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Contact patch size versus tire inflation

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25 November 2006
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  1. A common comment on RBT is that the contact patch for the same load on
    a bicycle tire increases and decreases linearly according to the tire
    pressure.

    That is, the contact patch under the same rider will double in size if
    the tire inflation is cut in half, and vice-versa.

    For example, a 100 lb load on a tire inflated to 50 psi is supposed to
    produce a contact patch of 2 square inches, while the same load on a
    tire inflated to 100 psi is supposed to produce a contact patch of
    only 1 square inch.

    Testing suggests that this theory is mistaken.

    I clamped a frame with a rear wheel and tire in a vise by the bottom
    bracket, leaving the whole frame free to pivot.

    A 2x4 attached to the top tube provided convenient leverage. I just
    hung enough weights from the end of the 2x4 sticking out past the rear
    tire to produce a 100-lb reading on a bathroom scale under the tire.

    After rubbing a worn 700x26 tire with a red ink pad, I lowered the
    tire onto a sheet of paper on some well-supported sheet metal, and
    then let the weight hang, pressing the inked tire down on the paper
    with about 100 lbs of force.

    The tire was locked in place by brake pads held tight with a zip tie.

    Using a floor pump with a dial gauge, I took the tire's fingerprint at
    40-60-80-100-120 psi:

    http://i7.tinypic.com/2z4wuib.jpg

    Even a glance shows that the red tire marks don't shrink at the same
    rate that the tire pressure rises.

    Here's the data:

    3x 2.5x 2x 1.5x 1x

    psi 120 100 080 060 040
    mm length 76 81 88 96 108
    mm width 11 11 11 12.5 13
    ---- ---- ---- ---- ----
    L x W 836 891 968 1200 1404

    1x 1.07x 1.16x 1.44x 1.68x

    Multiplying the longest diagonal of the mark by its widest section
    gives a high approximation of the area of the contact patch (a
    rectangle rather than a vague oval).

    Even if the extremes are discarded, the size of the contact patch
    appears to change significantly less than expected.

    For example, pumping the tire up 20% from 80 psi to 100 psi reduces
    the approximate size of the patch by only 8%.

    The asymmetrical ends of the tire print are likely due to the wheel
    not being perfectly vertical, the rope and weight not hanging
    perfectly in line with the tire, and the tread being worn. (There was
    plenty of ink smeared on the tread--the diagonal lower end of each
    mark just shows where the tire was reluctant to touch the paper.)

    One possible partial explanation is that pressure varies considerably
    in the contact patch, with much lighter pressure toward the edges.

    Another possible partial explanation is that casing tension
    complicates matters.

    Better explanations would be welcome.

    Cheers,

    Carl Fogel

  2. Quoted message said:

    A common comment on RBT is that the contact patch for the same load on
    a bicycle tire increases and decreases linearly according to the tire
    pressure.

    That is, the contact patch under the same rider will double in size if
    the tire inflation is cut in half, and vice-versa.

    For example, a 100 lb load on a tire inflated to 50 psi is supposed to
    produce a contact patch of 2 square inches, while the same load on a
    tire inflated to 100 psi is supposed to produce a contact patch of
    only 1 square inch.

    Testing suggests that this theory is mistaken.

    I clamped a frame with a rear wheel and tire in a vise by the bottom
    bracket, leaving the whole frame free to pivot.

    A 2x4 attached to the top tube provided convenient leverage. I just
    hung enough weights from the end of the 2x4 sticking out past the rear
    tire to produce a 100-lb reading on a bathroom scale under the tire.

    After rubbing a worn 700x26 tire with a red ink pad, I lowered the
    tire onto a sheet of paper on some well-supported sheet metal, and
    then let the weight hang, pressing the inked tire down on the paper
    with about 100 lbs of force.

    The tire was locked in place by brake pads held tight with a zip tie.

    Using a floor pump with a dial gauge, I took the tire's fingerprint at
    40-60-80-100-120 psi:

    http://i7.tinypic.com/2z4wuib.jpg

    Even a glance shows that the red tire marks don't shrink at the same
    rate that the tire pressure rises.

    Here's the data:

    3x 2.5x 2x 1.5x 1x

    psi 120 100 080 060 040
    mm length 76 81 88 96 108
    mm width 11 11 11 12.5 13
    ---- ---- ---- ---- ----
    L x W 836 891 968 1200 1404

    1x 1.07x 1.16x 1.44x 1.68x

    Multiplying the longest diagonal of the mark by its widest section
    gives a high approximation of the area of the contact patch (a
    rectangle rather than a vague oval).

    Even if the extremes are discarded, the size of the contact patch
    appears to change significantly less than expected.

    For example, pumping the tire up 20% from 80 psi to 100 psi reduces
    the approximate size of the patch by only 8%.

    The asymmetrical ends of the tire print are likely due to the wheel
    not being perfectly vertical, the rope and weight not hanging
    perfectly in line with the tire, and the tread being worn. (There was
    plenty of ink smeared on the tread--the diagonal lower end of each
    mark just shows where the tire was reluctant to touch the paper.)

    One possible partial explanation is that pressure varies considerably
    in the contact patch, with much lighter pressure toward the edges.

    Another possible partial explanation is that casing tension
    complicates matters.

    Better explanations would be welcome.

    Cheers,

    Carl Fogel

    a) a treaded tyre will not give the same result as a smooth one

    b) can you get & use a planimiter?

