# Thread: The placebo effect of light wheels (tech)

1. ## The placebo effect of light wheels (tech)

While I was browsing the web looking for some hindsight about “Aero vs Light wheelset debate” I came across this very interesting tech article about “real world” wheel theory.
http://www.biketechreview.com/archive/wheel_theory.htm

Even if the author focused on road bike racing I think the article could be interesting for anyone to read.
Note: I’ve often been told that MTB trails require more acceleration than road riding, but seeing the values of accelaration involved (0,1 to 0,2g on a criterium race) I’m thinking that we are splitting hairs – or looking for an excuse to burn 1200\$ for the new hyper light wheelset of the moment…

Quote:
Roughly, the average rider power requirements on a course with a zero net elevation gain is broken down into 60% rider drag, 8% wheel drag, 8% frame drag, 12% rolling resistance .5% wheel inertia forces and 8% bike/rider inertia. The uphill TT example given is a special case where the rider aerodynamics and the bike/rider weight have nearly equal contributions to power – somewhere around 35% each with wheel mass contributing around 1%. The steeper the hill, the more important mass becomes and the less important aerodynamics becomes.
(…)
How can it be that wheel inertial forces are nearly insignificant, when the advertisements say that inertia is so important? Quite simply, inertial forces are a function of acceleration. In bike racing this peak acceleration is about .1 to .2 g’s and is generally only seen when beginning from an initial velocity of 0 (see criterium race data in Appendix D ). Furthermore, the 0.3kg/0.66lb difference in wheels, even if this mass is out at the rim, is so small compared to your body mass that the differences in wheel inertia will be unperceivable. Any difference in acceleration due to bicycle wheels that is claimed by your riding buddies is primarily due to cognitive dissonance, or the placebo effect (they paid a lot of money for the wheels so there must be some perceivable gain).
(…)

Unquote

2. nice.

When my old wheels got beat down, I replaced them with something a little lighter, and I did not notice a difference in speed/acceleration - though I did notice the new wheels were a little stiffer due to a better build. When I swap between lighter and heavier tires, i notice the difference in traction and rolling resistance a lot more than any weight gain. I'm not saying saving weight on wheels makes no difference, but in my experience, small changes are difficult to detect.

I can't really comment too much on the scientific accuracy of the article, but in MTB a lot less resistance comes from drag, much more resistance comes from rolling resistance, and there is a lot more acceleration, which increases the effect of the two inertias mentioned. The sports really are quite different.

I do feel that people seem to overstate the effect of saving a few grams on your wheels. I guess that is the general nature of the weight weenie crowd though. People get excited about any gain, big or small.

3. From http://en.wikipedia.org/wiki/Bicycle...#Rotating_mass :
Due to the fact that wheels rotate as well as translate (move in a straight line) when a bicycle moves, more force is required to accelerate a unit of mass on the wheel than on the frame. It is not the center of gravity (mass) which matters, but the moment of inertia, which takes the rotation into account. Reducing the rotational inetria can be achieved by moving the spoke nipples to the hub or using lighter nipples such as aluminium. This can be pictured by imagining an ice skater spinning; when they pull in their arms, they rotate more quickly. A rule of thumb, in the case of acceleration only, is "the effect of a given mass on the wheels is almost twice that of the same mass on the frame."[8] But again, the distinction between rotating and non-rotating mass is only felt during acceleration (and braking to a lesser extent), and mass at the hub matters a lot less than mass at the rim.
Bear in mind that acceleration is a vector quantity meaning it not only has a rate of change but also a direction. So by that logic it means steering would also be affected. Lower rim weight equals less gyroscopic effect equals more responsive steering and a lot less for the brakes to slow down. What is called 'deceleration' is actually just negative acceleration, so basically braking is the opposite of sprinting but would likewise see an improvement from lighter wheels.

http://velonews.com/article/9727

It is also worth noting that on suspended frames, wheels count as unsprung mass. That means that your suspension would be MUCH more effective. Suspension effectively pushes down on the wheels so that it maintains traction with the trail surface after impact. If the tire gets bumped up off the ground, guess what, there goes your traction. A lighter weight wheel will return to its ideal position more quickly than a heavier wheel, delivering better traction and better handling.

4. Before this diverts into religious warfare can I point out that there are two views here that are correct, if only both sides would understand that they are compatible.

View 1: Saving weight in rotating mass gives better benefit than saving non-rotating mass

View 2: The differences are unimportant

View 2 does not infer that View 1 is not the case. It just disregards the result as being important.

