Help me understand jerk of a falling object hitting an "ideal foam"

[zsero]

TL;DR Summary

Questions about jerk of a quickly decelerating object

I'm trying to calculate the theoretical minimum thickness of an "ideal foam" for a given jerk and acceleration limit.

Say we have a ball in free fall from 1.83 meter, reaching 6 m/s. It then reaches an "ideal foam" for decelerating it. I'm trying to understand the connection between the thickness of the material and the connection to peak acceleration / peak jerk.

Basically: how thin a foam can theoretically be, while keeping acceleration / jerk under a given limit.

So, let's suppose our "ideal foam" is 5 cm thick.

A. If I'm correct, we can say that with constant acceleration, the lowest theoretical peak acceleration is:
a = v^2 / 2s, or 36 / 0.1 = 360 m/s2 or ~36 g.
The time in this case is t = sqrt(2s/a) or sqrt(0.1 / 360) = 0.016 sec

The problem with constant acceleration is that it means infinite jerk. How can I keep the same peak acceleration (~36g) while limiting jerk? I guess I need to raise thickness, there is no way around it, right?

Also, I'm trying to understand, how does this problem look if the object is falling on a linear spring? How does a linear spring (as in Hooke's law) behave in terms of jerk and acceleration?

---

I mean let's say I can create any material I want. It can be a "magic" spring or a "magic" foam, or any layer of variable density foams. I'm interested in the theoretical limits on such a material. Basically I'm trying to calculate a lower boundary on an ideal material, from which real world materials would have to be thicker.

So I'm interested in the direction of:
To have max. X peak acceleration and max. Y peak jerk, I need at least Z cm thickness. How can I calculate Z?

 

[erobz]

Also, I'm trying to understand, how does this problem look if the object is falling on a linear spring? How does a linear spring (as in Hooke's law) behave in terms of jerk and acceleration?

Probably should start your exploration here.

Can you write Newtons Second Law for a mass falling onto a (Hookean) spring? After you have that, how are jerk and acceleration related mathematically?

 

[jbriggs444]

Basically: how thin a foam can theoretically be, while keeping acceleration / jerk under a given limit.

So you have a limit to peak acceleration and a separate limit to peak jerk.

The optimal decelleration profile would clearly be one that maximizes jerk until the acceleration peak is reached and then delivers constant acceleration thereafter.

The acceleration profile is easy to calculate for both phases. Calculate it. Integrate it once to get velocity and again to get displacement. Read off the result for thickness.

 

[zsero]

Yes, I came to the same conclusion. A linear spring-mass system was not the best system as it has that harmonic movement with sin-cos functions for all derivatives.

The theoretical best system is thus something like:
- a progressive spring for the constant jerk part
- after the jerk limit is reached, I need a constant force, like friction or similar

Calculating this is indeed simple.

 

[erobz]

Yes, I came to the same conclusion. A linear spring-mass system was not the best system as it has that harmonic movement with sin-cos functions for all derivatives.

The theoretical best system is thus something like:
- a progressive spring for the constant jerk part
- after the jerk limit is reached, I need a constant force, like friction or similar

Calculating this is indeed simple.

So for my own curiosity with a constant jerk you get the ODE.

$$ \frac{d^3x}{dt^3} = k $$

which has the characteristic equation:

$$ m^3 - k = 0 $$

Factoring the LHS:

$$ \left( m- k^{1/3} \right) \left( m^2 + k^{1/3}m + k^{2/3} \right) = 0 $$

Is that the route you take?

 

[zsero]

So for my own curiosity with a constant jerk you get the ODE.

$$ \frac{d^3x}{dt^3} = k $$

which has the characteristic equation:

$$ m^3 - k = 0 $$

Factoring the LHS:

$$ \left( m- k^{1/3} \right) \left( m^2 + k^{1/3}m + k^{2/3} \right) = 0 $$

Is that the route you take?

I don't know if such a spring or multi-layer foam can be made or not. I just calculated the other way, constant jerk -> linear growing acceleration -> quadratic velocity -> cubic displacement

I mean all I was interested is what is the minimal distance for a theoretically perfect constant jerk - constant acceleration system to slow down a falling ball.

I came up with the following calculations:

import math

def cj_t():
    return max_a / max_j
    
def cj_v():
    return (max_a ** 2) / (2 * max_j)

def cj_d():
    return (max_a ** 3) / (6 * max_j ** 2)

def is_final_v_in_cj_phase():
    return final_v <= cj_v()

def cj_t_f():
    if is_final_v_in_cj_phase():
        # Calculate time to reach final_v during constant jerk phase
        return math.sqrt(2 * final_v / max_j)
    else:
        return cj_t()

def cj_d_f():
    t = cj_t_f()
    # Calculate distance traveled during constant jerk phase up to time t
    return (1/6) * max_j * t**3

def ca_t():
    if is_final_v_in_cj_phase():
        # No constant acceleration phase needed
        return 0
    else:
        # Time during constant acceleration phase
        return (final_v - cj_v()) / max_a

def ca_d():
    t = ca_t()
    # Distance during constant acceleration phase
    return cj_v() * t + 0.5 * max_a * t ** 2

def total_t():
    # Total time is sum of times from both phases
    return cj_t_f() + ca_t()

def total_d():
    # Total distance is sum of distances from both phases
    return cj_d_f() + ca_d()

Two interesting points:
1. jerk limit: 1500 G/s, acceleration limit: 33 G, velocity: 6 m/s => depth required 6.21 cm, compression time: 29 ms
2. jerk limit: 500 G/s, acceleration limit: 33 G, velocity: 6 m/s => depth required 9 cm, compression time: 45 ms

 

[erobz]

constant jerk -> linear growing acceleration -> quadratic velocity -> cubic displacement

Ahh! I guess that will fall out of the solution (clearly the hard way) if I continue on...

 

[erobz]

Ahh! I guess that will fall out of the solution (clearly the hard way) if I continue on...

I can't ( I can recently- but that another story) believe I didn't see this can just be consecutively integrated three times w.r.t. $t$... Still asking the mathematicians,

If I'm more careful we should be able to get there from instead squaring $k$ on the RHS:

$$ \frac{d^3x}{dt^3} - k^2 = 0 $$

and

$$ \frac{d^3x}{dt^3} + k^2 = 0 $$

So there are six zeros, two solutions to this set of ODE's? ,and only one of general solutions (the one you get from integrating consecutively) survives the process? Sorry for the tangent, but now I'm a bit frazzled.