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Vertical force generated by a wheel

  1. Mar 4, 2015 #1
    Hi all. I'm working on building a go kart, and it's been a very long time since my college trig based physics classes. I'm having trouble figuring out the proper stiffness and length of spring to buy for the suspension. In short I need to know the vertical force generated by a wheel and attached suspension encountering an obstacle. I can understand simple physics, but impact centripetal physics seem to be beyond the capabilities of this cobwebbed brain.

    Forward velocity = 18 m/s
    Obstacle = .05 m high
    Acceleration (linear degradation) = 0 m/s2 @ 18 m/s, 4.5 m/s2 @ 0 m/s
    Vehicle weight = 113.5 kg

    Front wheel diameter = .10 m
    Front tire diameter = .26 m
    Front Tire & Wheel weight (ea) = .91 kg
    Front unsprung weight (each side including tire/wheel) = 2.8 kg
    Rear wheel diameter = .15 m
    Rear tire diameter = .37 m
    Rear tire and wheel weight (ea) = 4.1 kg
    Rear unsprung weight (each side including tire/wheel) = 14.79 kg

    Sorry I don't know the compression rate for the tires. Each tire is inflated to 27.6 kPa. If it makes a difference they are pretty simple 2 ply tires.

    Here's what I've tried.
    Speed equilateral right triangle, hypotenuse 17.88 gives Vx=Vy=12.64 m/s
    Obstacle equilateral right triangle, sides .05m, hypotenuse .0707/17.88 = t = .004s
    Since Vy = 0 to start, ay = 3160 m/s2
    Using F = ma, forward force exerted on bottom of spring is 8848 N, rear 46736 N

    These seem high to me, even considering the impact speed. I'm sure that the low pressure tire absorbs partial impact, more force is lost since the tire rolls over the object instead of point impact. Obviously I've not found a reasonable spring which can accommodate this, which is why I need someone who can understand this better than I can. Thanks for the help.
  2. jcsd
  3. Mar 4, 2015 #2

    Simon Bridge

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    Your calculation need not be exact - basically you want to know how rugh the terrain is that the suspension has to cope with - i.e. how big is the typical obstacle?
    If the cart is to be steady when going over a 2cm high whatsit, then the spring must compress by 5cm in the time it takes for the wheel to climb the whatsit without slowing down. You also need the springs to be stiff enough that they don't compress too much under the passenger...

    Usually you'd build an adjustable suspension - do the physics to get the right ballpark figure.
  4. Mar 5, 2015 #3


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    Without breaking/bending wheels, you're not going to be putting much more force on anything than the tire inflation pressure times tire footprint area for full cart weight on a single wheel. How much stiffer you want beyond that ---- ?
  5. Mar 5, 2015 #4
    What I'm concerned about is getting springs too weak to adequately react to obstacles up to .05 m (2") at full speed as this will be almost exclusively driven off-road. Of course having springs too stiff is almost as bad. Initially I'd wanted to clear 4" obstacles at full speed, but with just 10.2" of front wheel I think that height would probably be more likely to blow the tire off the wheel before the suspension could react. I'd also rather not send the kart flying off on wild attitudes and tangents at every bump. Additionally the overall weight of the kart is really 113.5kg, as it will be radio control, there's no person to add weight to the kart. So an adjustable suspension shouldn't be required. I've already figured out the force on each spring by the weight above so it's really just the tire impact that I'm having difficulty with.
  6. Mar 5, 2015 #5

    Simon Bridge

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    All suspensions IRL are adjustable - it means you don't have to have as fine calculations in design and you can maintain better against wear and tear.

    A weak spring will go to max compression too easily while a strong spring will not react enough.

    It is not really possible to help in more detail without the details of the suspension design - these things don't just have springs in them.
    The best we can do here is point in likely directions ... you'll have to apply Newtons' laws etc to your design.

    Suspension design is quite complicated - but there are lots of helpful documents online.
  7. Mar 5, 2015 #6


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    What I don't see anywhere is the travel of the suspension.

