Does weight matter with electric go-karts?

Jeff Glucker

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Out here in Californy way, one of my favorite places to turn a few laps is at a K1 Speed. This is a chain of indoor karting locations. They have fun track layouts, nice amenities for when you’re done racing, and quality electric karts. The lap timing is clear and easy to read while you’re racing too.

I’ll admit that there have been times when I believe I could’ve beaten a competitor if I had a few less pounds on me. Or they weighed a bit more. It seems I’m wrong though, because the torque provided by an electric kart should negate that weight penalty.

Two drivers of differing size set a few laps. Each did so without a weight penalty, which pitted a 230-lb driver against his 180-lb friend. Then the two went out and set laps wearing a 40-lb weight vest. Neither driver saw worse times. It looks like the instant power of the electric kart overcomes the additional heft.

I’m curious if the story would be different if the drivers were using gas karts. Power-to-weight is a very real thing, and it should be a clear help or hindrance in a kart. I’m quite surprised to find that’s not the case with the karts used in the video above.

About

Jeff Glucker

Jeff Glucker

Jeff Glucker is the co-founder and Executive Editor of Hooniverse.com. He’s often seen getting passed as he hustles a 1991 Mitsubishi Montero up the 405 Freeway. IG: @HooniverseJeff

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23 responses to “Does weight matter with electric go-karts?”

  1. smalleyxb122 Avatar
    smalleyxb122

    Weight doesn’t matter (much) because it is a tight, low speed track. It’s more about how much speed you can carry through the corners than it is how fast you can get it up to on the straights. How much speed you can carry through the corner is mostly about traction, and traction doesn’t care about weight. Weight falls out of the calculations when calculating lateral grip.

    Gas karts on the same track would similarly show very little disadvantage to a few extra pounds.

    Enough weight would slow you down, but I’m not sure what would constitute “enough” weight. Clearly, it is more than 40 (or 80) pounds.

    Reply
  2. Eric Rucker Avatar
    Eric Rucker

    Another factor is the weight of the kart.

    A lot of these places use karts with lead acid batteries, and that means they’re heavy – looks like an OTL Storm (what K1 Speed uses – not sure about the batteries, though, as there’s a Li-ion option) with lead acid weighs 602 lbs.

    This means that driver weight is a lower percentage of total weight, and therefore less of an influence.

    Reply
    1. Troggy Avatar
      Troggy

      A battery won’t get lighter as it ’empties’ like a fuel tank would, so you’ll probably get those same results all day. It might make for a good measure of who the better driver is I’d imagine.

      Reply
      1. Fuhrman16 Avatar
        Fuhrman16

        Would that really be much of a factor though? A go cart shouldn’t be carrying all that much fuel for it to make that much of a difference.

        Reply
  3. smalleyxb122 Avatar
    smalleyxb122

    Weight doesn’t matter (much) because it is a tight, low speed track. It’s more about how much speed you can carry through the corners than it is how fast you can get it up to on the straights. How much speed you can carry through the corner is mostly about traction, and traction doesn’t care about weight. Weight falls out of the calculations when calculating lateral grip.

    Gas karts on the same track would similarly show very little disadvantage to a few extra pounds.

    Enough weight would slow you down, but I’m not sure what would constitute “enough” weight. Clearly, it is more than 40 (or 80) pounds.

    Reply
    1. danleym Avatar
      danleym

      “Weight falls out of the calculations when calculating lateral grip.”

      Can you explain this? I’m not saying you’re wrong, it just seems counterintuitive and I’m curious to see a more detailed explanation.

      I’m not an engineer- had a few engineering classes in college and quickly realized some of it was over my head.

      Reply
      1. smalleyxb122 Avatar
        smalleyxb122

        F=m*a, therefore, a=F/m
        F=mu*N (Accelerative force, coefficient of friction, Normal force)
        N=m*g (Normal force, mass, acceleration of gravity)
        a=mu*(m*g)/m (coefficient of friction, mass, acceleration of gravity, mass)
        a=mu*g*m/m
        a=mu*gm/m

        a=mu*g

        I tried to do it in steps here to show how mass cancels out. Let me know if you can’t follow it, and I can try to clarify.

        Reply
    2. danleym Avatar
      danleym

      “Weight falls out of the calculations when calculating lateral grip.”

      Can you explain this? I’m not saying you’re wrong, it just seems counterintuitive and I’m curious to see a more detailed explanation.

      I’m not an engineer- had a few engineering classes in college and quickly realized some of it was over my head.

      Reply
      1. Jason Lancaster Avatar
        Jason Lancaster

        When you’re calculating grip, it’s just friction coefficient times surface area. The calcs assume uniform contact, which is obviously a big assumption. A tire with no load on it will have the same lateral grip as a tire carrying 1,000 lbs, according to the calcs. But experience tells us that this isn’t so.

        Suffice to say, the calculations we learn in physics 101 aren’t wrong, but aren’t 100% correct in the real world either.

