A pound on the wheels is worth...

hwttdz said:

I saw someone mention that a "pound on the wheels is worth two on the frame" and someone else said, "no you're stupid" (slightly rephased the second one there). Well in the interest of bettering out knowledge I broke out some old equations, massively oversimplified things made gross assumptions, and got an answer (how much it's worth is questionable).

For an 8kg bike (17.6 lbs approx) traveling at 10m/s (22.36 mph) with a 1700 grams of weight at the rim (rims and tires, rims are 500ish tires 250ish, tubes 100ish) 700mm wheel diameter (pretty normal).

1 pound of mass at the rim has the same amount of energy as 1.025 pounds of non-rotating mass. (1:1.025 make the units whatever makes you happy, g:g kg:kg doesn't matter).

So it's not as much of a magic bullet as some might think, but hey it's something.

you're math is off, your friend is correct the ratio is two. see attached

I see what you did, in the v=rw you plugged in r and v. I'm actually not sure about this, haven't done any physics in so long. What I did was I figured the circumference of the wheel 2.2m the speed 10m/s and figured the wheel had to make 4.55 rotations a second, I then used 4.55 s-1 as my angular velocity. You say this is no good.

Yeah that's my mistake I thought for some reason we cared about rotations per second when it's actually radians per second. That's a neat result then exactly 2:1. Thanks for the catch.

Ok more musings assume we have two bikes for the sake of easy numbers
bike a: 7 kg (excluding wheels) 1 kg wheels
bike b: 6 kg (excluding wheels) 2 kg wheels
So at a given speed bike b has more kinetic energy. In theory even climbing a hill at constant speed they are equal because you're just adding mgh to both and you can replace bike/wheels with a point particle, right?

hwttdz said:

I see what you did, in the v=rw you plugged in r and v. I'm actually not sure about this, haven't done any physics in so long. What I did was I figured the circumference of the wheel 2.2m the speed 10m/s and figured the wheel had to make 4.55 rotations a second, I then used 4.55 s-1 as my angular velocity. You say this is no good.

Yeah that's my mistake I thought for some reason we cared about rotations per second when it's actually radians per second. That's a neat result then exactly 2:1. Thanks for the catch.

Ok more musings assume we have two bikes for the sake of easy numbers
bike a: 7 kg (excluding wheels) 1 kg wheels
bike b: 6 kg (excluding wheels) 2 kg wheels
So at a given speed bike b has more kinetic energy. In theory even climbing a hill at constant speed they are equal because you're just adding mgh to both and you can replace bike/wheels with a point particle, right?

kinda mixing concepts there. the kinetic energy equations pertain to acceleration, or how much work it takes to accelerate up to speed. The potential energy equation relates to quasi static lifting. emphasis on quasi static, meaning no kinetic energy gain or loss. So in the scenario you propose both bikes gain the same energy since they are both gaining the same elevation. But at the end of the climb, bike b, with the heavier wheels has moretotal energy since it started off with more, assuming you started the climb with both bikes going the same speed and they maintained it throughout the climb and ended at the same speed.

Mr_Potatohead said:

kinda mixing concepts there. the kinetic energy equations pertain to acceleration, or how much work it takes to accelerate up to speed. The potential energy equation relates to quasi static lifting. emphasis on quasi static, meaning no kinetic energy gain or loss. So in the scenario you propose both bikes gain the same energy since they are both gaining the same elevation. But at the end of the climb, bike b, with the heavier wheels has moretotal energy since it started off with more, assuming you started the climb with both bikes going the same speed and they maintained it throughout the climb and ended at the same speed.

Interesting discussion, and good to see the equations. Understand solving the kinetic equation gets you the factor of 2, but this is of course far from the whole story.

As you state, this kinetic energy difference only relates to acceleration. Putting this in context of overall energy/power required for a given ride seems difficult to me.

Also, have to keep the wheels in context of the overall bike and rider. The work required for linear acceleration of your body and frame, say from 0 to 10 m/sec, far outweighs the energy going into the angular acceleration of the wheels because your non-rotating body and bike weight is so much greater.

dhk said:

Interesting discussion, and good to see the equations. Understand solving the kinetic equation gets you the factor of 2, but this is of course far from the whole story.

As you state, this kinetic energy difference only relates to acceleration. Putting this in context of overall energy/power required for a given ride seems difficult to me.

