# Rollers math fun?

Discussion in 'Cycling Equipment' started by joseph.santaniello@gmail.com, Dec 1, 2005.

1. ### joseph.santaniello@gmail.com Guest

Hi Everyone,

Last week our cycling club had a spinning session at the local health
club. Dag-Otto Lauritzen was the leader and Steffen Kjærgaard was
there too. (Both ex pros, 7-Eleven, Motorola, US Postal) They showed
Tour de France video on the big screen, and it was big fun. At one
point I made a feeble joke about managing not to get dropped, and later
the conversation turned to hooking up odometers or similar to the
spinning bikes. I suggested a watt or work meter. Late while a few of
us got together to ride rollers, we discussed the effort involved in
using rollers.

So I started thinking about calculating the watts required to acheive a
given speed as measured by the computer. This might be a fun problem
for the math whizzes over on rec.bicycles.tech, I thought!

So going with somewhat normal wheels like Shimano WH-550 or Campy
Vento, how would be estimate the watts need to spin them in still air?
And how would we use rider weight to estimate rolling resistance for 3
contact points? And losses from the drive train? Assume 53x13.

Any takers?

Joseph

#1
Tags:

2. ### Phil, Squid-in-Training Guest

> So I started thinking about calculating the watts required to acheive
> a given speed as measured by the computer. This might be a fun problem
> for the math whizzes over on rec.bicycles.tech, I thought!

I'm not a math whiz, but I'll try.

> So going with somewhat normal wheels like Shimano WH-550 or Campy
> Vento, how would be estimate the watts need to spin them in still air?
> And how would we use rider weight to estimate rolling resistance for 3
> contact points? And losses from the drive train? Assume 53x13.

--
Phil, Squid-in-Training

#2
3. ### Kinky Cowboy Guest

On Thu, 1 Dec 2005 16:58:59 -0500, "Phil, Squid-in-Training"
<phil_leeIHEARTBASHGUARDS@hotmail.com> wrote:

>> So I started thinking about calculating the watts required to acheive
>> a given speed as measured by the computer. This might be a fun problem
>> for the math whizzes over on rec.bicycles.tech, I thought!

>
>I'm not a math whiz, but I'll try.
>
>> So going with somewhat normal wheels like Shimano WH-550 or Campy
>> Vento, how would be estimate the watts need to spin them in still air?
>> And how would we use rider weight to estimate rolling resistance for 3
>> contact points? And losses from the drive train? Assume 53x13.

>

Here's a slightly more sensible guess: 250W at 90rpm on 56x12 fixed
with Mavic Ellipse wheels, Conti GP Supersonics, Tacx 120mm roller
drums, 90kg rider

Kinky Cowboy*

*Batteries not included
May contain traces of nuts

#3
4. ### Skippy Guest

<joseph.santaniello@gmail.com> wrote in message
Hi Everyone,

>Last week our cycling club had a spinning session at the local health
>club. Dag-Otto Lauritzen was the leader and Steffen Kjærgaard was
>there too. (Both ex pros, 7-Eleven, Motorola, US Postal) They showed
>Tour de France video on the big screen, and it was big fun. At one
>point I made a feeble joke about managing not to get dropped, and later
>the conversation turned to hooking up odometers or similar to the
>spinning bikes. I suggested a watt or work meter. Late while a few of
>us got together to ride rollers, we discussed the effort involved in
>using rollers.

>So I started thinking about calculating the watts required to acheive a
>given speed as measured by the computer. This might be a fun problem
>for the math whizzes over on rec.bicycles.tech, I thought!

>So going with somewhat normal wheels like Shimano WH-550 or Campy
>Vento, how would be estimate the watts need to spin them in still air?
>And how would we use rider weight to estimate rolling resistance for 3
>contact points? And losses from the drive train? Assume 53x13.

>Any takers?

How about getting someone with a Powertap or SRM to ride at a few speeds and
recording it. I'm guessing that this might be the case if you get to ride
with ex pros. That will give you a curve of power against speed. Then it
should be reasonably easy to derive a formula (quadratic?) that is a close
match. Whatever equipment was being used (mass, tyres + pressure, wheels)
gives you a reference to match all the bikes against.

From memory, friction is always a function of the applied force, and not
area. The main force resulting in tyre friction here is gravity. It's
force is proportional to mass. So, on the rollers a rider+bike with twice
the mass of another requires double the force to produce the same speed. As
a rule-of-thumb, this would make adjusting the formula for different riders
easy.

