Re: An Indirect Measure of Tire Rolling Resistance

[Johnny Sunset]

Joe Riel wrote:
> Prompted by the recent discussion of the tire rolling-resistance (RR)
> measurements published by Bicycle Quarterly [1], I wondered whether
> there might be a simple method for estimating the relative RR of
> bicycle road tires, one that avoids the inherent difficulties of an
> experiment in the field and the expense and complexity of a more
> direct measurement using specialized tire-testing equipment,
> specifically a drum tester.
>
> The RR of a bicycle road tire is largely due to hysteresis losses in
> the tread and sidewalls of the tire as they flex while passing through
> the load zone. Instead of rolling a bicycle wheel, suppose that we
> bounce it. The coefficient of restitution (Cr), which can be readily
> measured, depends on the same hysteresis losses in the tire as the
> rolling resistance....

Is the coefficient of restitution constant for different loading rates
of the elastomer and/or casing cords? Dropping the tire onto a hard
surface would seemingly result in a significantly higher rate of
loading than rolling the tire along a surface at 150 to 250 rpm
(typical range for an ISO 622-mm tire).

--
Tom Sherman - Post Free or Die!

 

[Joe Riel]

"Johnny Sunset" <[email protected]> writes:

> Is the coefficient of restitution constant for different loading rates
> of the elastomer and/or casing cords? Dropping the tire onto a hard
> surface would seemingly result in a significantly higher rate of
> loading than rolling the tire along a surface at 150 to 250 rpm
> (typical range for an ISO 622-mm tire).

Good question, but it isn't clear that the rate of loading is higher
with the bounce. If I use my (crude) approximation, the spring
constant of the tire is

(1) K ~ 2*P*pi*sqrt(Rw*Rt).

The natural frequency of the system is then

(2) w0 = sqrt(K/m).

The bounce time (Tb) is half a period, so

(3) Tb = pi/w0

Plugging in values of

P = tire pressure = 100psi
Rw = wheel radius = 12.8in
Rt = tire radius = 0.5in
m = mass of wheel = 2.5lbf/g
g = 32ft/s^2

Assuming I did the arithmetic correctly (always questionable),
I get

(3a) Tb = 6.5 msec

At 20mph = 350in/s, so the rolling transition through a one inch long
load zone occurs in about 3 msec, quicker than the bounce.


--
Joe Riel

 

[Tim McNamara]

In article <[email protected]>, Joe Riel <[email protected]>
wrote:

> At 20mph = 350in/s, so the rolling transition through a one inch long
> load zone occurs in about 3 msec, quicker than the bounce.

The contact patch with a 100 pound load on a 700 x 25 mm tire at 100 psi
would be 1 in^2, of course, but the patch would be at least twice as
long as it is wide- .5 in wide and 2 in long. At least measuring the
flat area on the rear tires of my bike suggests that the contact patch
is about .5 in wide. So the transition would be probably about 6 msec.

 

[Joe Riel]

Tim McNamara <[email protected]> writes:

> In article <[email protected]>, Joe Riel <[email protected]>
> wrote:
>
>> At 20mph = 350in/s, so the rolling transition through a one inch long
>> load zone occurs in about 3 msec, quicker than the bounce.
>
> The contact patch with a 100 pound load on a 700 x 25 mm tire at 100 psi
> would be 1 in^2, of course, but the patch would be at least twice as
> long as it is wide- .5 in wide and 2 in long. At least measuring the
> flat area on the rear tires of my bike suggests that the contact patch
> is about .5 in wide. So the transition would be probably about 6 msec.

Agreed, I knew the 1in was too short, but close enough to show that
the rates for bouncing (unloaded) and rolling are reasonably close.

--
Joe Riel

 

[Phil Holman]

"Joe Riel" <[email protected]> wrote in message
news:[email protected]...
> Tim McNamara <[email protected]> writes:
>
>> In article <[email protected]>, Joe Riel
>> <[email protected]>
>> wrote:
>>
>>> At 20mph = 350in/s, so the rolling transition through a one inch
>>> long
>>> load zone occurs in about 3 msec, quicker than the bounce.
>>
>> The contact patch with a 100 pound load on a 700 x 25 mm tire at 100
>> psi
>> would be 1 in^2, of course, but the patch would be at least twice as
>> long as it is wide- .5 in wide and 2 in long. At least measuring the
>> flat area on the rear tires of my bike suggests that the contact
>> patch
>> is about .5 in wide. So the transition would be probably about 6
>> msec.
>
> Agreed, I knew the 1in was too short, but close enough to show that
> the rates for bouncing (unloaded) and rolling are reasonably close.
>
Joe, I've been thinking how the drop test could be revised to eliminate
the additional losses associated with high impulse loads on the whole
wheel/tire. Maybe a weight should be dropped onto a stationary
wheel/tire (swinging arm) and its rebound measured. I don't know if this
has already been suggested.

Phil H

 

[Joe Riel]

"Phil Holman" <piholmanc@yourservice> writes:

> Joe, I've been thinking how the drop test could be revised to eliminate
> the additional losses associated with high impulse loads on the whole
> wheel/tire. Maybe a weight should be dropped onto a stationary
> wheel/tire (swinging arm) and its rebound measured. I don't know if this
> has already been suggested.

Interesting idea. The trick might be to adequately support the back
of the rim. Not sure if this is easier than my current setup. I was
hoping to do a series of tests tomorrow, but just "popped" my sternum
doing some dips. Fortunately, it doesn't hurt much when riding, but
does so the rest of the time. I'll probably delay the testing for a
while.


--
Joe Riel