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Re: Archery test

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Cycling Equipment
Published
1 January 2007
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Ron Ruff
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  1. Quoted message said:

    Obviously, the measured actual tension will disagree with calculations
    using the t = (L x F) / ( d x 4) equation for fixed instead of springy
    anchor points.

    After all, the fixed anchor-point equation must predict a spoke
    tension increase as soon as the spoke begins to bend, while a real
    spoke attached to a springy wheel actually loses tension as it first
    bends.

    I guess you still don't get this... and I'm getting tired of trying to
    explain it. This equation does *not* require fixed anchor points at
    all. They can move wherever they will. You only need to be able to
    measure "L" with reasonable accuracy, and d and F, to calculate the
    tension.

    Since the bowstring is limp and very long we surely don't need to worry
    about a bending moment on the string. The equation will work
    *perfectly*. In this form the equation assumes small angles, so for the
    entire range of a bow string, you should use the sine of the angles
    instead of the approximation.

    As you draw the string back, if the deflection increases at a faster
    rate than the increase in force, then this means that the tension is
    *decreasing*. If you measure the force and deflection of the bow, this
    is exactly what you will see.

  2. Ron Ruff said:
    Quoted message said:

    Obviously, the measured actual tension will disagree with calculations
    using the t = (L x F) / ( d x 4) equation for fixed instead of springy
    anchor points.

    After all, the fixed anchor-point equation must predict a spoke
    tension increase as soon as the spoke begins to bend, while a real
    spoke attached to a springy wheel actually loses tension as it first
    bends.

    I guess you still don't get this... and I'm getting tired of trying to
    explain it. This equation does *not* require fixed anchor points at
    all. They can move wherever they will. You only need to be able to
    measure "L" with reasonable accuracy, and d and F, to calculate the
    tension.

    Since the bowstring is limp and very long we surely don't need to worry
    about a bending moment on the string. The equation will work
    *perfectly*. In this form the equation assumes small angles, so for the
    entire range of a bow string, you should use the sine of the angles
    instead of the approximation.

    As you draw the string back, if the deflection increases at a faster
    rate than the increase in force, then this means that the tension is
    *decreasing*. If you measure the force and deflection of the bow, this
    is exactly what you will see.

    you two seem to be arguing at crossed purposes. carl /is/ measuring
    decreasing tension. the question is /why/. fixed anchor points predict
    the tension will increase, as has been calculated here numerous times.
    elastic anchor points however, depending on their degree of flexibility
    and linearity, can see that tension drop. again, the question is /why/.
    does carl's archer cite not explain that?

  3. jim beam said:

    and linearity, can see that tension drop. again, the question is /why/.
    does carl's archer cite not explain that?

    It is because the ends are not rigid. If the displacement rises at a
    faster rate than the force applied, then we will calculate a decrease
    in tension. The thing I disagreed with was his statement that the
    equation was invalid because of this... it is not.

    If someone with the proper equipment to make accurate measurements
    would like to test for the deflection of a spoke due to a side load,
    calculate the tension, and compare it to values obtained with a
    tensiometer, I bet we'd find that the results are the same.

  4. On 31 Dec 2006 21:26:40 -0800, "Ron Ruff" <[email hidden]>

    Quoted message said:


    Quoted message said:

    Obviously, the measured actual tension will disagree with calculations
    using the t = (L x F) / ( d x 4) equation for fixed instead of springy
    anchor points.

    After all, the fixed anchor-point equation must predict a spoke
    tension increase as soon as the spoke begins to bend, while a real
    spoke attached to a springy wheel actually loses tension as it first
    bends.

    I guess you still don't get this... and I'm getting tired of trying to
    explain it. This equation does *not* require fixed anchor points at
    all. They can move wherever they will. You only need to be able to
    measure "L" with reasonable accuracy, and d and F, to calculate the
    tension.

    Since the bowstring is limp and very long we surely don't need to worry
    about a bending moment on the string. The equation will work
    *perfectly*. In this form the equation assumes small angles, so for the
    entire range of a bow string, you should use the sine of the angles
    instead of the approximation.

    As you draw the string back, if the deflection increases at a faster
    rate than the increase in force, then this means that the tension is
    *decreasing*. If you measure the force and deflection of the bow, this
    is exactly what you will see.

    Dear Ron,

    Possibly I've missed the crucial point because it's been so badly
    explained--repeatedly, without the necessary example.

    Let's use the simple equation that your final paragraph says will work
    "exactly"--I'm beginning to see how it might do just that.

    From where to where should D and L be measured?

    I may have missed it, but the general impression that I've gotten is
    that both D and L are being treated as being measured from their
    original positions.

