# Training & Nutrition - What is a watt and why is it important?

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HiYoSilver
11-30-04, 03:46 PM
A standard unit of measurement for energy output is a watt, but I never had physics.

1. What is a watt? Please speak English and not just physics. Physics is ok, but explain each jargon word/concept.

2. How do you tell how many watt's your body can generate?

3. Do any training programs use watt's as a measurement unit?

Just curious, thanks.

late
11-30-04, 04:05 PM
1) A watt is a unit of energy. It's just a number describing how much work you're doing.
2) You need a device that measures watts.
3) Yes

You can get estimated watt output from Polar or the HAC4. I think the Powertap isn't an estimate. Kurt Kinetics just released a cyclocomputer that estimates watts for it's trainer. At \$50, it's the cheapest way to wrangle a watt I know of.

Phatman
11-30-04, 04:51 PM
1) A watt is a unit of energy. It's just a number describing how much work you're doing.
2) You need a device that measures watts.
3) Yes

You can get estimated watt output from Polar or the HAC4. I think the Powertap isn't an estimate. Kurt Kinetics just released a cyclocomputer that estimates watts for it's trainer. At \$50, it's the cheapest way to wrangle a watt I know of.

a watt is measurement of power, not energy. 761 watts=1 horsepower. (its 700-something, that number might be the speed sound at sea level.)

power is the ability to do work over a given time. work is force over a given distance.

CdCf
11-30-04, 05:00 PM
Basic physics (SI units, but the principle is the same for any unit system).

1 N (Newton) is the unit of force and 1 metre (m) is the unit of length.
If you exert a force of 1 N over a distance of 1 m, you have spent 1 Nm.
1 Nm = 1 J (Joule), which is the unit of energy, or work.

Power is the rate of doing work. The unit of time is 1 s (second).
1 Nm/s = 1 J/s = 1 W (Watt)

If you put out 100 W for 30 seconds, you get:
100 J/s * 30 s = 100*30 = 3000 J
(The first unit has the energy divided by seconds and you then multiply by seconds, cancelling each other out, leaving only the energy.)

The relationship between force, distance, time and power is beautifully illustrated by the gearing mechanism of a bike.
Since force*speed=power, you can vary the force needed by varying the spin rate of the pedals.
A high speed gives a low force, while a low speed gives a high force, but the power can remain the same.

late
11-30-04, 05:03 PM
And neither one of us said why it's important.
It's important because it's a direct measurment of the work you're doing. Before watts we measured heart rate, which is an excellent training tool. But a given heart rate doesn't tell you how much power you're putting into the pedals. Measuring output directly is a terrific advance, if expensive.

CdCf
11-30-04, 05:07 PM
Actually, it's relatively easy to measure your cycling power.
Just install force sensors in the pedals, and a crank RPM sensor, and just record both variables in a computer.

operator
11-30-04, 06:39 PM
Actually, it's relatively easy to measure your cycling power.
Just install force sensors in the pedals, and a crank RPM sensor, and just record both variables in a computer.

Before I go ahead and redicule people for spending thousands on srm system. Could you elaborate on this?

lowracer1
11-30-04, 07:31 PM
Either computrainers or Tacx I-magic trainers both measure wattage, heartrate, cadence, speed, elapsed time, and distance. I use the i-magic for winter training. You can keep track of improvements. The key is over time to increase wattage with the same heartrate.

CdCf
11-30-04, 08:10 PM
Before I go ahead and redicule people for spending thousands on srm system. Could you elaborate on this?

Well, it's simple really.
As I said, force times speed equals power.
If you continuously measure the force you push the pedals with and at the same time measure the rotational speed (at the pedal pivot point), you can just put it all together and get the instantaneous power output at any given moment.

That's the mechanical power output. The actual power output of the body will be around five times higher, due to the body's inefficiency.

Now, to actually rig something like this up is probably beyond the average biker, but a techie with an electronics store nearby, plus any old computer, could have it up and running in a day, if all goes well.
And it would be a bit cumbersome to bring along for a ride outside. Best used with a trainer setup... :)

Dusk
11-30-04, 08:45 PM
Watt are you guys talking about.....LOL....oh I crack myself up at time....

Cheers,
Dusk

sscyco
11-30-04, 08:57 PM
746 = 1 HP I = ~ 1/2 HP (on a good day)

Pat
12-01-04, 02:00 AM
And neither one of us said why it's important.
It's important because it's a direct measurment of the work you're doing. Before watts we measured heart rate, which is an excellent training tool. But a given heart rate doesn't tell you how much power you're putting into the pedals. Measuring output directly is a terrific advance, if expensive.

