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Torque required to push an EV

  1. Apr 10, 2008 #1
    Hi,
    A normal query regarding moviement of the wheel of an EV.
    In the designing of an EV, I calculated the torque required to pull the wheel of my EV on normal surface as T=F*r (keeping in mind friction and other factors too),where 'r' is the radius of the wheel. Similarly, I calculate the torque for uphill and for clearing the obstacles. Now, the torque I calculated is the peak torque as I am using T=F*r, where F=M*acceleration, say during my normal operation of the EV on the normal surface, I require this much of torque to push or to provide a start to my EV, because its considering the acceleration here, now my problem is here that I want to consider the velocity of my EV, like after supplying the initial torque to the wheels, what is going to be the normal torque, say like my velocity is 1 m/s and radius is .3 m and acceleration is 1.5 m/s2. So basically, I want to know the torque require to push my vehicle for a certain time 't' and with a velocity of 1 m/s irrespective of the acceleration.

    Any help would be of great help.

    Thanks!!!
    Dave
     
  2. jcsd
  3. Apr 10, 2008 #2

    rbj

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    if you're not accelerating, the amount of torque needed is usually broken up into two components: rolling resistance due to bearing friction bending loss in the rubber tires and which is constant with speed. the other torque component is wind drag which the torque will increase with speed (i dunno, somewhere between linear and a square function). if i remember right, for most conventional cars, we needed 3 hp per 1000 lbs of vehicle weight to overcome rolling resistance to pull a car at 50 mph. with all those numbers, you should be able to figure out that constant torque. i don't remember the wind drag numbers but it was negligible for slow speeds (1 m/s) but much more than the rolling resistance at 50 mph.
     
  4. Apr 10, 2008 #3

    rbj

    User Avatar

    one thing you can do is get a reasonably new conventional car (hopefully with standard transmission) and on a straight and flat and level and smooth paved street (or parking lot) (maybe with tape markers every 5 meters), drive it up to 30 mph and then put it in neutral and let it coast to a stop, while filming it with a digital camera with known frame rate. that will tell you what the rolling resistance is.

    if the wind was very low, and you had a very long and empty parking lot, you could do the same thing to get some numbers on wind resistance but you would have to start at 50 mph and let it coast to a stop.

    you'll also need to weigh the car somehow.

    a way to experimentally get numbers.
     
  5. Apr 11, 2008 #4
    What I have done, to get a rough idea for a small vehicle, is attach a rope to the vehicle, over a pulley, tied to some weights--jugs of water to be precise. It's good for a rough estimate at low speed where wind drag is nil. If you have especially hard wheels, the surface smoothness has a substancial effect on rolling resistance.
     
  6. Apr 11, 2008 #5
    Hey rbj and Phrak thanks for the suggestions!

    rbj, you suggested to consider the rolling resistance and the aerodynamic drag, that is fine but i think that along with these factors i also have to consider the kinetic energy of my vehicle (here i am talking about the designing of a power wheelchair). What i am considering here is say that my vehicle move a distance of 100 m with a velocity of .5 m/s, so:
    1) the total work done in moving the vehicle from initial to final distance = integration of the product of force and displacement and this work done eventually comes out to be equal to the difference of the final and initial kinetic energy. Which gives that the work done is equal to the difference of final and initial kinetic energy.
    So, calculating the work done here for some distance and then relating it to w=F*d, and then finding Force from here having the values of 'd' and 'w' in hand. Now as this force is a factor of the velocity of the vehicle (i mean not related to the acceleration of the vehicle as what i asked in my previous question) and with this force plus the force require to overcome the aerodynamic drag plus force require to overcome rolling resistance will give me the total force require to drive my vehicle for a ceratin distance irrespective of the initial acceleration and depending on this force i can find the torque required. Is this approach correct? Please guide me in this regards.

    Any help would be of great help!

    Thanks!
    Dave
     
  7. Apr 11, 2008 #6

    rbj

    User Avatar

    well, since wind drag is not so much at issue because of the slow speed, then if it is flat, level, and smooth pavement, the only force to overcome once you are at the final velocity, is rolling resistance, which is a constant force no matter what speed. how much is that force? i dunno, but an experiment like the one i described with the car will work. put a person (not the handicapped person, but someone of the same weight) in the wheelchair, push it to at least twice the speed you expect and let it coast to a stop, all with a digital movie camera of known frame rate recording it. you could even put tape markers on the wheel of the chair so that you could see how much it turned in each adjacent frame.

    if the rolling resistance force is constant, which it should be to a good level of precision, then the rate of decleration should be constant and you can calculate that from the movie frames.
     