    If D is measured from a straight, fixed line to the bend, things don't
    work. But if D is re-measured from a moveable line drawn between the
    two moving points, then the equation will work.

    Similarly, if L is left at its original length, things don't work. But
    if L is re-measured as the end points come slightly together, then the
    equation will work.

    But every example that I recall from other posters used a fixed L and
    gave no indication of any practical heed being paid to the change in
    D.

    So here's a crude effort to illustrate what I think is the point that
    you're trying to explain.

    In the three ASCII drawings below, L and D change so that tension
    reduces as the deflecting force increases.

    The x's and .'s are a deforming springy structure bulging to the left.

    It's crucial to measure D from the line L, not from the x.

    F=0 F=1 F=2
    L=11 L=9 L=7
    D=0 D=2 D=3

    T=(LxF)/(4xD) T=(LxF)/(4xD)
    T=? T=(9x1)/(4x1) T=(7x2)/(4x3)
    unknown T = 9/4 T = 14/12
    T = 2.25 T = 1.17 (tension drops)
    1 L
    2 L .L
    3 L . L .L
    4 L . L . L
    5 L . L . L
    6 x x DD x DDD
    7 L |. L | . L
    8 L | . L | . L
    9 L | . L | .L
    10 L | .L |
    11 L | |
    | | |
    | | |
    x x DD x DDD
    | | 12 | 123
    0 0123456 0123456789

    Repeated exhortations to "just measure it" failed to explain that for
    a situation where tension decreases, it is probably crucial to measure
    the reduction of L and to measure D from a new starting point.

    Whenever practical examples were given, they explicitly assumed that L
    did not change length or position, and that D was measured from the
    same place, which is reasonable for extremely stiff structures, but
    extremely misleading for the puzzling drop in measured tension.

    It's unlikely that anyone measured the deflection from the new
    imaginary line drawn between the slightly changed new end points of
    the spoke.

    Only something as exaggeratedly springy as the archery example makes
    this visible for practical measuring.

    Now I'm curious if anyone can provide an example with figures of how D
    and L would change to produce a drop from say 200 to 190 pounds of
    tension for a spoke with a 295 mm original L, using a 5-lb and a 10-lb
    force,
    and the same again for the force staying steady at 200 pounds.

    I suspect that such figures will suggest that no practical measurement
    could have revealed what was going on.

    Does this seem like what you've been trying to explain? I think that
    you're exasperated because I missed what was obvious to you, while
    I've been exasperated because a single example of when to drop the
    assumption that the length of L and the physical starting point for
    measuring D were unchanging.

    On balance, I think that I'm in your debt--you got close enough to an
    example for me to work it out. So thanks, Ron.

    Happy New Year!

    Carl Fogel

  5. Quoted message said:

    On 31 Dec 2006 21:26:40 -0800, "Ron Ruff" <[email hidden]>

    Quoted message said:


    Quoted message said:

    Obviously, the measured actual tension will disagree with calculations
    using the t = (L x F) / ( d x 4) equation for fixed instead of springy
    anchor points.


    [snip]

    Quoted message said:
    Quoted message said:

    The equation will work *perfectly*. In this form the equation assumes
    small angles, so for the

    [snip]

    Quoted message said:

    Let's use the simple equation that your final paragraph says will work
    "exactly"--I'm beginning to see how it might do just that.

    From where to where should D and L be measured?

    Quoted message said:

    I may have missed it, but the general impression that I've gotten is
    that both D and L are being treated as being measured from their
    original positions.

    No, from the bent position, although L is the same in both cases if the
    supports are fixed as I'm sure you realize.

    The bent string forms a kind of V shape. Join the two ends of the V and
    you have a triangle. The top edge of the triangle is L. The
    perpendicular distance from that edge to the point of V is d.

    Quoted message said:

    If D is measured from a straight, fixed line to the bend, things don't
    work.

    That is how it should be measured.

    Quoted message said:

    But if D is re-measured from a moveable line drawn between the two
    moving points, then the equation will work.

    No, it should be the perpendicular distance.

    Quoted message said:

    Similarly, if L is left at its original length, things don't work. But
    if L is re-measured as the end points come slightly together, then the
    equation will work.

    Yes, L must be measured from new positions of the end points.

    Quoted message said:

    But every example that I recall from other posters used a fixed L and
    gave no indication of any practical heed being paid to the change in
    D.

    No, they didn't need to.

    I will have a go at explaining the principle here.

    Suppose the string is pulled vertically downwards in the centre with a
    force of 19lbs. The tension in the string either side has to match that
    19lbs downwards with 19lbs upwards. But these tension forces (one each
    side) do not act directly upwards, they are constrained to act along the
    string.