I think it is only practical to measure power output or watts directly on a stationary bike. If you are out riding on the road, your only real practical and convenient measure of power output is an indirect one like heart rate.

HiYoSilver
12-01-04, 08:25 AM
Heart rate is not a good motivation tool for me. I will have to check out some other cyclometers. Besides speed/distance/cadence, I would love to have the ability to measure watts on the road.

Good discussion so far, thanks!!!

FOG
12-01-04, 10:49 AM
Do you have the conversion from joules to calories?

MichaelW
12-01-04, 11:09 AM
SRM cranks measure power output during a ride. You can combine this info with your heart-rate montor to work out how many watts you are producing at a certain heart rate. This is a really important measure for racing athletes, it tells them how efficient they are. Athletes dont just want to work harder at a higher heart beat, they want to squeeze as many watts as possible form each heart and lung cycle.
http://www.srm.de/old/index.html

Derailed
12-01-04, 12:08 PM
In an effort to put this in terms you may see on your monthly utility bills: ever notice on your electric bill that you are charged for "kilowatt-hours" (kWh)? This is a unit of energy: the electric company charges you for the amount of energy your household uses over the course of a given month. You get a kilowatt-hour by multiplying the power used by something and the time it was used. The longer something is operates, the more total energy it uses.

If you think of lightbulbs, for example, you'll see that they are designated by their wattage (e.g. a 60 Watt bulb). This is the amount of energy they require to operate over an interval of time. The longer you have it on, the more total energy that it will use. If you have a 60 Watt bulb on for one minute that will use 60 Watt-minutes of energy. You can convert this to kilowatt-hours by taking account of the fact that a kilowatt is one thousand Watts and that there are 60 minutes in an hour: 0.001 kilowatt-hours for this example. This is not a large contribution to your monthly energy bill, but if you leave quite a few lights on throughout the day it does start to add up.

Stricly speaking, a Watt is defined as a Joule per second. A Joule is a unit of energy (another common energy unit is the calorie: 1 calorie is equal to about 4.2 Joules; another unit of energy you may come accross when shopping for appliances it the British Thermal Unit or Btu: there are about 250 calories in a Btu; physicists like to use an energy unit called the electron-volt or eV: there are a very large number of eV's in a Joule-- a 6 followed by eighteen zeros!).

The longer you produce a certain wattage on a bicycle, the more total energy you will use. This total energy can then be compared to the foods you eat. That is, if you know know the enegy you expended during a workout you can figure out how many PowerBars you need to eat to replace that energy. One caveat: there can be some confusion when using units of calories. The relationship I mentioned above between Joules and calories is for a "true" calorie. Foods often state the number of Calories they contain, but they are really refering to kilocalories!! Notice that I capitalized "Calorie" when referring to food? That was intentional; a unfortunante convention in physics is that 1 calorie (c) is equal to 1,000 Calories (C). Merely making this distinction based on a capital vs a lower case c has been the source of quite a few computational errors!

Wow.... I went on way too long. In the meantime I suppose others have already fished all the details of this discussion out. Furthermore, I can't imagine a single person will have made it this far through my post. To delete or not delete? Ah, what the heck.

a watt is measurement of power, not energy. 761 watts=1 horsepower. (its 700-something, that number might be the speed sound at sea level.)

power is the ability to do work over a given time. work is force over a given distance.

Dane
12-01-04, 08:30 PM
The above definitions of force, energy, and power are accurate. BTW, I don't think it was mentioned, but for those who don't know what a newton (N) is, it is the force required to make a 1 kilogram (2.2 lbs) object accelerate at a rate of 1 meter per second, per second. If you push on a stationary one kilogram object with 1 newton of force, after 1 second it will move with a speed of 1 meter/second. After 2 seconds of applying the force, it will go 2 meters/second, and so on. To give you a relative idea of what a newton "feels like", lifting a 1 kg (2.2 lb) object with constant speed reqiures 9.8 newtons of force.

Going back to power...why is it important for cycling? Its important because when you're cycling, you aren't simply moving forward with a velocity, you are pushing against a resistive force with a velocity. Since power = force x velocity, making more power means you can push against the force of air resistance with a greater velocity and ride faster. Making more power also means you can push against the force of gravity with more velocity, and therefore climb faster. This is why you want more power.

