Work done on a stationary bike

[mwright2cycle]

Our local YMCA is putting on an indoor triathlon and I am trying to come up with a formula to help them determine the winner of the cycle segment.

The indoor triathlon consists of a 10 minute pool swim, a 20 minute cycle on an indoor, stationary bike, and a 20 minute run on the treadmill. The winner of the swim segment is clearly the one that swims the furthest in 10 minutes, and the winner of the run segment the one that runs the furthest on the treadmill, but on the spin bikes the distance reported at the end of the session is not the distance traveled by revolutions of the fly wheel.

The spin bikes that are being used are the Keiser M3 (http://www.keiser.com/m3/mscience.html ). They have a gear selector that allows the user to set the resistance on the bike between a setting of 1 and 24 with 1 being the easiest and 24 the hardest amount of resistance. The bike also has a computer that displays current cadence (RPM), watts, Kcals, and “distance”. At the end of the biking session, the display is summarized for average cadence, average watts, and total “distance”.

From what I understand, for every 200 revolutions of the flywheel, the Keiser M3 will display a “distance” of 1.0 (http://www.powerbiketrain.com/Keiser-M3.html ) regardless of the gear selection. If you are riding a bike on the road this is not the case—the harder the gear, the further the bike travels per revolution of the crank.

The VP of Sales & Marketing at Keiser, Darrin Pelkey, has provided this explanation of how wattage is computed on the Keiser M3:

Watts are calculated from the gear setting. A potentiometer is attached to the magnet holder (the round cone shaped disc at the end of the shifter cable). As the shifter is moved, the cable rotates the magnet holder. A potentiometer is rotated by the rotation of the magnet holder, thus feeding information to the computer on the position of the magnet holder. The rotation of the potentiometer is broken down into 24 gear settings. A dynamometer was used to test and develop a table of wattage at various gear settings and speeds. The speed of the crank is determined by a magnet attached to the large pulley on the right crank arm and a magnetic switch attached to the circuit board in the magnet holder assembly. Each time the magnet on the pulley passes by the magnetic switch, a signal is sent to the computer to compute the RPM's of the crank arm. Power equals force times velocity. The force is determined by the magnet position and the speed by the crank speed. The lookup table is programmed into the computer and the computer simply looks at the gear setting and speed and goes to the lookup table to find and display the Watts for those two settings.

One way to solve the problem would be to require the participants to keep the bikes in a fixed gear, but it is hard to pick a gear that would suite the inexperienced and experienced participant alike; a fixed gear that is easy enough for a younger participant would be a disadvantage to a strong, more experienced cyclist. Also, the gear selector is very easy to move, and it would be hard to monitor the bikes to ensure that no one cheated even if just by accident.

Based on my understanding of this problem, it is not possible from the data supplied by the Keiser M3 computer to determine the actual distance traveled. If the bike were kept in a fixed gear, then this could be determined, but if the participant is allowed to vary the gear, then the only way to determine the actual distance would be to know what gear the bike is in per revolution of the fly wheel. If it is true that actual distance cannot be determined, then it seems that the only way to rank contestants on the cycle segment is by the amount of work that they put forth during the 20 minutes.

One theory currently being proposed to determine the amount of work is to take the average watts reported at the end of the 20 minute cycle session and divide that by the rider’s weight in kilograms, giving you the power to weight ratio. I disagree with using the rider’s weight in the calculation. I know that weight matters when riding a bike outdoors because you actually have to propel the bike forward, but on a stationary, indoor bike I do not believe that weight matters, though admittedly I really do not know. My idea for determining the amount of work done is to take average watts * distance (revolutions of the fly wheel) or just the average watts itself.

Here is some test data that we put together on 3 participants riding the Keiser M3 for 20 minutes:

rider,watts,distance,fly wheel revolutions,Kcal,weight (lbs),weight (kg),watts/weight (kg),watts * distance

A,330,10.8,2160,880,147,66.67807839,4.949152825,3564

B,360,10.2,2040,1102,161,73.02837157,4.929590956,3672

C,290,9.8,1960,1163,200,90.718474,3.196702802,2842

Using watts/weight (kg), rider A wins at 4.949 watts/kg; using watts * distance, rider B wins at 3672. And since watts itself is a function of the cadence, I think you could also just use average watts so that rider B wins at 360 avg watts.

If you have time to consider this interesting problem, what is your opinion on how to determine the rankings for the cycle segment of our indoor triathlon? Is the weight of the rider a factor on a stationary, indoor bike?

Thanks!

 

[Borek]

Can't you order them according to kcal spent? That's a good measure of the effort.

 

[Lsos]

Sorry I didn't read the whole thing. Either way, it seems to me the "best" way to calculate the distance traveled is to figure out the average power it takes to travel a certain speed on a real bike (by doing measurements of different cyclists traveling at different speeds) and then plotting this data into a computer, which then goes on the stationary bike.

