has reached the nadir plus any extra force that body might apply (assuming he has bent his knees prior to that point).

I'm interested in understanding the interaction between the trampoline's storage and release of energy and the body's own contribution (addition) to that energy with its own kinetic energy.

CLARIFICATION: In short, I want to know when the best time to crouch is (pull your knees up) and when the best time to release (jump) is. Anyone who's been on a trampoline knows you can get better height by crouching and jumping at the right time - you can also bring yourself to a complete stop by crouching and jumping at different times. I can't seem to find anything documented though about what these times actually are. Since I couldn't find the information from the sports / athletics side, I thought I'd come over here to the physics side and see if anyone knew.

I'm surely not being very clear, which only further reveals my own shortcomings in this matter. Any guidance would be appreciated. I understand that this forum typically operates at a much higher level than what I'm looking for, but I thought I'd ask here, since this is where the great minds seem to gather.

EDIT I can very easily do a "good enough" simulation of a mass "bouncing"off the trampoline by taking its incoming velocity and simply inverting it and multiplying by an arbitrary restitution amount to simulate the springiness of the material. Nothing more complicated really need from the perspective of a game.

The part that's killing me is getting the body to continue moving past the plane of the trampoline while decelerating to 0 and then accelerate again up to final "release velocity". The part that baffles me is the part that happens between the time the body touches the material to the time it leaves contact, taking into account any additional force added by the person through his legs (I've decided not to model any additional lift sure to arm motion etc, to keep both the simulation and the user controls thereof straightforward).

SECOND EDIT I've read through the answers here and they're very good! And surprisingly, I seem to be following about 75% of it.

The challenge I'm having is that I'm principally interested in Velocity, rather than Energy, even though it's obviously energy that is behind all of this. Every 30th of a second (every "step") I need to know the velocity of the body so that I can plot its new position. It's really easy to plot the position of the body once it has left "contact": starting at some arbitrary position and velocity, v = v + g, y = y + v. See, it's a very different way of representing the same facts when you're dealing with visual graphics than physics equations.

So now y = the plane of the trampoline. v = v * c, where 'c' is a factor of how far the springs will compress, less than 1. y = y + v until v approaches 0 at which point we start the "bounce back". Here I suppose v = v * (1/c) * s, where 's' represents whatever factor of 'loss' we want to simulate so that we don't get 100% bounce back. Once y > the plane of the trampoline, we can then forget the more complicated stuff and go back to letting gravity take over.

Obviously this whole bit ignores mass, which is unfortunate, but I have to keep it simple for now. The "mass" factor gets rolled into the numbers for 's' and 'c' that I fudge until it 'feels' right.

Okay, now that you guys have stopped shuddering in utter horror at this wishy-washy approach, the bit I'm trying to figure out is the part about the human body bending his legs and then performing the jumping motion to add his force to the rebound of the trampoline.

It seems to me, having spent a little time on a trampoline, that you can jump at the right time and at the wrong time.

So, from a physics perspective, when is the optimal time to crouch? When is the optimal time to jump / release?

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Related question (and answers) : physics.stackexchange.com/q/1467/392 – ja72 Jul 31 '11 at 4:05
The part of my question that I think people miss out on is the part about how the human body contributes to the bounce. Actually, the bit about swinging the arms was interesting, though I had already decided I wasn't going to model that part. But kerbs focus on the leg motion: how does bending the legs (and the timing of that action and its counterpart: the release) affect the buildup of potential energy? when is the optimal time to crouch and when is the optimal time to jump? – Tom Auger Aug 1 '11 at 2:14

I will make the following approximating assumptions:

1. The surface of a trampoline has zero mass density. This implies standing or crouching will not instantaneously affect the position of your center of gravity.
2. The upward restoring force is a monotonic function of vertical displacement.
3. Humans can straighten their legs in zero time. This is just a way to avoid extra optimization details.

