A Theory on Gearing: Part 2

Part 2: Rolling Resistance

Some astute readers might note there’s another factor besides drag to consider, and that’s rolling resistance from the tires.

The formula for rolling resistance is:

F = Crr*N

Coefficient of rolling resistance (Crr) and normal force (N) are both essentially constant for the point of this exercise, unless you plan on riding to the moon or losing quite a bit of weight during your ride or getting a new wheel with a new tire from the team car. If you recall from earlier:

p = F*V

Let us again, be lazy, and assign arbitrary values of 1 mystery units to Crr and N, so that simply:

p = V

Note that power here is denoted by p, because I don’t know where subscripts are. P from the previous post and p here are both power, but in different arbitrary units. In order to convert p to P, I’ll introduce a new conversion factor, constant a.

P = ap

Adding in the aerodynamic drag component, substitute V for p, we get:

P = aV + V^3

Where a is our conversion factor, V is speed in whatever units, and P is our arbitrary power units that are definitely not watts.

Now we need to determine the value of a to establish the relative relationship between the rolling resistance component (aV) and aerodynamic drag (V^3). We can do so by determining they they intersect and solving for a.

aV = V^3

a = V^2

Now we just need to know V where rolling resistance is equal to aerodynamic drag, and we can skip all the other math and estimating we would have had to do. There’s a few numbers thrown out for when tire rolling resistance and aerodynamic drag are even, oft repeated seems to be 15mph. I can only assume this is because someone misread a graph without unit labels that actually said 15km/h. Schwalbe puts it at around 18km/h (11.2mph). BikeCalculator seems to put it at around 15.1km/h (9.4mph) using default values. Both of these may be right, as the relationship between drag and rolling resistance depends on the bike, rider and tires. But they give a good ballpark figure.

Neither of these models includes drive train losses, but calculating them doesn’t really change the conclusion so I’ll skip it.

The equation for arbitrary power units for Schwalbe is:

a = (11.2)^2 = 125.44

P = 125.44V + V^3

The equation for arbitrary power units for default values for BikeCalculator is:

a = (9.4)^2 = 88.36

P = 88.36V + V^3

And since bike calculator gives us a speed of approximately 15mph for 100w, we can actually come up with a conversion factor to turn our arbitrary power units P into watts.

W ~= (V*9.4^2 + V^3)/47

Which should come within a watt of BikeCalculator with default values.

On flat land with no wind, at a speed of 7.5mph, which is equivalent to 34×32 at 90rpm, about the lowest gear you will find on a road bike, if you increase the gear ratio at a given cadence by 10% you require an additional 19.0% power. At 33.9mph, equivalent to 53×11 at 90rpm, if you increase the gear ratio by 10% you require 31.9% more power. This might be closer to the result you were expecting from the previous post on drag. That seems like quite the difference at first glance, until you realize these speeds are much slower and faster than most people will go on windless flats.

If we narrow it down to normal cruising speeds of amateur recreational riders to 15-20mph and run the same calculations with 15mph and 20mph, we get 26.6% and 28.9% respectively. Much much closer together.

Consider again, that our cog choices must be in integers and often times a limited selection of integers, and we can not arbitrarily increase gear ratio by 10% in real life unless you have a NuVinci.

A Theory on Gearing: Part 1

Part 1: Drag

You’ve probably heard that drag increases exponentially with speed, which is why aerodynamics only matter at fast speeds. Fast speeds variably starting anywhere from 15mph to 25mph. Some people treat it as a magical barrier that only kicks in once you’ve passed a certain speed threshold, then suddenly it takes exponentially more power to increase your speed because of aerodynamic drag. Indeed, anyone who has tried to go as fast as they can can attest to the effects of drag. Some people will say this applies to gearing, and this exponential drag increase is why cogs need to be clustered closely together at the high/small end of the cassette.

NASA provides us with the equation to calculate drag:

One way to deal with complex dependencies is to characterize the dependence by a single variable. For drag, this variable is called the drag coefficient, designated “Cd.” This allows us to collect all the effects, simple and complex, into a single equation. The drag equation states that drag D is equal to the drag coefficient Cd times the density r times half of the velocity V squared times the reference area A.

D = Cd * A * .5 * r * V^2

Cd is called a variable, and it is indeed variable and influenced by a host of variables, but your Cd but for a given rider, for given equipment, in a given position, for the purpose of calculating drag here it’s constant. You will see people doing aerodynamics tests calculating the Cd of a certain part and so on. r is also essentially a constant, given that you cycle at reasonable altitudes in reasonable temperatures, and so on, we can also treat r as a constant for the purpose of this exercise. A will also be constant for this exercise, as the hypothetical bicycle and rider in this example won’t change in area.

Don’t worry, I’ll skip the calculus and present this in simple algebra.

So we will represent the product of these factors as the constant c.

c = Cd * A * .5 * r

Now we simplify to:

D = c * V^2

Now I’m going to just assign arbitrary values to the factors so that

c = 1

and

D = V^2

because they really don’t matter for this exercise. If it really bothers you, you can multiple our arbitrary drag units (D) by c. We’re just focusing on the relation of an increase in velocity on the increase in drag.

So we begin by calculating our arbitrary drag units (D) for an arbitrary velocity (V) of 15 arbitrary velocity units that are not miles per hour.

D = (15)^2 = 225

We can do this again for 16 and so on.