From http://en.wikipedia.org/wiki/Bicycle...#Rotating_mass :

Bear in mind that acceleration is a vector quantity meaning it not only has a rate of change but also a direction. So by that logic it means steering would also be affected. Lower rim weight equals less gyroscopic effect equals more responsive steering and a lot less for the brakes to slow down. What is called 'deceleration' is actually just negative acceleration, so basically braking is the opposite of sprinting but would likewise see an improvement from lighter wheels.
What you've not taken into account of is that if the effects are marginally unimportant for acceleration, they are even less relevant to braking where human exertion does not directly provide the force vector. I disagree with your use of the phrase: "a lot less for the brakes to slow down".

6. Lighter wheels have less rotating mass working to continue rolling against the brake force applied. How is this unclear?

Lighter wheels have less rotating mass working to continue rolling against the brake force applied. How is this unclear?
Not unclear. The point is that you don't work harder to slow down, your brakes do. I guess you have to pull the lever a bit harder. Probably less fade too. I don't think it has an effect on your overall braking power though. Most people can lock their brakes and break traction pretty easily.

Lighter wheels have less rotating mass working to continue rolling against the brake force applied. How is this unclear?

In one post you;ve gone from "a lot less" to "less". You are heading in my direction.

9. With a long background in drag racing I'll tell you this: reducing unsprung weight is paramount in helping to fine-tune a suspension. And rotating mass is important in terms of acceleration, be it a car or bicycle. Less rotating mass is easier for a braking system to stop as well.

10. Some things are partially true..... magnitude is the key frase, lets see

Person body mass: 75 Kg (165 lbs)
bike weight 12,5 Kg (27,5 lbs) Wheels included (1,8 Kg or 4 pounds)

Total weight: 192,5 lbs
Rider %: 85,7
Bike %: 14,3
Wheels %: 2%

Now, if you are a Tour the France guy or are in the Olympics, 1% makes a difference and when only 0,1 sec counts between a glod and silver medal.... you get the point
For the average joe or racer wanna-be... not worth it unless you just want to burn money

remember that any of the guys in the olympics kick our a\$\$es in a Kmart bike

11. i removed my old, average build, soft 2.2kg wheelset and replaced them with a new, stiffer, better built 1.6kg wheelset. i observed an improvement in acceleration and climbing. my conclusion? i think that i, like many people, put a set of nice new wheels on and noticed an improvement and assumed it to be the weight when in fact part of it was also the stiffness. perhaps even the majority...

i should try the reverse experiment where i put set of very strong but very heavy wheels on and see how that goes...

lighter wheels are always going to be better than heavier wheels unless the lighter wheels are significantly weaker imho.

12. Originally Posted by BBW
Wheels %: 2%
Sure, as a percentage of overall weight, wheels don't account for much. But, you are ignoring the whole idea of rotational weight/moment of inertia, and all that fun stuff that makes wheel weight different from the other bits that weight your bike down.

I'm not disagreeing with the fact that the difference is small or insignificant to some, but I just feel that the numbers your have presented don't really prove the point I think you are intending to prove.

And no, I don't feel like busting out any physics equations and trying to quantify the differences.

13. The equation for the moment of rotational inertia is:

I = ∑ m.r^2

As an approximation of the second moment of inertia, you use the mass of the rim, tire and other rim hardware in the equation:

I(r) = m x (ETRTO/2+10) ^2 where m is the total mass of all components and I've approximated the effective radius as ETRTO and a bit.

The contribution of the mass of the spokes is given by:

I(s) = 1/3 x m x (ERD/2)^2, where m is the total mass of the spokes

The hub is ignored because of its central location.

I =I(r) + I (s)

With a 52/559 ETRTO tyre and all rim components and tyre weighing 'Mt' and Spokes weighing 'Ms' on a rim with ERD of 538mm, this gives:

I(r) = 83810 x Mt

I(s) = 24120 x Ms

The equivalent equation for rotational acceleration is:

T = I x w

Where T is torque and w is the rotational acceleration.