If these hold, then the best time to crouch is before landing, and the best time to straighten your legs is when your speed is zero (i.e., when your height is at a minimum). This behavior optimizes three quantities:

1. The kinetic energy you gain before reaching the trampoline is maximized. Because you are in a crouching position, your center of gravity falls further before the restoring force starts slowing you down.
2. The height of your center of gravity when your speed reaches zero is minimized. This is because for a fixed position of your center of gravity, the restoring force is minimized when you are in a crouching position.
3. The work done by the springs in releasing you is maximized. If you set $h_0$ to be the displacement of your center of gravity where you are stationary, the energy imparted is the integral $\int_{h_0}^0 F ds$ of the restoring force with respect to displacement. For any fixed position of your center of gravity, the restoring force is maximized when you are in a standing position.

If you watch experienced gymnasts gaining altitude, you will see that they straighten their legs roughly when they reach the bottom, and they let the trampoline carry them up.

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Brilliant! I'm not sure I follow the differential math, but the other answers will help me with the math. This info is really good for me to make the game user input be as realistic as possible. Thanks – Tom Auger Aug 4 '11 at 14:36

Can you be more specific about what you mean by 'reasonable'?

The simplest treatment, ignoring loss of energy to heat/friction/etc, would be to approximate the human body as a point mass falling on a spring (i.e. the trampoline) with some spring constant $k$. Here, we have the initial potential energy of the falling body - $PE = mgh$, this is entirely transformed to kinetic energy at the time of impact - $KE = \frac{1}{2}mv^2$, and you stretch the trampoline to convert this into spring potential energy - $SPE = \frac{1}{2}kz^2$, where $z$ is the compression distance. Since the sum of the potential energy in the spring (SPE) and the kinetic energy of the falling body (KE) is constant, you can solve for the velocity at a given spring compression: let $C = SPE + KE$, which implies $v = (\frac{2C-kz^2}{m})^\frac{1}{2}$. The derivative with respect to $z$ is $v' = \frac{-kz}{m(\frac{2C-kz^2}{m})^\frac{1}{2}}$.

As for the body bouncing upwards, start with the kinetic energy at impact and play the same game where the sum of the body's kinetic energy (KE) and the potential energy due to gravity (PE), again $PE = mgh$ (for some height $h$), is constant. As such, we have $C = PE + KE$, which implies $v= (2)^\frac{1}{2}(\frac{C-mgh}{m})^\frac{1}{2}$. The derivative w.r.t. $h$ is $v' = \frac{-g}{(-2gh+\frac{2C}{m})^\frac{1}{2}}$.

A more complex treatment would involve energy loss due to heat and irreversible stretching of the trampoline fabric, air resistance on the falling body, etc. You could add in some fudge terms to account for this. As for a more complex treatment of the human body, i.e. some kind of ragdoll model, nothing fundamental changes, but you'll have a much harder time finding the instantaneous velocity of the individual body parts, as well as the compression velocity of your trampoline at any given time. To make traction there, you'd probably have to do something like a ball-and-spring simulation of the body, and more precisely define how it will be falling on the trampoline.

Some suggested units for the terms:

$g$, or 'little $g$' denotes the acceleration due to gravity near the earth's surface (http://en.wikipedia.org/wiki/Earth%27s_gravity). It's value is approximately $~9.81 \frac{m}{s^2}$.

$v$, your velocity, can be measured in meters/second, $\frac{m}{s}$.

$m$, or your mass, can be measured in kilograms ($kg$).

These work well if you measure your terms representing energy (i.e. PE, SPE, KE, and 'C') in joules, since the units for a joule are: $\frac{kg*m^2}{s^{-2}}$.

As for how the trampoline should 'look' when depressed a distance $z$, perhaps you could model it as an upside-down cone of height $z$ (http://mathworld.wolfram.com/Cone.html)?

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This is awesome and thanks for the noobie vocab lesson too: very helpful! It's going to take me some time to grok what type equations mean and how I would translate them into machine code, but let me ask: do you also explain the part wether the body "sinks" below the plane of the trampoline and decelerates before starting its motion back up? – Tom Auger Aug 1 '11 at 2:27
@Tom Auger, I'm treating the body hitting the trampoline as a point mass hitting a spring (which occurs at the "time of impact" I refer to above). The distance the spring is compressed, 'z', is treated as equivalent to the distance 'z' the body "sinks" below the plane of the trampoline. – TheSheepMan Aug 1 '11 at 15:10

As the secretary of a trampoline club I spend too much time watching people bouncing on trampolines and thinking about the physics of what they are doing.

