My Crazy Idea to Measure Power without a Meter, sort of...

Hypothesis:

Watts are simply a measure of how much “work” was done to move an object, i.e., bike + rider, from point a to point b.

Power meters use a “torque tube” to measure the amount of torque being applied at any given time to measure watts. However, I don’t think this is the only way.

With gps technology growing more and more accurate it’s is probable that you could use gps with preloaded topography maps to measure watts accurately over the course of a ride. You simply need the correct equation that calculates current speed over the given terrain, taking into consideration the total weight being moved at that speed. The more rapidly this equation could be calculated the more accurate the data.

PROBLEM #1: In the discussed equation, watts would still be calculated as work being done, even while coasting. The equation could take into account that you are losing elevation due to the topography and gps, however, the times you coast being in the draft of another rider would still be calculated as a more constant production of work, where the torque tube method knows you are no longer pedaling.
ANSWER #1?: Most cyclist would want this computer to have cadence as a measure anyway. Could the equation take cadence into account? Not necessarily in the equation of watts production, but as a “if/then” type of code. If cadence is registering then the person is pedaling and producing watts, If not, then no work is being done.
PROBLEM 2: What if the person is descending and pedaling but there is no resistance. The computer may assume the land speed is being generated by the pedaling when it is not. How do you measure this level of resistance without a censor on the drive train and basically making it just like any other power meter that is currently on the market?
PROBLEM 3: Headwind! In this equation headwind wouldn’t be taken into consideration unless there was some way to factor in HR or perceived exertion into the equation. Without those factors and no way to measure the amount of extra work being done due to the headwind this calculation would simply show the rider going slower with an amount of work done being grossly lower than accurate. Is there way to take a variable like headwind into account?

Given the terrain and total weight, it makes sense that to go from point a to point b in a certain amount of time (speed) a certain amount of work must be done to cover that distance. This work can be expressed as watts and should be measurable without the need of a torque tube for instant feedback.

Variables for Calculation:

· Gravity

o Is this idea of gravity simply the same as weight?

o Do you need to calculate gravity or is a variable of total weight (person+bike=tw) good enough?

o Gravity should be a constant. (g=32.2ft/s^2) This refers to falling objects gaining velocity. However, the same theory should hold true for the way the earth holds a person to the ground, which is how we measure weight. Ie., when a fat person jumps, their weight still gets pulled back to earth at the same velocity as a skinny person. It does take more force to get off the ground, to defy gravity, but that isn’t a factor of gravity, simply a factor of weight.

· Weight

o Has to be included as total weight (TW=bike+person).

o This is the total weight that will be forced to accelerate and move forward in the equation.

o Same idea as the computrainer. For most accurate results you’re supposed to weigh the person holding the bike.

· Terrain

o Using GPS and topography maps the terrain function shouldn’t be that big of a challenge.

o If the calculation of watts is going to be accurate the terrain must be taken into consideration for every computation of the equation. A difference of the road being 3% and 4% would make a huge difference in the amount of watts required to cover that distance at whatever speed.

§ I feel like this is probably going to be a problem of integration from calculus 2. However,

§ When finding tangent lines of a graph it’s possible to take smaller and smaller slices of any segment so that you can get the most accurate measure. These measures can vary for the purposes of whatever you’re calculating. In this case I think we need to know the slope of the tangent line of each computation to accurately know the grade of the road that is being ridden over.

§ In order for our “slice” to be small enough, the limits would need to be constantly changing as well.

· Speed

o This is probably the easiest of the variables to calculate. If we’re using GPS for the topography of the route, it’d give you instant speed feedback as well.

· Other Variables?

Equation for Calculating

I have now realized my math skills may not be up to this challenge. I know how to calculate all of these aspects individually.

Examples:

The terrain is what generally makes the biggest difference in the difficulty of a ride. When the road goes up, most of the time it starts to get hard. Even for the purest climber, an increase in the grade of the road is more difficult than riding on a completely flat piece of land. (All other variables being equal).

The grade of the road is a simple calculation of the slope of the linear line that expresses elevation gained divided by distance gained. Calculating slope is not a difficult thing. Everybody pretty much learns how to do it in 7^th grade pre-algebra. And, for the sake of saying the jingle that makes it easy to remember:

“Slope is easy, it can even be fun, just remember, rise over run.”

I apologize! But, the rise is the part of the road going up (y-axis), the gain in elevation, and the run is the distance we’re traveling (x-axis). The final equation would need to be constantly calculating this slope in order to get instant feedback on the grade at which were riding over.

Grade = change in elevation / change in distance.

Easy enough.

Speed is also a very basic calculation. I suppose we’re technically calculating velocity and that’s simply the measure of distance divided by time. If you travel 60 miles and it took you 2 hours, then you were traveling, 30miles/hour. Again, simple. However, that’s just an average, so what we technically want is a calculation for instantaneous velocity (iv). It’s the measure of the absolute value of the change in (delta) distance divided by the change in time.

iv = |dx / dt |.

Also easy enough because dx and dt are simply measures taken at much smaller integrals of the whole trip. But, in order to get instant feedback of how fast you’re going over the various terrain, you’ll need to know you’re instantaneous velocity.

Weight is easy to calculate as long as you have a scale available. Simply pick up bike, stand on scale and face the music. I suppose there should be something mentioned of the fact that most of us lose weight over the course of any ride lasting longer than 1 hour. Maybe the formula in the end should also come with a reminder to drink more during the work out.

Total weight of the rider and the bike is, I think, extremely important given that it is going to take much more work for a rider of 200 lbs to get over a hill than it does for a rider of 150 lbs. This doesn’t mean that the 200 lb rider can’t beat the lighter rider over the hill, but he’ll have to produce more watts (do more work) than the 150 lb rider will over the same grade.

This is where all of the variables of the equation need to start coming together and I’m not exactly sure how that happens. There are lots of websites that give formulas and even calculators for an average estimated measure of watts after you have this date recorded in another fashion. That’s not really what I hope to accomplish. Any ideas of the missing steps of this equation would be greatly appreciated, as well as the advice to simply give up because the factors I point out as problems earlier are not going to be overcame very easily.