Today I have been playing around with some information from:
www.mayq.com/Best_European_trips/cycling_speed_math.htm

The whole page can be expressed thusly:
WattsRider =Cfriction X velocity X weight + Cair(V + Vwind)^2 X veloc. + Cslope X weight X %slope X V + acceleration

It looks like a lot of stuff but it simplifies into:

Watts necessary to overcome friction = Constant value of friction, which is around .1, X velocity X weight
For me personally, that comes to .1 times 5.8 meters per second, or 13 mph times 77.2 my weight with bike in kilograms
I get 45 watts of power that I need to produce just to overcome the friction of the road and my weight on the hub bearings. The additional friction of chains, idlers, leg hair, cranks etc. is measurable but insignificant overall.

45 Watts for friction plus 70 watts to overcome air resistance at 13 mph = 115 watts to go 13mph flat, no wind

OR: Watts necessary to overcome friction = Constant value of drag times velocity cubed, assuming no wind.
I had to interpret the Cair, the constant value. An upright rider on a Mountain bike is about .9. A road racer in lycra and well tucked in is about .36. For recumbents the only measured value was for a Easy Racer Gold Rush at .046. I'll use .36 because I can't get any hard numbers for high racers.
That is: Watts to overcome air resistance at 13 mph = Cair .36 x velocity cubed 5.8 m/sec = .36 x 195 = 70 watts

On a slope the wind and friction become secondary to gravity because the rider is going slowly uphill.
Watts to climb = Cgravity x weight x slope % x velocity = 9.8 m/s (77.2 kilograms) x .06 for a 6% slope x velocity in m/sec

To make this useful to myself I changed it to find how fast I might expect to go on a given slope when producing 160 watts. That happens to be my max watts working just below my lactate threshhold.

That means velocity = 160 divided by (756.6 x .06 ) = 8 mph So climbing up Yarnell Grade between time stations in Congress and Prescott Arizona will take me, say 1hour 15 minutes at 8 miles per hour. I will need to do four hours going at least 14.5 mph to make climbing that grade average out to 13 mph, which is my target average speed. Since road engineers try to keep road grades at 6% or less I can more or less plan climbs at that speed. My speed drops .7 mph for each 1% increase in grade over 6%. Below a 4% grade velocity increases to where air resistance becomes significant.
Notice that climbing a hill is purely a problem of power vs weight. Going down the other side weight doesn't help as much as it hurt going up. Going down aerodynamics comes into play with a vengence. Most people can coast down a steep grade fast enough that maintaining control is more important than small increases in speed. In most cases bicycles just aren't geared so that pedalling over 25 miles per hour makes any difference.
As an aside, my trike with a Rohloff hub and Schlumpf speed bracket allow me to pedal against resistance at 35 mph. Much beyond that pedalling makes the steering wobbly.

A head wind is computed with (velocity + wind) squared. So going 13 mph into a 10 mph headwind requires:

.36 X (5.8 m/s + 4.5 m/s) squared x 5.8 = 220 watts, versus 115 watts to do 13 mph with no wind.

Now you know what your legs have always tried to tell you. That is what I did with my day today. Feel free to use it.

Useful numbers: 1 kilogram = 2.2 pounds 1 horsepower = 1000 watts 1 meter per second = 2.237 mph
5 mph = 2.25 meters per second 3 liters of water = 6.6 lbs or one big camelback
Average filled large water bottle = 1.6 lbs


ChrisM