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Simplified power measurement(9 posts)

Simplified power measurementContinental
Apr 15, 2003 8:47 AM
In principal it is easy to measure the power in terms of watts required to maintain a steady speed on a bicycle. The bicyclist must be traveling at steady speed on a uniform grade with constant wind velocity (flat, uphill, or downhill, with the wind or against the wind, but no change in steepness or gusts of wind). The cyclist riding at steady speed begins to coast. When the speedometer updates the cyclist starts a stopwatch and remembers the velocity, uses the stopwatch to time a short interval of coasting, then stops the stopwatch when the speedometer updates at the end of the coasting interval, and remembers the final velocity. To calculate the power in watts at the average velocity during the coasting interval the cyclist will need to know the combined weight in pounds of the cyclist and the bicycle, the velocity in MPH at the beginning of the coasting interval, the velocity in MPH at the end of the coasting interval, and the duration in seconds of the coasting interval. The change of velocity during the coasting interval should be about 2mph or 3 mph. Example data from my test: Velocity when speedometer updates at beginning of coasting interval = 23.7 mph. Velocity when speedometer updates at end of coasting interval = 21.9 mph. Duration of coasting interval = 3.4 seconds. Combined weight of bike and rider = 202 lbs

Step 1--Calculate average velocity during the coasting interval. (initial velocity + final velocity)/2.
Example: (23.7 mph + 21.9 mph)/2 = 22.8 mph

Step 2--Calculate power required. Watts = 0.0954 * weight of bike plus rider in lbs * average velocity during coasting interval * (initial velocity in MPH - final velocity in MPH)/ duration of coasting interval in seconds.

Example: 0.0954 * 202 * 22.8 * (23.7 -21.9) / 3.4 = 233 Watts
Under these riding conditions I require 233 Watts to maintain a constant 22.8 MPH

The key to successful calculation of watts generated is to get accurate measurements of the velocities and times. And don't lie about your weight. It's better to get several sets of data at the exact same location on the road, at as close to the same initial speed as possible, then average the data. Averaging the data will minimize the effect of measurement errors.

The speed display on Cateyes seem to update more predictably than the updates on my new Flightdeck. If anyone knows the algorithms used to compute the displays and determine the update intervals on these computers, please send me a note.

Also, I am trying to determine the reproducibility and resolution of this power measurement technique. If anyone wants to collect data for me, please send the raw data to bikewatts@yahoo.com. I'd especially like to see data from someone who has a Powertap to compare.
Simplified power measurement for geeks onlyContinental
Apr 15, 2003 8:51 AM
Power is defined as force times velocity in the direction of the force. Force equals mass times acceleration. When traveling at uniform velocity, the net force is zero and the acceleration is zero. The gravitational force from change in altitude, frictional forces, and aerodynamic forces acting on the cyclist are offset by the force generated by the cyclist. When the cyclist begins to coast, the gravitational, frictional, and aerodynamic forces decelerate the cyclist. The rate of deceleration can be measured as the change in velocity divided by the change in time over a time interval. For ease of calculation the velocity in MPH should be converted to meters/sec and the acceleration calculated in meters/sec^2. Now, knowing the combined weight of the bicycle and cyclist and the rate of acceleration, the Force can be calculated. For the calculation it is easiest to convert the weight in pounds to mass in kilograms, and calculate the force in kg*m/sec^2 or Newtons. Then, multiply the force times the average velocity during the costing interval to obtain Newton*meters/sec or Watts. Two approximations are made in this equation. The first approximation is that the average force during the coasting interval occurs at the average speed during the coasting interval. In reality, the aerodynamic force increases as approximately a square of the velocity, and the frictional force increases approximately proportional to velocity, while gravitational force from changing altitude is independent of velocity. Therefore, the force at the average velocity is lower than the average force, so this approximation leads to an overestimation of the power. However for small changes in velocity the error introduced by this approximation is negligible. The second approximation is that the inertia of the spinning wheels is negligible compared to the total inertia. This approximation leads to an underestimation of the power. Unless someone is riding with very heavy wheels, the error introduced by this approximation is also negligible
You must be a top notch engineerbigrider
Apr 15, 2003 11:10 AM
Who else could delicately merge accuracy with confusion and then call it a simplified power measurement. You have won my highest esteem (like you care) and therefore I will repeat my favorite engineering joke.

