|how many watts?||BostonDave|
Nov 7, 2003 1:37 PM
|Is there any simple way to calculate a rough approximation of the amount of power a cyclist is generating? I know there are fancy monitoring systems for pros, but I'm just curious to know what I'm doing--even just on my trainer.|
|re: how many watts?||divve|
Nov 7, 2003 2:00 PM
|Analytic cycling ...||Humma Hah|
Nov 7, 2003 2:13 PM
Nov 7, 2003 2:20 PM
|I've had 2 trainers that displayed watts, including the Computrainer. My results with them has been pretty close to the calculations on analyticcycling.com. Climbing tends to be more accurate than faster speeds, though, because for some reason I get extremely aero, allowing me to go faster on the flats and down hills than my meager power would permit by the calculations.
|re: how many watts?||asgelle|
Nov 7, 2003 2:28 PM
|To estimate power on your trainer,
gives curves of measured power vs. speed for several brands and models. As other have said www.analyticcycling.com to estimate power on the road.
|Formulaeic approach||Kerry Irons|
Nov 7, 2003 5:04 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. 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."
|do you mean calories/hour ?||PeterRider|
Nov 7, 2003 5:15 PM
|hp = power
calories = energy
watt = power
Nov 7, 2003 5:30 PM
|1 hp = 746 Watt
1 kcal/hour = 1.163 Watt
1 hp = 641.4 kcal/hour
Calories define work and watts and Kcals/hour define power.
Also, apply the efficiancy coefficient at the end, otherwise your conversions do not make sense for other purposes.
|The rest of you equation looks good though. (nm)||gala7516|
Nov 7, 2003 5:46 PM
|It's not simple, but it requires no special equipment||KeeponTrekkin|
Nov 7, 2003 6:09 PM
|Find a good, long hill of constant slope that you can ride.
If you have a cyclocomputer with altimeter, ride the hill and measure your time, distance and altitude gained.
If you have an ordinary cyclocomputer, go to topozone.com and find your hill. Determine the altitude gained.
Go to analytic cycling; find the page to determine power, given speed, weight, slope, etc.; input the correct data.
The result is your watts of power. The result will be soberingly pathetic when you think of it in terms of light bulbs.
If you monitor your body carefully and if the hill is long enough, you can determine if you did it entirely aerobically or if you lapsed into anearobic respiration. If it's aerobic, that's a good baseline to know. It is your power for sustained efforts. You can also use analytic cycling to confirm the data for straight and level.
|Here's what worked for me, but it's not too simple||Continental|
Nov 7, 2003 7:21 PM
|Kerry's formula is easier and accurate, but my calculation below can show power change from changing position or equipment.
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.
Nov 8, 2003 5:23 AM
|Once you do this, you can go to analyticcycling and determine by trial and error what your effective frontal area is (assuming a value for rolling resistance). Then, all their tools can be used with your numbers that have all been measured.
Comment: As power and speed are not linearly related, I think this method requires short intervals to be reasonably accurate, as in your example.
You could also use your climbing determined watts and your sustainable steady speed to estimate your frontal area. It would be good to cross check.