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Calculating Maximum Velocity Electric Longboard

  1. Oct 8, 2016 #1
    Hello Physics Forums,

    I've been looking around on these forums for help with trying to calculate the max velocity of an electric longboard my group of friends are making. After reading a few posts, you guys have helped me understand these calculation a lot better.

    Right now we have a gear we are thinking about using, but we are considering getting a new gear depending on a maximum speed calculation.

    This longboard is powered by one electric motor with a pulley belt going from the motor shaft pulley to a pulley on a wheel of the longboard.

    Like this:

    Fastest-Electric-Longboard.jpg

    I am aware that the maximum speed of the longboard can be can be calculated from rpm of the wheel times the circumference:

    Velocity = rpm shaft * gear ratio * 2 * PI * radius wheel

    (gear ratio = shaft pulley / wheel pulley)
    (rpm shaft = Battery Volts * Kv Motor)

    But it we wanted to use a different calculation because this equation didn't take into consideration other forces acting on the longboard.

    Therefore we used the equations below that uses rolling friction, drag, and the force at the wheel of the longboard from the power and torque supplied by the motor.

    Free body diagrams:

    https://docs.google.com/presentation/d/1AQB0hobkEDjml_ogeAiS_b97AHGMdfuXgTlEd_Qzw64/edit?usp=sharing

    Google sheets calculations using the equations below:

    https://docs.google.com/spreadsheets/d/1_gZbtuwIs4DpJcVSWQHrwEuCpFddRLJ7mRFlSEbtPy0/edit?usp=sharing

    Power shaft = Torque shaft * Angular Velocity
    Power shaft = Voltage * Current
    Angular Velocity shaft = rpm shaft * (1/60) * (2*PI / revolution)
    Torque shaft = Power shaft / Angular Velocity shaft
    Force Pulley Belt = Torque shaft / radius shaft pulley
    Torque Wheel = Force Pulley Belt * Radius Wheel Pulley
    Force Wheel = Torque Wheel / radius wheel

    The only issue is that as the wheel pulley gets larger, the Force Wheel and maximum velocity of the longboard increases. Furthermore, if the wheel pulley gets smaller, the Force Wheel and maximum velocity of the longboard decreases. This doesn't make sense because a smaller pulley would be spinning much faster than a bigger pulley and should go faster.

    So we were wondering if there is something we are missing with our calculations that would account for the rpm of the wheel. Or does anyone have another suggestion for calculating max velocity?

    I saw people using:

    Power = Force*Velocity

    And "jack action" has the HP Wizard website that has the max velocity equation displayed here:

    http://hpwizard.com/accelerating-power-limit.html

    But I think I might run into the same problem as I did before because it doesn't account for rpm. Do you guys have any ideas?

    Thanks for your time.
     
  2. jcsd
  3. Oct 8, 2016 #2
    I was looking at a website recently which explained the energy required for a cyclist to go up hill:

    http://theclimbingcyclist.com/gradients-and-cycling-how-much-harder-are-steeper-climbs/

    While not exactly what you are looking for, I think it explained very clearly the different forces involved and gave some easy to use formulas for calculating the energy used in different circumstances. If you look at the maximum power rating of the motor and couple this with your gearing equations you should be able to come up with a sensible answer.
     
  4. Oct 8, 2016 #3

    billy_joule

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    The problem looks to be that you don't know the RPM the motor produces max shaft power. For whatever reason, most BLDC motor manufacturers don't provide characteristic motor curves like PMDC manufacturers do.

    The kV *V value tells you the no load RPM of the motor (zero shaft power out so zero percent motor efficiency) for a given applied voltage.
    That doesn't help much as you want to know the RPM when the motor is loaded, and producing max shaft power. A bit of googling may help you here.

    Once you know the rpm the motor produces it's max power at and the skateboards max velocity due to that max power, then the gear ratio is defined.
    The skateboards velocity and wheel radius defines the wheel RPM, and then
    ##GR = \frac {RPM_{Wheel~@~Vmax}} {RPM_{Motor~@~Pmax}} ##

    The problem is, this gear ratio will not give the best range. If you're also interested in range, you can do a similar process where you optimise the ratio for ##RPM_{Motor~@~max~efficiency}## Maybe a GR somewhere between these two extremes is the best compromise.
     
  5. Oct 8, 2016 #4

    jack action

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    It should read:

    as the wheel pulley gets larger, the Force Wheel of the longboard increases and maximum velocity decreases. Furthermore, if the wheel pulley gets smaller, the Force Wheel of the longboard decreases and maximum velocity increases.

