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Maximum velocity of a vehicle moving downhill

  1. Aug 1, 2009 #1
    Suppose some vehicle with mass [tex] m [/tex] is moving down a hill with angle [tex] \theta [/tex]. Say that the vehicle is exerting a force parallel to the ground of magnitude [tex] f [/tex]. Suppose that the maximum velocity of the vehicle on flat ground is [tex]
    v_{max} [/tex]. What is the max velocity of the vehicle down the hill, neglecting air resistance? I have a scooter, and I am trying to figure out it's max speed down any hill of an arbitrary grade.
     
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  3. Aug 1, 2009 #2

    Hootenanny

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    This is an extremely complex question. Even if we ignore drag (which isn't a reasonable assumption for bodies of this size, travelling at typical speeds) there are still several other factors to consider such as the rolling resistance of the tyres, the mechanical 'resistance' of the engine and transmission etc. To produce a reasonable estimate, one would need to take into account all of these factors.

    I would suggest that it would be better to try and find the relation empirically.
     
  4. Aug 1, 2009 #3

    Cyosis

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    Are you just ignoring air resistance or resistances all together? In the latter case there won't be a maximum speed (classically).
     
  5. Aug 1, 2009 #4

    Pengwuino

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    Also.... it's applying a force parallel to the ground? Do you mean that there's a driving force?
     
  6. Aug 1, 2009 #5
    If vmax is the maximum velocity on the flat ground, then ignoring air resistance and considering f to be the frictional force on flat ground(if that is what you mean) the force downhill is mgsin(theta) - f {where f is mu times mgcos(theta), mu being the coefficient of friction}theta is the angle to flat ground
    With the force giving the value of acc in the dir of theta use the kinematic exp
    v = u + at {u being zero, a being the sum of acc you get from the previous step and the acc of the scooter from v=0 to v=vmax} upto the point its velocity becomes vmax
    and afer that let the equation be
    v=u'+a't'{u' being vmax, a' being the acc from the first step only and t' starting from 0 at that point}
     
    Last edited: Aug 1, 2009
  7. Aug 1, 2009 #6

    jtbell

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    Why does it have a maximum velocity on flat ground?
     
  8. Aug 1, 2009 #7

    Cyosis

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    I guess he means that his speedometer can't exceed a certain value on flat ground.
     
  9. Aug 1, 2009 #8

    negitron

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    No, it's a clever question designed to get at the root of this problem. The reason a vehicle (any vehicle) has a maximum speed is because the force transmitted from the engine to the driving wheels becomes equal at some point to the forces acting in opposition to this, such as air resistance, friction, etc. This speed will also be different for various gearings because the engine's output power begins to decrease once the RPM rate is out of its power band. As someone noted upthread, it's a complex issue.
     
  10. Aug 1, 2009 #9
    The reason I put the max velocity on flat ground was because to my way of thinking, it gives some information about what it's max speed downhill might be. I appreciate the complexity of the issue, and have garnered some insight to the problem through your comments. If I find a reasonable solution, I shall post it for feedback.
     
  11. Aug 1, 2009 #10
    If you're under 30 mph, I would say you can safely ignore air resistance. The primary drag is rolling resistance which is easy enough to estimate. What is the top speed of your scooter?
     
  12. Aug 2, 2009 #11
    The rolling resistance coefficient for automobile tires is about 0.01, so the restraining force from tire "friction" on level ground is about 0.01 mg. For a slope angle theta, the gravitational accelerating force is mg sin(theta), while the tire rolling resistance restraining force is mg cos(theta). Because it is independent of velocity (below about 80 Km/hr for cars), it will not lead to a terminal velocity.
     
  13. Aug 3, 2009 #12
    That, is a very good post. I just coded a simulation that deals with friction and everything you said is dead on in terms of the numbers. That's a fine point about the sin and cosine. Now consider a realistic scenario. How big of a hill do you think he will be on? For all intents and purposes, you will probably run out of hill well before this approximation breaks down. It's something easy enough to graph and see though.
     
