larryh

For truly autonomous driving, the car needs to be aware of its surroundings. Is the car going to recognize a deer at the side of the road and understand that it needs to slow down. Will it recognize cattle that are inside a fence versus outside a fence. What if something unexpected happens, i.e. a bridge is out or their is a large sink hole in the road? What if a tire experiences a blow out? Will the car know what to do on a busy freeway? Is it going tor recognize children playing in a residential neighborhood? It needs to understand the environment and identify potentially dangerous conditions. What about the system itself? Computer software crashes. Compare hardware fails. Are there redundant systems to take over in the case of failures?

I think the author is being overly optimistic about autonomous driving. It requires intelligence to drive a car. I haven't seen rapid progress in artificial intelligence in the past few decades. I think the more likely result is people won't know how to properly use the new technology. People have already been in accidents because they do not understand how TACC works.

I did this same experiment a week earlier with an additional person in the car. I get the same results as before. When climbing the hill, the friction appears to be half of what I would expect. So either the transmission is more efficient climbing a hill or their is less friction for some unknown reason. Another explanation is the car is reporting the power output by the motor incorrectly when going uphill (as well as when going downhill). In that case, the efficiency of the motor is more than 88% when climbing a hill. It would have to be around 93%.

Most of the energy increase resulting from faster acceleration is due to increased friction with higher speeds. With faster acceleration you will be going faster over the same section of the road vs slower acceleration. A small amount is due to the motor being less efficient when supplying more power. Efficiency seems to range from about 84% at max power to 95% during regen (when traveling greater than 15 mph). When going less than 15 mph, efficiency plummets. Don't accelerate rapidly from a stop. When going up a hill, the only difference is that you now also have to provide the potential required to climb the hill. The car is going to be about 85% efficient in supplying the required potential energy. If you accelerate faster, it is going to be slight less efficient. However, as is the case with level roads, most the energy losses will be due to friction which is a function of the car's speed. If you accelerate faster, you will have more losses from friction. However, there is an anomaly, as pointed out in post 39, where friction is less than expected when climbing a hill. I don't fully understand what is happening when ascending hills.

Where Does All The Energy Go Ascending/Descending a Hill For the hill in the previous two posts, the change in potential energy for the trip was 0.325 kWh. To ascend the hill, 0.428 kWh of electrical energy from the HVB was required. The energy above the change in potential energy represents energy losses of 0.428 – 0.325 = 0.103 kWh (24% loss). The motor is 88% efficient, so it output 0.88*0.428 = 0.378 kWh of mechanical energy. That, of course, amounts to a 12% loss of energy. Friction internal to the motor results in a 0.019 kWh loss (4 % loss). The transmission is 97% efficient, so we now have 0.97*(.0378-0.019) = 0.348 kWh of energy remaining (2% loss). Friction to overcome aerodynamic drag, tire rolling resistance, and transmission friction was 0.032 kWh (7% loss). About a 15% loss (= 100% - 88%*97%) resulted from the motor and transmission not being 100% efficient converting electrical to mechanical energy and transmitting it to the road to propel the car. The remaining losses were from various sources of friction (which depend only on the car’s speed and not the grade of the hill). These friction losses would be present regardless if you are climbing a hill or not. In order words, the car is about 85% efficient in supplying the potential energy required to ascend a hill. When descending the hill, the HVB received 0.239 kWh of energy. The energy below the change in potential energy represents energy losses of 0.325 – 0.239 = 0.086 kWh (26% loss). Regen is about 95% efficient, so 0.239 / 0.95 = 0.252 kWh of mechanical power was supplied to the motor (4% loss). Approximately 0.019 kWh of energy was lost due to internal friction within the motor (6% loss). Friction to overcome aerodynamic drag, tire rolling resistance, and transmission friction resulted in a 0.044 kWh (13% loss). The transmission is 97% efficient, so 0.03 * 0.325 kWh = 0.010 kWh of the potential energy doesn’t make it to the motor (3% loss). About an 8% loss (=100% - 95%*97%) resulted from the motor and transmission not being 100% efficient in transmitting mechanical energy from the road to the motor and then converting the mechanical energy to electrical energy. The remaining losses were from various sources of friction (which depend only on the car’s speed and not the grade of the hill). These friction losses would be present regardless if you are descending a hill or not. In order words, the car is about 92% efficient in regening potential energy when descending a hill.

