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.

 

 

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