The computer can then approximate the distance you would have traveled in the real world based on how much power you are using.
In the end, it's a stationary bike. It can't replicate the fact that different gears require different forces, or the changes in air resistance due to speed. Even if it could it would be unnecessary, since looking at the power output essentially gives the computer the total combined picture of each of these components together. Therefore, if the computer tried to account for the gears, etc, the end result would be the same, but with probably lower accuracy.

 

[Borek]

approximate the distance you would have traveled in the real world based on how much power you are using

It is not power that counts, it is amount of work. These are related, but not equivalent. It is not clear to me if the power displayed is average, maximum or instantaneous. If kcal are real they are the best.

 

[mwright2cycle]

The watts displayed while cycling is instantaneous, but when you stop cycling it displays average watts over the duration of the session. Kcal is a straight calculation from the watts so it is no more or less useful that the watts itself. I am trying to determine the amount of work done, as you say, from the power (avg watts) and I don't know if body weight is a factor considering the bike does not move forward through space.

 

[Borek]

The watts displayed while cycling is instantaneous, but when you stop cycling it displays average watts over the duration of the session. Kcal is a straight calculation from the watts so it is no more or less useful that the watts itself. [q/uote]

time*power=energy, assuming all were riding for the same time ratio of watts to kcal should be constant, that's not the case in the data you have listed.

I am trying to determine the amount of work done

Work done = energy spent, that's is simple energy conservation.

as you say, from the power (avg watts) and I don't know if body weight is a factor considering the bike does not move forward through space.

If you were trying to convert to distance then yes, body mass counts. I believe I have seen calculators on the web that let you calculate amount of energy spent while biking as a function of body mass, distance and speed - but they were very approximate.

 

[Lsos]

It is not power that counts, it is amount of work. These are related, but not equivalent. It is not clear to me if the power displayed is average, maximum or instantaneous. If kcal are real they are the best.

I'm pretty sure power is instantaneous, at least that's from my experience.

I also believe calculating distance from power would be far more accurate, since speed vs power is not linear...and therefore distance isn't a function of work.

 

[mwright2cycle]

To be honest, I am not sure how the bikes are calculating the Kcal; it should just be a straight conversion from watts, but yes, the ratios are different for the 3 riders. The instantaneous feedback on the bikes switches from watts to Kcal, but at the end of the session is just shows average watts. I was just disregarding Kcal and using watts instead because they should be both based from gear & cadence. The manufacturer of the bikes provides the info about how the watts are computed, but nothing that I have found on the Kcal.

 

[Lsos]

Maybe they figured out that the fatter you are the less efficient?

Any other inputs? heart rate, height, etc?

 

[Borek]

I also believe calculating distance from power would be far more accurate

Would be. Sadly, as we already know - it is not, as it is calculated just from the number of crank revolutions.

 

[Curl]

You're comparing apples to apples so the best way to do it is compare the work put in.

 

[Rap]

I think you can make a model - pushing a weight across a surface with friction. The amount of force you need to make it move is k*f where k is the coefficient of friction, f is the weight you are moving. If you move it a distance D, then the work you have done is W=k*f*D. Suppose the person weighs fp and the bike weighs fb. Then the work to go distance D is W=k*D*(fp+fb). Suppose you know the weight of the bike fb. So now you specify the amount of work a 100 lb person must do to go the distance D. Call that Wo. Then Wo=k*D*(100+fb). You can solve this for k*D : k*D=Wo/(100+fb). The amount of work a person weighing fp must do to go the distance D is W(fp)=k*D*(fp+fb) and using the K*D that you solved for, you get

W(fp)=Wo*(fp+fb)/(100+fb).

Just plug in the persons weight into fp and that's how much work they must do to go the distance D. The person who wins is the one who did their required amount of work in the shortest time - in other words the person that went the distance in the shortest time.

 

[Borek]

pushing a weight across a surface with friction

Have you ever ride a bike alone and in a group? Amount of power required to ride alone and to follow someone else (in his aerodynamic shadow) is substantially different, we are talking about 10-20-30% difference. So your model ignoring drag is oversimplified.

 

[Rap]

Have you ever ride a bike alone and in a group? Amount of power required to ride alone and to follow someone else (in his aerodynamic shadow) is substantially different, we are talking about 10-20-30% difference. So your model ignoring drag is oversimplified.

The problem is, at any given time we only have the total work done by the rider (Kcal) and the time they have been on the bike. (The distance measurement is unreliable). Thats all we have to work with. Drag force is F=bV where b is a constant and V is the velocity, but we don't have the velocity.

I guess we could pretend the drag force was like the friction force, (which is wrong) and then the above model would be the same, except the k is due to friction and drag together. The other thing we could do is to pretend that the velocity is constant, then we could make a model that would be a little better, but we would have to guess at the values of the k and b constants. That might not be so bad. I could figure that out if need be.