I'd be curious to see how accurate this approach would be compared to
something derived from first principles.

Skippy

Joseph

#4
5. ### joseph.santaniello@gmail.com Guest

Kinky Cowboy wrote:
> On Thu, 1 Dec 2005 16:58:59 -0500, "Phil, Squid-in-Training"
> <phil_leeIHEARTBASHGUARDS@hotmail.com> wrote:
>
> >> So I started thinking about calculating the watts required to acheive
> >> a given speed as measured by the computer. This might be a fun problem
> >> for the math whizzes over on rec.bicycles.tech, I thought!

> >
> >I'm not a math whiz, but I'll try.
> >
> >> So going with somewhat normal wheels like Shimano WH-550 or Campy
> >> Vento, how would be estimate the watts need to spin them in still air?
> >> And how would we use rider weight to estimate rolling resistance for 3
> >> contact points? And losses from the drive train? Assume 53x13.

> >
> >Forty-two watts. Sounds about right.

>
> Here's a slightly more sensible guess: 250W at 90rpm on 56x12 fixed
> with Mavic Ellipse wheels, Conti GP Supersonics, Tacx 120mm roller
> drums, 90kg rider

My math teachers always used to say: "No credit, unless you SHOW YOUR
WORK!" ;-)

Jospeh

#5
6. ### joseph.santaniello@gmail.com Guest

Skippy wrote:
> <joseph.santaniello@gmail.com> wrote in message
> Hi Everyone,
>
> >Last week our cycling club had a spinning session at the local health
> >club. Dag-Otto Lauritzen was the leader and Steffen Kjærgaard was
> >there too. (Both ex pros, 7-Eleven, Motorola, US Postal) They showed
> >Tour de France video on the big screen, and it was big fun. At one
> >point I made a feeble joke about managing not to get dropped, and later
> >the conversation turned to hooking up odometers or similar to the
> >spinning bikes. I suggested a watt or work meter. Late while a few of
> >us got together to ride rollers, we discussed the effort involved in
> >using rollers.

>
> >So I started thinking about calculating the watts required to acheive a
> >given speed as measured by the computer. This might be a fun problem
> >for the math whizzes over on rec.bicycles.tech, I thought!

>
> >So going with somewhat normal wheels like Shimano WH-550 or Campy
> >Vento, how would be estimate the watts need to spin them in still air?
> >And how would we use rider weight to estimate rolling resistance for 3
> >contact points? And losses from the drive train? Assume 53x13.

>
> >Any takers?

>
> How about getting someone with a Powertap or SRM to ride at a few speeds and
> recording it. I'm guessing that this might be the case if you get to ride
> with ex pros. That will give you a curve of power against speed. Then it
> should be reasonably easy to derive a formula (quadratic?) that is a close
> match. Whatever equipment was being used (mass, tyres + pressure, wheels)
> gives you a reference to match all the bikes against.
>
> From memory, friction is always a function of the applied force, and not
> area. The main force resulting in tyre friction here is gravity. It's
> force is proportional to mass. So, on the rollers a rider+bike with twice
> the mass of another requires double the force to produce the same speed. As
> a rule-of-thumb, this would make adjusting the formula for different riders
> easy.

I don't know anyone with a power meter. My own fixation is making me
contemplate getting one for myself, but I can't justify the expense.
Plus experimenting first takes all the fun out of the exercise. (The
mathematical exercise, that is)

> I'd be curious to see how accurate this approach would be compared to
> something derived from first principles.

That's what I am interested in. I think someone here can figure this
out.

Joseph

#6
7. ### Ron Ruff Guest

joseph.santaniello@gmail.com wrote:
> Skippy wrote:
> > I'd be curious to see how accurate this approach would be compared to
> > something derived from first principles.

>
> That's what I am interested in. I think someone here can figure this
> out.

I doubt it.

Both the aerodynamic and the rolling resistances, and the losses in the
drivetrain could only be guesstimated from "first principals". It might
be possible to use previous test measurements of these components and
adjust them to your specific situation, but the results would not be
precise. In particular, since rolling resistance should dominate, you'd
need to know that with good accuracy... and it will vary a lot
depending on what tires you use, their condition, their temperature,
etc.

