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Wrong Bicycle equations?

  1. May 31, 2007 #1
    Dear Gurus,
    I'm coding a microprocessor controller for a 0.1hp motor.
    to assist going up hill on a bike. But the RR equation seems strange

    I found the equations relating the bike losses, mechanical, rolling resistance , airdrag... with speed and hence horsepower and calories.

    Here are the equations connecting losses, speed and Watts (hp)

    The motor has to overcome losses, and as I said I found these
    formulas for watts needed to keep bike at the chosen speed. .
    RR = M*g*Crr.................................(1)
    Airdrag = 0.5*Ac *Cd*Da*v2 ............(2)
    Totloss =(RR+Airdrag)*102/100..........(3) 102/100 term for bike mec losses
    P = v *Totloss ..............................(4)
    Where v is in m/s and w in kg ( they use mass for w????)

    They gave an example 20mph, rider +bike weight 190lbs, HP needed
    expressed in watts 329W. In metric units
    ....20mph = 8.89m/s and 190lb =86.18kg

    Table of coefficient values that were used:-
    Air Density coeff: Da =1.226 kg/m3
    Air Drag coefficient : Cd = 0.9
    Rider surface area:- Ac = 0.7 m3
    Rolling Resistance:- Crr = 0.007

    My question is :-

    Could you explain if equation (1) below is correct?
    RR = M*g*Crr........................(1)

    The author of the internet article directly multiplies
    0.007*9.81 * 86.18 *v= 52.55.
    But 86.18 is weight, not mass.

    On the other hand the magnitude of the result seems believable, about
    0.44hp.
    Most people in this example would need a motor as even a good cyclist would soon become tired.

    I'd also be most grateful if someone could explain if the whole set of the above equations are valid enough for engineering use?

    The actual values of course of the loss coefficients will vary depending on terrain, tires, rider frontal area. That is acceptable for practical engineering calculations..

    I know that the bicycle mechanical losses are not covered, but they are very small, perhaps 2% of the total losses because modern bikes are extremely efficient, hence the 102/100 term above
    My interest in the RR part, is because that RR part of the equation at low speeds predominates. (This is because the air resistance is a v^3 function.)
    Air losses becomes predominant at about 12km/hr, always supposing that the RR term is correct.

    Bye and thanks
    Fred
    Madrid
    Spain
     
  2. jcsd
  3. May 31, 2007 #2

    FredGarvin

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    Science Advisor

    The weight in English units is usually interchangeable with mass. You just have to know that it is what you are dealing with. For most applications on Earth, one usually assumes that 1 Lbf = 1 Lbm. You correctly converted the weight over to mass in SI units so the multiplication by 9.81 will result in a force unit of Newtons.

    For the rolling resistance I get:
    [tex]P = (86.18 kg)(9.81 \frac{m}{s^2})(.007)(8.9 \frac{m}{s})[/tex]

    [tex]P = 52.67 \frac{N*m}{s}[/tex]

    [tex]P = .0071 hp[/tex]

    By the looks of it, I would think the rolling resistance number is believable, but I would also think that it would be based on experience and not so much on theory.

    I think you have the right idea for your calculations.
     
    Last edited: Jun 1, 2007
  4. May 31, 2007 #3

    rcgldr

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    Homework Helper

    .44hp seems high. The top bicycle racers can go 30 miles just under an hour, and this correlates to about .4hp (that's 30+mph for almost an hour).
     
  5. Jun 1, 2007 #4

    FredGarvin

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    How does one correlate power by just having a speed and time?
     
  6. Jun 1, 2007 #5
    Dear Gurus
    First thank you Fred for getting me out of the muddle as to which pounds we are using. That has go me up and running with my assembly code.

    As to correlation I take it you are asking about what what Jeff Reid wrote in a previous answer.
    Maybe he had assumed as obvious that the third element in the equations is losses.
    Hence we are relating the power needed to overcome those losses. or have I missed the point?

    If you would allow me to throw a new spanner in the works, I'm mulling over how to actually make a measurement of the rolling loss coefficient, Crr.

    I'm going to try and put into an equation this scenario.
    I usually cycle on dirt roads and on a flat bit I'm going to do the following.

    1. Get up to a speed of 10km/hr ( the air resistance at this speed is small)
    2. Stop pedalling and start my stopwatch.
    3. Also note the distance on the distance counter
    4. The bike will start slowing and just before I fall off., read distance and time again.

    I did this once before and I think I did about 83m before coming to a halt.
    Obviously the amount of sand on the road is the important factor here, not the tire material.

    So I'd be grateful for any ideas if this has been done before. I don't believe in reinventing the wheel unless it becomes necessary. :-)

    Bye
    And once again , many thanks
    Fred
    Madrid, Spain
     
  7. Jun 2, 2007 #6

    rcgldr

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    Homework Helper

    Sorry, I left out the fact that the power output of top bicyclists has been measured numerous times. They can generate almost 1 hp for about 5 to 10 seconds, less power for longer periods of time and about .4 hp for several hours. My point was that .4hp corresponds to about 30mph, not 20mph.

    It's probably easier to find this information on the web by searching for "gossamer albatros", a human powered aircraft. The original version required .37hp from the "pilot".

    Regarding bicycle power requirements, wikipedia has an article, but cleary states it's not known how accurate the data is. What is known is the power output of the top bicyclists. I'm not sure how much wind tunnel testing has been done by actual bicycle manufactureres or race teams.

    An intereseting tid-bit here. The bicyclists that were prepping to "pilot" the gossamer albatros were each consuming about 5,000+ calories per day, even though they only weighed about 150lbs.
     
    Last edited: Jun 2, 2007
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