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Blog Smart exercise bike computer. Part #3: Power measurement
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  • Author Author: yuritikhonov
  • Date Created: 18 May 2018 8:01 AM Date Created
  • Views 2307 views
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  • Comments 17 comments
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Smart exercise bike computer. Part #3: Power measurement

yuritikhonov
yuritikhonov
18 May 2018

Good day!

 

This week I want to continue my mathematical calculations. Last time I made the assumption that the force needed to pedal the exercise bike is constant and equivalent to that of a real bike riding straight on High Gear. This statement is valid only if you ride an exercise bike always with the same tension.

 

Today I plan to do the following:

  • determine the actual force (or rather the torque) created by the exercise bike, depending on the position of the tension knob;
  • calculate the power that creates a rider rotating the pedal;
  • compare the received power with the power that is consumed when the real bicycle moves.

image

 

Let's start the experiments!

 

Experimental research

 

Initially, I wanted to conduct an experiment, spinning the exercise bike up to a certain speed using an electric drill, and then measure the power consumed by the drill. However, this did not work in my case for several reasons:

 

1. The bike comes with a flywheel and in order to quickly spin it up just a huge torque is needed, which literally breaks the attachment between the drill and the pedal assembly of the exercise bike.

image

 

2. I do not know exactly what power factor my drill has, but in my home workshop I can only measure the RMS values of current and voltage, so even if I could perform this experiment, I would not speak about any accuracy.

 

3. It turned out that the tension of my exercise bike depends not only on the position of the tension knob, but also on how quickly one or another "gear" was switched on and even of what "gear" was turned first, for example:

  • switching 6 – 7 – 8 – 7 gives the torque Te(A);
  • switching 6 – 7 gives the torque Te(B);
  • according to sensations, Te(A) > Te(B).

 

Disclaimer: further calculation uses an obviously simplified model of the dynamics of the exercise bike, the conclusions obtained as a result of this calculation do not pretend to be scientific truth!

 

I was saddened by the current situation and even thought not to do the power calculation at all, but I still managed to find one solution: if we assume that the torque at the central axis of the pedal assembly of the exercise bike does not depend on the speed of rotation of the pedals, we can conduct the following experiment:

  • set the desired "gear" (G) with the help of tension knob;
  • put an empty cylindrical jar on the pedal;
  • fill the jar with water until the pedals begin to rotate;
  • determine the weight of water (m) in the jar;
  • determine the torque Te(G) by the formula:

Te(G) = 10 * m / (S1 / 2000)

 

where S1 is the distance between the pedal axes of the exercise bike (we already know that S1 = 480mm).

 

For the experiment, I used a cylindrical jar with a diameter d = 85mm, knowing that 1 liter of water weighs 1 kg, the mass of water in the jar can be determined knowing the water level l by the formula:

m = 3,14 * (d  / 1000/ 2)2 * l

 

As a result of the experiment, I obtained the following data table:

image

 

As I said, and as it can be seen from the table, it is not possible to define Te(G) precisely, for this reason I needed to approximate the value obtained. I performed the approximation with the help of my favorite program [Curve Expert], a polynomial of the second order. As a result of the approximation, the following model was obtained:

Te_a(G) = 5,49 – 0,26*G + 0,13*G2

 

Power calculation

 

Power, as you surely know well, is the product of torque at speed. If everything is clear with the torque, then what would be the speed?

 

In the course of my training I found the optimal speed for myself – It is about 32 km/h or 20 mph. As for the rationale for choosing this speed, I propose to postpone it for one of the following articles, but for now let's try to understand what is that 32 km/h?

 

As I wrote in my [last blog], my old bicycle computer interprets one cycle of pedals of exercise bike as D3 = 6.67m. Knowing D3 and speed Vn = 32 km/h it is easy to calculate Wn – the speed of rotation of the pedals in RPM:

Wn = (Vn * 1000  / 60) / D3

Wn = (32 * 1000 / 60) / 6,67 = 80,0 rpm

 

Great, we have the angular velocity Wn! Now we can calculate the power

P(G, W = Wn) according to the following formula (power in watts):

P(G, W) = Te_a(G) * W / 9,549

 

In the highest gear G = 8, we get:

P(G=8, W=Wn) = (5,49-0,26*8+0,13*64) * 80,0 / 9,549 = 98,27 W

 

On the Internet, I managed to find a very interesting [link] following which you can find the table "Watts required to maintain a speed of 20 mph". If we make the assumption that we are really driving at a speed of 20 mph, the closest thing to our result is the Low racer bicycle. Let's take this as a reality.

 

But how can we estimate the figure obtained? To do this, you need to determine the calorie consumption.

 

Calculation and correlation of power

 

It is known that 1 kcal/h is equal to 1.16 W. However, one very important parameter is not taken into account here – the coefficient of efficiency (u) of rider. In our calculations, I am guided by the data from [the next article], which says that when riding a bike u = 0.25. Thus, rider’s power can be calculated by the formula:

Pc(G, W) = P(G, W) * 1,16 / 0,25

 

As a result, when riding a bicycle at the 8th "gear" at a speed of 32 km/h, we have:

Pc(G=8, W=Wn) = 98,27 * 1,16 / 0,25 = 455,97 kcal/h

 

My old bicycle computer counts Pc(G, W) according to the following formula:

Pc3(G=any, W=Wn) = 835,6 / 55,71 * 32,0 = 479,97 ckal/h

image

 

The values turned out to be very close (difference of ~ 5%, taking into account all the assumptions made in the article, it is possible to write off the error of the method), so we take the following: when G = 8, the speed of my old bicycle computer (V3) – is true.