  8. Apr 12, 2008 #7
    Now I'm really confused.
    Usually what you want to know is how fast you will get up to speed, or accelerate with a given motor and gear ratio under some appropriate rolling resistance and grade.

    You might be limited by the maxium amount of current your motor can sustain. In DC motors this translates directly to torque at the motor shaft. Every motor has a constant Kt, relating torque to current.

    In any case, I've found it usefull to graph some constant-torque curves. Remember, this is torque at the motor shaft, not the wheel. Usually you will gear down the motor, so the wheel spins slower than the motor. Call this the gear ratio, G. Then the force applied on the vehicle F = GRT, where T=motor torque.

    You might want to plot velocity as a function of time given some constant torque.
    This will look like a charging curve of a resistor-capacitor circuit. The initial torque goes to accelerating the vehicle. After some period of time all the torque goes into maintaining the final velocity.
     
  9. Apr 12, 2008 #8
    ! I got caught in a PF database crash before I could edit that last post, so some of it's misleading. I've looked over my graphs. The top speed will be limited under no-load conditions by the motor's Kv that relates motor voltage to shaft RPM. Under conditions where the drag is non-zero, the top speed will be less.

    To get acceleration and velocity curves as a function of time, I've assumed full voltage applied to the motor while assuming some particular drag coefficient. I also plot current drain, and instantanious power loss in the motor as a function of time.

    The equation F=GRT is still valid, where G=gear ratio, R=wheel radius, and F is the force applied to the vehicle. I think a good place to start, would be to assume, on level ground, F = 6% of the total gross weight of the passanger and vehicle; 2% for rolling friction and 4% for the power train.
     
  10. Apr 12, 2008 #9
    Make that:

    F=GT/R, where G=gear ratio, T=motor torque, and R is the wheel radius.
     
  11. Apr 12, 2008 #10
    Hey thanks for replying!

    Phark! along with the acceleration and velocity curve with respect to time, my major concern is "to plot the torque curve with time", I have initial torque required (i.e of acceleration) say of around 40N-m, I have torques available for climbing uphill and clearing obstacle.

    Now once my wheelchair accelerated at say around 40N-m initial torque (with an acceleration of around 1m/s2), now inorder to plot the torque-time curve, I am looking for the torque require to move at constant speed, Well as per rbj, it is related only to the amount of force require to overcome the rolling resistance. I am agree with that, but say if my rolling resistance is 0.001, then the force which i calculated with this value is very very less and there is a huge difference in the initial torque (at starting with an acceleration of 1 m/s2) and the torque at a particular speed of say 0.5 m/s. And these values are not looking practical. Please guide me.

    Thanks!!!
    Dave
     
  12. Apr 12, 2008 #11
    Hi dave. Are you sure you really want my help, with all the errors I've incurred above??

    But, be that as it may, ther're a few points that need addressing. 40 Nm is a way lotta torque. I think you mean either 0.040 Nm torque, or you're taking about a force of 40 mN, or 40 milliNewtons somewhere.

    I don't think in SI units at all well, so bare with me. An initial acceleration that will get you 15 fps in 3 seconds should be nice. That's an acceleration of about 4.5 m/s^2. Half of that might be sufficient.

    You quote a drag coefficient of .001. That's 1/10 of one percent--way lower than you can realistically expect. For really hard (Shore 80A), but small diameter wheels on a smooth surface I think I get 1%-2% using (quality) AEBC bearings. With inflatable, or softer (Shore 60A?) surfaced wheels you can expect less. I'd guess 2 to 3%. But you need to gear down maybe 15 to 1. Add another 5%. I'm not familiar with gear train losses or sprocket and chain losses so double-check this. This is an overall frictional coefficient of 0.008.

    For every kilogram of mass your vehicle+passanger has, you have 9.8 Newtons of downward force on the wheels, right? At 4.5 pounds = 1N .....I get an initial design mass of about of 55 Newtons gross. (add 5 newtons for fast food american riders)

    8% of 55 Newtons = 0.440 N is the force you're drive wheels have to exert tangentially to the ground at a first guess.

    Now the gear ratio and wheel radius need to be considered. I'll assume a 3.5" radius = 0.090m. The required torque on the wheels is .440N x .09m = .040 Nm. At a 15 to 1 gear ratio the required torque is .0027 Nm on the motor shaft.

    In summary, on level ground your motor must be capable of .0027 Newton-meters of torque. You will need to check my figures! SI is not at all natural to me.

    Now for the force to accelerate the vehicle.

    I got 4.5 m/s^2. What the hey is the mass?? 112 kilos?

    See if you can find the torque required on the motor to accelerate. I cant think in SI, convert back and forth between SI and FPS, get interrupted every 15 minutes by a 6 yr old, and have any natural feel for what seems right vs. two decimal places off.
     
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