  3. In article <[email hidden]>,

    Quoted message said:

    A common comment on RBT is that the contact patch for the same load
    on a bicycle tire increases and decreases linearly according to the
    tire pressure.

    That is, the contact patch under the same rider will double in size
    if the tire inflation is cut in half, and vice-versa.

    For example, a 100 lb load on a tire inflated to 50 psi is supposed
    to produce a contact patch of 2 square inches, while the same load on
    a tire inflated to 100 psi is supposed to produce a contact patch of
    only 1 square inch.

    Testing suggests that this theory is mistaken.

    <snip>

    Quoted message said:

    Multiplying the longest diagonal of the mark by its widest section
    gives a high approximation of the area of the contact patch (a
    rectangle rather than a vague oval).

    That may change the numbers quite a bit and might change the
    proportionality of the change in area related to the pressure- e.g., the
    contact patch may become more "blunt" at the ends with lower pressure
    which would increase the area of the contact patch more than would be
    indicated by a simple height x width multiplication.

    Quoted message said:

    Even if the extremes are discarded, the size of the contact patch
    appears to change significantly less than expected.

    For example, pumping the tire up 20% from 80 psi to 100 psi reduces
    the approximate size of the patch by only 8%.

    I plugged your numbers into a spreadsheet and charted them. The curve
    is not quite a curve, being nearly linear for the 40-60-80-100 pressures
    and flattening significantly between the 100 and 120 psi data points.
    More data points would probably smooth the curves out quite a bit.

    Interestingly, the curve does fairly well fit the curves that resulted
    from the IRC rolling resistance tire testing data from years ago.
    Remember that those curves saw largish reductions in rolling resistance
    towards the low pressure end of the graph, and flattened out at the high
    end with there being basically diminishing returns from increasing
    pressures above 100 psi.

    http://bike.terrymorse.com/rolres.html

  4. Quoted message said:
    Quoted message said:

    A common comment on RBT is that the contact patch for the same load on
    a bicycle tire increases and decreases linearly according to the tire
    pressure.

    That is, the contact patch under the same rider will double in size if
    the tire inflation is cut in half, and vice-versa.

    For example, a 100 lb load on a tire inflated to 50 psi is supposed to
    produce a contact patch of 2 square inches, while the same load on a
    tire inflated to 100 psi is supposed to produce a contact patch of
    only 1 square inch.

    Testing suggests that this theory is mistaken.

    I clamped a frame with a rear wheel and tire in a vise by the bottom
    bracket, leaving the whole frame free to pivot.

    A 2x4 attached to the top tube provided convenient leverage. I just
    hung enough weights from the end of the 2x4 sticking out past the rear
    tire to produce a 100-lb reading on a bathroom scale under the tire.

    After rubbing a worn 700x26 tire with a red ink pad, I lowered the
    tire onto a sheet of paper on some well-supported sheet metal, and
    then let the weight hang, pressing the inked tire down on the paper
    with about 100 lbs of force.

    The tire was locked in place by brake pads held tight with a zip tie.

    Using a floor pump with a dial gauge, I took the tire's fingerprint at
    40-60-80-100-120 psi:

    http://i7.tinypic.com/2z4wuib.jpg

    Even a glance shows that the red tire marks don't shrink at the same
    rate that the tire pressure rises.

    Here's the data:

    3x 2.5x 2x 1.5x 1x

    psi 120 100 080 060 040
    mm length 76 81 88 96 108
    mm width 11 11 11 12.5 13
    ---- ---- ---- ---- ----
    L x W 836 891 968 1200 1404

    1x 1.07x 1.16x 1.44x 1.68x

    Multiplying the longest diagonal of the mark by its widest section
    gives a high approximation of the area of the contact patch (a
    rectangle rather than a vague oval).

    Even if the extremes are discarded, the size of the contact patch
    appears to change significantly less than expected.

    For example, pumping the tire up 20% from 80 psi to 100 psi reduces
    the approximate size of the patch by only 8%.

    The asymmetrical ends of the tire print are likely due to the wheel
    not being perfectly vertical, the rope and weight not hanging
    perfectly in line with the tire, and the tread being worn. (There was
    plenty of ink smeared on the tread--the diagonal lower end of each
    mark just shows where the tire was reluctant to touch the paper.)

    One possible partial explanation is that pressure varies considerably
    in the contact patch, with much lighter pressure toward the edges.

    Another possible partial explanation is that casing tension
    complicates matters.

    Better explanations would be welcome.

    Cheers,

    Carl Fogel

    a) a treaded tyre will not give the same result as a smooth one

    b) can you get & use a planimiter?

    I'd suggest that the contact patch measured on the interior surface of
    the tyre follows the theoretical prediction, but translating this
    through a variable thickness of moderately rigid rubber is bound to
    give non-linear results. Try calibrating your experiment first with a
    thin inner tube with no tyre around it; you'll probably have to run
    pressures in the 0.5 - 1.5 bar range, with a corresponding reduction
    in load.

    Kinky Cowboy*

    *Batteries not included
    May contain traces of nuts
    Your milage may vary

  5. On Sat, 25 Nov 2006 10:53:58 -0600, Tim McNamara
    <[email hidden]> wrote:

    [Carl wrote:]

    Quoted message said:
    Quoted message said:

    Multiplying the longest diagonal of the mark by its widest section
    gives a high approximation of the area of the contact patch (a
    rectangle rather than a vague oval).