    We can work out those tension forces because we know both their
    direction (that's given by the sides of the triangle whose base is L and
    height is d) and that their vertical components must add up to exactly
    19lbs upwards.

    This is why L and d are involved in the calculation-- they determine the
    direction of the string either side of the point the weight is hanging
    from. So it is always L, d and F in the loaded configuration that we are
    interested in.

  6. Ben C said:
    Quoted message said:

    On 31 Dec 2006 21:26:40 -0800, "Ron Ruff" <[email hidden]>

    Quoted message said:

    [email hidden] wrote:
    > Obviously, the measured actual tension will disagree with calculations
    > using the t = (L x F) / ( d x 4) equation for fixed instead of springy
    > anchor points.


    [snip]

    Quoted message said:
    Quoted message said:

    The equation will work *perfectly*. In this form the equation assumes
    small angles, so for the

    [snip]

    Quoted message said:

    Let's use the simple equation that your final paragraph says will work
    "exactly"--I'm beginning to see how it might do just that.

    From where to where should D and L be measured?

    Quoted message said:

    I may have missed it, but the general impression that I've gotten is
    that both D and L are being treated as being measured from their
    original positions.

    No, from the bent position, although L is the same in both cases if the
    supports are fixed as I'm sure you realize.

    The bent string forms a kind of V shape. Join the two ends of the V and
    you have a triangle. The top edge of the triangle is L. The
    perpendicular distance from that edge to the point of V is d.

    Quoted message said:

    If D is measured from a straight, fixed line to the bend, things don't
    work.

    That is how it should be measured.

    Quoted message said:

    But if D is re-measured from a moveable line drawn between the two
    moving points, then the equation will work.

    No, it should be the perpendicular distance.

    Quoted message said:

    Similarly, if L is left at its original length, things don't work. But
    if L is re-measured as the end points come slightly together, then the
    equation will work.

    Yes, L must be measured from new positions of the end points.

    Quoted message said:

    But every example that I recall from other posters used a fixed L and
    gave no indication of any practical heed being paid to the change in
    D.

    No, they didn't need to.

    I will have a go at explaining the principle here.

    Suppose the string is pulled vertically downwards in the centre with a
    force of 19lbs. The tension in the string either side has to match that
    19lbs downwards with 19lbs upwards. But these tension forces (one each
    side) do not act directly upwards, they are constrained to act along the
    string.

    We can work out those tension forces because we know both their
    direction (that's given by the sides of the triangle whose base is L and
    height is d) and that their vertical components must add up to exactly
    19lbs upwards.

    This is why L and d are involved in the calculation-- they determine the
    direction of the string either side of the point the weight is hanging
    from. So it is always L, d and F in the loaded configuration that we are
    interested in.

    that's true from a simplistic vector viewpoint, but imagine instead that
    it's an i-beam showing that deflection, not a flexible wire. the force
    magnitudes are different [obviously], but the math is different too.
    that's all that's being said here - we just need the right math model to
    match what we can measure. saying the measurements are wrong [as has
    been argued by others] is not the solution!

  7. jim beam said:
    Ben C said:
    Quoted message said:

    On 31 Dec 2006 21:26:40 -0800, "Ron Ruff" <[email hidden]>
    wrote:

    > [email hidden] wrote:
    >> Obviously, the measured actual tension will disagree with calculations
    >> using the t = (L x F) / ( d x 4) equation for fixed instead of springy
    >> anchor points.


    [snip]

    Quoted message said:

    > The equation will work *perfectly*. In this form the equation assumes
    > small angles, so for the


    [snip]

    Quoted message said:
    Quoted message said:

    I will have a go at explaining the principle here.

    Suppose the string is pulled vertically downwards in the centre with a
    force of 19lbs. The tension in the string either side has to match that
    19lbs downwards with 19lbs upwards. But these tension forces (one each
    side) do not act directly upwards, they are constrained to act along the
    string.

    We can work out those tension forces because we know both their
    direction (that's given by the sides of the triangle whose base is L and
    height is d) and that their vertical components must add up to exactly
    19lbs upwards.

    This is why L and d are involved in the calculation-- they determine the
    direction of the string either side of the point the weight is hanging
    from. So it is always L, d and F in the loaded configuration that we are
    interested in.

    that's true from a simplistic vector viewpoint, but imagine instead that
    it's an i-beam showing that deflection, not a flexible wire. the force
    magnitudes are different [obviously], but the math is different too.

    I agree, and Carl has done a good job of demonstrating that the "string"
    model may not be so appropriate for spokes with his recent picture of a
    weight hanging from a spoke that wasn't anchored at the ends at all. I'm
    still trying to figure out what the maths is for a beam though...

