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Electric bicycle

  1. Nov 20, 2009 #1
    How do I go about calculating the required power of an electric motor (will be attached to bicycle) needed to accelerate a bicycle (about 200lbs) to a minimum of 25 mph?

    I'm not asking people to do all the work for me. I just want to know what I need to know in order to find the answer.
     
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  3. Nov 20, 2009 #2

    russ_watters

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    I'd probably just use the output a computerized stationary bike tells you you need and double it.

    Or I'd pick a hill climbing requirement and apply the potential energy equation.

    The problem here is that in constant speed, level motion, all of the power of a vehicles engine goes to overcomming friction and wind resistance losses and these are extremely difficult to calculate.
     
    Last edited: Nov 20, 2009
  4. Nov 20, 2009 #3

    jambaugh

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    Bikes roll with pretty low friction. The main concern is wind resistance or drag. You should be able to find resistance factors by searching the internet. (Actually looking I found in a discussion on recumbent bikes the calculations going into determining drag force from drag coefficient, cross sectional area, and dynamic pressure.)

    The basics will then be this. Power is the time rate of work and work is force times distance so for a moving object power equals force time velocity.

    Work out the force at 25mph due to wind resistance for a typical bike and rider. That force times the speed (converted to appropriate units such as meters /second) will give you the power requirement. Keep your units in your work throughout. They will guide you in the calculations and help catch mistakes.

    Power = work per unit time = force times velocity.
    1 Watt is one Joule per second which is one Newton meter per second.

    Now if you like you can try another tack. Decide how quickly you want to get from zero to 25mph neglecting wind resistance and work out the weight of the bike and rider. This will give you their kinetic energy which divided by the given time to reach 25mph is the power required.
     
  5. Nov 20, 2009 #4
    At 25 mph, the biggest power loss will be due to air drag. You cannot ignore air resistance, because the power required increases roughly as speed cubed. See
    http://users.frii.com/katana/biketext.html
    For a person sitting upright, the power loss from everything at 25 mph is about 1 HP. See plot "Power required (friction plus air resistance."
    Bob S
     
  6. Dec 17, 2009 #5
    I'm a bit confused. Lets leave out wind resistance and just calculate the HP required for the bike+me to go up to 25mph. In the equation, "Power = force times velocity," would force equals to the weight of me plus the bike? I don't understand how to find the force. As for velocity, it's just 25mph (i know you have to convert to the appropriate units)?

    Thanks for the help. I had nightmares about physics when I was in college.
     
  7. Dec 18, 2009 #6

    jambaugh

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    You also have to use force equals mass times acceleration. So the force depends on how quickly you want to get from zero to 25mph. If you neglect all friction then in principle you could use a teeny-tiny motor and a little solar cell to get the bike up to any speed given enough time. 25mph is 11.176 meters per second.

    Let's say you want to get up to that speed in 10 seconds using constant acceleration You must accelerate at 1.1176 meters per second per second. That times your mass equals your force. But notice that your power will be force times speed which is constantly changing so you are not talking about constant power. To figure constant power we work with kinetic energy.

    Recalling that power is rate of energy or rate of work notice that in this case (no friction) all the work goes into increasing the kinetic energy which is E=1/2 mass times velocity squared. In this case given the velocity of 11.176m/s the square of the velocity is 124.9 (m^2/s^2) Half that is 62.45 (m^/s^2).
    So 62.45 times the mass in kilograms is the energy in Joules = kg m^2/s^2 = watt seconds.

    Supposing the man plus bike have mass of 91kg (weigh just over 200 lbs.) They would then have a kinetic energy of 5683 Joules (watt seconds) or 5.683 kilowatt seconds. It would thus take 0.5683 kilowatts to get them up to speed in 10 seconds.
    (Divide kinetic energy by power to get time, or divide kinetic energy by time to get power).

    Let's say you want to get up to speed in 3 seconds then you need a power of 5.683/3 =1.9 kilowatts.