Regen Descending a Hill Unlike the previous post analyzing energy consumption when ascending a hill, the analysis used in the previous posts fully explains regen when descending the same hill. The following chart shows regen when descending the same 8% grade section of a hill as in the previous post at 30 mph. The bottom green line shows the potential that could be captured if the car were 100% efficient. I assume the mass of the car is 1895 kg as in the previous post. The purple line and light blue line (which you can’t see because it coincides with the dark blue and red lines showing the electrical energy input to the HVB) show the predicted and actual mechanical energy applied to the motor, respectively. Something is wrong with what the car is reporting. If the power reported by the car were correct, then the efficiency of regen would be 100% (the light blue line coincides with the dark blue and red lines). I don’t think regen is quite that efficient. The predicted mechanical energy input to the motor is the potential energy times 97% (transmission efficiency) minus the energy required to overcome friction going 30 mph (see post 2) and minus the energy loss from internal friction within the motor itself (post 6). The dark blue and purple lines show the predicted and actual electrical energy applied to the HVB. The predicted electrical energy input to the HVB is 95% of the mechanical energy input to the motor assuming the motor is 95% efficient during regen.

Energy Consumption Climbing a Hill The analysis presented in the previous posts doesn’t quite work when climbing hills. Something strange is going on. The following chart shows energy consumption when climbing an 8% grade at 30 mph. The bottom light blue line shows the potential energy required to climb the hill if the car were 100% efficient. This time I assume a lower mass for the car rather than a mass of 1940 kg as in the previous posts when computing kinetic energy. The rotating wheels and gears have rotational inertia that increases the apparent mass of the car when computing kinetic energy. This is not the case for potential energy. I had previously assumed the apparent mass of the car increased by 45 kg due to rotational inertia. I am unsure of the true mass of the car. In order to match the actual results, I can choose any one of the following from among an infinite number of possibilities: Car’s mass is 1895 kg, transmission efficiency is 97%, and friction is 50% of normal friction on a level road. Car’s mass is 1895 kg, transmission efficiency is 93%, and friction is 10% of the normal friction on a level road. Car’s mass is 1770 kg, transmission efficiency is 97%, and friction is 100% of the normal friction on a level roadCase 3 is not acceptable. That would make the car’s mass with me in it is less than what is listed in Ford’s specifications. Case 2 seems strange. How could friction be reduced by so much? We are forced to conclude that friction must be reduced when going uphill for some unknown reason. The best choice seems to be Case 1. Either there is less friction when climbing a hill or the transmission is more efficient when climbing a hill. The dark blue and red lines show the predicted and actual mechanical energy output of the motor, respectively. The predicted energy output of the motor is the potential energy divided by 97% plus 50% of the energy required to overcome friction going 30 mph (see post 2). The green and purple lines show the predicted and actual electrical energy output from the HVB. The predicted electrical energy output from the HVB is the mechanical energy output of the motor plus the energy loss due to internal friction inside the motor (post 6) all divided by the motor efficiency of 88% (post 8).

Energy Consumption During a Short Trip The following chart shows energy consumption during a short trip. I begin at rest, accelerate to 50 mph, maintain 50 mph for about 20 seconds, and then slow to a stop. If there were no friction, the total energy consumed for the trip would be 0 kWh--no energy would be consumed by friction and all the energy supplied by the HVB to provide the kinetic energy to accelerate the car would be recaptured during regen. Instead, the trip consumes about 0.104 kWh of energy from the HVB. Friction from aerodynamic drag, tire rolling resistance, and the transmission accounts for 0.084 kWh (81%) of the loss. Friction internal to the motor itself accounts for 0.016 kWh (15%). The motor is about 88% efficient converting electrical power to mechanical power for about a 0.003 kWh loss (3%). The motor is about 96% efficient converting mechanical to electrical power for about a 0.001 kWh loss (1%). By far, the most energy is lost due to friction. The power required to overcome friction increases with the cube of speed. The most effective way to improve mileage is to simply drive slower.