    Torque and rpm are always going in opposite directions with gear change: If one goes up, the other goes down. This is because power is constant and equal to Torque X RPM. So:

    Power Shaft = Power Wheel
    Torque Shaft X RPM Shaft = Torque Wheel X RPM Wheel
    Torque Shaft / Torque Wheel = RPM Wheel / RPM Shaft​

    Your equations are correct, you just read them wrong.
     
  6. Oct 8, 2016 #5
    I appreciate the reply, I took a look at this website and it has me thinking about the run time for our longboard. Furthermore, we have a hill at school that we can use the gradient equation on to determine the speed. Thanks!


    I hadn't considered that the Volts * Kv would calculate the rpm of the motor under no load. I will look for the motor curves tonight on the internet and hopefully I get lucky. If I find the motor curve, what you explained makes it easy to find max speed. Thanks for the information and response.

    Thank you for the response!

    So here is my issue with my calculations:

    Using 40 teeth wheel pulley, 15 teeth motor pulley,

    Fmotor = 195.9N (From calculations on Google Sheets. Fdrag and Frollingfriction also calculated).

    m*a = m* 0 = 0 = Fmotor - Fdrag -Frollingfriction

    0 = 195.9 - 0.258V^2 -8.829V

    As I increase the teeth on the wheel pulley, the Fmotor goes up. So when the Fmotor goes up, my max speed V goes up. This doesn't make sense for the reasons you listed above. So I am thinking I need to approach my calculations a different way.
     
  7. Oct 9, 2016 #6

    jack action

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    All of this is true. But something else is also true (I replaced «motor» with «board» for clarity):

    m*a*V = m* 0*V = 0 = Fboard*V - Fdrag*V -Frollingfriction*V = Pboard - Pdrag -Prollingfriction

    0 = Pboard - 0.258V^3 -8.829V^2

    Whenever Fboard increases, V must come down if Pboard is the same (Power = Force X Velocity). Given a certain amount of power, you can transform it into force (or torque) and velocity (or RPM) but never both at the same time, it is either one or the other. The only thing you know for sure is that power will always stay the same, no matter how you transform it:

    Pmotor = Pwheel = Pboard
    Tmotor X RPMmotor = Twheel X RPMwheel = Fboard X Vboard

    Example: If you have 10 W of power at your motor, what is the available force at the board, given the board' speed:

    Code (Text):
    V (m/s)    F (N)
        0.01  1000
        0.1    100
        1       10
        2        5
        3        3.333
        4        2.5
        5        2
        6        1.667
        7        1.429
        8        1.25
        9        1.111
       10        1
      100        0.1
     1000        0.01
    All of these are true because when you multiply one with the other, it always gives 10 W.
     
  8. Oct 15, 2016 #7
    Sorry for the late reply, I have been a bit busy.

    Thank you for explaining the relationship between velocity and force in terms of power. I hadn't considered it in that fashion before and it makes sense.

    So the Pboard that I calculated using various equations was 2072 watts. Therefore the equation you listed above becomes:

    0 = 2072 - 0.258V^3 -8.829V^2 (MAX VELOCITY)

    So this would give the max velocity for the longboard right?

    But one of my concerns is that this equation doesn't consider the gear ratios, only the power. If the equation above is correct, then if we put a 5 tooth pulley at the wheel, we would still have a the same max velocity as a 60 tooth pulley.

    I think I understand what you were talking about when you said that in terms of power, force (torque) and velocity (rpm) are inversely propositional. But I still don't know how to incorporate the gear ratios.

    Thanks for the great information!
     
  9. Oct 15, 2016 #8

    billy_joule

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    You need to gear the motor so it's producing it's max power at top speed.
     
  10. Oct 15, 2016 #9

    jack action

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    The equation gives you the maximum velocity for the given power. It doesn't matter if your power comes from an electric motor, a gas engine or a nuclear reactor.

    You can choose any gear ratio you want, you'll never be able to achieve a higher velocity unless you increase the power.

    That being said, it doesn't mean that you can achieve that speed with any gear ratio.

    Once you have your desired speed (which must be equal or less than the maximum velocity), you can find the wheel RPM (##\frac{30}{\pi}\frac{V}{r}##) and compare with the motor RPM (where you have the available power), such that ##GR = \frac{RPM_{motor}}{RPM_{wheel}}##. Re-read post #3.
     
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