One thing you should take into account is that the reaction forces from the trampoline are not well modeled by an ideal spring with force proportional to the depression $x$. When the trampoline bed and springs are detached they shrink to a width $L$ just a little less than the stretched width $W$ You can model this with a stiff rod or string of length $L$ attached to an ideal spring of stretched length $S = \sqrt{W^2+x^2}-L$

The upward component of force from the bed is therefore proportional to $T = x - \frac{xL}{\sqrt{W^2+x^2}}$

Here I am assuming that the tramplinist is bouncing in the middle of the trampoline and that it is circular. For rectangular trampolines you need a model that adds up contributions from parts of different length. Off centre bouncing will also create transverse forces.

Even if you ignore these things you should allow for some losses due to friction

The trampolinist gets lift from the bed by bending knees and ankles and moving arms during the contact with the bed. The combined effect is to move the position of his centre of gravity relative to the position of his feet. In the simplest case he keeps this constant and bounces just losing some height due to friction.

For an analysis including novement but no friction you can take the vertical position of the centre of mass as your primary variable and use a Lagrangian with kinetic energy given by $\frac{1}{2}mv^2$ from the veloicty of the centre of mass and potential energy from the bed and gravitation, then work out equation of motion. How you proceed from there depends on what you are trying to achieve.

If you want to model summersaults and twists with the trampolinist moving across the trampoline it could get complicated! Hopefully I have given some initial pointers at least.

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Thanks for the great diagram! That will definitely come in handy when it comes time to model the depression of the fabric: I intend to use a bezier curve for that. – Tom Auger Aug 1 '11 at 2:30
I was making the approximation that the surface of the bed will form a cone (circular trampoline) This is a good enough model for the forces, but the real shape would be different because of the cross tension, so that the sides curve down into the depression. If you are doing graphics and want it to look perfect you may want to add that curvature. – Philip Gibbs - inactive Aug 1 '11 at 13:35
Thanks for the great info. – Tom Auger Aug 4 '11 at 14:37

I don't what to trow a wench in the works BUT you can jump on the trampoline bay just moving your arms up and down. The shift of mass and the change of momentum bay swinging your arms down at the moment of contact increases the mass towards the trampoline and then trowing them upwards accelerates the body in the direction of travel. We do this instinctively the problem with calculating this is that the jumper does not keep there arms straight. Now if you take in account that we do a similar motion whit our lags you have one hell of a calculation just considering the shift in mass in the air. Calculating the elastic capabilities of the trampoline is fairly straight forward but the work capability of individual people is something all together different. An experienced jumper can achieve impressive height bay just shifting there mass a very precise way.

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This raises the point that, depending on application, you may have an easier time filming real humans jumping on a trampoline and constructing a simple model of the system using real data. – TheSheepMan Jul 30 '11 at 18:06
@SheepMan darn straight, of course I would need access to a: high speed camera, a trampoline, and a trained athlete... – Tom Auger Aug 1 '11 at 2:44
@Tom Auger, I think you could actually do pretty well with a cheap digital camera with a known framerate. You should be able to approximate how long it takes the jumper to complete different events (i.e. how long it takes to go up and come down, depress the trampoline a certain distance, etc). You could also probably figure out the angular deformation of the trampoline surface as a function of the jumper's distance to the ground. – TheSheepMan Aug 1 '11 at 15:04
@Tom Auger, "...and a trained athlete..." it might be a little hard to figure out the deformation angle of trampoline, but couldn't you use digital footage of Olympic events for your purpose? I'm sure you can find it online. – TheSheepMan Aug 1 '11 at 15:13
@SheepMan, interesting - never thought of looking for Olympic footage. I tried looking for training videos and there was nothing that dealt with the crouch and release - everything is about the arms and spins and stuff. The basics seem to be just too... intuitive to warrant any study. Which seems odd to me. – Tom Auger Aug 1 '11 at 15:17