|Speed as a function of power...||NewDayNewWay|
Aug 2, 2003 9:28 AM
|So we've all seen power = torque X cadence. That's nice and simple. But here's the next obvious question: What is the relationship between SPEED and power?
OK, I realize this is not so simple, so let's simplify the situation a little. Assume the following: 1) no loss of power transfer from you to the road 2) no rolling resistence, 3) completely flat road, 4) no wind resistance, 5) maybe a few other needed assumptions I'm leaving out.
So assuming the above simplified scenerio, what is the formula for the relationship between power and speed, assuming a bike and rider combined weight of X lbs?
2nd question: What is the formula for the relationship between calories burned and power? OK, I guess we need a simplifying assumption here as well, such as the calories are being converted directly into power to move the bike forward (opposed to calories needed to blink you eyelids, for example).
Tx in advance for any insights on understanding all this. NDNW
|no drag = near infinite speed, eventually||DougSloan|
Aug 2, 2003 9:34 AM
|I think you have assumed too much. You assume no drag whatsoever. That being the case, you are almost describing a free fall in a vacuum, and would be constantly accellerating.
Go here: http://www.analyticcycling.com
|I believe you need some units of measure||Live Steam|
Aug 2, 2003 9:35 AM
|Watts are the units cyclists measure their power output in. That much I know. You'd have to search some training sites on the web for a formula. I think you also need some type of instrument to measure the output.|
|By your (strange) criteria, speed = infinite||Kerry Irons|
Aug 2, 2003 4:18 PM
|You state no aero drag and no rolling friction, therefore no resistance whatsover. A constant application of power will yield infinite speed. F = ma (force = mass times acceleration). Apply a force F to a mass m, and a is fixed. Since (under your bizzare conditions) acceleration would go on forever, speed reached would be infinite. However, I think the following two formulas are what you are really asking for:
P = (Vg*W*(K1+G) + K2*(Va)^3)/375
Where P is in horsepower, Vg is ground speed (mph), W is bike/rider weight in pounds, G is the grade, and Va is the rider's speed through the air (mph). Grade is feet or altitude gain per foot of horizontal distance, and while often expressed in per cent, in this equation is used as a decimal (a 6% grade is 0.06). K1 is a lumped constant for all frictional losses (tires, bearings, chain) and units conversion, 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 0.0083. Note that power to overcome friction and gravity is proportional only to rider weight and ground speed. Power to overcome wind drag is proportional to the cube of the air speed. For reference, 1 hp-hr = 641 calories delivered to the pedals and 1 hp = 746 watts. Here, all calories are kg-calories, "big" calories, or "food calories." The human body runs at about 24% efficiency so to deliver 1 hp to the pedals requires the body to consume 2700 calories, more than 4 times the 641 calories delivered to the pedals.
Obviously, both of the lumped constants in this equation depend on many variables, such drive train efficiency, the rider's position, tire pressure, road surface, etc. Also, recognize that air speed is not constant in speed or direction nor easily measured. It's certainly reasonable that the aerodynamic lumped constant would be different in cross winds or tail winds than in direct head winds, as the profile the bike/rider presents to the wind is different in each situation. Also, wind speed as seen by the bike/rider is not uniform except in zero wind conditions. "Weather report" wind speed is measured at some distance above the ground in free air with no obstructing trees or buildings nearby. Yet, by definition, the wind speed is always zero right at the road surface. Assuming a single wind velocity and a single lumped drag constant are just two of the dreaded "simplifying assumptions" of this equation. You should know that many high powered Computational Fluid Dynamics modelers have looked at the "bicycle problem" and pronounced it "really hard." In layman's terms, this means that much more sophisticated models can be developed, but they will still have simplifying assumptions.
Given this simplified equation, however, you can calculate some values of interest. For example, assuming zero wind, you get the following results for calories required and power delivered to the pedals (watts):
- 630 calories per hr (174 watt-hrs) for a 200 lb. bike+rider to go 20 mph on the flats (76% of effort to overcome aerodynamic drag), or 5.7 mph on a 7% grade (3% of effort to overcome aerodynamic drag).
- 1125 calories per hr (310 watt-hrs) for a 200 lb. bike+rider at 25 mph on the flats (83% of effort to overcome aerodynamic drag) or 9.8 mph on a 7% grade (7% of effort to overcome aerodynamic drag).
- 585 calories per hr (161 watt-hrs) for a 140 lb. bike+rider to go 20 mph on the flats (82% of effort to overcome aerodynamic drag), or 7.4 mph on a 7% grade (3% of effort to overcome aerodynamic drag).
- 1070 calories per hr (295 watt-hrs) for a 140 lb. bike/rider at 25 mph on the flats (87% of effort to overcome aerodynamic drag) or 12.6 mph on a 7% grade (7% of effort to overcome aerodynamic drag).
|By your (strange) criteria, speed = infinite||NewDayNewWay|
Aug 2, 2003 6:58 PM
|Very insightful. I realized after I made my post that I was making too many assumptions! Maybe I'm just an oddball for even being interested in such stuff, but I find it connects me a little more to my riding in some twisted manner. Thanks for the info!|
Aug 3, 2003 3:50 PM
|You are not alone. You really should go to www.analyticcycling.com and see a detailed description and derivation of the formulas relating power to speed (or speed to power) as well as other factors controlling bicycle motion.|| |