Now we just need to link rotational acceleration to linear acceleration. This is simply a matter of:

a = w x (559/2 + 52)

F = T / (559/2+52)

Because we know F=ma, we can use all these equations to give us the mass equivalent for the rotational inertia:

m= F/a

m = T / ( w x (559/2+52)^2 )

I = T / w

m = I / ( (559/2+52) ^2 )

Now, as (559/2+52)^2 = 109892 the result is....

m = 83810/109892 x Mt + 24120/109892 x Ms

ie. m = 0.762 x Mt + 0.219 x Ms

This means that the rim components contribute ~76% extra mass and the spokes contribute ~22% extra mass

Now supposing the tyre and rim components represent 2kg out of an all up mass of ~75kg for a reasonably light bike and a light rider (i.e. heavy wheels, light everything else), the rotational contribution to inertia is ~2%.

14. Just to give an easier to understand perspective...

Say you swap out a 600g tyre (effective inertial mass = 1057g) for a 480g tyre (effective inertial mass = 846g) representing a 120g mass saving.

The bike will be as easy to accelerate as though you had in fact saved 211g. i.e. the real world weight weenieing contribution is 91g.

15. ## It's nice to see somebody put forth some numbers.

Originally Posted by petercarm
ie. m = 0.121 x Mt + 0.035 x Ms
I'm unsure of the following line:

a = 2 x pi x w x (559/2 + 52)

maybe I'm wrong (it's getting late), but I don't think the 2xpi should be there. for a rolling circular object, the forward linear acceleration of the centre (and hence the bike) is given by the angular acceleration x the radius.

16. Furthermore, the 0.3kg/0.66lb difference in wheels, even if this mass is out at the rim, is so small compared to your body mass that the differences in wheel inertia will be unperceivable. Any difference in acceleration due to bicycle wheels that is claimed by your riding buddies is primarily due to cognitive dissonance, or the placebo effect (they paid a lot of money for the wheels so there must be some perceivable gain).

This is an interesting question, so I decided to do some calculations. Obviously, the lighter your wheel already is the more of an impact adding 300g at the rim will be. Adding 300g to the wheel rim accurately simulates the effect of running a 'superlight'
tire (~300g) vs. a standard kevlar tire (~600g).

The first part is a tabulation of weights and distances required to accurately calculate the moment of inertia of my actual wheel. I also decided to throw in a dream lightweight wheel where cost is no object for comparison purposes. All distances are measured and rounded to the nearest millimeter. For simplicity, all parts were considered as a 'solid disc', and the distance to the mass of the disc is half of the radius to the edge of the disc. For example, the hub is considered a solid disc with a radius of 30mm based on a flange diameter of 60mm. Thus, r/2 = 15mm. While this method isn't perfect, it should give us very good results.

This calculation finds the difference between the angular acceleration of the specified wheel vs. the specified wheel with a 300g weight added at the rim. Because the angular acceleration for the heavier wheel will be less than the unladen wheel, the ratio of angular accelerations will be less than one. We will then subtract this from unity to find the percentage decrease in angular acceleration due to the addition of the 300g weight.

Conclusions:

Adding 300g to my actual wheel results in a 22.2% decrease in angular acceleration for a given torque.

Adding 300g to my dream lightweight wheel results in a 34.3% decrease in angular acceleration.

Obviously, these are significant percentages. While it is difficult to estimate what percentage difference somebody will actually feel, it is hard to imagine that somebody would not feel a 20% difference in angular acceleration.

In fact, I think most professional riders would probably feel a ~5% difference quite easily. A 5% difference in angular acceleration accounts for adding approximately 55g to the wheel rim. This would be the difference between running a lightweight innertube and a standard one.

17. Originally Posted by ginsu2k
For simplicity, all parts were considered as a 'solid disc', and the distance to the mass of the disc is half of the radius to the edge of the disc. For example, the hub is considered a solid disc with a radius of 30mm based on a flange diameter of 60mm. Thus, r/2 = 15mm. While this method isn't perfect, it should give us very good results.
I'm not happy with your assumption. For an item such as the rim, it would make much more sense to approximate it as a thin ring, for which the moment of inertia is mr^2.

18. Originally Posted by rkj__
I'm unsure of the following line:

a = 2 x pi x w x (559/2 + 52)

maybe I'm wrong (it's getting late), but I don't think the 2xpi should be there. for a rolling circular object, the forward linear acceleration of the centre (and hence the bike) is given by the angular acceleration x the radius.

Good point and I'm glad someone is paying attention. I posted that at something like 04:30, so I'm pleading early hours fatigue. Original posts corrected.

The numbers now stack up and explain why rotating weight is a good thing to save.