        Source: sparsely used engineering degree. Take it with a grain of salt. 🙂

        EDIT: I said “friction coefficient times surface area”, and that is incorrect. I was thinking “surface area” when I should have been typing acceleration of gravity. My apologies. This is why I don’t usually dive into these conversations, as I inevitably make a dumb error…

        Reply
        1. smalleyxb122 Avatar
          smalleyxb122

          Surface area doesn’t matter in a physics 101 calculation of traction. It affects the resultant mu when factoring in the shear strength of the materials.

          Reply
          1. Jason Lancaster Avatar
            Jason Lancaster

            True! But friction coefficient itself is a shortcut…when friction coefficients are measured, they’re based on a set of assumptions that include contact patch size and shape. There’s a whole rabbit hole we can dive into – and perhaps you’ve already done so – but there are a bunch of different measurement processes for mu, and all of them make certain assumptions.

            EDIT: I re-read my comment and realized the mistake you were pointing out. My bad. Thank you – I fixed it.

          2. smalleyxb122 Avatar
            smalleyxb122

            Granted. This is not the place to get into the weeds. I’m also not a tire engineer. This is just basics, and the basics are that if you can control for a constant mu, all of the other stuff drops out of the equation. I’m sure we could have an in-depth conversation on the particulars of tire engineering that would be interesting to… nobody else.

            Unless one is actually engineering tires, or developing non-newtonian traction fluids, a constant mu is usually good enough.

          3. Jason Lancaster Avatar
            Jason Lancaster

            Ya I did a little reading on this and some tire companies use lasers to map various surfaces and use some math to “model” the coefficient at various speeds/conditions, some use a glorified drill press with an attachment for spinning little pucks of rubber, there were old-school pendulum tests that tried to approximate friction with precise adjustments to the length of the wire, so that the rubber tip would just barely touch, etc.

            All seems sort of sub-optimal to me. With all the onboard data tools available, it seems like it would be pretty easy to put a set of tires on a “standard” vehicle, run some laps on a test track, and then swap out tires and re-do the last lap. But what do I know.. 🙂

      2. smalleyxb122 Avatar
        smalleyxb122

        F=m*a, therefore, a=F/m
        F=mu*N (Accelerative force, coefficient of friction, Normal force)
        N=m*g (Normal force, mass, acceleration of gravity)
        a=mu*(m*g)/m (coefficient of friction, mass, acceleration of gravity, mass)
        a=mu*g*m/m
        a=mu*gm/m

        a=mu*g

        I tried to do it in steps here to show how mass cancels out. Let me know if you can’t follow it, and I can try to clarify.

        Reply
        1. danleym Avatar
          danleym

          Took me a minute, but I got through it.

          Thanks for the physics lesson!

          Reply
        2. outback_ute Avatar
          outback_ute

          The assumption that mu is constant has to be the problem area here, because the fact is that the real world does not bear this out. Do you propose that a 4000lb car can corner at the same speed as a 2000lb car on say a 185 width tyre?

          Reply
          1. smalleyxb122 Avatar
            smalleyxb122

            For the purposes of this conversation, yes. If that 185 section tire has a sufficiently low mu to slip, rather than shear, the 2000lb car and the 4000lb car will corner at the same speed before slipping.

            The reason this doesn’t hold up in most real world applications is that the effective mu is limited by the shear strength of the tire’s rubber, leaving rubber on the road. To balance the shear forces, the heavier car will need a wider tire. Continuing my oversimplification of tire design, you’d want a 265 section on the 4000lb car to match the 185 on the 2000lb car.

          2. outback_ute Avatar
            outback_ute

            Sufficiently low mu, eg a wet road. Then you have other factors such as hydroplaning complicating things.

    3. Jason Lancaster Avatar
      Jason Lancaster

      It’s true that weight doesn’t matter when calculating lateral grip.

      However, the basic calculations assume a round, flat contact patch between the tire and driving surface. Weight can impact the shape and size of that contact patch.

      Weight is very important, but in this particular situation not so much so as to change lap time in a measurable way. But increase speeds, the number of laps, or other factors, and I guarantee the difference will become more obvious.

      Reply
    4. nanoop Avatar
      nanoop

      Great thread so far!

      So for “everything linear, no screaming tires” weight isn’t a factor (literally) for lateral grip. Now I wonder, as long as we’re in the same regimen: weight shouldn’t be in the equation for longitudinal grip, too, right? Still talking grip, not 0-60 times.

      Reply
  4. neight428 Avatar
    neight428

    Also curious how much braking these karts on these tracks need. My last experience on a rednecky/carny kart circuit involved planting the throttle at the green light and never letting it off the stop. The brake was only for not ramming the person that beat you when you pulled into the staging lane at the end, unless they were your brother, in which case, you would want to ram them.

    Reply
  5. Vairship Avatar
    Vairship

    “Weight is useful only in a steam roller” – Uffa Fox.

    Reply
    1. outback_ute Avatar
      outback_ute

      Don’t forget “road-hugging weight”, which must have given the road a nice warm feeling?!?

      Reply