#7
8. ### Kinky Cowboy Guest

On 1 Dec 2005 15:24:00 -0800, joseph.santaniello@gmail.com wrote:

>
>Kinky Cowboy wrote:
>> On Thu, 1 Dec 2005 16:58:59 -0500, "Phil, Squid-in-Training"
>> <phil_leeIHEARTBASHGUARDS@hotmail.com> wrote:
>>
>> >> So I started thinking about calculating the watts required to acheive
>> >> a given speed as measured by the computer. This might be a fun problem
>> >> for the math whizzes over on rec.bicycles.tech, I thought!
>> >
>> >I'm not a math whiz, but I'll try.
>> >
>> >> So going with somewhat normal wheels like Shimano WH-550 or Campy
>> >> Vento, how would be estimate the watts need to spin them in still air?
>> >> And how would we use rider weight to estimate rolling resistance for 3
>> >> contact points? And losses from the drive train? Assume 53x13.
>> >
>> >Forty-two watts. Sounds about right.

>>
>> Here's a slightly more sensible guess: 250W at 90rpm on 56x12 fixed
>> with Mavic Ellipse wheels, Conti GP Supersonics, Tacx 120mm roller
>> drums, 90kg rider

>
>My math teachers always used to say: "No credit, unless you SHOW YOUR
>WORK!" ;-)

Here's my work; that's how fast I go on the rollers for the same HR as
riding the same set up on the road except for 49x14 gearing. this site
http://sportech.online.fr/spen_esy.html
gives 259W at 38km/h for my set-up

Kinky Cowboy*

*Batteries not included
May contain traces of nuts

#8
9. ### dianne_1234@yahoo.com Guest

joseph.santaniello@gmail.com wrote:
> Hi Everyone,
>
> Last week our cycling club had a spinning session at the local health
> club. Dag-Otto Lauritzen was the leader and Steffen Kjærgaard was
> there too. (Both ex pros, 7-Eleven, Motorola, US Postal) They showed
> Tour de France video on the big screen, and it was big fun. At one
> point I made a feeble joke about managing not to get dropped, and later
> the conversation turned to hooking up odometers or similar to the
> spinning bikes. I suggested a watt or work meter. Late while a few of
> us got together to ride rollers, we discussed the effort involved in
> using rollers.
>
> So I started thinking about calculating the watts required to acheive a
> given speed as measured by the computer. This might be a fun problem
> for the math whizzes over on rec.bicycles.tech, I thought!
>
> So going with somewhat normal wheels like Shimano WH-550 or Campy
> Vento, how would be estimate the watts need to spin them in still air?
> And how would we use rider weight to estimate rolling resistance for 3
> contact points? And losses from the drive train? Assume 53x13.
>
> Any takers?
>
> Joseph

Have a look here, and scroll down:
http://www.geocities.com/almost_fast/trainerpower/

#9
10. ### Phil, Squid-in-Training Guest

> So going with somewhat normal wheels like Shimano WH-550 or Campy
> Vento, how would be estimate the watts need to spin them in still air?

I'm curious. What would you consider not "somewhat normal" wheels? Discs?

--
Phil, Squid-in-Training

#10
11. ### joseph.santaniello@gmail.com Guest

Phil, Squid-in-Training wrote:
> > So going with somewhat normal wheels like Shimano WH-550 or Campy
> > Vento, how would be estimate the watts need to spin them in still air?

>
> I'm curious. What would you consider not "somewhat normal" wheels? Discs?

I was thinking along the lines of 20-28 spokes, slightly "aero" rim
cross-section. Something that approximates a standard set of wheels in
terms of performance. Not deep, no bladed spokes.

Joseph

#11
12. ### joseph.santaniello@gmail.com Guest

Ron Ruff wrote:
> joseph.santaniello@gmail.com wrote:
> > Skippy wrote:
> > > I'd be curious to see how accurate this approach would be compared to
> > > something derived from first principles.

> >
> > That's what I am interested in. I think someone here can figure this
> > out.

>
> I doubt it.
>
> Both the aerodynamic and the rolling resistances, and the losses in the
> drivetrain could only be guesstimated from "first principals". It might
> be possible to use previous test measurements of these components and
> adjust them to your specific situation, but the results would not be
> precise. In particular, since rolling resistance should dominate, you'd
> need to know that with good accuracy... and it will vary a lot
> depending on what tires you use, their condition, their temperature,
> etc.

Perhaps it cannot be figured out completely, but I'll bet someone can
figure out pretty close. At most needing only one or two big guess
factors that can be measured with something like a timed spin-down on
the rollers or similar.

Joseph

#12
13. ### Dan Connelly Guest

dianne_1234@yahoo.com wrote:
> Have a look here, and scroll down:
> http://www.geocities.com/almost_fast/trainerpower/
>

This confirms my appraisal of the situation....

