Max tip speed of a spinning cable

[charlesrwest]

A company called SpinLaunch claims it can get something to 2200 m/s by spinning it up on a carbon fiber composite arm. I'm trying to figure out the limit of that approach. How fast can you go with existing materials?

I tried to work it out for a constant with cable with no payload (result below, I think it's right?). However, a constant stress design with variable cable width would definitely do better. That said, the math for that is beyond me. If I may ask, would anyone care to take a crack at it? It's probably similar to the equations for space elevator cable thickness (https://en.wikipedia.org/wiki/Space_elevator#Cable_section), but I'm not sure how to adapt it.

Thanks!

Dyneema specific strength = 3711000 N*m/kg
Carbon Epoxy composite: 785 N*m/kg

Payload mass = Pa
Cable specific strength = Css
Centripetal Acceleration = Ca = V^2/r
Cable Linear Mass = CLM = F/Css
Cable Linear Speed = Cls = Cev*r/Cl
Cable Length = Cl
Cable Edge Velocity = Cev

Force at cable center with just constant thickness cable:
F = (integral over r from 0 to Cl (CLM*V^2/r))

F = (integral over r from 0 to Cl (CLM*Cls^2/r))

F = (integral over r from 0 to Cl ((F/Css)*(Cev*r/Cl)^2/r))

F = (integral over r from 0 to Cl ((F/Css)*(Cev*r/Cl)^2/r))

F = (F/Css)*(integral over r from 0 to Cl ((Cev*r/Cl)^2/r))

F = (F/Css)*(integral over r from 0 to Cl ((Cev^2*r^2/Cl^2)/r))

F = (F/Css)*Cev^2*(integral over r from 0 to Cl ((r^2)/r))/Cl^2

F = (F/Css)*Cev^2*(integral over r from 0 to Cl (r))/Cl^2

F = (F/Css)*Cev^2*(Cl^2/2)/Cl^2

F = (F/Css)*Cev^2*(1/2)

1 = (1/Css)*Cev^2*(1/2)

1 = (1/Css)*Cev^2*(1/2)

2*Css = Cev^2

Cev^2 = 2*Css

Cev = sqrt(2*Css)

With Dyneema:
Cev = sqrt(2*3711000)
Cev = 2724 m/s

With Carbon Composite:
Cev = sqrt(2*785)
Cev = 396.2 m/s

 

[jrmichler]

The specific strength of Dyneema is no way almost 5000 times greater than carbon epoxy composite. And starting with constant thickness cable with no payload is not useful.

Spinlaunch is designing for 5000 MPH at a 150 foot radius, which is 11,000 G's acceleration, and a 200 lb payload (rocket). That's a force of 11,000 X 200 = 2,200,000 lbs at the payload attachment.

That's your starting point. I like to use numerical methods to solve this type of problem. Choose a cable material and allowable stress, calculate the size, and calculate the lineal mass in kg per meter of length. Move a distance $\Delta L$ up the cable, calculate the acceleration, calculate the additional force due to the acceleration times the mass of cable length $\Delta L$, and add to the total force. Use that new force to calculate the necessary cross sectional area of the cable, along with the mass per length. Repeat until you get to the axis of rotation. The actual program is only a few lines.

Note that you work with actual strength, actual density, and an appropriate safety factor, not with specific strength.

 

[charlesrwest]

You are right. I did the Dyneema calculation and then went back to do the composite calculation. In the process, I forgot to convert the specific strength kN to N,

I followed your advice. The resulting python program is below (please let me know if anything seems wrong). The good news is that launching stuff from the moon to L5 with a spinner seems potentially doable!

Thanks!

