Potential good news for Mt. Washington access.

Discussion in 'Road Cycling' started by Wayne Pein, Dec 30, 2004.

  1. "Joe Riel" <[email protected]> wrote in message
    news:[email protected]
    > Joe Riel <[email protected]> writes:
    >
    >> where
    >> 1/Feff = 1/Fs + (n/2/pi)/Fr
    >> ~ 30,000kgF

    >
    > That should be
    >
    > Feff ~ 30,000kgF
    >
    > Nothing else changes.
    >
    > Joe


    Joe, should (n/2/pi) be (n/2pi). Depending on how you ran the
    calculation this "could" make a difference by a factor of ~10.
    0.75kgf/deg C for all 36 spokes seems really high..... don't ya think!

    Phil H
     


  2. Joe Riel

    Joe Riel Guest

    "Philip Holman" <[email protected]> writes:

    > Joe, should (n/2/pi) be (n/2pi).


    Division is left-associative. n/2/pi = (n/2)/pi = n/(2*pi).

    Joe
     
  3. Joe Riel

    Joe Riel Guest

    Joe Riel <[email protected]> writes:

    > Here is an approximation. Assume that we can model this by computing
    > the transverse bending of a circular plate of radius a with a load F
    > uniformly distributed around a concentric circle of radius b. The
    > diameter of the plate is set to the width of the rim, the diameter of
    > the loading circle to that of the spoke hole. From [1,article 314
    > (vi)] the displacement, w, is
    >
    > w = F/8/Pi/D*((a^4-b^4)/2/a^2 - (a^2+b^2)*ln(a/b))
    >
    > where
    >
    > D = 2/3*E*(T/2)^3/(1-sigma^2)
    > T = material thickness = 1mm
    > sigma = Poisson ratio = 1/3
    > E = elasticity of aluminum = 75kN/mm^2
    >
    > a = half rim width = 25mm/2
    > b = radius of spoke hole = 3.2mm
    >
    > This gives
    >
    > w/F = 8.3um/kgf
    >
    > So a change in the spoke tension causes a deflection in the nipple
    > hole of approximately (75kgF)(8.3um/kgF) = 0.6mm. That represents
    > 0.2% of the spoke length. For a 100degC rise the diameter of the rim
    > increased by 0.25%. So it appears as though much of the compliance is
    > in the bottom of the rim. The proper technique is to combine this
    > compliance with the previous computation---but its late and I'm tired.
    >
    >
    > Reference
    > ---------
    > [1] A. E. H. Love "A Treatise on the Mathematical Theory of Elasticity"


    I assumed the entire load was supported by one wall (the inner
    surface) of the rim. With sockets distributing the load to both
    walls, the compliance should be about half that computed.

    Joe Riel
     
  4. Joe Riel

    Joe Riel Guest

    Joe Riel <[email protected]> writes:

    > Here is an approximation. Assume that we can model this by computing
    > the transverse bending of a circular plate of radius a with a load F
    > uniformly distributed around a concentric circle of radius b. The
    > diameter of the plate is set to the width of the rim, the diameter of
    > the loading circle to that of the spoke hole. From [1,article 314
    > (vi)] the displacement, w, is
    >
    > w = F/8/Pi/D*((a^4-b^4)/2/a^2 - (a^2+b^2)*ln(a/b))
    >
    > where
    >
    > D = 2/3*E*(T/2)^3/(1-sigma^2)
    > T = material thickness = 1mm
    > sigma = Poisson ratio = 1/3
    > E = elasticity of aluminum = 75kN/mm^2
    >
    > a = half rim width = 25mm/2
    > b = radius of spoke hole = 3.2mm
    >
    > This gives
    >
    > w/F = 8.3um/kgf
    >
    > So a change in the spoke tension causes a deflection in the nipple
    > hole of approximately (75kgF)(8.3um/kgF) = 0.6mm. That represents
    > 0.2% of the spoke length. For a 100degC rise the diameter of the rim
    > increased by 0.25%. So it appears as though much of the compliance is
    > in the bottom of the rim. The proper technique is to combine this
    > compliance with the previous computation---but its late and I'm tired.
    >
    >
    > Reference
    > ---------
    > [1] A. E. H. Love "A Treatise on the Mathematical Theory of Elasticity"


    I assumed the entire load was supported by one wall (the inner
    surface) of the rim. With sockets distributing the load to both
    walls, the compliance should be about half that computed.

