http://www.analyticcycling.com/

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Christopher Kautz
Director of Technology, Product Development, and Education
GURU Sports, a division of Cannondale Sports Unlimited

Watts is a measure of energy expended per a given unit of time... when you're riding 200 watts, you're expending energy at 200 joules per second. Are you wanting to keep the rider's finishing time constant? Or are you wanting to know how much extra work is required of the rider regardless of the time it takes him/her? Without knowing a whole host of other variables, like rider weight (to figure out work needed to overcome gravity), consistency of the elevation gain, etc., it's hard to say.

How much extra work is what I am looking for. So if I am 185 lbs and can ride 10 miles no gain at an average of 200 watts. I can compute the total # of watts expended for the ride. Now if that same 10 miles had an elevation gain of 1000 ft how many more total watts would be needed to accomplish. I am not looking to keep speed or duration constant.

If you are not looking to keep speed and duration constant - you have your answer.

200 watts is required, it would just take longer.


Elevation changes does not change the amount of raw power required, only the amount of power required to cover a certain distance in a certain duration.

The extra work is mgh where m is mass, g is gravitational force and h is elevation change. As an example, if bike+rider weigh 205 lbs (~93 kg), g is 9.8 m/sec**2 (earth at sea level) and h is 1000 ft (305 m), the extra work for the 2nd ride would be 93*9.8*305 = ~278000 Joules.

Watts is an instantaneous number, i.e. how quickly the work is being done. It doesn't make sense to talk about total Watts. An analogy is asking "What was your total miles/hour on that ride?" What you calculate for a ride is not total Watts but average.

We should start here.
Power is a rate of energy use. Much like speed is a rate of distance covered.
How many miles are you going in an hour = speed.
How many joules are expended in a second = power.

The standard measurement of power is the watt which is a joule divided by a second - (J/s see I told you it was a rate).
A joule is, of course, the measurement of energy. It is defined as the amount of energy required to apply one newton of force over a distance of one meter.
A newton is, of course, the measurement of force in SI. The force exerted on an object to lift it vertically at a constant speed is about 9.8 Netwons. Lifting 1 kg 1 meter (at a constant speed) takes approximately 9.8 Joules.

If you want to do that in one second it requires 9.8 watts.
If you want to do that in .1 second it requires 98 watts.
If you want to do that in 10 seconds it requires .98 watts.

You see - the amount of power you apply to something doesn't change how far it will go, just how fast it will get there.

If you ignore friction for a moment... on flat ground you can actually go any distance with any amount of power.

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Erik
Strava

To simplify a bit, let's say you finished those 10 miles at 200 watts in 45 minutes. THat would give an energy output of 200 j/s*(45mins*60sec/min) = 540000 J - since watts is a measure over a given time, if you know the elapsed time you can figure out the total work.

the extra work to overcome the elevation change is mass*acceleration of gravity*elevation.

So, quick conversion to metric gives 84.1 kg*9.8 m/s/s/*304.9m=251292.482 joules of extra work to overcome the elevation.

Then to get watts you would divide by time in seconds. That gets you in the ballpark. There's some mechanical advantage to riding up the hill (it's a giant inclined plane, and I've calculated it here as a straight lift).

sorry to those whom have already posted above, i left the window open for a while before reposting :)

Mayday thank you so much for the physics lessons. So to summarize am I correct in computing that the 1000 ft gain requires 46% more effort (251,000 / 540000). Or in other words the same effort it would take to ride at at 200W avg for 14.6 miles on level grade in this example??

Yes - the real world is always a bit different than things work out on paper, but you do have the idea. We have to simplify things a great deal to make an approximation here, but it's a good starting point. I would think that the 'real' number is a little less than the 46% due to things we've left out for simplicity's sake, such as factoring in mechanical advantage for the climb and bike, as well as assuming that the grade of the climb is uniformly steep.

I forgot to add that you can convert joules to Calories by dividing by 4187 - for me anyways, it helps to look at things in familiar terms (what the hell is a joule anyway?).

So, the added elevation in this example only requires another ~60 Calories of output.

OK So you will probably figure out where I am heading with this. If a course is 84 miles with 8,121 ft of climbing this would be equal for the 185 pound rider to 122 miles on a flat course all else being equal?

All else being equal, yes, it sounds like you did the math correctly. But that is a lot 'else' to be equal.

You have to remember that these equations only describe perfectly efficient systems operating in frictionless vacuums. In the real world, the work done to propel bicycle and rider is only a small percentage of the story - you have to factor the work needed to overcome aerodynamic drag, rolling resistance, other sources of friction, and the inefficiencies of the human body (especially as rider's mass increases, but that's covered by the 'all else being equal' caveat).

It may be that your flat course rider is going faster than the your rider on the mountain - the flat course rider is losing much more to drag than your mountain rider as drag increases at the square of speed (i.e., it takes 4x more work to go twice as fast).

You would also need to take out all the downhill sections of the mountain course profile to be more accurate - you would only want to measure the actual distance on the hilly course where you're doing work before making a mathematical comparison.

Let's assume an 85kg rider and bicycle weight, CdA of 0.35, average rolling resistance, and constant air density.

Then a rider in order to maintain the same average speed on a flat course when putting out 200 watts on average would need to average 346 watts on a straight incline of 1.89% grade (1000 feet per 10miles).