    That may change the numbers quite a bit and might change the
    proportionality of the change in area related to the pressure- e.g., the
    contact patch may become more "blunt" at the ends with lower pressure
    which would increase the area of the contact patch more than would be
    indicated by a simple height x width multiplication.

    [snip]

    Dear Tim,

    Actually, the picture shows that the blunter and more "rectangular",
    and therefore "larger" contact patches are the smaller, higher
    pressure examples, not the lower pressure patches. As pressure
    increases, the lower dagger-like ends of the patches become blunter
    and shorter.

    http://i7.tinypic.com/2z4wuib.jpg

    So the smaller patches fill out a rectangle more than the larger
    patches, which reduces the relative size change even further.

    That is, the bottoms of the marks become more and more dagger-like as
    pressure decreases and the marks lengthen. Rearrange the long,
    dagger-like lower end of the longest 40 psi mark to resemble the
    blunter lower end of the 120 psi mark, and you'll get a length and
    width of about 97 x 13 mm, for an even smaller rough (high) estimate
    of 1261 mm^2 instead of 1404 mm^2.

    (The 60 psi mark would shorten a little, too, if its lower end were
    re-shaped more bluntly, and so on as pressures increase.)

    So again, re-shaping the ends to match the shape of the highest
    pressure would only reduce the rough estimates of the area
    differences.

    Given the extreme length to steady width ratio (about 7.5 to 1) of the
    streaks, the somewhat different ends aren't going to change the total
    area very much.

    If the pressure to contact area relationship were purely linear, the
    40 psi mark should have been roughly 3 times the length of the 76 mm
    120 psi mark, over 200 mm. But it was only 108 mm, about half the
    length predicted by theory.

    Cheers,

    Carl Fogel

  6. Quoted message said:
    Quoted message said:

    A common comment on RBT is that the contact patch for the same load on
    a bicycle tire increases and decreases linearly according to the tire
    pressure.

    That is, the contact patch under the same rider will double in size if
    the tire inflation is cut in half, and vice-versa.

    For example, a 100 lb load on a tire inflated to 50 psi is supposed to
    produce a contact patch of 2 square inches, while the same load on a
    tire inflated to 100 psi is supposed to produce a contact patch of
    only 1 square inch.

    Testing suggests that this theory is mistaken.

    I clamped a frame with a rear wheel and tire in a vise by the bottom
    bracket, leaving the whole frame free to pivot.

    A 2x4 attached to the top tube provided convenient leverage. I just
    hung enough weights from the end of the 2x4 sticking out past the rear
    tire to produce a 100-lb reading on a bathroom scale under the tire.

    After rubbing a worn 700x26 tire with a red ink pad, I lowered the
    tire onto a sheet of paper on some well-supported sheet metal, and
    then let the weight hang, pressing the inked tire down on the paper
    with about 100 lbs of force.

    The tire was locked in place by brake pads held tight with a zip tie.

    Using a floor pump with a dial gauge, I took the tire's fingerprint at
    40-60-80-100-120 psi:

    http://i7.tinypic.com/2z4wuib.jpg

    Even a glance shows that the red tire marks don't shrink at the same
    rate that the tire pressure rises.

    Here's the data:

    3x 2.5x 2x 1.5x 1x

    psi 120 100 080 060 040
    mm length 76 81 88 96 108
    mm width 11 11 11 12.5 13
    ---- ---- ---- ---- ----
    L x W 836 891 968 1200 1404

    1x 1.07x 1.16x 1.44x 1.68x

    Multiplying the longest diagonal of the mark by its widest section
    gives a high approximation of the area of the contact patch (a
    rectangle rather than a vague oval).

    Even if the extremes are discarded, the size of the contact patch
    appears to change significantly less than expected.

    For example, pumping the tire up 20% from 80 psi to 100 psi reduces
    the approximate size of the patch by only 8%.

    The asymmetrical ends of the tire print are likely due to the wheel
    not being perfectly vertical, the rope and weight not hanging
    perfectly in line with the tire, and the tread being worn. (There was
    plenty of ink smeared on the tread--the diagonal lower end of each
    mark just shows where the tire was reluctant to touch the paper.)

    One possible partial explanation is that pressure varies considerably
    in the contact patch, with much lighter pressure toward the edges.

    Another possible partial explanation is that casing tension
    complicates matters.

    Better explanations would be welcome.

    Cheers,

    Carl Fogel

    a) a treaded tyre will not give the same result as a smooth one

    b) can you get & use a planimiter?

    Dear J.,

    I thought about using 1x1 mm graph paper, but it wouldn't really
    improve the accuracy to any useful extent.

    The air pressure, after all, is only eye-balled on the dial gauge of a
    floor pump, and then a little air is lost as the chuck is wrestled off
    the Presta valve.

    Luckily, the trend is so gross and contradicts simple theory so
    obviously that there's no need for greater precision in measurement.

    If contact patch area had a purely linear relationship with air
    pressure, then the 40 psi patch would be three times the size of the
    120 psi patch.

    It's plain that the 40 psi patch isn't even twice as big.

    http://i7.tinypic.com/2z4wuib.jpg

    The 120 psi patch is on the left, the 40 psi is on the right.