A Crr of 100psi tires is approx 0.4% (Wilson, Bicycling Science 3rd edition
2004, Fig 6.9, pg 227). Riding at 11 m/sec (40 kph) with a total load
of 80 kg (weight 780N) yields a load of 34 W/. This is around 11% of the
total load, given a typical time trial power of 300 W.

Wind resistance from wheels at 27mph (10% higher than this assumed speed) are
shown:
http://www.diablocyclists.com/aerowheels.htm

A typical pair of aero wheels has a wind resistance power of 25 to 40 watts.
Wind resistance power is roughly proportional to speed^3, so at 40kph (25mph)
this is 20-32 watts. However, this both includes the rotating resistance of
the spokes, and the translational resistance. Translational resistance is
eliminated with rollers. Clearly the translational resistance is the greater
force, as spokes contribute to both, while the rims contribute just to translational
resistance.

Suppose, for sake of argument, the wheel resistance is reduced by 2/3 when translational
resistance is eliminated (this obviously depends on spoke count, spoke shape, rim profile,
etc). Then wheel resistance is only 8 to 13 watts. Rolling resistance remains 34 watts.

Rollers, of course, increase rolling resistance. Not only do they increase the
tire deformation, but there is the additional load of the roller bearings.
Including these factors, the wind resistance of the spokes becomes even a smaller
fraction of the total.

If the rotational power of the wheels is 8 to 13 watts, increasing the power by 9x
(to 70 to 120 watts) would require a 3x increase in speed -- 120 kph. Clearly such
speed is not going to be achieved on typical rollers, with typical gears. So I
conclude wind resistance is a small factor in rolling riding -- it's mostly rolling
resistance.

Thus a linear power-versus-speed relationship, roughly proportional to mass load, is expected.

Dan

#13
14. ### Dan Connelly Guest

joseph.santaniello@gmail.com wrote:

>
> But given the same tire, with more or less the same pressure, you think
> rider weight increases the power demands in a linear fashion? By what
> factor?
>
> The guy I do roller riding with weighs 40kg less than I do, and he uses
> his 53x12 the whole time, while I struggle with a much lighter gear,
> like 53x19 or 17. On the road (flat of course) he is only marginally
> faster than me.

Yes. Rolling resistance is typically load-proportional.

For coast-down tests, you need to relationship of power versus speed.
See my last post (responding to the Tom Compton's data) -- the power is
proportional to speed. In other words, retarding torque is independent
of speed: speed = speed(t=0) - a t, for some decelleration a, where t is
time.

But the big problem is to get power from the speed. Coastdown on the road
is relatively simple -- it's dominated by rider weight + bike weight.
Neither is decelerating in the roller test -- the load is far more complicated.
You need to measure it somehow.

Dan

#14
15. ### Ron Ruff Guest

joseph.santaniello@gmail.com wrote:
>
> But given the same tire, with more or less the same pressure, you think
> rider weight increases the power demands in a linear fashion? By what
> factor?
>

Because of the small rollers, the rolling resistance dominates... the
only other thing to cause resistance is the aero drag of the wheels
spinning, but that is relatively small... especially at lower speeds.

I'm not sure what you mean by "what factor?". If the weight (mass)
doubles, then the power requirement doubles... if the speed doubles,
then the power requirement doubles... and it passes through 0,0 of
course.

P=Crr*M*V

But... we don't know what Crr is...

> The guy I do roller riding with weighs 40kg less than I do, and he uses
> his 53x12 the whole time, while I struggle with a much lighter gear,
> like 53x19 or 17. On the road (flat of course) he is only marginally
> faster than me.

That makes sense. On a flat road aero resistance dominates, and rolling
resistance (and weight) is small. On the rollers, this is reversed...
power to weight ratio is the important factor for determining speed...
kinda like climbing a steep hill.

#15
16. ### Chris Zacho The Wheelman Guest

Where's Sheldon when yo really need him? §:-3)>

- -

Chris Zacho ~ "Your Friendly Neighborhood Wheelman"

"May you have the winds at your back,
And a really low gear for the hills!"