    Joe Riel
     
  5. Joe Riel

    Joe Riel Guest

    Joe Riel <[email protected]> writes:

    > Here is an approximation. Assume that we can model this by computing
    > the transverse bending of a circular plate of radius a with a load F
    > uniformly distributed around a concentric circle of radius b. The
    > diameter of the plate is set to the width of the rim, the diameter of
    > the loading circle to that of the spoke hole. From [1,article 314
    > (vi)] the displacement, w, is
    >
    > w = F/8/Pi/D*((a^4-b^4)/2/a^2 - (a^2+b^2)*ln(a/b))
    >
    > where
    >
    > D = 2/3*E*(T/2)^3/(1-sigma^2)
    > T = material thickness = 1mm
    > sigma = Poisson ratio = 1/3
    > E = elasticity of aluminum = 75kN/mm^2
    >
    > a = half rim width = 25mm/2
    > b = radius of spoke hole = 3.2mm
    >
    > This gives
    >
    > w/F = 8.3um/kgf
    >
    > So a change in the spoke tension causes a deflection in the nipple
    > hole of approximately (75kgF)(8.3um/kgF) = 0.6mm. That represents
    > 0.2% of the spoke length. For a 100degC rise the diameter of the rim
    > increased by 0.25%. So it appears as though much of the compliance is
    > in the bottom of the rim. The proper technique is to combine this
    > compliance with the previous computation---but its late and I'm tired.
    >
    >
    > Reference
    > ---------
    > [1] A. E. H. Love "A Treatise on the Mathematical Theory of Elasticity"


    I assumed the entire load was supported by one wall (the inner
    surface) of the rim. With sockets distributing the load to both
    walls, the compliance should be about half that computed.

    Joe Riel
     
  6. Joe Riel

    Joe Riel Guest

    Joe Riel <[email protected]> writes:

    > Here is an approximation. Assume that we can model this by computing
    > the transverse bending of a circular plate of radius a with a load F
    > uniformly distributed around a concentric circle of radius b. The
    > diameter of the plate is set to the width of the rim, the diameter of
    > the loading circle to that of the spoke hole. From [1,article 314
    > (vi)] the displacement, w, is
    >
    > w = F/8/Pi/D*((a^4-b^4)/2/a^2 - (a^2+b^2)*ln(a/b))
    >
    > where
    >
    > D = 2/3*E*(T/2)^3/(1-sigma^2)
    > T = material thickness = 1mm
    > sigma = Poisson ratio = 1/3
    > E = elasticity of aluminum = 75kN/mm^2
    >
    > a = half rim width = 25mm/2
    > b = radius of spoke hole = 3.2mm
    >
    > This gives
    >
    > w/F = 8.3um/kgf
    >
    > So a change in the spoke tension causes a deflection in the nipple
    > hole of approximately (75kgF)(8.3um/kgF) = 0.6mm. That represents
    > 0.2% of the spoke length. For a 100degC rise the diameter of the rim
    > increased by 0.25%. So it appears as though much of the compliance is
    > in the bottom of the rim. The proper technique is to combine this
    > compliance with the previous computation---but its late and I'm tired.
    >
    >
    > Reference
    > ---------
    > [1] A. E. H. Love "A Treatise on the Mathematical Theory of Elasticity"


    I assumed the entire load was supported by one wall (the inner
    surface) of the rim. With sockets distributing the load to both
    walls, the compliance should be about half that computed.

    Joe Riel
     
  7. Joe Riel

    Joe Riel Guest

    Joe Riel <[email protected]> writes:

    > Here is an approximation. Assume that we can model this by computing
    > the transverse bending of a circular plate of radius a with a load F
    > uniformly distributed around a concentric circle of radius b. The
    > diameter of the plate is set to the width of the rim, the diameter of
    > the loading circle to that of the spoke hole. From [1,article 314
    > (vi)] the displacement, w, is
    >
    > w = F/8/Pi/D*((a^4-b^4)/2/a^2 - (a^2+b^2)*ln(a/b))
    >
    > where
    >
    > D = 2/3*E*(T/2)^3/(1-sigma^2)
    > T = material thickness = 1mm
    > sigma = Poisson ratio = 1/3
    > E = elasticity of aluminum = 75kN/mm^2
    >
    > a = half rim width = 25mm/2
    > b = radius of spoke hole = 3.2mm
    >
    > This gives
    >
    > w/F = 8.3um/kgf
    >
    > So a change in the spoke tension causes a deflection in the nipple
    > hole of approximately (75kgF)(8.3um/kgF) = 0.6mm. That represents
    > 0.2% of the spoke length. For a 100degC rise the diameter of the rim
    > increased by 0.25%. So it appears as though much of the compliance is
    > in the bottom of the rim. The proper technique is to combine this
    > compliance with the previous computation---but its late and I'm tired.
    >
    >
    > Reference
    > ---------
    > [1] A. E. H. Love "A Treatise on the Mathematical Theory of Elasticity"


    I assumed the entire load was supported by one wall (the inner
    surface) of the rim. With sockets distributing the load to both
    walls, the compliance should be about half that computed.

    Joe Riel
     
  8. Joe Riel

    Joe Riel Guest

    Joe Riel <[email protected]> writes:

    > Here is an approximation. Assume that we can model this by computing
    > the transverse bending of a circular plate of radius a with a load F
    > uniformly distributed around a concentric circle of radius b. The
    > diameter of the plate is set to the width of the rim, the diameter of
    > the loading circle to that of the spoke hole. From [1,article 314
    > (vi)] the displacement, w, is
    >
    > w = F/8/Pi/D*((a^4-b^4)/2/a^2 - (a^2+b^2)*ln(a/b))
    >
    > where
    >
    > D = 2/3*E*(T/2)^3/(1-sigma^2)
    > T = material thickness = 1mm
    > sigma = Poisson ratio = 1/3
    > E = elasticity of aluminum = 75kN/mm^2
    >
    > a = half rim width = 25mm/2
    > b = radius of spoke hole = 3.2mm
    >
    > This gives
    >
    > w/F = 8.3um/kgf
    >
    > So a change in the spoke tension causes a deflection in the nipple
    > hole of approximately (75kgF)(8.3um/kgF) = 0.6mm. That represents
    > 0.2% of the spoke length. For a 100degC rise the diameter of the rim
    > increased by 0.25%. So it appears as though much of the compliance is
    > in the bottom of the rim. The proper technique is to combine this
    > compliance with the previous computation---but its late and I'm tired.
    >
    >
    > Reference
    > ---------
    > [1] A. E. H. Love "A Treatise on the Mathematical Theory of Elasticity"


    I assumed the entire load was supported by one wall (the inner
    surface) of the rim. With sockets distributing the load to both
    walls, the compliance should be about half that computed.

    Joe Riel
     
  9. Joe Riel

    Joe Riel Guest

    Joe Riel <[email protected]> writes:

    > Here is an approximation. Assume that we can model this by computing
    > the transverse bending of a circular plate of radius a with a load F
    > uniformly distributed around a concentric circle of radius b. The
    > diameter of the plate is set to the width of the rim, the diameter of
    > the loading circle to that of the spoke hole. From [1,article 314
    > (vi)] the displacement, w, is
    >
    > w = F/8/Pi/D*((a^4-b^4)/2/a^2 - (a^2+b^2)*ln(a/b))
    >
    > where
    >
    > D = 2/3*E*(T/2)^3/(1-sigma^2)
    > T = material thickness = 1mm
    > sigma = Poisson ratio = 1/3
    > E = elasticity of aluminum = 75kN/mm^2
    >
    > a = half rim width = 25mm/2
    > b = radius of spoke hole = 3.2mm
    >
    > This gives
    >
    > w/F = 8.3um/kgf
    >
    > So a change in the spoke tension causes a deflection in the nipple
    > hole of approximately (75kgF)(8.3um/kgF) = 0.6mm. That represents
    > 0.2% of the spoke length. For a 100degC rise the diameter of the rim
    > increased by 0.25%. So it appears as though much of the compliance is
    > in the bottom of the rim. The proper technique is to combine this
    > compliance with the previous computation---but its late and I'm tired.
    >
    >
    > Reference
    > ---------
    > [1] A. E. H. Love "A Treatise on the Mathematical Theory of Elasticity"


    I assumed the entire load was supported by one wall (the inner
    surface) of the rim. With sockets distributing the load to both
    walls, the compliance should be about half that computed.

    Joe Riel
     
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