There actually are several ways to estimate power without a computer that calculates watts. One method is to time how long it takes you to accelerate to certain speed. If you know the change in speed and the total weight (your body weight plus weight of the bike), you can calculate your change in kinetic energy in a given amount of time. (This will require a speedometer or some other method to measure speed.)

For example, lets say you weight 170 lbs and your bike weighs 20 lbs. This is a total weight of 190 lbs, or 86.36 kg. You accelerate to 25 mph in 20 seconds. 25 mph is equal to 11.176 meters per second.

Kinetic Energy = 1/2 x Mass x (Velocity^2)

Change in Kinetic Energy = Final Kinetic Energy - Initial Kinetic Energy

Power = Change in Kinetic Energy/Change in Time

Power = {[1/2 x Mass x (Final Velocity^2)] - [1/2 x Mass x (Initial Velocity^2)]** / Time

Power = {[1/2 x 86.36 kg x (11.176 m/s ^2)] - [1/2 x 86.36 x (0 m/s ^2)** / 20 sec

Power = 269.67 Watts*

*-Please note that this is the average power during acceleration, and is actually less because it doesn't take into account air resistance, tire rolling resitance, wheel rotational inertia, and drivetrain friction. A person accelerating on a bike like this would actually produce more power.

A more accurate figure for this model would be:

Power > 269.67 Watts

The other way to measure power is to calculate the time it takes to climb a hill of known slope (or known height) at constant speed. In this situation, we are calculating power by the change in potential energy over a given time. Again, lets assume the total weight of cyclist and bike is 190 lbs (86.36 kg). Lets assume the hill slopes upward at an angle of 10 degrees, and that the total distance covered is 800 meters. (You need to know trigonometry for this one) The hill is climbed at a speed of 12 mph (5.36 meters per second).

Elevation Gain = sin (slope angle) x Distance

Potential Energy = 9.8 x Weight x Elevation Gain

Change in Potential Energy = Final Potential Energy - Initial Potential Energy

Change in Time = Distance / Velocity

Power = Change in Potential Energy / Change in Time

Power = [(9.8 x 86.36 kg x sin 10 degrees x 800 m) - (9.8 x 86.36 kg x sin 10 degrees x 0 m)] / (800 m / 5.36 meters per second)

Power = 787.72 Watts*

*-Power is actually greater than this because it doesn't take air resistance, tire rolling resistance, and drivetrain friction into account.

Power > 787.72 Watts

This example is a lot of power, probably only professional cyclists would be able to put out an effort like that...but feel free to prove me wrong...if you have the numbers to back it up.

Anyways, just showing that you don't need a bike computer to estimate power, although this is a pretty low estimation of power since it doesn't take everything into account. It can still give you a good idea though.

HiYoSilver
12-03-04, 08:25 AM
Ok so far. Now what if want to measure watts generated constant speed ride on level ground:

Variables: distance, weight, avg speed, time

Change in kinetic energy would be 0, so first formula won't work. Slope would be 0, so would sin calcs generate anything significant, or just garbage.

I would think there would be a formula to calc watts at avg speed over set distance. Thus I could use time trial approach to see if improving.

Thanks

CdCf
12-03-04, 11:18 AM
Sure, if you know your rolling friction coefficient, your drag coefficient, your frontal area and all power losses in the mechanical parts, you could use your speed over a perfectly flat ground to estimate your power output.

You could probably use a roll down a uniformly sloped hill to estimate your total power requirement for a given speed, and use that to derive a "power coefficient" that you can then apply to the flat-ground case. If you hold the same position, wear the same clothes, load the bike the same way, inflate your tyres to the same pressure, and so on...

SSP
12-03-04, 12:19 PM
Do you have the conversion from joules to calories?

4.1868 joules = 1 calorie

Dane
12-04-04, 10:24 PM
HiYoSilver,

Yes, the formulas I gave would not work for calculating power output while riding at constant speed on flat ground. (But used correctly, one of them would)

As far as estimating power while riding on flat ground goes, I've got an idea.

Think of it this way: When you ride at constant speed on level ground, you are performing work. However, you aren't speeding up or slowing down. You could think of this situation as adding energy to the bike through work at the same rate that the bike losses energy. Since energy is added at the same rate that it is lost, the bike is in equilibrium and moves at constant speed.

Since "Energy Added" and "Energy Lost" are equal, if we know "Energy Lost" we also know "Energy Added". If you could calculate "Energy Lost" over a very short time interval, then you could know the theoretical "Energy Added" over that short time interval...or in other words you could calculate power.