D = (15)^2 = 225

D = (16)^2 = 256

D = (17)^2 = 289

D = (18)^2 = 324

D = (19)^2 = 361

D = (20)^2 = 400

When we increase V from 15 to 16, D is increased by 31, from 16 to 17, and increase of 33. Then 35, 37 and finally 39. Each additional unit of V results an increasing amount of D. This is the common understanding of drag be proportional to velocity squared. From this, some come to the conclusion, you need smaller gaps between gears to compensate for this non-linearity, so each increase in speed is proportional to a certain amount of power.

However, we can rephrase the problem into a question of gearing. Let’s assume out cassette isn’t bound by the fact that the number of teeth have to be integers. Each gear results in a 10% greater gear ratio, and therefore a 10% increase in speed at a given cadence. That is after all, how we look at gearing when examining gear calculators.

Starting again with 15.

D = (15)^2 = 225

For our next gear

V = 15*1.1 = 16.5

D = (16.5)^2 = 272.25

And again

V = 16.5*1.1 = 18.15

D = (18.15)^2 = 329.4225

We’ll stop there. Maybe you noticed a pattern in the way it increments?

272.5/225 = 329.4225/272.25 = 1.21

That is, every time we change gearing and therefore speed by 10%, the amount of drag is increased by 21%. In fact, plugging in our relative change in gearing results in the relative change in drag.

D = (1.1)^2 = 1.21

Let’s not stop there though. The equation to calculate the power needed to counteract a drag is

P = D*V

D = V^2

P = V^2*V = V^3

Now calculating arbitrary power units (P) that definitely aren’t watts if V is measured in mph, which it isn’t.

P = (15)^3 = 3375

P = (16.5)^3 = 4492.125

P = (18.15)^3 = 5979.018375

Do you notice the relationship this time?

P = (1.1)^3 = 1.331

Conclusions:

An increase of 10% gearing increases drag by 21% and requires 33.1% more power, or force at a given cadence to counteract that drag, at any speed, fast or slow. Or decreasing the gear by 9.1% (the same gears, the change just measured from a different point of reference, 1-1/1.1) is 24.9% easier, at least the power component to counteract the increased drag.

 

 

GUB SI Seatpost

gub si
GUB SI Seatpost (27.2 x 240mm)

This seatpost weighs only 146g when cut down to 240mm, saving up to 128g over a generic 300mm seatpost. In the original form at 350mm, it weighed 188g. It resembles numberous light weight seatposts and some of the lightest seatposts share the cross-pin design. The yokes are flat, making them easier to align with the rails. It has an oval bore to save weight and the cross-pin has a flange to prevent slipping and is also angled to keep the screws straighter. However, it does not have spherical screw-heads although the need is lessened because the cross-pin is angled. The clamp design offers less support to rails, imparts a bending load instead of a clamping load to grip them, and is very wide offering little adjustment, features common to this design. Larger riders and riders with lightweight rail materials should exercise caution when using this post. For each centimeter removed, the weight will be reduced about 3.8g, but remember to leave enough post for safe usage.  In stock form, at $12 and saving 8.8g per dollar, it represents an excellent value, moderate cost upgrade. This increases to an exceptional 10.7g per dollar when cut down. It can be found on eBay.

generic sp
Generic Seatpost (27.2 x 300mm)

 

Overview:

Price: $12 [Moderate Cost]

Value: 8.8g/$ ($0.09/g) [Exceptional Value]

✔ Recommended with caveats

These seatposts are zero setback and offer even less adjustment range than the GUB GS Seatpost, so they are not for everyone. These are both cheaper and lighter, but the clamp design is also less friendly to exotic rail materials. There is little to lose by cutting them down for further weight savings.

Example Bike Build Spreadsheet

If you want to build a light weight bike on a budget, you have to keep track of every part to figure out upgrading which parts saves you the most weight for the least money. You should budget $10 for a cheap kitchen gram scale to verify weights as well.

I’ve done most of the hard work for you, and modified and simplified one of my personal spreadsheets for your use. Attached is a spreadsheet with two examples. One is an example build on a $1,500 $1,400 budget that weighs 16 lbs with pedals and cages. The other is is an example of a sub-UCI build for less than $2,000 $1,900.

Continue reading

Forte Light/Nashbar Lite Tube (48mm)

forte tube
Forte Light/Nashbar Lite Tube

This butyl inner tube weigh only 63g, saving up to 45g over generic inner tubes. A pair of them saves 90g of rotating mass. When used as a spare, they are also more compact and reduce the weight of your bag saving even more weight. Advertised as 67g, I weighed 5 of them without caps or nuts, and they weighed 62g, 62g, 63g, 63g and 65g, averaging 63g. Even if you patch them up, they are still significantly lighter than a normal tube. They are less expensive that other light tubes and don’t require the high maintenance of latex tubes but are just as light and have rolling resistance somewhere in between. They frequently go on sale as well. The value is even more impressive when only considering the price on the margin. Performance Bike and Nashbar are the same company, but Performance bike seems to have a better deal when buying bulk, 4 for $21, barely more than normal tubes, but charge more for the long stem versions because they have retail locations and people with deep carbon wheels are willing to pay when they get a flat. Be warned, they have threaded stems for those that dislike that. At $5 and saving 9g per dollar, it represents an exceptional value, moderate cost upgrade. It can be found at Performance Bike or Nashbar.

kenda tube
Generic Butyl Tube

 

Overview:

Price: $5 [Low Cost]

Value: 9.0g/$ ($0.11/g) [Exceptional Value]

✔ Recommended