    Quoted message said:

    that's all that's being said here - we just need the right math model
    to match what we can measure. saying the measurements are wrong [as
    has been argued by others] is not the solution!

    It still seemed that we hadn't explained adequately where "T=FL/4d"
    comes from though and what it's supposed to mean, even if it turns out
    not to be useful in this case.

  8. Quoted message said:

    From where to where should D and L be measured?

    I may have missed it, but the general impression that I've gotten is
    that both D and L are being treated as being measured from their
    original positions.

    If D is measured from a straight, fixed line to the bend, things don't
    work. But if D is re-measured from a moveable line drawn between the
    two moving points, then the equation will work.

    Similarly, if L is left at its original length, things don't work. But
    if L is re-measured as the end points come slightly together, then the
    equation will work.

    You are correct! It is very simple once you see it. You don't use
    "original" positions for anything... you take measurements after the
    load is applied so you can calculate the force balance. You are always
    working with the geometry that *exists* in order to compute the
    tension. L will generally not change enough to make a difference, but
    we can just as easily measure it after the load is applied. Use a ruler
    to measure the distance between "supports" (the ends of the triangle),
    and then a caliper to measure the deflection from a straight line
    (straightedge) between these supports, to the point where the load is
    applied.

    The ends of the triangle can be anywhere on the unsupported length of
    spoke... you don't have to use the ends of the spoke or even the
    crossing point... in fact it would probably be better if you didn't.
    You just need to make sure that both of the points you select are
    equidistant from where the load is applied. For instance you could make
    a mark 80mm from the load on either side, then lay your straightedge
    across those two points, and measure L and d.

    Yes, getting more precise than +- 0.5mm will not be easy.

  9. Quoted message said:

    Possibly I've missed the crucial point because it's been so badly
    explained--repeatedly, without the necessary example.

    Quoted message said:

    Repeated exhortations to "just measure it" failed to explain that for
    a situation where tension decreases, it is probably crucial to measure
    the reduction of L and to measure D from a new starting point.

    Whenever practical examples were given, they explicitly assumed that L
    did not change length or position, and that D was measured from the
    same place, which is reasonable for extremely stiff structures, but
    extremely misleading for the puzzling drop in measured tension.

    It's unlikely that anyone measured the deflection from the new
    imaginary line drawn between the slightly changed new end points of
    the spoke.

    Only something as exaggeratedly springy as the archery example makes
    this visible for practical measuring.

    Now I'm curious if anyone can provide an example with figures of how D
    and L would change to produce a drop from say 200 to 190 pounds of
    tension for a spoke with a 295 mm original L, using a 5-lb and a 10-lb
    force,
    and the same again for the force staying steady at 200 pounds.

    I suspect that such figures will suggest that no practical measurement
    could have revealed what was going on.

    I think you're still missing the point somewhat.

    Newton says that any point that's not accelerating must have all forces
    add to zero.

    Forces have magnitude and direction. This construct is a "vector".

    Vectors can be broken down into "orthogonal" (at right angles) components.

    The number of orthogonal components depends on the number of dimensions
    of the problem/model space - 2 for planar, 3 for 3D.

    To solve the tension in a cable/wire/bowstring, we view it as a force
    vector with 2 independent (in planar model) vectors - one in the
    direction of the pull/load force and one at right angles. This is called
    decomposition.

    Summing vectors is simple geometry, the resultant magnitude is the
    square root of the sum of the squares of the orthogonal components. The
    resultant direction is a trig function (tangent) of the orthogonal
    components.

    All of the above stipulates that the tension to load ratio is given by
    the angle of the tension vector (which must be the same as the
    cable/wire/string angle) to the load vector.

    All the bow example illustrates (as Ron described) is that, as the
    string is pulled, the angle changes faster than some critical rate, such
    that the tension must drop to balance the forces.

    This is not an obscure scenario. A very common scenario is to have the
    "spring" force provided by gravity (simple weight hanging from
    cable/wire/string, or suspended over pulleys). In those cases the
    tension is required to be constant.

    If, when drawing a bowstring, the tension drops, it just means that the
    string angle and spring (stave) force are changing at certain critical
    rates. The geometry forces that.

    In all pf this, the angle is what's critical, the measurements are just
    ways of computing that. The angle is needed to decompose the tension
    vector. The tension vector can be decomposed directly by the geometry by
    visualizing a right triangle where the hypotenuse is the tensioned
    element. If you visualize that, it should be obvious what "L" and "D" are.