[CODE lang="python" title="GetCableMassAndCenterCrossSectionDiameters"]
import math

def GetForce(mass, radius, velocity):
return mass*(velocity**2)/radius

def GetCableMassAndCenterCrossSectionDiameter(payloadSpeed, payloadMass, cableRadius, specificStrength, density, numberOfSegments = 1000000):

#Get force required for initial payload
previous_force_sum = GetForce(payloadMass, cableRadius, payloadSpeed)
cable_mass = 0.0
cross_section_diameter = 0.0

distance_increment = cableRadius/numberOfSegments
for index in range(0, numberOfSegments):
tip_distance = distance_increment*(index+.5)
radius = cableRadius - tip_distance
required_mass = previous_force_sum*distance_increment/specificStrength
velocity = (radius/cableRadius)*payloadSpeed
force_added_from_mass = GetForce(required_mass, radius, velocity)

cross_section_area = previous_force_sum/(specificStrength*density)
cross_section_radius = (cross_section_area/math.pi)**.5
cross_section_diameter = 2*cross_section_radius

previous_force_sum += force_added_from_mass
cable_mass += required_mass

return cable_mass, cross_section_diameter

safety_factor = 1.5
payload_mass = 100 #kg
cable_radius = 100 #Meters
print("Safety Factor: " + str(safety_factor))
for speed in [1000, 2000, 3000, 4000, 5000]:
composite_mass, composite_cross_section_diameter = GetCableMassAndCenterCrossSectionDiameter(speed, payload_mass, cable_radius, 785000/safety_factor, 1580)
dyneema_mass, dyneema_cross_section_diameter = GetCableMassAndCenterCrossSectionDiameter(speed, payload_mass, cable_radius, 3711000/safety_factor, 970)
print("Speed: " + str(speed))
print("Composite mass: " + str(composite_mass))
print("Composite cross section diameter (cm): " + str(composite_cross_section_diameter*100))

print("Dyneema mass: " + str(dyneema_mass))
print("Dyneema cross section diameter (cm): " + str(dyneema_cross_section_diameter*100))
print("")
[/CODE]

 

[jrmichler]

Check your program with some hand calculations:

1) Cross sectional area and stress at the tip, where the only force is from the payload.
2) Is the cross sectional area and stress correct for the calculated total force at the center of rotation?
3) Estimate the force due to the weight of the cable by assuming the entire mass of the cable is at the center of mass of the cable. Does that agree with the total force from your program?
4) What is the taper ratio of the cable?

Suggestions:
The program will be more understandable if you use stress and density, and avoid specific strength.
Depending on the taper ratio, as few as 100 segments may give acceptable accuracy.
It's good practice to plot the cable cross sectional area or diameter as a sanity check.

 

[mfb]

It's surprising but you can do all this with pen and paper. All we need is material constants and the effective potential difference. For the space elevator this is the sum of gravitational potential and centrifugal potential, for SpinLaunch it's only the centrifugal potential $V = \int F dr = \frac 1 2 \omega^2 r^2 = \frac 1 2 v^2$ (looks familiar?).

Consider an infinitesimal element of a tapered design that needs to support a tension T pulling from the outside. We allow a material- and engineering-dependent maximal tensile strength S, so we need a cross section of A=T/S which comes with a material density of $A\rho$. That means tension increases (with decreasing radius) by $\frac {dT}{-dr} = A \rho \omega^2 r = A \rho \frac{dV}{dr}$ or $\frac{dT}{dr} = -\frac{T \rho}{S} \frac{dV}{dr}$. Mathematicians will hate this step, but in physics we can rewrite this as $\frac{dT}{dV} = -\frac{T \rho}{S}$ and solve: $T(V) = T_0\exp(-V\rho/S)$.

Here $\frac{\rho}{S}$ is just the inverse specific strength. Let's plug in numbers: v=2000 m/s, 3700 kJ/kg specific strength of Zylon without safety factor -> tapering ratio of 1.7. If we load it with 2500 kJ/kg only to have some safety factor the tapering ratio increases to 2.2. These are perfectly reasonable values and you can even go with larger safety factors (or higher speed, but you still have the atmosphere as problem).