Dot Physics

I just realized that something has been bothering me. It is this KillaCycle – the electric zoom-fast motorcycle that I posted about previously. It clearly is super fast. However there are two claims that seem iffy.

  • It can be recharged in 4 minutes.
  • It is recharged by wind power.

It may be possible that these individually could be true – but could it be recharged in 4 minutes by the wind? I am not sure. Let me do some estimations to see if this is possible. What am I starting with:

  • From the review on gas2.org, uses cordless drill batteries.
  • Uses 1200 batteries
  • Produces 500 bhp – not sure what a bhp is
  • Bike can reach 168 mph.
  • The bike did the quarter mile in under 8 seconds (I assume just under or they would have said 7 sec)
  • Uses A123 System batteries? NanoPosphate batteries? Again, not sure what those are.
  • Recharge in 4 minutes.

Ok. bhp is brake horsepower (according to wikipedia). This is the horsepower without loses to transmission and stuff (which I assume does not apply to an electric bike? Why use bhp – not sure)

What am I going to calculate? First, how fast was this thing going? How much energy would that take? How long would it take to put that amount of energy into the batteries. If I get reasonable answers, then I guess it could happen.

How fast? How much kinetic energy?

The original article said it could go 168 mph. But does this agree with the 1/4 mile time? First, I will assume a constant acceleration (that way, I can use the kinematic equations). I know: (this is a 1-d problem, so no vector notation)

i-16656da332b0b5c881c472863e846718-knows-v.jpg

Here are two useful kinematic equations – for more on kinematic equations, see this

i-98f839560dc4565d52295f3fddbacd57-2-kinematics.jpg

Solving the second one for the acceleration and plugging into the first equation, I get:

i-9c45e44da8394fadae9cb6f027bc34ea-final-v.jpg

Notice that I put 0 m/s in for the initial velocity. This gives a final velocity of:

i-7e685beb57e094e4cd52bd98ece81ce2-final-speed.jpg

Ok, slower than 168 mph. Maybe that was the max speed, not the speed in the quarter mile. I will go with the lower speed for now. In order to calculate the kinetic energy, I need just one thing. The mass. Here is a picture of the bike.

i-31d65cb7940967832ff9ed4ff18a2d0f-pic-bike.jpg

It is not clear if that is all batteries or just stuff to improve aerodynamics. Here is the problem. The article says 1,200 cordless drill batteries, but it also says it uses these A123 batteries. I have a DeWalt cordless battery, but it doesn’t say what kind of battery it is. It is not very massive, maybe 0.5 – 1.0 kg. I will go with 0.5 kg. This would mean that 1,200 of them would have a mass of 600 kg. I will completely off the wall guess that the rest of the bike is 150-300 kg, and go with 200 kg. I am sure someone will set me straight here, but this gives a total mass of 800 kg or 870 kg if I include the rider (which I should). This site lists the mass of a normal drag motorcycle as 265 kg, so maybe I over estimated. I will go with 600 kg for the total mass plus rider (this might be an underestimate). This would give the kinetic energy at the end as:

i-5e4586cd220e215f0118db603e015eb4-ke-of-bike.jpg

Yes, that is a lot of energy. What is the energy density of a lithium ion battery (which is what I think the A123 is)? Wikipedia to the rescue again (I should totally buy wikipedia a christmas present or something). This lists Lithium ion nanowire as 2.54 MJ/kg. Wow. That means that a 1 kg battery has enough energy to get this thing up to speed? Ok, that really doesn’t seem right. Here is what I will do. I will just use variables and put the numbers in a zoho spreadsheet with calculations. That way if you don’t like any of the numbers, you can just change them yourself.

Oh, let me do a check on the power. If it is 500 hp, that would be 3.72 x 105 watts. So, if it is running at 500 hp for 8 seconds, that would be:

i-0d0eb096327c784420f11ea6d2e872e7-energy-from-power.jpg

This is comparable to my kinetic energy calculation. Ok. I will proceed anyway (spreadsheet at the end). Even if more energy is stored in the battery, this is how much must be recharged after one race. Actually, the amount of energy in the battery would be a little higher because it is not 100% efficient. Probably something like 70% efficient. Since my numbers are not certain anyway, I will just leave this part off.

Recharging the batteries

I know the energy going into the batteries and I know the time. From this I can calculate the power of the charger. I will assume that the charge is also only 70% efficient. This would make the energy input into the charger:

i-3a6ea9c373fb6156aeaff3c0b6fd6eac-charger-input.jpg

This is the energy needed in 4 minutes. I can calculate the power.

i-611ca4edbe9f37c2b1e6603683f4f077-power-charge.jpg

Even if this is half as much, it is still a large power requirement. My little portable generator (gas) is 2 kilowatts. Could this be provided by a wind generator? I could easily look up some values for power output of a generator – but isn’t it more fun to calculate it? Basically, the wind generator gets energy by slowing down air. Here is a diagram.

i-61d43e2bd44ad1f138330dcaacd5da38-generator-wind.jpg

In this diagram, the air comes in with a speed of v1 and leaves at a speed v2. I will assume the fan blades has a total diameter of d. So, in a time delta t what is the change in energy from the air? Well, this depends on the air. Imagine a cylinder of diameter d with a length of v1t long. This is the amount of air that will hit the generator in this time. (really it should be delta t, but I am too lazy). The density of air is about 1.2 kg/m3. From this I can get the mass of air:

i-b9037c08cbb96cb73a9a4f99b311a662-mass-air-1.jpg

Maybe you already see a problem, I do. If the velocity of the air changes, the density on the back side of the generator would be different. I guess this all gets fixed by some fluid dynamics stuff or something. Anyway, I can now get the change in kinetic energy of the air.

i-5ce2d30605b0fa055d0ffae84532a582-delta-k-air.jpg

This is a negative value because the air slows down. I will assume that most of this is turned into electrical energy. Maybe the efficiency is like 70%, I will call this efficiency e. This means that the energy produced by the generator is:

i-e884f210bc19859cc06a6427c16fa246-e-gen-delta-t.jpg

And the power produced:

i-ea2475211351c796f40180c802bdc014-power-gen.jpg

I can guess some of these values, but I will put this calculation in a spread so that you can change them if you like.