    I don't see anything in the geometry of a wheel that would cause the
    same rapid change of angle with load that causes the slight detension of
    a bow. I do see several "couplings" that could achieve the same effect.
    Imbalances of test forces could cause lateral rim movement and/or
    detensioning of one spoke by another through rim deflections. As a
    practical matter, it would seem very difficult to control these things,
    particularly in the realm of small applied loads in a "spoke squeezing"
    simulation.

    Looking at large load tension changes (the real subject of all this),
    it's not hard to measure the angle of the tension vector. Rim
    deflections, both radial and lateral are large enough to see/measure,
    but don't affect the angle much.

  10. jim beam said:

    Ben C wrote:

    Quoted message said:
    Quoted message said:

    This is why L and d are involved in the calculation-- they determine the
    direction of the string either side of the point the weight is hanging
    from. So it is always L, d and F in the loaded configuration that we are
    interested in.

    that's true from a simplistic vector viewpoint, but imagine instead that
    it's an i-beam showing that deflection, not a flexible wire. the force
    magnitudes are different [obviously], but the math is different too.
    that's all that's being said here - we just need the right math model to
    match what we can measure. saying the measurements are wrong [as has
    been argued by others] is not the solution!

    What do I-beams have to do with spokes?

    Are you ready to retract?

  11. Ben C said:

    I agree, and Carl has done a good job of demonstrating that the "string"
    model may not be so appropriate for spokes with his recent picture of a
    weight hanging from a spoke that wasn't anchored at the ends at all. I'm
    still trying to figure out what the maths is for a beam though...

    I dug up my old statics book and found the
    following relation for a beam that is rigidly fixed on the ends (no
    rotation allowed on the ends) subjected to a vertical displacement.

    M= 6*E*I*d/L^2

    If E = 200,000 N/mm^2 (steel)
    I= pi*r^4/4 = .785 for a 2mm diameter spoke
    d= 10mm
    L= ~105mm = the distance from one of the end supports to the center

    M= 854N-mm x2 (for both sides) = 1708 N-mm

    With a lever arm of 105mm this is a 16.3N force, or 3.7 lbs. This is a
    "worst case" since it assumes that both ends of the test region are not

    allowed to pivot, which is probably close to being right for the end
    near the hub, but not the end near the rim.

    You can demonstrate this yourself by taking a spoke and holding the
    ends parallel, while pushing up in the center with your thumbs. The
    resistance due to bending in the spoke is not that small... but
    compared to a side load of 50lb, a 3.7lb "error" due to bending
    stiffness isn't very big.

    You can measure it in a manner similar to what Carl showed. Place
    supports a little more than L apart, place marks on the spoke L/2 from
    the center, rest a straight spoke on it with one side clamped
    perpendicular to F and the other just resting on the support, and hang
    weights until the deflection equals d, and then you will have the error
    due to bending stiffness.

    Quoted message said:

    It still seemed that we hadn't explained adequately where "T=FL/4d"
    comes from though and what it's supposed to mean, even if it turns out
    not to be useful in this case.

    It is a simplification of the force balance, assuming that the spoke
    has negligible bending stiffness, and the angles are small. For bigger
    angles, the full equation is:

    T=F/(2*sin(atan(2*d/L))

  12. Ron Ruff said:


    Ben C said:

    I agree, and Carl has done a good job of demonstrating that the "string"
    model may not be so appropriate for spokes with his recent picture of a
    weight hanging from a spoke that wasn't anchored at the ends at all. I'm
    still trying to figure out what the maths is for a beam though...

    I dug up my old statics book and found the
    following relation for a beam that is rigidly fixed on the ends (no
    rotation allowed on the ends) subjected to a vertical displacement.


    [snip]

    Thanks, I'll take a look at that.

    Quoted message said:
    Quoted message said:

    It still seemed that we hadn't explained adequately where "T=FL/4d"
    comes from though and what it's supposed to mean, even if it turns out
    not to be useful in this case.

    It is a simplification of the force balance, assuming that the spoke
    has negligible bending stiffness, and the angles are small. For bigger
    angles, the full equation is:

    T=F/(2*sin(atan(2*d/L))

    Yes, this gives the same results as my own calculations.

    If you work out the actual vectors for the tension in the string, FL/4d
    is actually the horizontal component, which is close to the magnitude if
    the vertical component is small (i.e. for small angles).

    The direction of the string is the vector [L/2, d]. Call it a. Now we
    just need to find what coefficient to scale a by to get the right
    magnitude.

    We can state that the tension in one side of the string must balance
    half the force with this equation:

    dot(lambda*a, f) = F/2

    where f is a vector for the normalized direction of the applied force
    (i.e. [0, 1]) and F is its magnitude.