Chris'Z Corner
http://www.geocities.com/czcorner

#16
17. ### joseph.santaniello@gmail.com Guest

Ron Ruff wrote:
> joseph.santaniello@gmail.com wrote:
> >
> > But given the same tire, with more or less the same pressure, you think
> > rider weight increases the power demands in a linear fashion? By what
> > factor?
> >

> Because of the small rollers, the rolling resistance dominates... the
> only other thing to cause resistance is the aero drag of the wheels
> spinning, but that is relatively small... especially at lower speeds.
>
> I'm not sure what you mean by "what factor?". If the weight (mass)
> doubles, then the power requirement doubles... if the speed doubles,
> then the power requirement doubles... and it passes through 0,0 of
> course.
>
> P=Crr*M*V
>
> But... we don't know what Crr is...
>
> > The guy I do roller riding with weighs 40kg less than I do, and he uses
> > his 53x12 the whole time, while I struggle with a much lighter gear,
> > like 53x19 or 17. On the road (flat of course) he is only marginally
> > faster than me.

>
> That makes sense. On a flat road aero resistance dominates, and rolling
> resistance (and weight) is small. On the rollers, this is reversed...
> power to weight ratio is the important factor for determining speed...
> kinda like climbing a steep hill.

On the road, one must overcome a small percentage RR and the remainder
in wind resistance. On rollers one must more or less only overcome RR,
as wind resistance of the wheels is pretty small. So why is RR so
drastically increased on rollers? I can't see it being the bearings in
the rollers. And the contact patch is obvioulsy the same size, but most
likely a different shape. So what is it that makes RR on rollers so
much greater than on the road?

Joseph

#17
18. ### Ron Ruff Guest

joseph.santaniello@gmail.com wrote:

> as wind resistance of the wheels is pretty small. So why is RR so
> drastically increased on rollers? I can't see it being the bearings in
> the rollers. And the contact patch is obvioulsy the same size, but most
> likely a different shape. So what is it that makes RR on rollers so
> much greater than on the road?
>

The contact patch is very different because the rollers are so small,
sort of putting a "point" load on the tire instead of a long thin
patch.

According to "Bicycling Science", the equivalent radius is:

1/((1/r1)+(1/r2))

So, for a 27inch wheel and a 3inch roller, the effective diameter is
2.7inches.

Since the rolling resistance is roughly proportional to 1/d^0.7, the
difference between your 27in wheel on a smooth road vs the rollers is:

27^.7/2.7^.7= 5.0... so you'd expect the rolling resistance to be about
5 times higher on a 3in roller.

The back tire has 2 rollers, though they are also spaced apart which
would increase the pressure on the tire a bit. Overall the ratio is
closer to 4.

So, if your Crr is .006 on a smooth road, it will be about .024 on the
rollers... or the equivalent of climbing a 1.8% grade (.024-.006) on

#18
19. ### Kinky Cowboy Guest

On 3 Dec 2005 11:53:52 -0800, joseph.santaniello@gmail.com wrote:

>On the road, one must overcome a small percentage RR and the remainder
>in wind resistance. On rollers one must more or less only overcome RR,
>as wind resistance of the wheels is pretty small. So why is RR so
>drastically increased on rollers? I can't see it being the bearings in
>the rollers. And the contact patch is obvioulsy the same size, but most
>likely a different shape. So what is it that makes RR on rollers so
>much greater than on the road?
>

The tyre has to bend round quite a small radius roller three times, so
there is much more deformation and therefore hysteresis and
interstitial (between carcass and tube) friction loss. The front
reaction loads are not in line with the imposed weight load. I doubt
that there is a huge loss in either the roller bearings or the drive
belt; PVC has quite a low softening point, and much more than about
10W per bearing would probably be problematic.

Kinky Cowboy*

*Batteries not included
May contain traces of nuts

#19
20. ### Ron Ruff Guest

Dan Connelly wrote:
>
> Suppose, for sake of argument, the wheel resistance is reduced by 2/3 when translational
> resistance is eliminated (this obviously depends on spoke count, spoke shape, rim profile,
> etc). Then wheel resistance is only 8 to 13 watts. Rolling resistance remains 34 watts.
>

When a wheel is being ridden, the power needed to spin it is much
higher, because some of the spokes are seeing a wind that is double
forward velocity... which results in a factor of 8 in power on these
spokes. Of course some are seeing essentially zero wind as well.
Eliminating both the translational drag and the extra power to spin on
the road would likely drop the drag even more than 2/3.

On rollers the rolling resistance goes up a lot, as I showed earlier.
Using your values for weight and speed, with a Crr of .024, the power
would be:

..024*780N*11m/s= 206W

Even if the aero resistance is 10W it is only 5% of the total at this
speed.

#20