I know a way that you could calculate "Energy Lost". You do this by riding at a constant speed, and then stop pedaling and time the rate of deceleration (slowing down). For example, you ride at 20 mph. You stop pedaling, and with a watch or other timing device, count the seconds until you slow down to 19 mph.

Now you can use the Change in Kinetic Energy/Change in Time equation to calculate your power output. Note that in this situation, it is average power output between 19 and 20 MPH, not instantaneous power output. The smaller your speed/time interval is, the more accurate you will be, not taking into account human error such as your reflexes when clicking the stop watch, or the precision of your speedometer or stopwatch.

Lets do a practice calculation. We'll use the same numbers that I used in my previous post. The combined weight between cyclist and bike is 190 lbs (86.36 kg). Riding at 20 mph (8.94 meters per second) on flat ground, the cyclist stops pedaling. It takes 3 seconds for the bike to slow down to 19 mph (8.49 meters per second).

Rate of Energy Loss = (Final Kinetic Energy - Initial Kinetic Energy) / Change in Time

Kinetic Energy = 1/2 x Mass x Velocity^2

Rate of Energy Loss = {[1/2 x 86.36 kg x (8.49 m/s ^2)] - [1/2 x 86.36 x (8.94 m/s ^2)** / 3 sec

Rate of Energy Loss = -112.89 Watts

Remember that this is an average rate of energy loss between 19 and 20 mph, so:

Rate of Energy Loss ~ -112.89 Watts

Since "Rate of Energy Loss" is equal and opposite to "Rate of Energy Added" while at constant speed:

Rate of Energy Added ~ 112.89 Watts

Rate of Energy Added is equal to Power, so:

Power ~ 112.89 Watts

SSP
12-05-04, 07:56 AM
FWIW, my CycliStats (http://www.CycliStats.com) ride logger and training diary estimates your Average Watts and Calories Burned for each bicycle ride you log. Both calculations are dependent on a number of input variables, which you can see in this screenshot (http://www.shastasoftware.com/CycliStats/caloriecalculator.htm). While not as accurate as an on-the-bike power meter, this method of estimatiing average watts should be in the ballpark and costs a whole lot less.

JavaMan
12-05-04, 01:04 PM
Hey Dane,
Thanks for the formulas. I've always wondered how many watts I produce on a hard climb. I wanted inputs of %Grade, Weight(lbs), and Speed(mph), so I revised things a little. Given the proper units, Watts = total weight of bike and rider multiplied by vertical rise in 1 second.

Weigh you and your bike with everything - helmet, shoes, pump, water bottles, etc. No need to convert to Kg, the formula below will automatically do it.

The vertical height h you gain in 1 second is the distance d you go in 1 second times the sin of the road angle. The road angle is the inverse tangent of the Grade divided by 100 (6% Grade becomes 6/100 or .06). The distance you go in 1 second is your Speed(miles/hour) X (1600 meters/mile) X (hour/3600 seconds) or .444 meters for every mile per hour. So h = Speed(mph) X .444 X sin(inv tan(Grade)).

After combining constants and re-arranging to make it calculator friendly I get this. Punch it into your scientific calculator just as written. Function keys are in brackets.

%Grade [divided by] 100 [equals] [2nd or inverse] [tan] [sin] [times] 1.98 [times] Weight(lbs) [times] Speed(mph) [equals]

Example:
There is a 2 mile long hill at 6% grade that I climb regularly. I can just stay above 10 mph all the way up. My bike and I weigh 216 pounds.

6 [divided by] 100 [equals] [2nd] [tan] [sin] [times] 1.98 [times] 216 [times] 10 [equals]

I get 256 watts. From what I've read, that's about half as much as the pros generate on a long climb. Would you mind verifying my work, Dane? I'd appreciate it.

It also occurs to me that if you coast down a known grade at a constant speed, then use the same formula to calculate watts, that is the wattage required to maintain the same speed on level ground. Rolling resistance and wind resistance are already accounted for. Is that correct?

Tom

CdCf
12-05-04, 05:48 PM
It also occurs to me that if you coast down a known grade at a constant speed, then use the same formula to calculate watts, that is the wattage required to maintain the same speed on level ground. Rolling resistance and wind resistance are already accounted for. Is that correct?

That's what I said...:

You could probably use a roll down a uniformly sloped hill to estimate your total power requirement for a given speed, and use that to derive a "power coefficient" that you can then apply to the flat-ground case. If you hold the same position, wear the same clothes, load the bike the same way, inflate your tyres to the same pressure, and so on...