Where Does All The Energy Go During Regen In the previous charts, 0.129 kWh of kinetic energy was consumed as the car decelerated from 50 mph to 8 mph. 0.025 kWh of the kinetic energy, or 19%, was consumed to overcome friction (aerodynamic drag, tire rolling resistance, and internal friction in the transmission). 0.005 kWh, or 4%, was used to overcome internal friction within the motor itself. 0.001 kWh, or 1%, was lost during the conversion from mechanical to electrical energy. So of the total available kinetic energy, 75% made it to the HVB. Had I decelerated a little faster, there would have been less energy lost due to friction (the car stops sooner and friction has less time to consume power) and I could maybe have converted 80% of the kinetic energy to electricity.

Regen (Electrical Energy) The following chart shows the electrical energy supplied to the HVB during the same deceleration as the previous post from 50 mph to 8 mph. The red line is the predicted electrical energy supplied to the HVB assuming that 96% of the mechanical energy supplied to the motor (the red and blue lines in the chart of the previous post) is converted to electricity. Ford claims the motor is 95% efficient during regen. The blue line is the actual energy supplied to the HVB. Only about 0.001 kWh of energy is lost during the conversion. The green line is the kinetic energy—the same as in the previous chart.

Regen (Mechanical Energy) The same analysis that was used during acceleration can be used during regen. The following chart shows the energy supplied to the motor during regen decelerating from 50 mph to 8 mph. The red line is the actual energy supplied to the motor and the blue line is the predicted energy supplied to the motor based on the previous posts. The green line is the kinetic energy of the car that would be supplied to the motor if regen were 100% efficient (in which case the red and blue lines would coincide with the green line). However, not all the kinetic energy from the car makes it to the motor. Some of it is consumed to overcome friction (post 2), 0.025 kWh, and some if it is consumed to overcome the internal friction in the motor itself (post 6), 0.005 kWh. Of the remaining kinetic energy, 97% percent of that makes it to the motor assuming the transmission is 97% efficient. Hence of the total 0.129 kWh of available kinetic energy, only 0.097 kWh makes it to the motor, i.e. 75%.

Absolutely not. When you make highly proprietary products (like Apple) and restrict who can sell it, who can service it, what modifications you are allowed to make to it, can't obtain service manual to do your own repairs, and, in general, make it a very closed system, you will get a very expensive product that costs a lot to buy, a lot to service, and a lot to repair. Consider Tesla. You are required to pay $600/year for annual inspection. That is far more than a couple of oil changes per year for a conventional car. If you need service, you are required to take the car to an authorized service center--you can't take it to just any shop. They are going to charge you a lot more than other repair places. If you hire an authorized electrician to install a charger for a Tesla, you are going to pay several times more than other electricians. If you need to replace the drive unit, that is $15,000. If you need simple body repairs, that will cost you $22,000 (more than many cars cost to begin wth). Insurance is going to cost a lot more. When you create a virtual monopoly, you can charge whatever you want.

Where Does All the Energy Go During Acceleration? Consider two trips, one where the car accelerates slowly from 0 to 40 mph and then maintains a constant speed of 40 mph. The total length of the trip is 0.5 miles. In the second trip, the car accelerates quickly from 0 to 40 mph and then maintains constant speed for the remainder of the 0.5 mile trip. The following is a detailed analysis of the energy consumed during the two trips. In both cases, the motor needs to provide the kinetic energy to accelerate the car from 0 mph to 40 mph. The transmission is about 97% efficient. That means the motor must supply 0.5*m*v^2/0.97 = 0.083 kWh of mechanical energy. 3% of that energy, or 0.002 kWh, is lost transmitting the power from the motor to the road to propel the car. The motor also needs to provide the energy to overcome friction (aerodynamic drag, tire rolling resistance, and internal transmission frictions). With fast acceleration, the average speed of the car during the trip will be faster than with slow acceleration. The faster the car goes, the more energy is required to overcome friction. Hence the motor will have to supply more energy to overcome friction for the trip with fast acceleration. The amount of energy required can be computed using the information from Post 2. For the trip with fast acceleration, 0.062 kWh of mechanical energy is required. For the trip with slow acceleration, 0.053 kWh of mechanical energy is required. Finally, the motor needs to provide the energy to overcome the internal friction within the motor itself. For fast acceleration, 0.017 kWh of energy is required. For slow acceleration, 0.017 kWh of energy is also required. Note that for slow acceleration, the trip takes longer (the trip length overwhelms less friction at lower speeds and hence the energy required for the trip with slow acceleration is slightly more than the trip with fast acceleration). The total mechanical energy required for the fast acceleration trip is 0.083 + 0.062 + 0.017 = 0.162 kWh. For the slow acceleration trip, 0.083 + 0.053 + 0.017 = 0.153 kWh is required. 0.162/0.153 = 1.06 times as much mechanical energy is required for fast acceleration, i.e. 6% more. The difference in mechanical energy between the trips, 0.162 – 0.153 = 0.009 kWh, is completely accounted for as the additional energy to overcome greater friction with higher average speed—the 0.5 mile trip with fast acceleration takes less time than the one with slow acceleration so your average speed is greater as is friction. Using the motor efficiency results from the previous post, we can calculate the amount of energy consumed from the HVB. For fast acceleration, 0.187 kWh is consumed. For slow acceleration, 0.174 kWh is consumed. 0.187 / 0.174 = 1.07 times as much energy is consumed from the HVB, or 7% more. The difference in electrical energy between the trips, 0.187 – 0.174 = 0.013 kWh, is greater than 0.009 kWh mechanical energy difference. The efficiency of the motor is generally around 88%, so I would expect at least a 0.009/0.88 = 0.010 kWh difference. The remaining 0.003 kWh difference is because the motor is simply less efficient during fast acceleration. Most of the energy difference for fast vs. slow acceleration is simply due to higher average speed, and consequently, greater friction for the fast acceleration trip. 25% of the difference is due to operating the motor less efficiently.