    Cheers,

    Carl Fogel

  7. On Sat, 25 Nov 2006 17:09:35 +0000, Kinky Cowboy <[email hidden]>

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

    A common comment on RBT is that the contact patch for the same load on
    a bicycle tire increases and decreases linearly according to the tire
    pressure.

    That is, the contact patch under the same rider will double in size if
    the tire inflation is cut in half, and vice-versa.

    For example, a 100 lb load on a tire inflated to 50 psi is supposed to
    produce a contact patch of 2 square inches, while the same load on a
    tire inflated to 100 psi is supposed to produce a contact patch of
    only 1 square inch.

    Testing suggests that this theory is mistaken.

    I clamped a frame with a rear wheel and tire in a vise by the bottom
    bracket, leaving the whole frame free to pivot.

    A 2x4 attached to the top tube provided convenient leverage. I just
    hung enough weights from the end of the 2x4 sticking out past the rear
    tire to produce a 100-lb reading on a bathroom scale under the tire.

    After rubbing a worn 700x26 tire with a red ink pad, I lowered the
    tire onto a sheet of paper on some well-supported sheet metal, and
    then let the weight hang, pressing the inked tire down on the paper
    with about 100 lbs of force.

    The tire was locked in place by brake pads held tight with a zip tie.

    Using a floor pump with a dial gauge, I took the tire's fingerprint at
    40-60-80-100-120 psi:

    http://i7.tinypic.com/2z4wuib.jpg

    Even a glance shows that the red tire marks don't shrink at the same
    rate that the tire pressure rises.

    Here's the data:

    3x 2.5x 2x 1.5x 1x

    psi 120 100 080 060 040
    mm length 76 81 88 96 108
    mm width 11 11 11 12.5 13
    ---- ---- ---- ---- ----
    L x W 836 891 968 1200 1404

    1x 1.07x 1.16x 1.44x 1.68x

    Multiplying the longest diagonal of the mark by its widest section
    gives a high approximation of the area of the contact patch (a
    rectangle rather than a vague oval).

    Even if the extremes are discarded, the size of the contact patch
    appears to change significantly less than expected.

    For example, pumping the tire up 20% from 80 psi to 100 psi reduces
    the approximate size of the patch by only 8%.

    The asymmetrical ends of the tire print are likely due to the wheel
    not being perfectly vertical, the rope and weight not hanging
    perfectly in line with the tire, and the tread being worn. (There was
    plenty of ink smeared on the tread--the diagonal lower end of each
    mark just shows where the tire was reluctant to touch the paper.)

    One possible partial explanation is that pressure varies considerably
    in the contact patch, with much lighter pressure toward the edges.

    Another possible partial explanation is that casing tension
    complicates matters.

    Better explanations would be welcome.

    Cheers,

    Carl Fogel

    a) a treaded tyre will not give the same result as a smooth one

    b) can you get & use a planimiter?

    I'd suggest that the contact patch measured on the interior surface of
    the tyre follows the theoretical prediction, but translating this
    through a variable thickness of moderately rigid rubber is bound to
    give non-linear results. Try calibrating your experiment first with a
    thin inner tube with no tyre around it; you'll probably have to run
    pressures in the 0.5 - 1.5 bar range, with a corresponding reduction
    in load.

    Kinky Cowboy*

    *Batteries not included
    May contain traces of nuts
    Your milage may vary

    Dear Kinky,

    If you experiment with an inner tube, a pump, and a gauge, I'll be
    astonished if you can produce even 3 psi, 0 0.2 bar:

    http://groups.google.com/group/rec.bicycles.tech/msg/de34b4e303a152d1

    My gauge stubbornly read 0 psi in that post and picture.

    Cheers,

    Carl Fogel

  8. Quoted message said:

    A common comment on RBT is that the contact patch for the same load on
    a bicycle tire increases and decreases linearly according to the tire
    pressure.

    That is, the contact patch under the same rider will double in size if
    the tire inflation is cut in half, and vice-versa.

    For example, a 100 lb load on a tire inflated to 50 psi is supposed to
    produce a contact patch of 2 square inches, while the same load on a
    tire inflated to 100 psi is supposed to produce a contact patch of
    only 1 square inch.

    Testing suggests that this theory is mistaken.

    I clamped a frame with a rear wheel and tire in a vise by the bottom
    bracket, leaving the whole frame free to pivot.

    A 2x4 attached to the top tube provided convenient leverage. I just
    hung enough weights from the end of the 2x4 sticking out past the rear
    tire to produce a 100-lb reading on a bathroom scale under the tire.

    After rubbing a worn 700x26 tire with a red ink pad, I lowered the
    tire onto a sheet of paper on some well-supported sheet metal, and
    then let the weight hang, pressing the inked tire down on the paper
    with about 100 lbs of force.

    The tire was locked in place by brake pads held tight with a zip tie.

    Using a floor pump with a dial gauge, I took the tire's fingerprint at
    40-60-80-100-120 psi:

    http://i7.tinypic.com/2z4wuib.jpg

    Even a glance shows that the red tire marks don't shrink at the same
    rate that the tire pressure rises.

    Here's the data:

    3x 2.5x 2x 1.5x 1x

    psi 120 100 080 060 040
    mm length 76 81 88 96 108
    mm width 11 11 11 12.5 13
    ---- ---- ---- ---- ----
    L x W 836 891 968 1200 1404

    1x 1.07x 1.16x 1.44x 1.68x

    Multiplying the longest diagonal of the mark by its widest section
    gives a high approximation of the area of the contact patch (a
    rectangle rather than a vague oval).