    We can rearrange that into:

    lambda = F / (2 * dot(a, f))

    since f is just [0, 1], this simplifies to:

    lambda = F / (2 * a(2))

    [I'm using 1 for the first index of a vector]

    lambda*a gives you the tension vector T, with the correct magnitude to
    oppose half of F (the other half is opposed by the other side of the
    string).

    The magnitude of lambda*a (sqrt of sum of its components squared) gives
    the same result as your formula-- the proper result without the small
    angle approximation-- when I tested it on a few values.

    Since a(2) is just d, lambda is F/2d, and the first component of T is
    given by:

    F/2d * L/2 = FL/4d

    which is how we arrive at the formula FL/4d.

  13. Ron Ruff said:


    Quoted message said:

    From where to where should D and L be measured?

    I may have missed it, but the general impression that I've gotten is
    that both D and L are being treated as being measured from their
    original positions.

    If D is measured from a straight, fixed line to the bend, things don't
    work. But if D is re-measured from a moveable line drawn between the
    two moving points, then the equation will work.

    Similarly, if L is left at its original length, things don't work. But
    if L is re-measured as the end points come slightly together, then the
    equation will work.

    You are correct! It is very simple once you see it. You don't use
    "original" positions for anything... you take measurements after the
    load is applied so you can calculate the force balance. You are always
    working with the geometry that *exists* in order to compute the
    tension. L will generally not change enough to make a difference, but
    we can just as easily measure it after the load is applied. Use a ruler
    to measure the distance between "supports" (the ends of the triangle),
    and then a caliper to measure the deflection from a straight line
    (straightedge) between these supports, to the point where the load is
    applied.

    The ends of the triangle can be anywhere on the unsupported length of
    spoke... you don't have to use the ends of the spoke or even the
    crossing point... in fact it would probably be better if you didn't.
    You just need to make sure that both of the points you select are
    equidistant from where the load is applied. For instance you could make
    a mark 80mm from the load on either side, then lay your straightedge
    across those two points, and measure L and d.

    Yes, getting more precise than +- 0.5mm will not be easy.

    Dear Ron,

    You've explained the point even better with the rulers.

    We agree on the margin-of-error problem, though I think that your
    ruler approach would help.

    But a crucial point, I think, is how we measure D.

    With crossed spokes, one end of the original line of L keeps moving to
    a new position, following the bend where the weight hangs, as if the
    triangle is rotating around the hub end of the spoke.

    If we mistakenly keep measuring D from the original straight spoke's
    midpoint position, we'll get an exaggerated value of D, an
    underestimated value of tension, and a paradoxically smaller value

    Really crude illustration:

    http://i18.tinypic.com/40ax507.jpg

    It is _always_ a mistake to measure D by starting from A. We need to
    measure D from a new position each time we hang a new weight, a
    position that exists only in thin air at the midpoint of a line drawn
    between the new ends of the spoke. If we measure from A (or any
    previous starting point), we'll get an exaggerated value for D.

    How much does the exaggeration matter?

    Because D is so small with a 23-lb weight hung on a 280 mm spoke
    (maybe 280 mm to rim, maybe 280 mm to crossing), the apparently minor
    movement of the spoke as it tugs on the crossing turns out to be
    significant.

    First, you should confirm how much the spoke moves. Squeeze a spoke
    pair with your hand and watch the crossing. It's easy to pull the
    spoke a spoke's width, about 2 mm, in the direction that you're
    squeezing.

    The middle of the new L (a line in thin air) should move half as far
    as the crossing moved, about 1 mm. (There are several other movements,
    but let's stick to just the crossing.)

    If a 6 mm D is mismeasured as a 7 mm D because the movement at the
    crossing wasn't considered and the ruler started from the wrong spot,
    here's what happens with a 280 mm L and a 23-lb force:

    ( F x L ) / (4 x D) = T

    (23 x 280) / (4 x 6) = 268 lbs tension (real is higher)

    (23 x 280) / (4 x 7) = 230 lbs tension (lower than real)

    Even if D is measured from the right spot, the measurements are just
    too darned small and tricky to inspire confidence in the results,
    particularly if we start out with a target tension in mind.

    The ruler-across-the-spoke-bend might get around the problems of the
    margin-of-error and where-to-measure-D-from.

    But we'd end up building a tool more and more like a tension gauge--a
    tool that attaches easily to the spoke, automatically centers a
    consistent force on a precisely measured span, always measures D from
    the correct spot, keeps L closer to its original length, and gives an
    accurate result in a few moments.

