General Cycling Discussion - Bike math: Is 25+10+35?

Suppose I am going to cycle at a speed of 10 mph. It will take me x units of force to go 10 mph. Now...I am on the same bike, which, by the way, has an electric motor. I am using the motor to move the bike at 25 mph. If I add X amount of force to it, i.e. the amount of force needed to make the bike travel at 10 mph, will the result be 35 mph -assuming my gear ratio has changed?

No, wind resistance isn't linear with respect to speed - it's a cubed relationship. If you add the same amount of power to the bike as you did at 10 mph when traveling at 25 mph, you won't go 35 mph.
See the power formula at the bottom of this link --> http://en.wikipedia.org/wiki/Bicycle_performance

No, wind resistance isn't linear with respect to speed - it's a cubed relationship. If you add the same amount of power to the bike as you did at 10 mph when traveling at 25 mph, you won't go 35 mph.
See the power formula at the bottom of this link --> http://en.wikipedia.org/wiki/Bicycle_performance

Correct.

That's our experience with cars too. Cars today are more aero. The faster the car goes, the more power it needs, more gasoline. I can't imagine the amount of power it takes the race cars to hit 200 mph.

Cubed? my ballistics programs use squared!
Anyway, the effort increases dramatically as the speed increases.

Air resistance increases exponentially. So it will take twice as much power to go 32MPH as going 25MPH.

Only way to overcome air resistance is to install a fairing.

Air resistance increases exponentially.

Quadratically, not exponentially.

A significant difference...thanks!

Only way to overcome air resistance is to install a fairing. You mean, to lower air resistance. There only two ways to truly overcome air resistance: stop moving or get rid of air. Neither is desirable for a cyclist. :)

Don't confuse force with power. Air resistance (a force) goes up with the square of the speed, but power is force times speed, so the power requirement goes up with the cube of the speed. To double the speed takes 8 times the power.

Don't confuse force with power. Air resistance (a force) goes up with the square of the speed, but power is force times speed, so the power requirement goes up with the cube of the speed. To double the speed takes 8 times the power.

Correct, but for air resistance only.

A car, and a bike, has rolling resistance as well, and rolling resistance power is roughly linear with speed.

For a typical bike+rider, the combined effect of air resistance power and rolling resistance power increases with the speed raised to the power of ~2.6, as an approximation, at least for typical speeds (say 15-35 km/h). It's close enough to be within a few percent of the actual value, as calculated by combining the true components.

To answer the OP, the difference in power used between 10 and 25 mph should be around (25/10)^2.6, which is roughly 11 times the power. That means that the power required to travel at 10 mph is 1/11 of that required for 25 mph, so adding the same amount of power to 25 mph, will only add just over 1 mph.

Right...I was considering only air resistance. I didn't have a good feel for the relative magnitudes of rolling resistance and air resistance, but doesn't rolling resistance vary wildly depending on the type of bike? Compare a mountain bike with fat, 26", low pressure, knobby tires with a road bike with skinny, slick, high pressure 700's for example.

Right...I was considering only air resistance. I didn't have a good feel for the relative magnitudes of rolling resistance and air resistance, but doesn't rolling resistance vary wildly depending on the type of bike? Compare a mountain bike with fat, 26", low pressure, knobby tires with a road bike with skinny, slick, high pressure 700's for example.

Right, that would be encapsulated in the first term of the equation below. Specifically the constant K1. If rolling resistance is less, K1 would be smaller. Rolling resistance power has a linear relationship to speed. Air resistance has a cubed relationship.

P = g * m * Vg(K1+S) + K2 * Va^3

Where P is in watts, g is Earth's gravity, Vg is ground speed (m/s), m is bike/rider mass in kg, s is the grade (m/m), and Va is the rider's speed through the air (m/s). K1 is a lumped constant for all frictional losses (tires, bearings, chain), and is generally reported with a value of 0.0053. K2 is a lumped constant for aerodynamic drag and is generally reported with a value of 0.185 kg/m [5].
Note that the power required to overcome friction and gravity is proportional only to rider weight and ground speed. The power required to overcome wind drag is proportional to the cube of the air speed.

I have been reading through this thhread with great interest and I am thoroughly thankful that my pretty good schooling in mathematics has helped me to actually understand it a little (might even be tempted to slap together a neat little power output calculator in Excel so people can know what speed they can roughly expect with any given number of gear inches). I just wanted to say one thing.

I love bike nerds

"I love bike nerds"

Me too :) Thanks for all of your help!