Also, have to keep the wheels in context of the overall bike and rider. The work required for linear acceleration of your body and frame, say from 0 to 10 m/sec, far outweighs the energy going into the angular acceleration of the wheels because your non-rotating body and bike weight is so much greater.

I think what you are after is this. Assume constant power output, P = Fv. So when you drop a pound of weight from your bike in a climb, the force required to lift you and your bike up the hill goes down and your speed can go up proportionally.

So for example drop 2 pounds off a 180 pound rider + 20 pound bike in a climb and you will increase your speed by 1 percent, with no increase in effort, dropping your time up the hill by 1 percent. Which comes out to saving about 36 seconds per hour of climbing.

But 90 percent of a riders effort in climbing goes towards lifting their own body weight.

The rotational inertia of the wheel only comes into play during acceleration. Otherwise the weight of the wheel is the same as the weight of the spare tire around your waist or in your saddle pack.

But now you can see why the heavy sprinters struggle so much in the mountain stages of the Tour de France. Compare the great Marco Pantani who weighed about 135 lbs riding a 15 lbs bike to another rider weighing 185 lbs riding the same bike. Assuming equal power output, Marco Pantani goes up the mountain 25% faster than the heavy guy or maybe about 13 minutes up Alp d'huez.

Actually the old (sixties) axiom was "a gram off the wheels is worth FOUR off the frame" which I think had more to do with the "feel" of things -- especially acceleration out of corners, etc. Not arguing with the math, just a little history to go with the science.......

Bititanio said:

Actually the old (sixties) axiom was "a gram off the wheels is worth FOUR off the frame" which I think had more to do with the "feel" of things -- especially acceleration out of corners, etc. Not arguing with the math, just a little history to go with the science.......

The backpacker's mantra goes "A pound on the foot is worth five in the pack." My experience tells me this is true. "A gram off the wheel is worth four off the frame" seems to have the ring of truth from my experience too.

At 155 lbs., I figure each pound off my gut is worth a tenth of a mile per hour. Each of these pounds come a off whole lot cheaper than losing 110g at a time from the wheels. I can also lose ten of them whereas losing 1100g from my Ksyrium Elites would be prohibitively expensive and potentially dangerous.

It seems to me that the smoothness of the hubs, the quality of the build, and the suitability for intended use are each worth at least 110g worth of consideration.

Also even if you think you're maintaining a constant speed you're not. Constantly accelerating and decelerating costs more energy with more inertia.

What I was trying to say earlier is that the two bikes should climb the same at constant speed (if we could do constant).

Does the pound off the body is an extra tenth of a mile per hour hold? That sounds good.

cachehiker said:

... At 155 lbs., I figure each pound off my gut is worth a tenth of a mile per hour. Each of these pounds come a off whole lot cheaper than losing 110g at a time from the wheels. I can also lose ten of them whereas losing 1100g from my Ksyrium Elites would be prohibitively expensive and potentially dangerous...

Is this a formula and is it for the flats? Or is it for climbing? The reason I ask is because I could stand to lose ... well, much. If that is for the flats then my speeds would be almost normal. I'm not too concerned with climbing for the time being. TIA.

The formula about "a pound on the wheels is worth two on the frame" is correct, as many have observed, when working out the energy required to accelerate the bike in a straight line, subject to the proviso that the pound on the wheels is all right at the edge. In practice this doesn't quite happen, so the factor is always a little shy of 2.

As an aside, the factor of 2 comes from the fact that the wheels rotate in such a way that their lowest points are always stationary relative to the ground. This is not the case for the cranks (except perhaps in perverse gear ratios), so it is generally NOT true that a pound on the cranks is worth two on the frame (i.e. not all "rotating weight" is the same). At realistic road-cycling speeds and cadences, a pound on the cranks is worth only a little more than a pound on the frame.

So maybe we'd be a little more accurate if we said a pound at the rims, but this is about the same as what you think the weight of your wheels is after tires and tubes.

You're right about other rotating weight not being "weighted" as much. The moment arms are shorter and the rotational velocity is less.

The formula for a pound off the body is worth 1/10 mph has more to do with physical conditioning than physics. Although you will get some benefit from the physics as well.