19. Now if we can just link in the discussion about the effect of the mass of the air in the tire and its effect on rotational momentum we could likely establish this thread as a college level physics course.

20. ## speaking of fluid mechanics...

Originally Posted by rockyuphill
Now if we can just link in the discussion about the effect of the mass of the air in the tire and its effect on rotational momentum we could likely establish this thread as a college level physics course.
I just realized i missed my fluid mechanics course today. I had a 9:30 class yesterday to start, and an 8:30 to start today. I never changed my alarm.

21. Wow, you guys are really getting into the calculations! Don't you have to choose a velocity to make those calculations? What are you choosing? Most of my climbing is only about 5mph, so there's no angular velocity built up at that speed to worry about. So, for general climbing, I'd agree it's not quite as important as some make it out to be.

On the other hand, concerning the OP's article, road bikes turn smoothly. Mountain bikes, otoh, may need to quickly change lines while going down a 20mph hill. Or maybe you want to pull off a little table-top or tail-whip over a jump? That stuff will all be better with lighter rims. Ever hold a wheel by the axles and have a buddy spin it, and then try to move it like you're steering or leaning the bike? You can really feel the gyroscopic weight in those types of changes. I'd think that effect isn't important in road biking but more important for mtbg fun.

So, I'm agreeing that a wheel weight difference may not make such a big difference in energy spent to climb a hill, but should make a difference giving your bike a responsive, agile feeling during tighter maneuvers.

22. Originally Posted by Wheelspeed
Wow, you guys are really getting into the calculations! Don't you have to choose a velocity to make those calculations? What are you choosing? Most of my climbing is only about 5mph, so there's no angular velocity built up at that speed to worry about. So, for general climbing, I'd agree it's not quite as important as some make it out to be.
Exactly: from a very quick calculation, accelerating from 0 to 30Km/h (18mph) in 15 seconds results in an acceleration value of only 0,05g
That's why the inertia term of the equation of motion is very, very close to zero.
Considering tha most of the climbs are done at a semi-constant speed (i.e. very limited speeed changes) this is even more true.
That's why in road riding the term "climbing wheels" may be incorrect, as much as the term "climbing seatpost" o "climbing stem" is incorrect.

In mtn biking however, the biggest advantage of light wheels is the reduction of unsprung weight and the impreved suspension sensitivity.

Fab

23. Originally Posted by rkj__
I'm not happy with your assumption. For an item such as the rim, it would make much more sense to approximate it as a thin ring, for which the moment of inertia is mr^2.
oh, actually that's exactly what I have done. the entire problem is 2d only, and i use radial distances to the center of the rim profile (it's about an inch wide). only the spokes are actually
a very wide ring.

So, I'm agreeing that a wheel weight difference may not make such a big difference in energy spent to climb a hill, but should make a difference giving your bike a responsive, agile feeling during tighter maneuvers.
The fact that we do not pedal at the same cadence all the time is what makes rotational inertia so important. You are constantly changing your cadence and thus your legs are constantly providing torques (negative or positive) that are influencing the angular acceleration of the wheels. Given the same torque, a light wheel will accelerate much faster. You will feel this as the bike being 'responsive' and 'agile'.

This effect is also seen when you lighten your seat, seatpost and any components far above the ground. The bike is much more maneuverable because there is a significant polar moment of inertia effect as the distance from the ground (center of rotation) is very big (can be a meter or more).

In mtn biking however, the biggest advantage of light wheels is the reduction of unsprung weight and the impreved suspension sensitivity.
Well, I don't think this is true at all. Personally, I don't climb at the same speed for an hour or more. Anytime the trail changes grade or direction you will also change speed. I've had friends take a GPS and graph the elevation change on an average ride. The elevation where I ride is constantly changing.

I think most people are probably arguing whether or not the FEEL a difference in mass at the wheel. Obviously, this is a completely different question. Mathematical equations don't give a crap whether humans can feel a numerical difference. The fact is there is a numerical difference.

24. Originally Posted by Ausable
That's why in road riding the term "climbing wheels" may be incorrect, as much as the term "climbing seatpost" o "climbing stem" is incorrect.
Fab
I would not say the term climbing wheels in incorrect. For a flatter course, many riders will take a weight penalty in favour of a more aero rim. But, at climbing speeds, there is no advantage to the heavier more aero wheels, so riders choose to go for a lighter set, since changing wheels is a relatively easy swap. It makes sense to save weight. I see your point though, that at a perfectly constant speed, going up a road climb, saving weight on your wheels is not too different from saving weight from other areas of the bike.