    Even if the extremes are discarded, the size of the contact patch
    appears to change significantly less than expected.

    For example, pumping the tire up 20% from 80 psi to 100 psi reduces
    the approximate size of the patch by only 8%.

    The asymmetrical ends of the tire print are likely due to the wheel
    not being perfectly vertical, the rope and weight not hanging
    perfectly in line with the tire, and the tread being worn. (There was
    plenty of ink smeared on the tread--the diagonal lower end of each
    mark just shows where the tire was reluctant to touch the paper.)

    One possible partial explanation is that pressure varies considerably
    in the contact patch, with much lighter pressure toward the edges.

    Another possible partial explanation is that casing tension
    complicates matters.

    Better explanations would be welcome.

    Cheers,

    Carl Fogel

    Tom Schmitz wrote to point out that he did similar tests with similar
    results. Here's a link to Tom's much more detailed data:

    http://www.tomschmitz.org/Contact%20PatchFrame1Source1.htm

    The obvious point of Tom's graph with area calculated two different
    ways versus tire inflation is that roughly doubling the tire pressure
    from 60 to 130 psi reduced contact patch area from about 1.80 to about
    1.40, a 22% decrease instead of the expected 50% decrease.

    Here's where Tom posted his data:

    http://groups.google.com/group/rec.bicycles.tech/msg/1e8256ee25dbe90d

    Unsurprisingly, Tom was informed that his test results were wrong
    because they disagreed with untested theories.

    Cheers,

    Carl Fogel

  9. On Sat, 25 Nov 2006 00:46:03 -0700, [email hidden] wrote:

    [snip]

    http://i7.tinypic.com/2z4wuib.jpg

    An email criticized my handwriting in the picture above ("badly
    scrawled figures"😉 and then complained that my description of the
    contact-patch-test setup was hard to follow.

    So here's a picture of the setup. A short rod runs through the hollow
    bottom bracket and is clamped in the vise, leaving the bike frame free
    to pivot:

    http://i7.tinypic.com/47ilsaw.jpg

    Holes in the 2x4 let it slip over the head and seat tubes, and a
    U-bolt visible near the head-tube anchors the board.

    The $2 red ink pad (somewhat dirty now) and the white zip tie locking
    the rear brake are visible.

    The inside of an old computer case sitting on a portable workbench
    provides a smooth surface for the paper.

    The four gray 15-lb weights sitting on the upside-down blue plastic
    bin (plus the weight of the frame pivoting at the bottom bracket)
    produce ~100 lbs on a floor scale under the tire when the weights are
    hanging by the rope from the end of the 2x4.

    The yellow color of the floor pump reduces pumping effort and
    increases psi accuracy by at least an order of magnitude.

    CF

  10. Carl Fogel said:
    Quoted message said:
    Quoted message said:

    Multiplying the longest diagonal of the mark by its widest section
    gives a high approximation of the area of the contact patch (a
    rectangle rather than a vague oval).

    Quoted message said:
    Quoted message said:

    That may change the numbers quite a bit and might change the
    proportionality of the change in area related to the pressure-
    e.g., the contact patch may become more "blunt" at the ends with
    lower pressure which would increase the area of the contact patch
    more than would be indicated by a simple height x width
    multiplication.

    Quoted message said:

    Actually, the picture shows that the blunter and more "rectangular",
    and therefore "larger" contact patches are the smaller, higher
    pressure examples, not the lower pressure patches. As pressure
    increases, the lower dagger-like ends of the patches become blunter
    and shorter.

    Quoted message said:

    http://i7.tinypic.com/2z4wuib.jpg

    Quoted message said:

    So the smaller patches fill out a rectangle more than the larger
    patches, which reduces the relative size change even further.

    I don't understand what sort of tire was used for these ink pad tests.
    Why are they parallelograms instead of canoe shaped as my tires make?
    What are the diagonal pale stripes in the printed area? Was the tire
    worn so that it had a flat zone, as worn rear tires usually have?

    Jobst Brandt

  11. Carl Fogel said:

    Tom Schmitz wrote to point out that he did similar tests with
    similar results. Here's a link to Tom's much more detailed data:

    Quoted message said:

    http://www.tomschmitz.org/Contact%20PatchFrame1Source1.htm

    Quoted message said:

    The obvious point of Tom's graph with area calculated two different
    ways versus tire inflation is that roughly doubling the tire
    pressure from 60 to 130 psi reduced contact patch area from about
    1.80 to about 1.40, a 22% decrease instead of the expected 50%
    decrease.

    I think Tom's test makes the results more apparent. These are tires
    with fairly thick tread and therefore, resist conforming to the ideal
    infinitely flexible model of contact patch and inflation pressure. I
    believe if you could get a light weight track tubular or better yet,
    pull the tread strip off a castoff but inflatable tubular, you would
    get closer to the theoretical contact area.

    Jobst Brandt

  12. Quoted message said:
    Carl Fogel said:
    Quoted message said:

    > Multiplying the longest diagonal of the mark by its widest section
    > gives a high approximation of the area of the contact patch (a
    > rectangle rather than a vague oval).

    Quoted message said:
    Quoted message said:

    That may change the numbers quite a bit and might change the
    proportionality of the change in area related to the pressure-
    e.g., the contact patch may become more "blunt" at the ends with
    lower pressure which would increase the area of the contact patch
    more than would be indicated by a simple height x width
    multiplication.