Note that I weigh 160 currently and the lowest I could possibly go without loosing a lot of power would be 145. If I really wanted to I could go down to maybe high 130's but my speed on the flats would suffer so much that any marginal gains climbing would be more than offset. So don't loose mass at the expense of power. None the less I'm thinking of shedding some extra mass over the next year.

mjw_byrne said:

As an aside, the factor of 2 comes from the fact that the wheels rotate in such a way that their lowest points are always stationary relative to the ground. This is not the case for the cranks (except perhaps in perverse gear ratios), so it is generally NOT true that a pound on the cranks is worth two on the frame (i.e. not all "rotating weight" is the same). At realistic road-cycling speeds and cadences, a pound on the cranks is worth only a little more than a pound on the frame.

Does that apply for hubs too then? Or are they considered to be part of the frame, since they are attached to it and the relative distances are always the same?

The reason I ask is because they rotate too, but around an axle whose separation from the centre of gravity of the bike is constant.....

I'm a bit confused..... 🙁

Hubs don't count, sorry, only weight at the rim. The formula for moment of inertia is the integral of the square of the moment arm with respect to mass. The moment arm on the hub is super small so it's not going to have significantly more energy than a hub moving at the same speed but not spinning.

hwttdz said:

I saw someone mention that a "pound on the wheels is worth two on the frame" and someone else said, "no you're stupid" (slightly rephased the second one there). Well in the interest of bettering out knowledge I broke out some old equations, massively oversimplified things made gross assumptions, and got an answer (how much it's worth is questionable).

For an 8kg bike (17.6 lbs approx) traveling at 10m/s (22.36 mph) with a 1700 grams of weight at the rim (rims and tires, rims are 500ish tires 250ish, tubes 100ish) 700mm wheel diameter (pretty normal).

1 pound of mass at the rim has the same amount of energy as 1.025 pounds of non-rotating mass. (1:1.025 make the units whatever makes you happy, g:g kg:kg doesn't matter).

So it's not as much of a magic bullet as some might think, but hey it's something.

You left out spoke weight...this is a lot of the mass. Difference between straight gauge and butted is huge.

All I know is a set of very light wheels makes a HUGE difference in climbing. I can definitely feel lighter wheels more than I can a lighter bike.

Spokes count but not quite as much as rims, because some of their mass is close to the rim and some is close to the hub. But steel is super dense so that material near the rim has some impact. Makes alloy nipples sound good doesn't it.

hwttdz said:

Spokes count but not quite as much as rims, because some of their mass is close to the rim and some is close to the hub. But steel is super dense so that material near the rim has some impact. Makes alloy nipples sound good doesn't it.

a gram of rotating mass in the spoke has 1.33 times more energy than a gram of mass on the frame. the difference between the ratio of 2 for rim mass and 1.33 for spoke mass arises from the difference in the mass moment of inertia for mass that is distributed at at constant radius versus mass that is distributed evenly along the radius.

Mr_Potatohead said:

I think what you are after is this. Assume constant power output, P = Fv. So when you drop a pound of weight from your bike in a climb, the force required to lift you and your bike up the hill goes down and your speed can go up proportionally.

So for example drop 2 pounds off a 180 pound rider + 20 pound bike in a climb and you will increase your speed by 1 percent, with no increase in effort, dropping your time up the hill by 1 percent. Which comes out to saving about 36 seconds per hour of climbing.

But 90 percent of a riders effort in climbing goes towards lifting their own body weight.

The rotational inertia of the wheel only comes into play during acceleration. Otherwise the weight of the wheel is the same as the weight of the spare tire around your waist or in your saddle pack.

Agree. Would add that on level-ground acceleration, the effect of losing say 500 grams on the wheels isn't huge to most of us. Using the factor of two, our 200 lb bike/rider would accelerate 0.05% faster with the lighter wheels.

To illustrate, I looked at a constant-acceleration example. Say with your standard wheels you can accelerate at a constant 1 m/s^2, going from 0 to 10 m/sec (22 mph) in 10 seconds. Your identical twin, on the 500 g lighter race wheels, would then accelerate at 1.005 m/s^2. You'd cover 50 meters from a standstill in the process, while your twin next to you on race wheels would hit the same 50 m line in 9.975 sec....about 0.25 meters ahead.

That's a clear winning advantage for a sprinter, but not important to me as a club rider.