Motor Efficiency During Acceleration The following graph is about the best I can do to create a map of the motor’s efficiency. It shows the efficiency of the motor vs. rpm for three different rates of acceleration. The green markers are measurements from slow acceleration using 11 kW of power from the HVB (less than 2 bars on the Empower screen). The red markers are from moderate acceleration using 20 kW of power from the HVB (2 bars on the Empower screen). The blue markers are from fast acceleration using 55 kW of power from the HVB (more than 4 bars on the Empower screen.). Efficiency is poor when motor rpm is less than 2000 (15 mph). Efficiency starts out at 0% at 0 rpm and gradual builds to 80% - 85% at 2000 rpm. You don’t want to use a lot of power from the HVB when going less than 15 mph—much of the energy is simply wasted. Efficiency peaks around 5000 rpm (35 mph) at 87% to 91% and then slowly falls again with increasing rpm. Efficiency is about 4% higher for slow acceleration vs. fast acceleration up to about 5000 rpm. Moderate acceleration is more efficient than fast acceleration up to about 7000 rpm (50 mph). Note that when computing efficiency of the motor, I am not including the internal resistance of the motor. I am plotting the function e(rpm,torque) from the previous posts.

Fast acceleration is above four bars (HVB output 57 kW), enough to turn the indicator yellow. Slow acceleration is less than 2 bars (HVB output about 10 kW).

The measurements are taken from different modules. I don't know the update rate of the modules or how long it takes a measurement to be read. The other problem is that I would have to do a lot of driving pushing the motor to high power output to get a complete map. I don't usually don't consume more than about two bars on the empower screen, so I have less than half the map.

I don't know the origin of the map for the Prius. It was presented at an engineering conference. They must have had special equipment required to make the measurements. I don't know that I can make such a map very accurately. I can only collect measurements opportunistically while driving. The measurements have lots of errors, mainly because they are not synchronized. The power measurements from the HVB is taken at a different time than from the power measurements for the motor. A lot can change in a fraction of a second. I will have to understand how the car works better before I can determine the best way to accelerate. But there are three components of energy: kinetic (acceleration), potential (hills), and that required to overcome friction. I have to understand better how they interplay with motor efficiency.

Why Fast Acceleration is Inefficient The plot below is the Electric Motor Map for a Prius. The x axis indicates rpms. The y axis indicates torque. The contour lines indicate efficiency. So at 20 Nm of torque and 3000 rpms, the motor is about 93% efficient in converting electrical power to mechanical power. If you accelerate quickly, you operate the motor along the top line labeled fast acceleration. You are operating the motor for a long time in some of its most inefficient regions. You start out at less than 73% efficiency and you will inefficiently consume high power from the battery for quite some time before efficiency reaches 90%. You have consumed most of the energy required to accelerate while the motor is operating inefficiently. If you accelerate more slowly, you operate the motor along the line labeled slow acceleration. You again start out at less than 73% efficiency. But this time, you inefficiently consume less power from the battery for less time before efficiency reaches 90%. You are consuming more of the energy required to accelerate while the motor is operating more efficiently. Although this map is for a Prius, the principle is the same for an Energi.