    Quoted message said:

    Actually, the picture shows that the blunter and more "rectangular",
    and therefore "larger" contact patches are the smaller, higher
    pressure examples, not the lower pressure patches. As pressure
    increases, the lower dagger-like ends of the patches become blunter
    and shorter.

    Quoted message said:

    http://i7.tinypic.com/2z4wuib.jpg

    Quoted message said:

    So the smaller patches fill out a rectangle more than the larger
    patches, which reduces the relative size change even further.

    I don't understand what sort of tire was used for these ink pad tests.
    Why are they parallelograms instead of canoe shaped as my tires make?
    What are the diagonal pale stripes in the printed area? Was the tire
    worn so that it had a flat zone, as worn rear tires usually have?

    Jobst Brandt

    Dear Jobbst,

    I used a worn 700x26 tire. Like most worn tires, it's noticeably
    flattened, but if anything that would simplify matters.

    The upper ends that were toward the front of the bike tend to be more
    rounded, possibly due to the locked wheel descending in an arc onto
    the paper.

    The trailing ends have a distinct diagonal shape that I suspect is due
    to the bike frame being rigidly clamped in a not quite perfectly
    square vise-to-workbench-to-cement-floor-to-portable-bench and the
    weights hanging not perfectly square and centered from the 2x4
    extending from the top-tube of the frame.

    Here's the setup:

    http://i7.tinypic.com/47ilsaw.jpg

    A close-fitting rod through the hollow crank is clamped in the vise.
    If things aren't perfectly square, the somewhat flattened tire may
    well be tilting slightly, so that the left rear of the tire is very
    slightly lifted, enough to show up on ink and paper, with a
    corresponding but smaller effect on the upper end, the right front of
    the tire:

    http://i7.tinypic.com/2z4wuib.jpg

    Much of the streakiness and lines visible on the sides is actually
    where the tread turns into the sidewall. A very fine chevron pattern
    marks the edge of the worn-smooth tread and is visible at much higher
    magnifications. The right-hand sides of the marks show more of this
    very fine pattern, confirming that the test rig is tilting very
    slightly to the right.

    An improved setup would probably raise the tire closer to level and
    eliminate some of these artifacts, but it's probably not worth the
    trouble. At 40 psi, theory predicts that the contact patch should be
    about 3 times the size of the contact patch at 120 psi, but it's
    obviously nowhere near that large.

    Again, it's possible that there's a considerable pressure gradient,
    with the raw smear on paper not showing that the edges have only 30
    psi while the center has 100 psi, and that the proportion of low to
    high pressure area changes with increasing inflation.

    Or it could be that there's more than just simple air pressure
    involved.

    Tom Schmitz showed that the contact patch is considerably larger than
    predicted. My crude test suggests the same thing, but I originally
    assumed that my figures were too rough to be significant.

    At ~120 psi with ~100 lbs load, my contact patch smear is about 76 x
    11 mm, roughly 836 mm^2, or 1.30 square inches instead of 0.83, about
    56% larger than expected.

    At ~100 psi with ~100 lbs load, my contact patch smear is about 81 x
    11 mm, roughly 891 mm^2, or 1.38 square inches instead of 1.00, about
    38% larger than expected.

    At ~60 psi with ~100 lbs load, my contact patch smear is about 96 x
    12.5 mm, roughly 1200 mm^2 (the figures just happened to work out with
    spurious precision to an even 1200), or 1.86 square inches instead of
    1.67, still about 11% larger than expected.

    Again, my L x W estimates, pressure measurements, and weight on a
    scale are rough, but they do tend to confirm Tom's observations.

    Cheers,

    Carl Fogel

  13. Carl Fogel said:

    I used a worn 700x26 tire. Like most worn tires, it's noticeably
    flattened, but if anything that would simplify matters.

    Quoted message said:

    The upper ends that were toward the front of the bike tend to be
    more rounded, possibly due to the locked wheel descending in an arc
    onto the paper.

    Quoted message said:

    The trailing ends have a distinct diagonal shape that I suspect is
    due to the bike frame being rigidly clamped in a not quite perfectly
    square vise-to-workbench-to-cement-floor-to-portable-bench and the
    weights hanging not perfectly square and centered from the 2x4
    extending from the top-tube of the frame.

    Quoted message said:

    Here's the setup:

    Quoted message said:

    http://i7.tinypic.com/47ilsaw.jpg

    Quoted message said:

    A close-fitting rod through the hollow crank is clamped in the vise.
    If things aren't perfectly square, the somewhat flattened tire may
    well be tilting slightly, so that the left rear of the tire is very
    slightly lifted, enough to show up on ink and paper, with a
    corresponding but smaller effect on the upper end, the right front
    of the tire:

    Quoted message said:

    http://i7.tinypic.com/2z4wuib.jpg

    Quoted message said:

    Much of the streakiness and lines visible on the sides is actually
    where the tread turns into the sidewall. A very fine chevron
    pattern marks the edge of the worn-smooth tread and is visible at
    much higher magnifications. The right-hand sides of the marks show
    more of this very fine pattern, confirming that the test rig is
    tilting very slightly to the right.

    I think you overlook that tires wear with the crown of the road and
    that the flattened area of the tread has a slant. Just the same, as I
    mentioned in another reply, tread shape and thickness has too much
    influence on contact patch to approach theoretical values.