    An ideal suspension would have a pre-stressed spring that is at an equilibrium in holding-up the vehicle, yet provides very little resistance to up and down motion. That way, when hitting an obstacle, the wheel moves up and down, but the vehicle doesn't. So if the travel of the suspension is larger than the largest obstacle you want to drive over, the spring and shocks just exist to dampen the induced vibration as quickly as possible.
  8. Mar 5, 2015 #7


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    'Real' cars have dampers that are arranged to let the wheel go up fast, when you hit a bump but return relatively slowly to the equilibrium position. The energy on the return is dissipated as the wheel moves downwards. This asymmetrical damping gives a much more comfortable ride. I believe that some old sports cars tended to use the frame as part of the suspension but that is not a good or comfortable system. I don't know just how sophisticated the highest performance karts are but some damping can do an awful lot to help stop a car hopping sideways when it hits a bump on a bend - by making sure that the wheel has contact for the maximum possible time.
    The object that provides most damping on a very simple kart is the body of the driver!! Your poor spine.
  9. Mar 5, 2015 #8
    OK here's some rough sketches of the suspension designs, obviously not to scale. The front has changed a little from this drawing in that the spring assembly will not go directly through the strut bar, but through an eye bolt trailing behind it to allow for some rotation. The spring assembly is a simple thing, a spring (length and stiffness tbd) encapsulated between washers pressing against eye bolts on either end. The entire assembly will be fixed at the bottom by nuts on a bolt. The bolt will act as a guide rod for the assembly and the head at the top will act as a stop point to prevent over extension. So in theory I can pre-load the spring to a certain amount by shortening the guide rod, however, it would not be able to extend beyond the pre-load length. There is load on the springs even with the car at rest from gravity, total 767.5N spread over the 4 springs, or about 191.9N if I can obtain 50/50 weight distribution. Obviously since the length and stiffness of the springs is still in question, the travel of the suspension is in question. The stiffer the spring used the less it will have to travel, but the spring free length will be <..15m, and travel <.05m. Obviously the longer the spring the higher the ground clearance, but also higher center of gravity. I'm more concerned with the CG as the wheel track is only .76m, hence trying to use shorter stiffer springs. There will be no damper or oil filled shock to smooth things out, finding such a device in my scale and load usage has proven nigh impossible. I've tried using coil over shocks from other go karts, but because they are designed to carry 1 or more people, they are too stiff and don't react properly to my load.

    Were my previous calculations done correctly? I thought they seemed high, but maybe I'm not accounting for something. The force of g on such a small weight doesn't off set the vertical force very much. I'd read somewhere in another thread about figuring the obstacle point of contact on the wheel as a lever, would that actually be appropriate?

  10. Mar 6, 2015 #9
    I've tried looking at this another way. Previously I had assumed a 45o angle for my triangles to the obstacle. Today I started to address the fact that I have a circular wheel to roll over the obstacle. Using the diameter for each tire (front 10.2" and rear 14.5") I obtained the actual angle appropriate to use for the triangle, which is smaller as the tire will rotate down on top of the obstacle instead of slide into it. Because the front and rear use different size and mass tires and unsprung weight they each result in different values for nearly everything. Hold the phone.....

    In typing this I just realized that I've been performing the calculations using the entire mass of unsprung weight as the mass which the force is applied to. In fact some of this mass is rotational about the suspension, and it is only in fact the weight of the tire and wheel which the force of impact is applied to. The unsprung mass is important because that entire mass (including the tire and wheel) is still subject to gravity and hence an opposite force to the vertical acceleration. So I've been getting such high force number because I was wrongly applying the acceleration to the entire mass of suspension for each wheel and not just the rotating mass of tire and wheel. It's 2am, and I'm very much hoping this has been my error all along, can someone please validate this new concept?
    Fy = aymtire - gmuw

    I also found a more appropriate weight for the front tire and wheel at 2.26kg, meaning front unsprung weight (each side incl tire & wheel) is now 4.16kg.
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