25. Originally Posted by ginsu2k
oh, actually that's exactly what I have done. the entire problem is 2d only, and i use radial distances to the center of the rim profile (it's about an inch wide). only the spokes are actually
a very wide ring.
Ok. It was just the wording. It sounded like you were approximating a rim as a solid disc, meaning it had mass distributed near the centre of the circle. I did not check the numbers.

Originally Posted by ginsu2k
This effect is also seen when you lighten your seat, seatpost and any components far above the ground. The bike is much more maneuverable because there is a significant polar moment of inertia effect as the distance from the ground (center of rotation) is very big (can be a meter or more).
Originally Posted by ginsu2k
I think most people are probably arguing whether or not the FEEL a difference in mass at the wheel. Obviously, this is a completely different question. Mathematical equations don't give a crap whether humans can feel a numerical difference. The fact is there is a numerical difference.
I stated in an earlier post that ...

Originally Posted by rkj__
When I swap between lighter and heavier tires, i notice the difference in traction and rolling resistance a lot more than any weight gain. I'm not saying saving weight on wheels makes no difference, but in my experience, small changes are difficult to detect.
Conclusions. No doubt, there is a numerical difference. Saving weight on wheels does make more of a difference than saving weight on a frame for example. This is the main result we are after I think. But, everybody perceives changes differently. Saving x grams on wheels may make a noticeable difference to one rider, while the change may go undetected by another. You can't draw a firm line saying "this much" of a difference is significant while "that much" is insignificant. This difference in perception is the cause of many debates here in the WW forum.

26. As some have pointed out, there is a huge difference between MTB and Road biking.

MTB requires constant change of pace and lots of quick acceleration/decelerations (unless you ride a lot of fire roads).

Also, I think people that ride a lot "feel" these weight differences, especially wheels. However, the perceived differences might not equate to a significant performance enhancement. But it sure feels a lot better pushing light weight wheels on a MTB than heavy wheels.

27. Ok. It was just the wording. It sounded like you were approximating a rim as a solid disc, meaning it had mass distributed near the centre of the circle. I did not check the numbers.
You make a good point though because in the first set of calculations i approximated everything as a thin ring and that is not accurate (although the results are the same regarding the difference in angular acceleration because i used the same formula for both wheels).

In the new calculations for wheel inertia i approximated the hub as a disc (cylinder), the spokes as a very thick ring, and the rim and tire as very thin rings. Each have their appropriate equations. The change from the previous calculation is only around 5%, so my approximation as thin rings wasn't really all that bad (certainly saved some time).

I looked up some weights on weightweenies for my dream lightweight wheel to see what it's inertia would turn out to be. Interestingly, the inertia is half of the inertia of my actual wheel, which strangely matches up with being half the weight of my actual wheel. I guess this isn't all that weird seeing as a lightweight wheel would have the weight removed proportionally, meaning everything is about half the weight of what it is on the heavy wheel (although the rim certainly is not).

I guess I would conclude from this that the best wheel you can get is only going to have about half the inertia of a normal XC wheel. Oh well, you're not going to see any hoops that are ridiculously easy to spin until we can get rim weights down to around 150g.

28. Some of the weight on your dream lightweight wheel are off. The ZTR Olympics are around 345 grams, the Tune Princess is 110 grams, Maxxis Maxxlite 285 are 3-4 grams over 285 gr. All these weight are verified as I have the exact same wheels/ tires. Hope this helps.

You also need to add either tubes / rim strips or veloplugs (+90 grams) or 2 oz of Stans with rim strips/valves/tapes (from 80 grams up per wheel)

29. Originally Posted by ginsu2k
You make a good point though because in the first set of calculations i approximated everything as a thin ring and that is not accurate (although the results are the same regarding the difference in angular acceleration because i used the same formula for both wheels).

In the new calculations for wheel inertia i approximated the hub as a disc (cylinder), the spokes as a very thick ring, and the rim and tire as very thin rings. Each have their appropriate equations. The change from the previous calculation is only around 5%, so my approximation as thin rings wasn't really all that bad (certainly saved some time).

I looked up some weights on weightweenies for my dream lightweight wheel to see what it's inertia would turn out to be. Interestingly, the inertia is half of the inertia of my actual wheel, which strangely matches up with being half the weight of my actual wheel. I guess this isn't all that weird seeing as a lightweight wheel would have the weight removed proportionally, meaning everything is about half the weight of what it is on the heavy wheel (although the rim certainly is not).