    Cheers,

    Carl Fogel

  14. Peter Cole said:
    jim beam said:

    Ben C wrote:

    Quoted message said:
    Quoted message said:

    This is why L and d are involved in the calculation-- they determine the
    direction of the string either side of the point the weight is hanging
    from. So it is always L, d and F in the loaded configuration that we are
    interested in.

    that's true from a simplistic vector viewpoint, but imagine instead
    that it's an i-beam showing that deflection, not a flexible wire. the
    force magnitudes are different [obviously], but the math is different
    too. that's all that's being said here - we just need the right math
    model to match what we can measure. saying the measurements are wrong
    [as has been argued by others] is not the solution!

    What do I-beams have to do with spokes?

    Are you ready to retract?

    retract what peter? that your simplistic assumptions don't paint the
    whole picture?

  15. Quoted message said:

    The ruler-across-the-spoke-bend might get around the problems of the
    margin-of-error and where-to-measure-D-from.

    I'd recommend making L a bit shorter than the unsupported span... ie
    inside of the spoke crossing and the nipple. Mark a point near the
    center where you will attach the weight, then two more the same
    distance away on either side. The straight edge will intersect the two
    end points. The weight needs to hang perpendicular to the straightedge,
    so adjust the tilt if necessary... but a slight error in this is not
    critical.

    Yes, hard to measure accurately... but I think this is the best
    approach for figuring tension without a meter. I bet the results will
    be very close to what you get with the Park gauge.

  16. Ron Ruff said:
    Quoted message said:

    The ruler-across-the-spoke-bend might get around the problems of the
    margin-of-error and where-to-measure-D-from.

    I'd recommend making L a bit shorter than the unsupported span... ie
    inside of the spoke crossing and the nipple. Mark a point near the
    center where you will attach the weight, then two more the same
    distance away on either side. The straight edge will intersect the two
    end points. The weight needs to hang perpendicular to the straightedge,
    so adjust the tilt if necessary... but a slight error in this is not
    critical.

    Yes, hard to measure accurately... but I think this is the best
    approach for figuring tension without a meter. I bet the results will
    be very close to what you get with the Park gauge.

    Dear Ron,

    Until you actually hang a lot of weights and try to measure the
    tension on squeezed spokes, it's hard to appreciate the annoying
    practical problems involved. A few pictures and pointers may help (or
    scare people off).

    Here's a quick setup with a handy wheel in a fork, ready for a 60-lb
    weight to be dangled, but with no complicating clamp on the spoke pair
    on the far side of the wheel:

    http://i18.tinypic.com/2ykc3o0.jpg

    Click on the lower right in Explorer for the full-size views.

    Notice that the ropes cross each other. The upper spoke points up to
    the right at 21~22 degrees, and that the lower spoke also points up to
    the right at 8~9 degrees.

    (Or so the onscreen protractor says. The camera distorts angles a bit.
    The pictures were taken in the dark to illuminate the spokes better.
    Some level edges can be seen in the computer nonsense beyond the
    wheel.)

    Now let's dangle the 60-lb weight:

    http://i14.tinypic.com/34t7961.jpg

    The wheel has rotated clockwise about 20 degrees. The spoke ropes no
    longer curve around each other. You use some Kentucky windage to get
    the offset close and then adjust a little more until you're happy.
    Unlike the shooting range, however, each shot involves heaving a 60-lb
    weight back onto a stand.

    With the weight, the upper spoke's hub-half points down at 5~6 degrees
    below level, and its rim-half points up at 6~7 degrees.

    The lower spoke's hub-half points down at 8~9 degrees, while its rim
    side points down even more steeply, about 20 degrees below the level.

    The rim-halves of the two spokes now diverge at 26~27 degrees.

    There's just room for the Park gauge, though the camera distortion
    makes it look as if the post is in the rope.

    That's a DT spoke ruler. Let's look at it more closely:

    http://i16.tinypic.com/33uygk1.jpg

    The bends are somewhere inside the white rope's knots. The spokes are
    2 mm thick. If you look up above the glare on the ruler, you can see
    how small a millimeter is relative to the knots hiding the bends.

    It would be darned hard to get 0.5 mm accuracy with this approach,
    which seemed likely to yield only half the accuracy of a Park gauge

    A single knot instead of a pair of half-hitches would let you get
    closer to imagining where the bend is, but of course a real spoke
    squeeze is spread out across several fingers. (I fooled around with a
    Terminator-style hand-rig inside a glove, but there's no room for a
    spoke gauge, and the silly thing never worked well, anyway.)

    Real-life spoke-squeezing by hand might produce a different tension
    than a thin rope because the squeeze is spread out over a broader
    curve. My purely ignorant guess is that a broad hand-squeeze of 60 lbs
    might produce a smaller tension increase than a narrow rope applying
    the same force, but it might be the other way around, or even have no
    effect at all.

