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Help me with some equations on torque & acceleration!

  1. Dec 7, 2007 #1
    I'm working on an electric car model and I need some help figuring out what I should set the gear ratio at it (it will be a fixed ratio that does not shift). I need to find what the options are on accelertion verses top speed.
    The motor being used maintains a steady torque of approx 670Nm from 0-1500 RPM, and then begins to drop off at a level rate to 100Nmm at 5000 RPM (top motor speed). The wheel diameter is approx 28 inches, (which I believe is about 88 inches in circumference). The vehicle weight is approx. 7000 lbs. Say that drag and friction are negligible (these are very loose approx. figures). How do I find acceleration and top speed of the vehicle? Mainly I need the acceleration on a level surface, so that I can figure out what the best gear ratio to set the car at.
    Mainly what I am asking for is an equation that would allow me to convert torque into acceleration given a vehicle weight.
  2. jcsd
  3. Dec 8, 2007 #2
    Well, first we're going to need a little more information, specifically about this wheel.
    - How many wheels? Doesn't matter if they're all being driven, just that they rotate as the car moves.
    - What's the mass of the wheel(s)?
    - What's the shape of the wheel(s)?

    These questions are important because what you really need to consider along with the total mass of the car is the rotational inertia of the wheels. Maybe you know this, but just in case: think about spinning a wheel on an axle. It should be pretty clear that for a given torque on the axle, a heavier wheel will not accelerate as quickly as a lighter wheel. Next consider a solid disk vs. a wheel with light spokes and a heavy rim. Even if the both have the same total mass and radius, the solid disk will accelerate at a higher rate, since its mass is distributed evenly over the radius, whereas the spoke & rim wheel has most of its mass out at the largest radial distance.

    The difference is quantified in the moment of inertia, which is ultimately what we're after, here. http://en.wikipedia.org/wiki/List_of_moments_of_inertia" [Broken] is a list of moments of inertia for objects of various shapes about various axes.
    Last edited by a moderator: May 3, 2017
  4. Dec 8, 2007 #3
    The simple answer to your main question would be Newtons second law: acceleration = force / mass.

    First, lets convert all measurments to metric. 7000 lbs = 3175 kg. 14 inches (the radius of your tires) = .3556 meters. 88 inches (circumference of tires) = 2.235 meters.

    Lets first assume a 1:1 ratio. Torque/ radius will equal the amount of force applied to the road by your tires. 670 Nm / .3556 m = 1884N. Back to Newtons second law, 1884N/ 3175kg = .5933 m/s/s of acceleration. (rather slow, but we'll fix that later.)

    Now lets jump to the top speed aspect. Multiply the circumference of the tires by the RPM of the tires. At a 1:1 ratio, that's 11,175 m/minute (670.5 km/h, a bit quick).

    Now that you have these figures all based on a 1:1 ratio, you can start adjusting your ratio to come up with your target top speed or acceleration, whatever they may be.

    You ought to start with a top speed, and THEN figure out what the acceleration will be. For instance, let's start with a top speed of 100 km/h. This of corse would require a gear reduction of about 6.7:1. Using that ratio your torque would increase to 4,489 Nm, which translates to 12,624 N of force applied. Again, using Newtons second law, that results in an acceleration of 3.976 m/s/s. So in ten seconds, you would achieve a speed of 143 km/h.

    I realize that 143 km/h is higher than the top speed was to begin with, and that all these calculations were based on a constant 670 Nm of torque. And of corse NO friction, resistance (indeed, moment of inertia being one of them), or other inefficiencies were factored in. But these are the BASICS you need to do your rough calculations. I'll leave the specifics to you. Good Luck!
    Last edited: Dec 8, 2007
  5. Dec 8, 2007 #4
    Hey, utimmer43, are you assuming that the inertia of the wheels is negligible? Why is that? I do a lot of bicycling, and we always worry about the effects of wheel mass on acceleration due to the moment of inertia of the wheels.
  6. Dec 8, 2007 #5
    If you read my last paragraph, I do acknowledge that NO friction, resistances (MOMENT OF INERTIA BEING ONE OF THEM) or inefficiencies were factored in. However, now that you mention it... While I would not necessarily consider moment of inertia "negligable", I certainly think it is far less a factor than the torque of the motor and the weight of the vehicle. Would you agree that your body weight+ the weight of your bike and the strength of your legs more so effect your acceleration than does the moment of inertia of your bicycle wheels?
  7. Dec 8, 2007 #6
    Well, I don't think it's quite right to include moment of inertia as a resistance, the latter being a term leading to dissipative losses, i.e. things that slow you down no matter what. Moments of inertia, like mass, oppose changes to motion, including deceleration.

    But more to your point - the rotational inertia of the wheel(s) of this car might well be negligible compared to the total mass, but I can't say with certainty without knowing more about the car, which is why I was asking. Experimental electric vehicles seem to have very light, thin wheels, so probably their inertia could be neglected, but if this is more a "normal" car with wide tires and steel rims, then I wouldn't think so.

    As for the bike wheel stuff, yes, it is pretty small compared to the total weight of the bike+rider, but given the choice, cyclists generally choose to lighten their wheels first, as opposed to getting, say, lighter handlebars, because of the benefits of lower rotating mass. Hey, they even worry about the weight of pedals and shoes for this reason, but then again, they're a weird bunch.
  8. Dec 8, 2007 #7
    Ok, so the moment of inertia is opposing the acceleration of the vehicle. "Resistance" or not, I'm not going to argue semantics.