    Jobst Brandt

  14. Quoted message said:

    Better explanations would be welcome.

    You're past the point of linear returns on this almost before you
    start, due to the stiffness of the casing and the limits on lateral
    expansion of the contact patch. Try it again with a fat, non-knobby
    balloon tire, and you'll get closer to the predicted results for at
    least a portion of the range of pressures. And always remember that
    in nearly any real-world experiment, there will be variables which
    have a bearing on the results but which are not explicitly described
    by the theory under test.
    --
    Typoes are a feature, not a bug.
    Some gardening required to reply via email.
    Words processed in a facility that contains nuts.

  15. Quoted message said:

    The yellow color of the floor pump reduces pumping effort and
    increases psi accuracy by at least an order of magnitude.

    Pfui. Paint it black to maximize heat dissipation and minimize
    thermally-induced dimensional change.
    --
    Typoes are a feature, not a bug.
    Some gardening required to reply via email.
    Words processed in a facility that contains nuts.

  16. <[email hidden]> wrote in message
    news:[email hidden]...

    Quoted message said:

    A common comment on RBT is that the contact patch for the same load on
    a bicycle tire increases and decreases linearly according to the tire
    pressure.

    That is, the contact patch under the same rider will double in size if
    the tire inflation is cut in half, and vice-versa.

    For example, a 100 lb load on a tire inflated to 50 psi is supposed to
    produce a contact patch of 2 square inches, while the same load on a
    tire inflated to 100 psi is supposed to produce a contact patch of
    only 1 square inch.

    Testing suggests that this theory is mistaken.

    Reality includes a bunch of other variables you haven't accounted for.
    For one, what area of contact patch do you get with a zero pound load? I
    suggest you subtract that area from your results. You should do a
    regression of different loads as well as different pressures.

    Other variables include the fact that contact pressure varies within the
    contact patch. In reality it's an integration of pressure over the area
    that must be equal to your 100 lbf.

    Phil H

  17. Quoted message said:

    Carl Fogel writes:

    <snip>

    Quoted message said:
    Quoted message said:

    http://i7.tinypic.com/2z4wuib.jpg

    Quoted message said:

    So the smaller patches fill out a rectangle more than the larger
    patches, which reduces the relative size change even further.

    I don't understand what sort of tire was used for these ink pad tests.
    Why are they parallelograms instead of canoe shaped as my tires make?
    What are the diagonal pale stripes in the printed area? Was the tire
    worn so that it had a flat zone, as worn rear tires usually have?

    Jobst, it would be helpful if you could post a jpeg of your tyres under
    a similar setup/load so we could compare the two.

  18. In article <[email hidden]>,

    Quoted message said:

    On Sat, 25 Nov 2006 10:53:58 -0600, Tim McNamara
    <[email hidden]> wrote:

    [Carl wrote:]

    Quoted message said:
    Quoted message said:

    Multiplying the longest diagonal of the mark by its widest section
    gives a high approximation of the area of the contact patch (a
    rectangle rather than a vague oval).

    That may change the numbers quite a bit and might change the
    proportionality of the change in area related to the pressure- e.g.,
    the contact patch may become more "blunt" at the ends with lower
    pressure which would increase the area of the contact patch more
    than would be indicated by a simple height x width multiplication.

    [snip]

    Dear Tim,

    Actually, the picture shows that the blunter and more "rectangular",
    and therefore "larger" contact patches are the smaller, higher
    pressure examples, not the lower pressure patches. As pressure
    increases, the lower dagger-like ends of the patches become blunter
    and shorter.

    http://i7.tinypic.com/2z4wuib.jpg

    So the smaller patches fill out a rectangle more than the larger
    patches, which reduces the relative size change even further.

    That is, the bottoms of the marks become more and more dagger-like as
    pressure decreases and the marks lengthen. Rearrange the long,
    dagger-like lower end of the longest 40 psi mark to resemble the
    blunter lower end of the 120 psi mark, and you'll get a length and
    width of about 97 x 13 mm, for an even smaller rough (high) estimate
    of 1261 mm^2 instead of 1404 mm^2.

    Yes, I note that. How worn is this tire- does it have a significant
    "flat spot" from wear that most rear tires exhibit? I wonder if that
    makes a difference compared to a basically new tire. It is interesting
    that your contact patch shapes are different from those shown in
    _Bicycling Science_ 3rd Ed., p 218 esp the second example.

  19. In article <[email hidden]>,

    Quoted message said:

    One possible partial explanation is that pressure varies considerably
    in the contact patch, with much lighter pressure toward the edges.

    I think this is important to what you're seeing. It takes no
    appreciable pressure to transfer ink from your tire tread to the paper.
    The curve of the tire outside of the contact patch meets up
    tangentially, and the tire beyond the edge of the patch makes grazing
    contact with the flat surface. This does not have significant contact
    pressure, but the inkblot marks it anyway.

    -Luns

  20. Quoted message said:

    A common comment on RBT is that the contact patch for the same load on
    a bicycle tire increases and decreases linearly according to the tire
    pressure.

    That is, the contact patch under the same rider will double in size if
    the tire inflation is cut in half, and vice-versa.

    For example, a 100 lb load on a tire inflated to 50 psi is supposed to
    produce a contact patch of 2 square inches, while the same load on a
    tire inflated to 100 psi is supposed to produce a contact patch of
    only 1 square inch.