JavaMan
12-05-04, 06:35 PM
That's what I said...:

You certainly said it first. Since you said it already, I guess that answers my question about verification. Too bad you did not say how the average person could apply it.

The purpose of my post was to put the equations into a form anyone could punch into a calculator, then show exactly how to do it using everyday concepts like Speed, Weight, and Grade. Not everyone is savvy enough to take a the technical information from these posts and apply it properly to their situation. For example, did anyone bother to explain the difference between the grade and the angle of incline? Most people I know are kind of shaky converting pounds to kilograms!

So now if i can find what grade I can coast down at 10 mph, I can calculate how many watts I must generate to overcome rolling friction and air resistance and add it to the 256 watts needed to lift me and my bike.

Tom

Dane
12-05-04, 08:46 PM
JavaMan,

I calculated your power on that hill and got the same number you did.

Percent Grade = (Rise/Run) x 100

Tan^-1 (Rise/Run) = Angle of Slope

CdCf,

As far as determining a power coefficient goes, what I would do is find a hill of known slope, coast down it and find your terminal velocity going down the hill, since terminal velocity is the point that your gravitational force parallel to the hill is equal to the sum of the resistive forces. I would use the equation:

Force = Force Coefficient x Velocity^2

You could determine the force with a vector diagram and a little trig. The gravitational force parallel to the hill is:

Force = Weight x Sin Angle

Since you already know Velocity, the only variable left to solve for is the Force Coefficient. Then you just plug that Force Coefficient into the power equation:

Power = Force Coefficient x Velocity^3

Then you could calculate your power output at any speed. It would be interesting to see if we could get some more exact formulas for tire rolling resistance as well as wheel rotational air resistance. I'll work on those ones...I've got some concepts in my head but need to refine them.

CdCf
12-05-04, 09:08 PM
The nice thing about finding that out is that it yields, what I would call, a "power coefficient".

Using grades instead of degrees is better for our purposes, as that directly links vertical and horizontal speeds. There is a small discrepancy, but for moderate slopes, the difference is negligible. For a 10-degree slope, you're less than half a percent off...

So, let's say you have a measured slope of 4%.
Suppose you can roll down at a constant 10 m/s in your normal riding position, legs stopped.
Your vertical speed will then be 0.4 m/s. The force is your total mass times the gravitational acceleration.
Let's say then, that your mass (with bike) is 90 kg. That makes the force (it's not really a force in this case, but it's easier to think of it that way) 90 * 9.8 = 882 N.
Power is force times speed, so the power is the vertical speed times the force. 882 * 0.4 = 353 W.
But it's really power gained in this case, as you're converting potential energy to keep moving forward.

Once you've come this far, you know the power it takes for you to maintain 10 m/s (~22 mph).
Divide the power required with the speed cubed. In this case, it's 353 / 1000 = 0.353.
You can then use this result as a "power coefficient".
If you want to know the power required to move at 13 m/s (29 mph), multiply the "power coefficient" by the speed cubed. In this case 13^3 = 2197, which gives 2197 * 0.353 =776 W.

In truth, the power required will be somewhat less than that. Perhaps, in this example, more like 650-700.
This is because at lower speeds, the rolling resistance is a larger power component than at higher speeds, so when we divide by the speed squared, we're assuming that all of the power is due to air drag.
Since power to overcome rolling resistance increases linearly with speed, it becomes less significant at higher speeds, but the larger portion it made up at the lower speed gets scaled up at the same rate as the power to overcome air resistance.

Still, it should be accurate enough to give a decent estimate.
If your normal speed range on flat ground is between, say, 15 and 25 mph, then the ideal would be a hill that set you rolling at 20 mph - in the middle of the range. The errors would be kept smaller that way.

HiYoSilver
12-06-04, 07:57 AM
Dana and all,

Great illustrations and explanations. This is exactly what I was looking for.

One more question: what is the difference between grade and slope? I seem to recall someone mentioning this past year that the X% grade signs are roads are not accurate, but I can't recall why or what the method was to determine the true angle of descent.

whitemax
12-06-04, 07:07 PM
I hereby confer upon all of you a degree in physics from Power University

jwrbikeman
12-13-04, 08:12 PM
Do you have the conversion from joules to calories?

1 Joule=0.239 caloris

jwrbikeman
12-13-04, 08:22 PM
I will add more detail later, but one needs to consider overcoming wind resistance
as a major use of power, more than 80 percent even at reasonable club ride speeds.
Power is important because it involves time and everything else being the same,
the rider that can generate the most power gets to the end first and wins the race!!