Solving Ropewalk Equations - Help from Niko

[niko2000]

Hi,
I am trying to do a model for a ropewalk.
I would like to describe it with two first order differential equations. If we consider a person walking on a rope there are two quantities changing; the speed of a rope in horizontal direction and the angle of a person. I have to write this in a correct mathematical form. Any advice would be helpful.
Regards,
Niko

 

[niko2000]

If we think about the relations between quantities:
Let's look an angle between vertical line and the man on the rope and we assume equilibrium at x=0.
A=angle, v=speed on x direction, k and g are some coefficients
dA(t)/dt=k*A(t)-g*v(t)
My explanation:
If the angle is greater, the angle is changing faster and if the speed in x direction is greater, the angle is changing slower thus I put -.
If I try to write an equation for dv(t)/dt I have some problems:
dv(t)/dt=h*A(t)-g*v(t)
Explanation:
If the angle is greater we have to accelerate faster to gain balance and if the speed is greater we have to accelerate slower, because we already have some speed. But the problem is that the right sides of equations are same, so I'm not sure about my illustrations.

 

[M98Ranger]

Hi Niko,

Thank you for reaching out for help with your ropewalk equations. It sounds like you have a clear understanding of the variables involved in your model. To write the equations in a correct mathematical form, you will need to use the appropriate mathematical notation and symbols.

First, let's define our variables. Let v represent the speed of the rope in the horizontal direction and θ represent the angle of the person walking on the rope. These variables will both change over time, so we can express them as functions of time, v(t) and θ(t).

Next, we need to consider the forces acting on the person and the rope. The person's weight and the tension in the rope will affect their movement. We can express these forces as Fp and Fr, respectively.

Using Newton's second law, we can write the equation for the horizontal motion of the person as Fp = m*d^2x/dt^2, where m is the mass of the person and dx/dt represents the acceleration in the horizontal direction. Since the person is walking on the rope, their motion is constrained to follow the rope's path, so we can write dx/dt = v(t). This gives us Fp = m*dv/dt.

Next, we need to consider the vertical motion of the person. The tension in the rope will affect their angle of inclination, θ. We can express this as Fr*sin(θ) = m*d^2y/dt^2, where dy/dt represents the vertical acceleration. Since the person is walking on the rope, their motion is constrained to follow the rope's path, so we can write dy/dt = v(t)*cos(θ). This gives us Fr*sin(θ) = m*v(t)*cos(θ)*dθ/dt.

Combining these two equations, we can write our system of first order differential equations as:
dv/dt = (Fp/m)
dθ/dt = (Fr*sin(θ))/(m*v(t)*cos(θ))

I hope this helps you get started with your ropewalk equations. If you need further clarification or assistance, don't hesitate to reach out. Good luck with your model!