I've been doing a little bit of tire testing lately. But, before I reveal any results, I thought it would be good to go over some math. I know, I know...(I can hear the groans already), but I think it's important to review so people understand why it's reasonable to equate power to move a tire on a roller to power on flat ground.

It's long been known that bicycle rollers act as a sort of rolling resistance "amplifier". In other words, the differences in the rolling resistance between tires is magnified when riding on the rollers. It's usually a fairly subtle thing to try to "feel" the difference in rolling resistance in tires when riding outside, but it's pretty easy to tell the fast tires from the slow tires on rollers just by the exaggerated effort it takes. But, the question has long been "how much" of an amplifier are they? Well, back in 2006 I was discussing this with a few folks and realized that the equations to make that comparison between rollers and a flat surface Crr (Coefficient of rolling resistance) were already available...they just needed to be combined. Then, it was pointed out to me that the particular geometry of a typical roller setup needed to be accounted for as well. The normal "dual roller" setup on the rear of a roller set results in a geometric effect that actually increases the normal force on each roller. In other words, you can't just take the rear wheel load as if it was a single roller. So, I added that to the equations as well.

Anyway, what you see below is the short "paper" I sketched up back then on the subject:

**Flat Surface RR from Roller Testing – Tom Anhalt – 5/2/06**

The power required to turn a wheel on a
drum at a specific speed is governed by the equation:

P

_{Drum }= Crr_{Drum}x V_{Drum}x M x g (a)
Where,

P

_{Drum }= Power required to turn drum (Watts)
Crr

_{Drum}= Coefficient of Rolling Resistance of the tire on the drum (unitless)
V

_{Drum}= The tangential velocity of the drum (m/s)
M = The mass load of the wheel on the
drum (kg)

g = gravitational constant = 9.81 m/s

^{2}
Rearranging equation (a) to solve for
the Crr of the tire on the drum results in:

Crr

_{Drum}= P_{Drum}/ (V_{Drum}x M x g) (b)
Then the contact patch deformation of a
tire of a specific diameter and a roller of a specific diameter can
be equated to the deformation of an equivalent diameter tire on a
flat surface using the following equation [Bicycling Science, 3

^{rd}edition, pg 211]:
1/r

_{eq}= 1/r_{1}+ 1/r_{2}(c)
Where,

r

_{eq}= equivalent wheel radius
r

_{1}= tested wheel radius
r

_{2}= tested drum radius
For convenience purposes, this equation
can be rewritten using the appropriate diameters (r x 2) and is then:

1/D

_{eq}= 1/D_{wheel}+ 1/D_{Drum}(d)
For a tire of a given construction, it
has been shown that the Crr varies inversely proportionally to the
wheel radius, and thus the wheel diameter, in the range of D

_{wheel}^{0.66 }to D_{wheel}^{0.75}[Bicycling Science, 3^{rd}edition, pg. 226]. To simplify for this purpose, the assumption is made that the Crr varies inversely proportionally to D_{wheel}^{0.7}
From this, it can be then written that:

Crr

_{flat}/ Crr_{Drum}= D_{eq}^{0.7}/ D_{wheel}^{0.7}(e)
Equation (e) can be combined with (d)
and rearranged to give:

Crr

_{flat}= Crr_{Drum}x [ 1 / (1 + D_{wheel}/D_{Drum})]^{0.7}(f)
Substituting equation (b) for Crr

_{Drum}in equation (f) results in:**Crr**

_{flat}

**= [P**

_{Drum}

**/ (V**

_{Drum}

**x M x g)] x [ 1 / (1 + D**

_{wheel}

**/D**

_{Drum}

**)]**

^{0.7}

**(g)**

**Mass Correction Factor:**

When doing Crr testing on rollers, the
mass loading of the wheel or wheels will need to be corrected due to
front-rear loading ratio and the fact that 2 offset rollers contact
the rear wheel, thereby increasing the normal force on the rollers
due to geometry effects.

