Dot Physics

This was on my ‘to do’ list, but Tom at Swans on Tea beat me to it. Basically, this grocery store has these plates that when depressed produce electrical energy. Tom does a good job pointing out that this is not free energy (the original article says this also). Clearly, the energy comes from the cars. How much would this cost the cars?

As always, let me start with some assumptions.

  • The original article says that the bumps will generate 30 kW of energy every hour. That is an odd thing to say. I am going to interpret that as 30 kW of power for all hours (every hour). They couldn’t have meant 30kW/hour. That wouldn’t be power OR energy. 30 kW of power for 1 hour would be 30 kW*hr of energy. Ok, so power is 30 kW.
  • I am going to assume that 1 car passes over this every 5 minutes. That is a complete guess. I would imagine the average rate that a car goes over it be anywhere from 1 a minute to 10 a minute.
  • The efficiency of the device. This is a wild guess, but I am going to say 20% to 50%.
  • The energy density of gasoline is 1.21 x 108 Joules/gallon.
  • The price of gasoline is $2.50 per gallon.
  • A car engine is 15%-25% efficient at converting energy into mechanical energy

Let me start off with the assumption that it is just one car doing all this driving over the device. If the car gets 25 mpg, how much extra gas would be expended from this device being in the way? First, I can calculate the estimated amount of energy per use it would get from the car.


If that is the energy loss by the car, how much gasoline would that take?


That is the extra gas needed for one car going over once. Let me put this all together in my normal spreadsheet reader changeable form.

Something doesn’t look quite right. This says it would take about 9 gallons of gas an hour. Ok, that may be ok. If I wanted to run a 30 kWatt generator, it would take some stuff. But what if people came in their cars and just hit the device while coasting? Then the electrical energy would come from the decrease in kinetic energy of the car. How fast would a car be going so that it is stopped after hitting that device? I added this in my calculation sheet. From that, I get 120 m/s? No way. That is with 30 cars hitting it an hour. Let me look at each piece and see if I can find a mistake.

In 1 hour, the energy from the device would be: (assuming a power of 30 kW):


Of course, the device is not 100% efficient. If it is 50% efficient, it would need twice that amount of energy coming in to get that much out. That means that during that hour, there must be a loss of 2.16 x 108 Joules of KE from the cars. If there were 30 cars during that hour, that would be 7.2 x 106 Joules per car. If this is all KE then:


I don’t see where I made my mistake. I estimated 1000 kg for the mass of the car, that seems ok.

Maybe this thing doesn’t really generate 30 kWatts of power?


  1. #1 Uncle Al
    June 23, 2009

    Dr. Schund, how can we supply America with all its electricity, too cheap to meter, using technology no more advanced than a tethered jackass?

  2. #3 Glen Thomas
    June 24, 2009

    The original article links to here, which suggests that

    “If it [the energy] wasn’t harnessed by the speed bumps, it would go to waste.”

    The ramps – which cost between £20,000 and £55,000, depending on size – consist of a series of panels set in a pad virtually flush to the road. As the traffic passes over it, the panels go up and down, setting a cog in motion under the road. This then turns a motor, which produces mechanical energy. A steady stream of traffic passing over the bump can generate 10-36kW of power.

    The bumps can each produce between £1 and £3.60 of energy an hour for up to 16 hours a day, or between £5,840 and £21,024 a year.

    What is ‘a steady stream of traffic? A dozen cars a minute?

    Even so, if the total energy, E = no. of cars per hour(N) * change in kinetic energy, then v^2-u^2 = 2E/Nm = acceleration * length of plate.

    If the plate is 1 metre across, two wheels travel 2 m (=x) over it, then a = 2E/Nmx = 300 metres per second squared, or 30g deceleration. If v^2=2ax, then any car traveling less than about 80 mph will be brought to a halt.

    The article says “The ramp is silent, comfortable and safe for vehicles … drivers don’t realise they are passing over it.”


  3. #4 Tom
    June 24, 2009

    Yes, a gallon of gas contains about 33 kWh of energy, so it will take several gallons of gas per hour to generate 30 kW, depending on how many layers of inefficiency we add. I suspect the unit foulup has to mean 30 kWh of energy over some longer span of time, perhaps all day. That’s about what an average US household uses per day.

  4. #5 Rob
    June 25, 2009

    The “mainstream” reporting on power will make a unit obsessive individual crazy. I’ve heard power plants described as producing “800 megawatts per year” on numerous occasions.

  5. #6 Rhett
    June 25, 2009


    Maybe they were really talking about the rate that the power changes? (dP/dt). Then it could be 30 kW/hr.

  6. #7 Rob
    June 28, 2009

    Yes, could be power acceleration, but over any significant length of time, an acceleration of 800 megawatts/year would lead to some prodigious delivery numbers

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