    Note that in this case of constant power you'll not get constant acceleration. Only one of the two may be constant in a given scenario.
     
  8. Dec 19, 2009 #7
    hi,

    i might be all wrong, so heed my words with caution...

    here is how i would go on about this:
    [tex]\sum[/tex]F = ma (Newtons 2nd law)
    so what is [tex]\sum[/tex]F? Its the torque on the wheels divided by wheel radius (adjusted according to any intermediate gears etc less force to overcome drag, rolling resistance and grade.

    You get a value of torque using engine power the vehicle speed (variable speed)
    you get force to overcome drag F=0.5 x rho x A x Cd x v^2
    obviously, not easy to calculate these, but maybe you can get data from experiments
    you get force to overcome rolling resistance and grade as F = (WxCf) + (Wxsin(grade))
    where W is the weight of the vehicle + driver
    Cf might be available from the tire manufacturer (dunno, its not in my part of the world)

    now, you put all these together, and you get a variable acceleration on one side and a variable velocity on the other

    from elementary dynamics, we know that:
    a = dv/dt
    so axdt=dv
    you get a differential equation. You solve it (integrate both sides) and get a velocity-time relationship

    you can put the equation in some equation solver etc. then you can easily see the variation of time to increase speed (acceleration) when motor power is changed. in fact, you can see the effect of changing other factors like weight etc on acceleration as well

    hope i might have made sense above
     
  9. Dec 19, 2009 #8

    RonL

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    Lots of good info above, in the past I found a Womack engineering handbook that has formulas on lots of mechanical applications, then there is Mechanical Engineering handbook. These are great for rolling resistance and wind resistance.
    I did the lazy route and went down to the local bike shop and bought a Mountain Bike with a 24Volt 7.2ah battery and a 430 watt 24 volt motor, it is geared 3:1 for the electric power and has a 5 speed gear for peddeling.

    Speed of electric is 20 mph, for approx 11 miles, first time I turned into a stiff wind, I was forever sold on electric. $400.00 and no math or building and I have been a lazy bike rider with a big smile on my face.

    Ron
     
  10. Dec 21, 2009 #9
    what kind of battery does the bike use and how long do you think is the battery's life? Battery has always been the weakness of electric vehicles.
     
  11. Dec 21, 2009 #10

    mgb_phys

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    A bit pessimistic power requirement.
    The average power for the 4000m individual pursuit record is about 500W ( 2/3hp) at almost 58km/h on a normal race bike (no fairing).

    As you said at this speed the majority of the power is air resistance, so if you wanted to go a more sensible 15mph you wouldn't need much electrical power - 200W should be fine.

    We use a 14.8V 12.6 Ah Li-ion cell for our portable equipment that weighs under 1kg in it's case = 190Wh so it would run a 200W motor for almost an hour.
    Not impractical, especially since you can still peddle if it runs down (easier than pushing a hybrid) and you would only use it uphill.
     
  12. Dec 21, 2009 #11

    RonL

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    The batteries are gel cell lead acid, I have not maintained a proper charging pattern so I suspect they will not last much longer, I have had the bike about 3 years.
    I have only gone out twice and had the low battery light come on, but I never go so far away from home that I would wear myself out to peddle back.

    I believe batteries being a weakness in electric powered vehicles is a myth or state of mind. A depleted battery is not much different than an empty fuel tank, the time to replenish the power might take longer, but most daily use will not result in a stranded situation as most driving needs in a 24 hour day leave a vehicle parked most of those 24 hours.

    People driving many miles or out of town trips will require fuel powered vehicles, but the vast majority of transportation needs fall in the " less than 25 miles a day" catagory, might be old data but I don't think it has changed much.

    Making present technology work for most needs, would not be much of a step, as I said " state of mind " is the biggest problem to address.

    Ron
     
  13. Dec 30, 2009 #12
    My friend, what you talk about is called perpetual motion, that is, motion that keeps going on forever and so on, without any requirement of an external energy source.

