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Physics
Classical Physics
Mechanics
Max tip speed of a spinning cable
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[QUOTE="charlesrwest, post: 6467213, member: 688350"] 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 ([URL]https://en.wikipedia.org/wiki/Space_elevator#Cable_section[/URL]), but I'm not sure how to adapt it. Thanks! [CODE] 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 [/CODE] [/QUOTE]
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Max tip speed of a spinning cable
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