I guess I would conclude from this that the best wheel you can get is only going to have about half the inertia of a normal XC wheel. Oh well, you're not going to see any hoops that are ridiculously easy to spin until we can get rim weights down to around 150g.
The spokes element in your calculation is wrong. The correct term is closer to:

1/3 x m x (R2^3 - R1^3)/(R2-R1)

Your thick ring model is incorrect. A disc implies the same mass/area density at all radii, whereas spokes are the same cross section (barring butting) so the mass/area density becomes lower as radius increases.

The term I suggest above doesn't take into account the tangential section of the spoke as it exits the hub flange. I'm having a think about that at the moment.

30. Ginsu2K
it's true that a dreamlight wheelset will have half of the moment of inertia of a standard wheelset.
But if you multiply the inertia values by the (very) limited acceleration occuring during a ride - something like 0.05 or 0.1g- you can see that the resulting difference in force is negligible.
For the nerds among us, here is the link to the equation of motion.
http://www.biketechreview.com/archive/images/appa.pdf

31. Got it.

The term for the spokes is...

I = m x (R2^2 + R1^2 - 1/3 .L^2 ) /2

Where L is the length of the spoke, R2 is the effective rim radius (ERD/2) and R1 is the flange radius. To get this you need to use the parallel axis theorem, using the cosine rule to work out the offset of the axis.

The working is this:

Consider a triangle formed by the vertices:

A- the hub centre
B- the flange end of the spoke
C- the rim end of the spoke

AB= R1
AC= R2
BC= L

Add a dividing line from the mid point of the spoke (M) to the hub centre. We want to know the length AM.

The moment of inertia of the spoke about its midpoint CoG is 1/12 x m x L^2.

The parallel axis theorem tells us that the moment of inertia for the spoke about the hub centre is given by adding m x AM^2 to the moment of inertia about its CoG.

The cosine rule can be used to get the length AM.

Looking at triangle ABC:

AB^2 = AC^2 +BC^2 -2.BC.AC.cos (c)
ie. R1^2 = R2^2 +L^2 - 2 L R2.cos (c)

L R2 cos (c) = 1/2 x (R2^2 +L^2 - R1^2 )

Looking at triangle AMC:

MC= L/2

AM^2 = AC^2 + MC^2 - 2 MC AC cos (c)

i.e. AM^2 = R2^2 + 1/4 L^2 - L R2 cos(c)

AM^2 = R2^2 +1/4 L^2 - 1/2 (R2^2 + L^2 - R1^2)

AM^2 = 1/2 R2^2 - 1/4 L^2 + 1/2 R1^2

Therefore, I = m ( 1/12 L^2 -1/4 L^2 + 1/2R2^2 + 1/2R1^2)
or I = m ( R2^2 +R1^2 - 1/3 L^2 ) / 2

32. Then whats the conclusion of all this formulas?

when I swapped my old wheelset to the crossmax plus light tubeless tires I felt a day to night difference in aceleration and cornering ability and that was not placebo. maybe the small variations calculated are very noticable.

33. Originally Posted by sergio_pt
Then whats the conclusion of all this formulas?
I posted MY conclusions (In bold) in the last paragraph of post #25.

34. Taking Ginus2k's Dream spec...

For the sake of saving 803g, the bike will accelerate as though 1.375kg had been saved.

Taking the dream spec, but keeping the Geax Beaumont tyres...

For the sake of saving 421g, the bike will accelerate as though 607g had been saved.

My conclusion is that the rolling inertia is dominated by the tyres. If you fit the tyres you need for the conditions you ride, you will be reaping the most benefit. Weight weenieing the wheel build does provide extra equivalence in weight savings, but not so much that the strength/stiffness of the build should be compromised. The further away from the rim you save the weight, the less "bonus acceleration" you get.

35. Originally Posted by petercarm
Taking Ginus2k's Dream spec...

For the sake of saving 803g, the bike will accelerate as though 1.375kg had been saved.

Taking the dream spec, but keeping the Geax Beaumont tyres...

For the sake of saving 421g, the bike will accelerate as though 607g had been saved.