    That fact is, that this is not as simple as it sounds. Perpetual motion violates the first and second laws of the formulated Thermodynamics. In which case, you mean to say that you power up a dynamo by pedalling the cycle, which in turn powers the wheel, you in fact forget that the energy that powers the wheel is always lesser that what you generate by the dynamo. The energy can be lost by the factors of, as in this case, listed below. The energy lost means that energy is in fact converted and thrown off like an overload, but can never be destroyed.

    • Conductivity of wire (thicker wires conduct better, but superconductive wires conduct without any transmission loss, eliminating the problem)
    • Friction of air (can be removed in a vacuum)
    • Friction of dynamo (no way this can be eliminated, for friction is required if you need to power it up)

    As you saw, energy in the first case is lost by conductivity, then by air friction, by maintaining the super-cold environment of the superconducting wires, and finally by the friction of the dynamo, doing so which it also loses more energy as heat.

    So you see, vacuumed and supercooled environments will simply make your cycle impractical for use, but can be exploited as a scientific curiosity.
     
  14. Dec 30, 2009 #13

    mgb_phys

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    it's not perpetual motion.
    The electric bicycles are recharged at night from the grid.
    Some of the proposed designs do use regnerative braking to recover some of the energy as you go down hill, but for a few reasons these aren't as effective as in an electric car.
    You could have an electric bike that was recharged solely by you peddling - but the weigth and efficency losses of the extra components probably wouldn't make it worthwhile.
    In this case it still isn't PM - the extra energy comes from breakfast!
     
  15. Jan 3, 2010 #14
    One of the variant of the HP equation is "ft*lb / min" or "work / time". Is it also correct to say that "hp = force / velocity" by rearranging the units in the equation to f * (ft/min)?

    What do you think of the drive system where the motor shaft with a small rubber wheel is pressed against the side wall of the bicycle wheel? I had an electric razor scooter with this drive system and it works pretty good. No chain required and it's very simple.
     
    Last edited: Jan 3, 2010
  16. Jan 4, 2010 #15

    jambaugh

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    Power would be force times velocity not force over velocity.

    Power is work done per unit time which is force times distance per unit time or force times velocity.

    If it works it works. The simpler the better. I've seen some electric assist attachments for regular bikes which were similar.

    But if you look online you'll find there are hub motors you can buy http://www.goldenmotor.com/" [Broken]
    which would probably be ideal.
     
    Last edited by a moderator: May 4, 2017
  17. Jan 4, 2010 #16
    Ignoring drag and rolling resistance, for street riding one needs take into account the local terrain.

    200 pound gross weight, 5% grade and 15 mph will require 0.400 HP (295 Watts) delivered to the wheels.
     
  18. Jan 11, 2010 #17
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  19. Jan 11, 2010 #18

    RonL

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    If you do a search of "friction drive systems" there is a lot of history and different types going back to the early days of transportation.
    I have been considering this as a method for one or both of my pickups, the one ton with dual wheels seems most logical, as it has a nice taper between the rubber tires. It might be good only in dry weather times and making a suitable pressure application method, is my current reason of delay.

    An interesting find while doing my searching, was a man in one of the southern states, makes friction wheels using cardboard compressed between two plates, I think these are used in old steam engine systems and farm tractors and attachments.

    What's the old saying " what's old will become new again" :wink:

    RonL
     
  20. Jan 11, 2010 #19
    I had a razor type scooter with friction drive. The manual said not to ride it in the rain. I'm guess the friction drive system will slip if it gets wet.

    One of the biggest problem with this project is the fact that I don't have machining equipments to fabricate parts (motor mounts etc.). Is there a "metal lego like" kit where I can assemble parts together to make what I want to make?
     
  21. Jan 11, 2010 #20

    jambaugh

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    Your local hardware store should sell stock metal with pre-punched holes (Like the old Erector Set pieces). With a hacksaw and some bolts and maybe some hose clamps you should be able to whip something together...maybe not pretty but functional.
     
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