Trying reasonable values, I don’t get anywhere near 10 kilowatts. Maybe the wind would be much higher, but I don’t think so. Perhaps they use the wind generator to store energy in a capacitor and then charge it in 4 minutes. Actually, I suspect the two claims are not meant to be at the same time (charging via wind AND charging in 4 minutes).

Comments

  1. #1 Uncle Al
    May 6, 2009

    If I get reasonable answers, then I guess it could happen.

    If you get incredibly unreasonable answers, then so much greater is the miracle. We are abandoning the Age of Science and entering the Age of Splendor. Mystics are baffled by the obvious yet possess a complete understanding of the nonexistent.

  2. #2 peter w
    May 16, 2009

    You seem to have lost a v in the expression for power in a vind turbine.
    It should be proportional to v^3, (the mass of air passing the turbine/s is proportional to v (as you write) and the kinetic energy is proportional to mass and v^2). Why do you think they use 600 kg of batteries if 1 kg is enough?

    charging 3 Mj in 4 min would require 12 kw hm.. assume charging at 4V means the current will be 3kA, something is fishy. storing the energy in capacitors is even less efficient than batteries. if you have 100 farad carged to 4 volt you can store 400 columb ie 1 amp i 400 sek would discharge it. I dont see how they make it.

  3. #3 voicecoils
    May 23, 2009

    hehe, you’ve gone about this in quite the funny way. What about keeping things simple.

    How much energy does the Killacycle store: 9.1 kWh
    (http://www.killacycle.com/about/)

    How much power does a large wind turbine output: 6 MW (6,000 kW)
    (http://www.thewindpower.net/wind-turbine-datasheet-223-enercon-e126-6000.php)

    How long would it take for such a wind turbine to provide 9.1 kWh: ~5.5 seconds

    So, from an energy perspective, 4 minutes is an easily achievable target. Also you must consider that the wind turbines are not on the bike or likely even near by. I think the actual situation is that the Killacycle team charges a trailer full of batteries at home from renewable energy they purchase from the grid. They are also working on a biodiesel trailer option for charging IIRC.

    Finally, on their ‘About’ page, they state that each 1/4 mile run consumes 0.6 kWh. You can assume an electrical power to mechanical power efficiency loss and work backwards from that, to check/redo your own calculations.

    Perhaps it’s time to remove the ‘fake’ from your post tag?

  4. #4 Rhett
    May 23, 2009

    @voicecoils,

    Yes, clearly the bike is not fake. That tag really should say “fake?”. Thanks for the info, I really didn’t look for many specifications for this bike because I was just using it as an example to talk about some physics.

  5. #5 voicecoils
    May 23, 2009

    @Rhett

    No worries. Another fast and interesting vehicle that might make for some fun physics are hot water rocket powered ones.

    Such as here:
    http://www.gizmag.com/214-mph-in-25-sec-water-powered-record-stands-the-test-of-time/11622/

    0 to 344 kph in 2.5 seconds! Perhaps not so much for high school physics but interesting for those studying thermodynamics later on…

  6. #6 voicecoils
    May 23, 2009

    p.s. those embedded spreadsheets are very cool!

  7. #7 Motorbike Helmets
    June 10, 2009

    I am not very good at physics but you explain it very well. Are you some sort of professor at a university? or just an average biker. Good info and good thermodynamics

  8. #8 Rhett
    June 10, 2009

    @Motorbike Helmets,

    Thanks for the compliment. Yes, I am a physics professor and I enjoy explaining stuff.

  9. #9 Rhett
    June 10, 2009

    @voicecoils

    The embedded spreadsheets are from ZoHo. Good stuff. You can do it with google docs also.

  10. #10 HJ
    June 20, 2009

    Being the un-mathematically inclined rider you could have baffled me enough to prove it was or wasn’t possible. I suspect you’re right and they were meant to be individually true. Either way, I look forward to hearing more.

  11. #11 Detox John
    October 2, 2009

    Wind power is good although it looks a bit bulkier compared to solar cells. i am trying to build a small wind generator at home too.

  12. #12 Bacnet
    November 4, 2009

    Wind Power is one of the best alternative energy sources that we should utilize, it is very clean and non-polluting. I built a small wind generator at home which can power small appliances.

  13. #13 Anthony
    February 5, 2011

    I don’t know what you are trying to prove with your theory? Killa Cycle has already proven its data by using, first, theory, then through “real world” empirical testing. There has been hundreds of man hours poured into this experiment. This is not a fact finding mission they were on, just simply a what would happen if we did this? They on there data by using a dynamometer. The best way to know HP to weight ratio is by knowing two basic things. One, is the total mass of the object under acceleration and second, it’s top speed at the end of the(1/4 mile)440 meter run.

  14. #14 Anthony
    February 5, 2011

    I don’t know what you are trying to prove with your theory? Killa Cycle has already proven its data by using, first, theory, then through “real world” empirical testing. There has been hundreds of man hours poured into this experiment. This is not a fact finding mission they were on, just simply a what would happen if we did this? They got there hp data by using a dynamometer. The best way to know HP to weight ratio is by knowing two basic things. One, is the total mass of the object under acceleration and second, it’s top speed at the end of the(1/4 mile)440 meter run. Mass/time and top speed.