__Rear Wheel Only Case__- When the test is done using a front fork mount and only the rear wheel contacting the rear rollers of the test setup, the following “effective mass” (M

_{eff}) needs to be calculated and substituted for M in equation (g) :

M

_{eff}= M_{rear }/ cos [arcsin (X/(D_{wheel}+ D_{Drum}))] (h)
Where:

X = separation distance of rear roller
axles (consistent units with D

_{wheel}and D_{Drum})
M

_{rear}= vertical mass load on rear wheel (kg)__Front and Rear Rollers__- When the test is performed using both the front and rear rollers, the following Meff needs to be calculated and substituted for M in equation (g) :

M

_{eff}= M_{front}+ M_{rear}/ cos [arcsin (X/(D_{wheel}+ D_{Drum}))] (i)
Where:

M

_{front}= vertical mass load on the front wheel (kg)**Power Correction:**

Depending on the method of power
measurement, the following offsets can be used to account for
drivetrain and drum rotation losses in the calculation of P

_{Drum}for use in equation (g):__For Powertap__- P

_{Drum}= P

_{Powertap}– 5W (accounts for drum bearing losses) (j)

__For SRM__- P

_{Drum}= P

_{SRM}– 15W (accounts for drum bearings and driveline losses) (k)

Where:

P

_{Powertap}and P_{SRM}are the power readouts (W) from the appropriate power meters.
These power offsets are somewhat
arbitrary and should be modified if better data is known about the
particular test setup.

That's basically it. I'd like to note a couple things about the last section on "Power Correction". First, after doing a bunch of testing since then, I don't bother accounting for the drum bearing losses. Also, when using a crank-based power meter, like an SRM or a Quarq Cinqo, I've found (after using a PT in conjunction for quite a few tests) that it makes more sense to account for the drivetrain friction with just a straight percentage. A typical value taken for drivetrain losses on bicycles is ~2.5%, but that is typical at higher power levels (i.e. higher chain tensions). For the lower power levels I usually see in tire testing with a rear only roller setup (usually ~100W or less, with the better tires closer to ~50W) I typically see ~5% drivetrain losses, so that's the figure I use.

It's important to remember that the point of this is to get a "ballpark" feel for the difference between tires, not necessarily an absolutely accurate value. It's been shown that percent difference in power requirements on the rollers equate very well to percent differences on the road. What we're really looking for is a sort of "scaling factor" to put the differences seen on the rollers in perspective as to what to expect on the road.

There you go...equation (g) is easily written into a spreadsheet. After that, it just takes a few measurements of the roller setup, weighing the rear wheel load, and some time on the rollers with a power meter equipped bike and nearly anyone can "test tires".

It's going to be interesting to see your results.

ReplyDeleteI've already done some comparative testing of my training wheels versus competition wheels.

(Mavic Ksyrium Elite with Michelin Pro Race 3 Service Course and Michelin latex tubes versus Enve 45 with Vittoria Corsa Evo CX II tubulars).

There's quite a clear difference in rolling resistance between these two wheelsets at the same tire pressure and the pressure limit on the Michelins actually make for a bigger difference!

Be careful with assuming that because you can pump tires up to >120psi on the rollers and they then will show lower rolling resistance on the road at those same pressures. A roller drum is effectively a "perfectly smooth" surface. A road is decidedly not. I've got data (probably material for another blog post) which suggests that at tire pressures above 120 psi or so, the "resistance to forward motion" actually goes up dramatically. This seems to correspond to pumping up the tires hard enough that they stop effectively acting as a suspension and start transmitting energy into the rider. When the tire flexes, most of the road surface energy can be "returned" in the trailing half of the contact patch. When it doesn't flex, that energy gets transmitted through the bike structure and into the rider's "squishy bits".

ReplyDeleteCorrect me if I'm wrong but shouldn't there be a factor of 1/2 the separation distance of the rollers in the effective mass calculations? I'm sitting here drawing out the geometry and not really sure how it would work out otherwise. Well sorry to bother you. Just a note you and Jack were right about water having higher rolling resistance. Its come out an average of about 10watts higher on the rollers.

ReplyDeleteHi Jason,

DeleteNo, because as you'll see in the equation I'm using the diameters of the rollers instead of the radii, so when setting up the vector triangle to calculate the angles, all the 1/2 terms cancel out. Also, make sure that when you apply the geometry to the loading calculations, that you do a free-body diagram and evaluate them at the roller centers...and then make sure to double the result. You'll see how it all turns out ;-)

I'm glad you were able to finally get a measure of your setup. Yeah, just intuitively it didn't make sense that filling the tires (even if just partially) with an incompressible, dense fluid would be a good thing...now we know!

Yeah that makes sense I knew there was something I was overlooking. I was using the radius. Just a note the 10watts delta was for 40kph. It still might be hard to detect on the road but I'm going to run 2 sets of trial runs a week for the month of March to get enough data for a good statistical analysis. I plan on a low wattage so rolling resistance is a large player in the drag. A fun piece of thermal physics came out of the roller tests as well. Where air took around 3min to reach its asymptotic power the water took around 8 minutes.

Delete