My conclusion is that the rolling inertia is dominated by the tyres. If you fit the tyres you need for the conditions you ride, you will be reaping the most benefit. Weight weenieing the wheel build does provide extra equivalence in weight savings, but not so much that the strength/stiffness of the build should be compromised. The further away from the rim you save the weight, the less "bonus acceleration" you get.
I must really be missing something because I don't see how you can get numbers like that without saying "at this speed" or "at such and such rate of acceleration"...

Go down a paved hill at 35 mph and slow down to 30mph watching your cyclocomputer and pay attention to how much lever force or how long it took you to slow down like that. Next, go down the same hill but only at 5mph then slow to 0. It's tons easier. (Especially obvious with v-brakes.) That's because of so much less work the brakes need to do to stop the spinning wheel at only 5mph instead of 35 mph. (At 35mph, you even have wind helping you to stop.)

36. Originally Posted by Wheelspeed
I must really be missing something because I don't see how you can get numbers like that without saying "at this speed" or "at such and such rate of acceleration"...
You got the point of the whole thread.
To say that a lighter wheel (especialli a lighter tire, etc) will have less rotating inertia is not properly new stuff.
But the important point is to relate the different inertia of lighter weels with the minimum rate of rate of acceleration experienced in real world cycling.

(Oh yes, I am selling my new Xsyrium road wheels for some 1350g American Classic .. PM if interested )

37. Originally Posted by Wheelspeed
I must really be missing something because I don't see how you can get numbers like that without saying "at this speed" or "at such and such rate of acceleration"...
Force = mass x acceleration.

Note that there is no speed involved in the above definition of Newton's law. All I have done is relate the rotational terms to the linear acceleration. As my first post pointed out, the maths is entirely correct in defining the effect, but opinions will vary on whether or not it is relevant.

38. Taking the math into practical application, I should get a lighter wheelset for my SS! There is a lot of speed change when I'm climbing, every pedal stroke is a wave of speed!

39. Taking Ginsu2k's standard and dream specs:

A rider weighs 160lbs.

Bike 1 has the standard wheel, weighs 24lb (all up 184lbs, 83.5kg, 84.6kg effective)

Bike 2 has the dream wheel, but heavier frame/components mean the bike still weighs 24lbs (all up 184lbs, 83.5kg, 84.1kg effective)

Bike 3 is the same frame components as Bike 3 but runs the dream wheel.(all up 182.2lbs, 82.66kg, 83.2kg effective)

Bike 4 uses the standard wheel, but non-rotating component weight savings mean it weighs the same as Bike 3. (all up 182.2lbs, 82.66kg, 83.8 kg effective)

Bike 5 uses the dream wheel with the standard tyre. (all up 183.1lbs, 83.1kg, 84kg effective)

Acceleration:

Bike 2 compared to Bike 1 accelerates 0.59% quicker for any given input.

Bike 3 compared to Bike 1 accelerates 1.6% quicker for any given input.

Bike 4 compared to Bike 1 accelerates 0.94% quicker for any given input.

Bike 5 compared to Bike 1 accelerates 0.71% quicker for any given input.

Energy to climb a hill at steady speed:

Bike 1 and Bike 2 are the same.

Bike 3 and Bike 4 save 1%

Bike 5 saves 0.47%

The above calculations are just for changing one of the wheels, so potentially the savings could be doubled.

I'm most interested in the Bike 5 result because I will never choose a tyre on grounds of weight alone. I also weigh 200lbs, so I'm not ever going to run Olympic Race rims. Bike 5 (with my riding weight) would be down to 0.59% advantage (1.2% acceleration advantage for both wheels). I currently run 819s, so the upper bracket of improvement I can make in my wheel choice is ~1%. For me, this isn't big enough to care. However, this wouldn't be the Weight Weenies forum if there weren't people for whom 1% is significant.

40. why da heck then pro cycling riders use the lightest possible wheels and bikes to climb?

Everytime you pedal down is energy you are saving for the next strokes with lighter wheels/ lighter bikes. I dont know what you are calculating, or if you are calculating it right. What about trying to calculate the energy saved in each pedal stroke?

41. Originally Posted by sergio_pt
why da heck then pro cycling riders use the lightest possible wheels and bikes to climb?
Because for the given power they will climb 1% faster. 1% over 10 kilometers is 100m. That's a decent margin.

Now - do you care for 100m over 10km of climbing? I do not. So there is a point of rapidly diminishing returns for me with light wheels - and a durability cutoff, as I am 190+ pounds. I will still go for a lighter wheel given the choice, but cost goes up exponentially with decreasing weight, while performance improves linearly - and then drops to zero once it is not durable enough.