An engineering student is walking across campus when he sees his fellow engineering student buddy riding a bike. He asks " where did you get the bike?". His friend replied " You won't believe my luck. Yesterday after class this beautiful blonde rode this bike up to me wearing a full length mink coat. She stopped opened up the coat to expose her naked body and said " take what you want""

Good choice the friend replied, the coat probably wouldn't have fit you.
I don't get it. Who would wear a mink coat while riding? nmContinental
Apr 15, 2003 11:37 AM
LOL, That verifies the engineer theory..... nmbigrider
Apr 15, 2003 12:32 PM
geek suggestion: for each trial, repeat in opposite direction.curtybirdychopper
Apr 15, 2003 12:05 PM
Average the two results in opposite direction to normalize the effects of the slope of the road.
another geek suggestion:bicyclerepairman
Apr 15, 2003 3:03 PM
First, perform the trial at noon during a total eclipse in an equatorial country, to eliminate tangential tidal forces and the coriolis effect. Average the two results in opposite direction to normalize the effects of the magnetic field of the earth.
Why not calculate directly?Kerry
Apr 15, 2003 4:34 PM
calories/hr = [V*W(.0053 + %G/100) + .0083(V^3)]*7.2

where V is speed, W is bike + rider weight in lbs., and %G is grade in per cent. Certainly no more complicated than your formula and doesn't require field measurements. Guess what - this formula would say 236 watts for your 22.8 mph! The factors listed here (0.0053 for friction + rolling resistance and 0.0083 for aerodynamic drag) are obviously not absolute. They will vary with efficiency of the tires and drive train, and with the aerodynamics of the bike + rider combination. Both of these assume a racing position on a racing bike. A clunker bike or a more efficient riding position will change these numbers, which are averages anyway. 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 = 2700 calories (because of human metabolic efficiency of 24%); 1 calorie = 0.276 watts; 1 hp = 746 watts. Here, all calories are kg-calories, or "food calories."

Recognize that true air speed is a bit difficult to estimate in anything but calm conditions, so substituting "speed through the air" for the second velocity term is problematic. You have to take into consideration wind angle (and the change in drag coefficient that comes with it) and the fact that the air speed near the ground is not the same as it is at head level. A simple head wind faced by a cyclist can be considered to be roughly 1/3 of the "posted" wind speed from a weather report (that wind speed is taken 30 feet above ground and away from any buildings, trees, etc.) By definition, the wind speed right at the ground is zero, so a bike sees a zone where the speed is heading toward nothing. Because of this, you should always be feeling a head wind and the more aero your position, the faster you will go. That doesn't prevent a lot of us from sitting up and relaxing a bit when we get pushed by the Giant Hand, but we would be faster if we stayed on the drops. No one has a model that can adequately represent these variables. For this reason, aerodynamic testing outdoors must be done in calm conditions.

In English, calories per hour equals speed times weight times the quantity 0.0053 plus grade plus 0.0083 times speed through the air cubed, all times 7.2. The Va term takes into account the added work due to working against a head wind, but recognize that the wind you actually face on the road is probably only 1/3 of the "posted" wind speed. For example, if you are riding 20 mph into a "15 mph head wind" you probably are only experiencing 25 mph for calculating work. The reason for this is that the reported wind speed is taken 30 feet above ground and well away from buildings, hills, trees, etc. When you're riding, you often have things that block the wind and wind speed near the ground is much less.
Want to experiment and find effects of things likeContinental
Apr 16, 2003 6:48 AM
tire pressure, riding position, different bikes.