“Portuguese drivers in the greater Lisbon area last monday were surprised by a strange scenario: a crazy dude with a camera and tripod in the backseat of the car, going back and forth on the same road, over and over again, accelerating to 80 kph and then letting the car stop by itself. He spent a lot of time doing that, making the other occasional road users slightly unhappy (although he had spent all morning and over 100km to find a suitable road).
Fig.1 – Camera with tripod locked in back seat
When asked about it, he said he was videotaping the speedometer of the car during the experiment. He also had a number of strange wires and electrical devices on the floor of the car, which he explained were necessary to keep the camera rolling without danger of running out of battery power.”
Fig. 2 – power inverter to feed camera.
This could have been the headlines, if the police had caught me. But they didn’t. And I’ve got my data. 🙂
Some time ago, I calculated the absolute maximum requirements for the motors, taking into account the car’s present maximum performance. That calculation is important because we have to be sure of what the motors can do.
But that’s quite different from what the motors usually will do; now I need a more realistic evaluation of what the vehicle must endure in normal everyday situations, in order to design a sensible system.
The first question that came to mind was: how much power does the car consume while rolling at constant speed on the highway? (Among the most common scenarios, the highway will be the worst place of performance for a serial hybrid, because the battery pack is being drained continuously at a relatively high rate, usually with very little chance of getting some charge back from regenerative braking.)
For this I needed to know the Cd (aerodynamic drag coefficient) and Crr (rolling resistance coefficient) of the car. Since I couldn’t get those values from any document, I teamed up with Njay and we executed the experimental method described in “IWillTry.org“‘s very neat instructable. (You gotta love the Internet!…). Our first quest for a long, flat, deserted road did not go well. We tried 2 different road stretches, at night, but the information was too bad (short and bumpy road) and no coefficients could be calculated from it. A few months went by, and I decided to hunt for the flat road again. This time I didn’t bother NJay; instead I took a video camera to film the speedometer. It’s got a digital stopwatch inside it, so all you need is a remote control and a tripod, and you can do it all alone (no more bothering a friend at unspeakable hours to go drive around like looneys). It also helps if you have a power inverter (fig. 2). That way you can keep the camera running without worrying about battery time.
So, after 140 km of road-chasing, this is what i got:
Fig. 3 – Graph of speed vs. time when in neutral gear, starting from 80 km/h (22,2 m/s).
Here are the calculated “1998 Mercedes-Benz A140” physical coefficients that fit that curve:
- Cd = 0,141 (EDIT: this is obviously wrong. should be around 0,30~0,40. I’ll have to repeat the measurements under better conditions…)
- Crr = 0,020
The spreadsheet is over here (Open Document format), if you care about the calculations. Change the extension to “.ods” if it doesn’t open correctly.
Now I can calculate the power requirements for the motors/controllers/batteries, depending on speed, inclination, cargo weight, etc.
Supposing a flat highway speed of 120 km/h (the maximum legally accepted), with 2 occupants (120kg) and corresponding luggage or a small child (30kg) in a 1080kg car (with a full tank of gas), the motors would have to sustain a constant forward force of:
- Velocity = 120 km/h = 33,3 m/s
- F(air) = – Cd * Frontal_Area * 0,5 * Air_Density * V^2
- = -0,14 * 2,41 * 0,5 * 1,22 * 1108,89 = – 228,22 N
- F(road) = – Crr * Mass * Gravity
- = -0,02 * (1080+120+30) * 9,81 = -241,32 N
- F(air+road)@120 = – 469,546 N
The minus sign just means this force is against the movement of the car. The motors have to counter that force to keep the car moving at the same speed.
Then the necessary power would be:
- P(120km/h) = F * V
- = 469,55 * 33,3 = 15,63 kW (~21 HP)
And the equivalent necessary torque:
- T(120km/h) = F * Wheel_Radius
- = 469,55 * 0,273 = 128,19 Nm
Which have to come from the 2 motorized wheels, so each motor must be capable of continuously working 7,81 kW and pushing 64,1 Nm, without heating up. These are now my target motor ratings for continuous usage, not withstanding the fact that they will have to exceed these values for short periods of time to overcome transient situations or obstacles (i.e. accelerating and braking).
With this in hand, we can now estimate an initial size for the battery pack. How long should the pack feed the motors without recharging? Remember this is the worst-case scenario for an electric vehicle, and we don’t want to carry around an extra 500kg of batteries, so let’s keep the demands down… let’s say 5 minutes. I’m assuming the generator can recharge the battery pack in under that time (either by start-stop, or by constant recharging).
Energy to hold in the battery pack:
- t = 5 min = 0,085 h
- E = P * t = 15.630 * 0,085 = 1.328,55 Wh ( 110,7 Ah for 12V batteries)
Assuming the textbook limits of the Lead-Acid (“Pb-H”) technology:
- “Pb-H” specific energy = 35 Wh/Kg
- Required “Pb-H” weight = 1.328 / 35 = 37,94 Kg
For a battery pack of Nickel-Metal-Hydride (“Ni-MH”) technology:
- “NiMH” specific energy = 80 Wh/Kg
- Required “Ni-MH” weight = 1.328 / 80 = 16,6 Kg
For a Lithium-Ion (“Li-I”) battery:
- “Li-I” specific energy = 160 Wh/Kg
- Required “Li-I” weight = 1.328 / 160 = 8,3 Kg
Quite a difference between 40, 20, and 8 kg, isn’t it? 🙂 Battery technology is probably worth the investment.
If you spot any errors in my math or logic, please let me know in the comments!! Thank you!! 🙂