I have calculated that my optimal point for price/performance/durability is about ~1500g for XC and ~1900g for AM wheelset for todays technology.

42. Originally Posted by ginsu2k
This is an interesting question, so I decided to do some calculations. Obviously, the lighter your wheel already is the more of an impact adding 300g at the rim will be. Adding 300g to the wheel rim accurately simulates the effect of running a 'superlight'
tire (~300g) vs. a standard kevlar tire (~600g).

The first part is a tabulation of weights and distances required to accurately calculate the moment of inertia of my actual wheel. I also decided to throw in a dream lightweight wheel where cost is no object for comparison purposes. All distances are measured and rounded to the nearest millimeter. For simplicity, all parts were considered as a 'solid disc', and the distance to the mass of the disc is half of the radius to the edge of the disc. For example, the hub is considered a solid disc with a radius of 30mm based on a flange diameter of 60mm. Thus, r/2 = 15mm. While this method isn't perfect, it should give us very good results.

This calculation finds the difference between the angular acceleration of the specified wheel vs. the specified wheel with a 300g weight added at the rim. Because the angular acceleration for the heavier wheel will be less than the unladen wheel, the ratio of angular accelerations will be less than one. We will then subtract this from unity to find the percentage decrease in angular acceleration due to the addition of the 300g weight.

Conclusions:

Adding 300g to my actual wheel results in a 22.2% decrease in angular acceleration for a given torque.

Adding 300g to my dream lightweight wheel results in a 34.3% decrease in angular acceleration.

Obviously, these are significant percentages. While it is difficult to estimate what percentage difference somebody will actually feel, it is hard to imagine that somebody would not feel a 20% difference in angular acceleration.

In fact, I think most professional riders would probably feel a ~5% difference quite easily. A 5% difference in angular acceleration accounts for adding approximately 55g to the wheel rim. This would be the difference between running a lightweight innertube and a standard one.

thr funny thing is that if you put the geax tire on the dream wheelset its inertia climbs to a whopping 0.08514 which is closer to the common wheelset 0.09785418 than the dream wheelset with the maxxis of 0.05315688.

so in the real world what determines the most the inertia are the tires along with the tubes if you run them.

not all the tracks or trails allow us to use 300 gram tires so in the end light wheelsets are useful but their effect would be negated with heavy tires.

i know i wont spend 900 bucks on a wheelset when a xtr hub dt swiss 4.2d rim and dt swiss double butted spokes should have rougly an inertia of 0.09something having in mmind i used a 600 gram tire on either of them.

so its 0.08514007 vs 0.09something or at the most 15 % more inertia. when it was already said wheelsets contribute to 2% of the inertia of a bike and rider system. it translates to 0.3% of the whole bike + rider system. in the end its about 80 rider and 20 bike given that the bikes are suitable.

43. Nice to see this thread is still going strong

Speaking of moment of inertia, check out this related discussion that popped up on the other wheel size forum:

Dumb (but legit) hypathetical Q about rotating weight

44. Originally Posted by petercarm
Got it.

The term for the spokes is...

I = m x (R2^2 + R1^2 - 1/3 .L^2 ) /2

I will put this in my excel spreadsheet. I'm far too busy to go through your proof, but I am
certainly impressed. It would certainly be more readable if you wrote with mathtype and posted the jpeg.

I don't really like to go through people's equations, especially since I'm in school and I have a bunch of homework all the time, so my time is better spent elsewhere. Besides, I trust your results if you spend the time to write it all out.

Besides, I've been wanting to do these calculations a bit differently after seeing a test on
roues artisanales. I have not considered both the rotational and kinetic energies together as a comparision, and by doing this you should easily see a substantial difference in the 'standard' vs. 'dream lightweight' setup.

BTW, I will be using data from the world's fastest sprinter.

45. ## My experience

Is that higher hubs and lighter wheelsets don't make feel faster because of weight loss but just the quality of the ride. I like high-end wheelsets! I typically run DT Swiss hubs in the rear and Hope front hubs. I just have had good luck with them....but they don't make me any faster. That is up to my pedalling speed

Jaybo

46. many people here speaks about percentage(%) for eg. there is 100% difference between 1mm and 2mm, but who cares it?
it's soo small - but guys... it IS 100% difference and this counts most - this is silly for me.

Soo look into numbers, this counts most. And than you will see that these numbers counts marginaly as we take whole bike + rider in to account.

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