 

Based on the foregoing, I compiled the formula for the dependence of the speed of the exercise bike (V) on the "gear" (G) and on the speed of rotation of the pedals (W):

V(G, W) = Te_a(G) / Te_a(8) * 32 * W / Wn

 

By analogy, you can determine the distance (here N – is a count of rotations):

D(G, N) = Te_a(G) / Te_a(8) * N * D3 / 1000

 

I plan to use these formulas in my new exercise bike computer.

 

Conclusion

 

I believe that I managed to achieve good results! I finally managed to connect the power (P) with the gear number (G)! The results obtained by me do not pretend to be scientific truth, however, for home use – this is quite enough (especially at a G=6..8). The main thing that I was able to learn (and this result is quite scientific) is the non-linearity of the tension variation when changing “gears”, which I previously only vaguely suspected. Now I just can not wait to bring the formulas I received into a new bicycle computer and to conduct the first real tests!

 

Oh yeah I still do not have an exercise bike computer Unfortunately I have not received my  FRDM-KW41ZFRDM-KW41Z board yet and I do not even know when to expect it Well if the board does not come this week I will have nothing to do but start working on another board that is close in functionality  FRDM-K64FFRDM-K64F which I got in one of the previous [Road Test]. I hope you will support me in this decision!

 

Thanks for reading and have a nice day!

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Top Comments

  • mcb1
    mcb1 over 7 years ago +4
    A real issue indeed. I was looking at your earlier post and the images provided. I resume the round item nearest the front is what your pedalling rotates, and the band just behind is connected to the tensioner…
  • mcb1
    mcb1 over 7 years ago in reply to mcb1 +3
    It seems it doesn't like my image ...
  • yuritikhonov
    yuritikhonov over 7 years ago in reply to mcb1 +3
    Hello mcb1! 1. This thing that looks like a standard bicycle brake is a permanent magnet that creates a constant torque on the flywheel. 2. Thanks for image! In my plan maximum, I want to install the electric…
Parents
  • jw0752
    jw0752 over 7 years ago

    Hi Yuri,

    I like your experiment and your approach. It is possible that you want radians per second in your equation for power instead of revolutions per minute. This should give you 1 RPS = 2 Pi radians per second. Notice you also want the angular velocity in a time frame of seconds not minutes. The work done by the pedal is torque times the change in angular distance. Power is work done per second or as you stated torque(in newton meters) times angular velocity (in radians per second). The method you are using is a pretty good way to estimate the power but also keep in mind the static friction causes the torque needed to start the pedal to be higher than the amount needed to keep it turning. Even if you can put a generator on the bike and measure its power output you will still be ignoring the parasitic losses due to friction before the power gets to the generator. The importance of the power calculations should allow you to make comparisons though even if they are not perfect. I will enjoy watching your progress.

    John

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  • jw0752
    jw0752 over 7 years ago

    Hi Yuri,

    I like your experiment and your approach. It is possible that you want radians per second in your equation for power instead of revolutions per minute. This should give you 1 RPS = 2 Pi radians per second. Notice you also want the angular velocity in a time frame of seconds not minutes. The work done by the pedal is torque times the change in angular distance. Power is work done per second or as you stated torque(in newton meters) times angular velocity (in radians per second). The method you are using is a pretty good way to estimate the power but also keep in mind the static friction causes the torque needed to start the pedal to be higher than the amount needed to keep it turning. Even if you can put a generator on the bike and measure its power output you will still be ignoring the parasitic losses due to friction before the power gets to the generator. The importance of the power calculations should allow you to make comparisons though even if they are not perfect. I will enjoy watching your progress.

    John

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  • yuritikhonov
    yuritikhonov over 7 years ago in reply to jw0752

    Hello John!

     

    1. I checked my formula for power P(G,W) it looks correct [link to formula].

    2. I'm using RPM rather than RPS because my reed-sensor can only determine the integer number of revolutions so RPM looks more comfortable.

    3. About static friction. You are right in 100%. But I don't know how to measure it correct. The most interesting thing here is that the torque is created not only by bearings and chain transmission, but also by a magnetic brake (controlled by tension knob). So static friction calculation looks complex.

    4. About measurements by generator: yes it's not correct method but if we use a motor to rotate pedals then we get "nearly" the right result (I wrote about it on experimental research section).

     

    Really thanks John, people like you help me better understand the nature of this world!

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  • jw0752
    jw0752 over 7 years ago in reply to yuritikhonov

    Hi Yuri,

    Thanks for the link to the source for your formula. I agree, the 9549 factor in the denominator is a conversion factor between units (kilowatts, revolutions per minute) in your formula and the standard (watts, radians per second) of the standard physics formula. It always amazes me how simple these things look on paper and how complex they become when we move into the real world. Keep up the good work. Your project has given me some fun trying to figure out a good way to accomplish what you are trying to do.

    John

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  • yuritikhonov
    yuritikhonov over 7 years ago in reply to jw0752

    Thanks again John!

    The most amazing thing for me was that this article was made from one cup of water which I put on pedal image

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