    My original reply was a direct answer to the very specific question posed, "Mainly what I am asking for is an equation that would allow me to convert torque into acceleration given a vehicle weight." and also "that drag and friction are negligible." One would assume that moment of inertia is to be considered either as important, or as negligible, as drag and friction. I never dismissed moment of inertia (in fact I mentioned it as something I left out, but that would have some effect).

    This forum is full of people who are way smarter than most questions necessitate. When a person asks a specific question about a specific formula, and factors like drag and resistance aren't important to him, people reply with a load of other factors, and never bother giving the poor guy the formula he asked for to begin with. I could have simply answered HIS question with a=F/m and let him do all the conversions and figure out the rest. At least that would be a starting point. The last thing I said was "I'll leave the specifics to you", meaning that after applying Newton's law, he can get as technical or be as vague as he needs to be.
  9. Dec 8, 2007 #8


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    Homework Helper

    Note, for this motor, peak power occurs at 2807 rpm:

    let w = rpm

    torque (nm) = (-570/3500) x w + (3200000/3500)

    power (nm x rpm) = (-570/3500) x w^2 + (3200000/3500) x w

    peak power occurs when the derivative is zero:

    0 = (-570/3500) x w + (3200000/3500)
    w = 2807 rpm.

    If you're ignoring drag factors then there is no limit to top speed (infinite gearing).
    If you assume infinite friction, then there is no limit to acceleration.

    torque (ft lb) = (.737561) x ((-570/3500) x w + (3200000/3500))

    let r = radius in feet:
    force (lbs) = torque / radius = ((.737561) x ((-570/3500) x w + (3200000/3500)))/r
    acceleration = force / 7000
    Last edited: Dec 8, 2007
  10. Dec 8, 2007 #9
    Okay, I don't want to make a big thing out of this, but this is not a semantic difference - we have a conceptual disagreement. Dissipative forces like friction may be neglected in a first approximation because they can be made small through design. Inertia cannot be changed, either linear or rotational, and neither can be neglected until you've shown that its contribution is clearly smaller than that of the other. The point, which I'm not sure you've gotten, is that moments of inertia are in the same category as mass - together they are what determine the acceleration, given the torque (or force). They are on the same footing and are completely different from drag or mechanical friction.
    Why would one make this assumption? They're in totally different categories. I'd believe you if you said that "one would assume that the moment of inertia is negligible compared to the total mass" - that's a valid comparison. I was just asking to verify this; I agree it's probably true.

    That's my problem - that equation is incorrect. It should be a = torque/(I/r +rm) . (I is the MOI.) Given the large mass of this car (7000 lbs), the rm term almost certainly dominates over the I/r term so that this equation reduces to yours, but that's something to confirm. You can't just say "moment of inertia is negligible like friction" - it's not true.
  11. Dec 9, 2007 #10
    Look, this is silly to argue about. You're acting as though by not factoring in MOI in my explaination that I am discounting MOI altogether and calling you a fool for suggesting it.

    If MOI isn't classified as a resistance, then so be it. It IS opposing the acceleration of the vehicle, which you and I both agree. The degree of the effect is where we disagree, and that is because right now both of us are guessing. If electricali is trying to hit a specific rate of acceleration, with no tolerance for error, then I would absolutly agree with you that he should factor in MOI before choosing his gear ratio. But if that is the case, then I would also argue that wind resistance, rolling resistance, energy loss in the form of heat, friction between all gears and linkage, and any other inefficiencies must also be factored before selecting a ratio. Perhaps the MOST important factor I left out is the horsepower curve, which will increase steadily with increased RPM until 1500 RPM when the torque begins to drop. At which point the horsepower will begin to level off, and eventually start to drop (which is the point Jeff Reid is making).

    ALL of these factors will have an effect on the final product. Again, when I said "I'll leave the specifics to you", I said that because I don't know what his tolerances are. I assumed them to be high because HE SAID "Say that drag and friction are negligible" and all of his other figures were approximate. When it's time for him to go to the drawing board and make some final decisions, HE will make the call as to what his tolerances are and how important all these factors are to him.
    Last edited: Dec 9, 2007
  12. Dec 9, 2007 #11
    When I said "one would assume..." I meant that, given his consideration that drag and friction are negligible, I would assume that he also considers MOI negligible. Again, only he knows what his tolerances are.
  13. Dec 9, 2007 #12
    Well, I agree that we passed the point of becoming silly some time ago ... but my point here was that I don't see why
    - one doesn't follow from the other. Friction and drag could be reduced as close to zero as you like, but MOI wouldn't change - it is what it is. There are perfectly good reasons for neglecting MOI, but yours isn't one of them.

    But I'd be happy to drop this since by now the OP has long since left us quibbling in the corner while he drives his electric car to the convenience store to get a snack.
  14. Dec 9, 2007 #13
    WOW!! What sir, if I may ask, do you believe IS my reason? I don't know how many different ways I could possibly say this: I never even suggested MOI BE neglected! The only reason I breezed over it was, first of all, you had already mentioned it. And second, The fact that HE considered drag and friction to be negligible gave me impression that HE would also consider MOI to be negligible. Now if I got the wrong impression there, then I guess that is a downside to text rather than an in-person conversation.

    You're right, it is what it is! It WILL have some effect. And if he considers it to be important enough to figure out, then I'm sure that he will use the link that you posted. And I wouldn't suggest that he not do it. THAT is up to him.
  15. Dec 9, 2007 #14
    Jeez - calm down!! I believed your reason to be just what you said - because drag and friction were said to be negligible, you then concluded that MOI was, as well. That's all! You've said it again, so it looks like you still believe it.

    As I've said, it most likely is negligible, but my original response was just to ask the OP about it so that we could know, since he stated only that drag and friction were negligible, which has no bearing on whether or not MOI will be.
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