    Testing suggests that this theory is mistaken.

    I clamped a frame with a rear wheel and tire in a vise by the bottom
    bracket, leaving the whole frame free to pivot.

    A 2x4 attached to the top tube provided convenient leverage. I just
    hung enough weights from the end of the 2x4 sticking out past the rear
    tire to produce a 100-lb reading on a bathroom scale under the tire.

    After rubbing a worn 700x26 tire with a red ink pad, I lowered the
    tire onto a sheet of paper on some well-supported sheet metal, and
    then let the weight hang, pressing the inked tire down on the paper
    with about 100 lbs of force.

    The tire was locked in place by brake pads held tight with a zip tie.

    Using a floor pump with a dial gauge, I took the tire's fingerprint at
    40-60-80-100-120 psi:

    http://i7.tinypic.com/2z4wuib.jpg

    Even a glance shows that the red tire marks don't shrink at the same
    rate that the tire pressure rises.

    Here's the data:

    3x 2.5x 2x 1.5x 1x

    psi 120 100 080 060 040
    mm length 76 81 88 96 108
    mm width 11 11 11 12.5 13
    ---- ---- ---- ---- ----
    L x W 836 891 968 1200 1404

    1x 1.07x 1.16x 1.44x 1.68x

    Multiplying the longest diagonal of the mark by its widest section
    gives a high approximation of the area of the contact patch (a
    rectangle rather than a vague oval).

    Even if the extremes are discarded, the size of the contact patch
    appears to change significantly less than expected.

    For example, pumping the tire up 20% from 80 psi to 100 psi reduces
    the approximate size of the patch by only 8%.

    The asymmetrical ends of the tire print are likely due to the wheel
    not being perfectly vertical, the rope and weight not hanging
    perfectly in line with the tire, and the tread being worn. (There was
    plenty of ink smeared on the tread--the diagonal lower end of each
    mark just shows where the tire was reluctant to touch the paper.)

    One possible partial explanation is that pressure varies considerably
    in the contact patch, with much lighter pressure toward the edges.

    Another possible partial explanation is that casing tension
    complicates matters.

    Better explanations would be welcome.

    Cheers,

    Carl Fogel

    Here are some more contact patch tests with a ~100 lb force on another
    tire, same model 700x26, almost new instead of worn, still shows tiny
    pebble/cross-hatch surface too small to be dignified as a tread
    pattern.

    The classic canoe-shape is much clearer, which probably says more
    about how real tires that have been used behave than anything else.

    The sides of the ink marks are a little harder to measure because of
    the faint pebbling.

    I did a series at 40-60-80-100-120 psi, and then another series at
    30-50-70-90-110 psi, partly to give a little blindness to the test and
    partly because I goofed and went to 60 instead of 50 (note the
    correction from 50 to 60 on the paper).

    I left the chuck on the Presta valve and noticed that what was
    supposed to be 40 psi was showing only 38 when I finished (note
    that correction, too).

    Being easily confused, I thought that the second series was showing
    unexpected results when I began multiplying length by width, but I
    persevered and found that I'm just dim-witted.

    Only the last figure for 110 psi seemed wrong, which it was--for some
    reason, I read the scale wrong, using 1 instead of 0 on the scale, so
    that's the last correction.

    Obviously, the widths are so small that they're much less accurate
    than the lengths, but they seemed fairly regular.

    http://i15.tinypic.com/4cdn0xi.jpg

    Here's the handwritten data in more legible form and in psi order:

    estimate
    mm mm mm^2
    psi widest length L x W
    030 11 [1] 122 1342
    038 [2] 11 118 1298
    050 10 110 1100
    060 10 103 1030
    070 9.5 98 931
    080 9 95 855
    090 9 92 828
    100 8.5 90 765
    110 8.5 86 [3] 731
    120 8.5 85 722.5

    [1] At 30 psi, widest is 11, but the ink on the left is very faint.
    And the paper seemed to wrinkle a bit on the side.

    [2] Gauge showed 38 psi afterward, not 40 psi.

    [3] First mismeasured as 90, plainly wrong mark on caliper.

    The results aren't significantly different than the first test.

    At ~40 psi, the mark was ~118 mm long and ~11 mm wide, suggesting a
    high rectangular area estimate of ~1300 mm^2.

    At ~120 psi, the mark shortened to ~85 mm, the width shrank to ~8.5
    mm, suggesting a high rectangular area estimate of around ~723 mm^2.

    This is obviously a much smaller change than predicted by the simple
    purely straight-line theory of contact patch size versus inflation.

    For example, the 60 versus 120 psi estimated areas are 1030 versus 723
    mm^2. The patch shrank about 300 mm^2 instead of the expected 515
    mm^2.

    Again, it could be that the pressure trails off toward the edges, but
    something funny is still going on.

    At ~100 lbs load, the ~100 psi high area estimate is 765 mm^2, or
    1.185 square inches, about 18% larger.

    Possibly the 0 psi edge has a border quickly increasing to 100 psi,
    which would account for the larger area.

    But the effect seems to reverse itself at lower pressures.

    Instead of a larger than expected area, there is a smaller than
    expected area at low pressure.

    At ~50 psi with the same ~100 lb load, the high area estimate of 1100
    mm^2 is 1.705 square inches, about 15% smaller (not 18% larger) than
    the expected 2 square inches.

    Cheers,

    Carl Fogel

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