  14. Aug 3, 2009 #13
    The force downhill should be written as mgsin(theta)-fcos(theta)
    where f is (mu)mg
    a will be (mu)gcos(theta) plus frictionless acceleration
    a' will be (mu)gcos(theta)
    The first post is faulty
     
  15. Aug 4, 2009 #14
    What I am trying to show is that given a hill with a 2% grade, a scooter with a passenger (a combined weight of roughly 500 lbs) can exceed 40mph, where the scooter has a 2 horsepower engine, and has a flatland top speed of 30mph. Now I believe that there may be no max velocity given an infinite hill, however let us say that this scooter is moving down a 1 mile stretch of road.
    Now knowing that the flatland top speed is 30mph, i suppose one can estimate the coefficient of friction. Since horsepower is 550 ft lbs per second then a 2 horsepower motor will exert
    [tex]
    \[
    1100 = \frac{fd}{t}
    \]
    [/tex]
    in one second at 30mph one travels 44 feet so
    [tex]
    \[
    1100 = 44f \Rightarrow f = 25
    \]
    [/tex]
    So the motor exerts 25 lbs of force while at top speed on flat ground
    now 25 lbs is about 111 newtons, and 500lbs is about 226 kg so then since we are at topspeed (zero net force) we have
    [tex]
    \[
    \mu m g = 111 \Rightarrow \mu(226)(9.8) = 111 \Rightarrow \mu = .05
    \]
    [/tex]
    Would my calculations be a reasonable estimate of [tex]\mu[/tex]?
     
  16. Aug 4, 2009 #15
    You are not solving this correctly. Find the drag coefficient from a level run, and us 0.01 for rolling resistance. Then you're done.
     
  17. Aug 4, 2009 #16
    What about speeds greater than 80Km/hr?
     
  18. Aug 4, 2009 #17

    rcgldr

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    That implies it would take about 40 lbs of force to keep a 4000 lb car moving on level ground at slow speed. I don't know how much of an issue the drive train is if the car is in neutral.
    I would assume that rolling resistance is related to deformation which is related to the total load on the tire, regardless of direction. If the vehicle is not accelerating vertically, then the load on the tires is m g, regardless of angle (within reason).

    Also rolling resistance increases a bit with speed (see figure 4):

    http://www.tut.fi/plastics/tyreschool/moduulit/moduuli_8/hypertext_1/3/3_3.html

    However, assuming constant drag greater than the 10 lbs of force that gravity generates in the direction of travel, there is a top speed. The engine produces 1100/speed (speed in ft / sec) lbs of force. If the drag force remained constant at 25 lbs, then top speed would be 50 mph (73.33 ft/sec), the speed at which the 2 hp engine generates 15 lbs of force, so that total force = 10 + 15 == total drag = 25.
     
    Last edited: Aug 4, 2009
  19. Aug 4, 2009 #18
    Rolling resistance arises due to the offset of the normal force resulting in it exerting a braking torque on the wheel. So the grade does reduce the rolling resistance in the same way as it reduces the normal force.

    I don't know understand how you arrived at values 10 lbs for gravity and 25 lbs for drag. :confused:
     
  20. Aug 4, 2009 #19
    If the vehicle is not accelerating vertically, load on tires is mg, but it is, with load of mgcos(theta)
     
  21. Aug 4, 2009 #20

    rcgldr

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    25 lbs was the stated amount of drag, based on a 30 mph top speed with 2 hp. force = power / speed = (2 x 550) / 44 = 25 lbs.

    Weight of vehicle with riders was stated to be 500 lbs. The component of this weight in the direction of travel = 500 lbs x sin(tan-1(.02)) ~= 10 lbs.

    As the vehicle approches it's top speed, the rate of acceleration decreases, and the load increases. Even if the slope was frictionless, the load on the tires would be m g cos((tan-1(.02)) ~= m g 0.9998 = 9.998 lbs, close enough. If the vehicle isn't accelerating, (and assuming no aerodynamic drag), then all of it's weight is supported by the tires, regardless of the slope angle.
     
    Last edited: Aug 4, 2009
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