    The advantage of the two half-hitches is that they're easier to adjust
    when you're setting things up (twenty times, if you measure up to 100
    lbs in 5-lb increments). The wider knot provides more friction to
    discourage the weights from stealthily slithering toward the hub and
    are also less likely to leave permanent bends in the spokes.

    Some kind of small, stiff Y-rig could hook over the spokes and expose
    the bend hidden inside the knot. Then some fancy caliper work would
    get around the rest of the rope to whatever ruler you were trying to
    fiddle into the mess.

    Or you could use strong, thin wire or even fishing line to expose the
    bend.

    But it seems enormously easier to use the normal industry tools
    designed to make this task easy, repeatable, and reasonably accurate.

    The Park tension gauge fits in ordinary 700c wheels with this rope
    trick. I don't know if the ropes would leave enough room for the
    Wheelsmith tension gauge (which works the same way as the Park, but
    looks as if its dimensions are different), or enough room for the DT
    Swiss and FSA designs (which use a different approach to achieve the
    same result as the Park and Wheelsmith gauges).

    Like you, I now expect that if D is measured accurately from the
    correct starting position (somewhere in the thin air between the two
    current end points of L) to the bend, then the equation should agree
    quite closely with an accurate tension gauge.

    Thanks again for the exasperating work of getting around my
    misunderstanding of the results.

    Cheers,

    Carl Fogel

  17. jim beam said:
    Peter Cole said:
    jim beam said:

    Ben C wrote:

    Quoted message said:

    > This is why L and d are involved in the calculation-- they determine
    > the
    > direction of the string either side of the point the weight is hanging
    > from. So it is always L, d and F in the loaded configuration that we
    > are
    > interested in.

    that's true from a simplistic vector viewpoint, but imagine instead
    that it's an i-beam showing that deflection, not a flexible wire.
    the force magnitudes are different [obviously], but the math is
    different too. that's all that's being said here - we just need the
    right math model to match what we can measure. saying the
    measurements are wrong [as has been argued by others] is not the
    solution!

    What do I-beams have to do with spokes?

    Are you ready to retract?

    retract what peter? that your simplistic assumptions don't paint the
    whole picture?

    Funny.

  18. Quoted message said:

    The bends are somewhere inside the white rope's knots. The spokes are
    2 mm thick. If you look up above the glare on the ruler, you can see
    how small a millimeter is relative to the knots hiding the bends.

    It would be darned hard to get 0.5 mm accuracy with this approach,
    which seemed likely to yield only half the accuracy of a Park gauge

    Sure, but if you simply use baling wire and a caliper, I think you can
    do substantially better.

    Quoted message said:

    Like you, I now expect that if D is measured accurately from the
    correct starting position (somewhere in the thin air between the two
    current end points of L)

    Or, just measure the gap between the spokes.

    Quoted message said:

    to the bend, then the equation should agree
    quite closely with an accurate tension gauge.

    Thanks again for the exasperating work of getting around my
    misunderstanding of the results.

    OK, now explain it to "jim".

  19. Peter Cole said:
    jim beam said:
    Peter Cole said:

    jim beam wrote:
    > Ben C wrote:

    >> This is why L and d are involved in the calculation-- they
    >> determine the
    >> direction of the string either side of the point the weight is hanging
    >> from. So it is always L, d and F in the loaded configuration that
    >> we are
    >> interested in.
    >
    > that's true from a simplistic vector viewpoint, but imagine instead
    > that it's an i-beam showing that deflection, not a flexible wire.
    > the force magnitudes are different [obviously], but the math is
    > different too. that's all that's being said here - we just need the
    > right math model to match what we can measure. saying the
    > measurements are wrong [as has been argued by others] is not the
    > solution!

    What do I-beams have to do with spokes?

    Are you ready to retract?

    retract what peter? that your simplistic assumptions don't paint the
    whole picture?

    Funny.

    why is that funny? since measured tension and calculated tension using
    T = LF/4D don't agree, there's something else that needs to be taken
    into account. you wouldn't send troops into battle with only parabolic
    artillery trajectory calculators would you? why wouldn't you be
    interested in getting this right either?

  20. jim beam said:
    Peter Cole said:
    jim beam said:

    Peter Cole wrote:

    Quoted message said:
    Quoted message said:
    Quoted message said:

    > Are you ready to retract?

    retract what peter? that your simplistic assumptions don't paint the
    whole picture?

    Funny.

    why is that funny? since measured tension and calculated tension using
    T = LF/4D don't agree, there's something else that needs to be taken
    into account. you wouldn't send troops into battle with only parabolic
    artillery trajectory calculators would you? why wouldn't you be
    interested in getting this right either?

    I'll keep that in mind the next battle I command.

    Maybe Carl can fill you in on the whole vector thing.

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