Revised Model for Electric Motor In post 8, I provided a very simple model of an electric motor. Given the discrepancies above, I am revising that model slightly. The new equation describing the operation of the motor is: Po = e(torque,rpm) * Pi – Pl(rpm), where Po is the mechanical output power of the motor; Pi is the electrical input power to the motor; e(torque,rpm) is the efficiency of the inverter/motor converting electrical to mechanical power, which is no longer a constant, but instead a function of both torque and rpms; and Pl(rpm) is the mechanical power loss of the motor due to internal friction, which is a function of the rpms of the motor. I have already shown Pl(rpm) in post 6. Previously, I assumed e(torque,rpm) was a constant equal to 88%. That is not true. In general, the higher the torque, or the lower the rpm, the lower the efficiency e. The following graph plots e as a function of rpm. The markers are spread out because the torque varies from one measurement to the next and efficiency depends on torque. The lower markers are measurements with higher torque. The x axis indicates rpms. The y axis indicates efficiency e. Efficiency increases with increasing rpms from 0 and 2000 rpms (15 mph), and then flattens out. You want to minimize power output of the motor when traveling less than 15 mph, as this is where the motor is least efficient. In particular, you want to accelerate gently until 15 mph. After that, the motor will be operating near peak efficiency of 88%. If you use maximum acceleration, the motor will be initially operating at less than 60% efficiency. Actually, the efficiency does not flatten out after 2000 rpms at a given torque. For a specified torque, efficiency peaks at some rpm and then begins to fall again. The peak occurs at higher rpms for lower torques. Efficiency just appears to flatten out in the graph below because the torque is not constant from one measurement to the next--there are a series of peaks corresponding to each torque.

The reason for the additional energy losses during fast acceleration is that the efficiency of the motor is also a function of torque (and not just rpms). The model I proposed, Po = e * Pi – Pl(rpm), is too simplistic. Torque affects the efficiency e of the motor, so e is not really a constant. At high torque, the motor is less efficient. It appears the motor is not as efficient at low rpms and high torques, which is what you have when you accelerate quickly from rest. Fast acceleration wastes energy.

Modelling HVB Energy Consumption During Fast Acceleration The following graph is similar to the one in the previous post, except this time the car accelerates quickly from 0 mph to 50 mph in 13 seconds and then maintains constant speed. Again the red and blue lines, showing predicted and actual energy output of the motor, coincide rather closely. However, this time the purple and green lines, showing predicted and actual energy consumption from the HVB, do not coincide so well. The light blue shows how much more energy is consumed from the HVB than predicted by the model, about 0.013 kWh. Again, all of this error occurs in the first 4 seconds. Something strange happens when accelerating from rest during the first few seconds. That is when most the additional actual vs. predicted energy is consumed from the HVB. To avoid wasting energy, you would want to accelerate slowly from rest, at least initially, to avoid this loss. I am going to guess that abrupt changes in power cause excessive energy losses above that predicted by the model. Since the red and blue lines coincide closely, this does not seem to be a mechanical energy loss, but rather an electrical one. But I’m not sure that the car is actually measuring the output torque of the motor rather than estimating it from the input power supplied to the motor. In that case, it could well be a mechanical energy loss.

Modelling HVB Energy Consumption During Acceleration From the information gathered in the previous posts, we can model energy consumption from the HVB during acceleration from rest to traveling at constant speed. Consider a short trip, where the car starts at rest, accelerates slowly to 40 mph, and then continues at a constant 40 mph. The graph below shows the predicted energy consumption and actual energy consumption for such a trip. The x axis is time in seconds. The y axis is energy consumed in kWh. The car accelerates from time 0 to 40 mph in 55 seconds and then maintains constant speed. The blue line shows the actual energy output from the motor. The red line shows the predicted output energy. The predicted power output from the motor is the power needed to overcome friction (aerodynamic drag, tire rolling resistance, and internal frictions—see post 2), plus the power needed to supply the Kinetic and Potential Energy assuming the transmission is 97% efficient. The blue and red lines coincide quite closely. The green line shows the actual energy consumption from the HVB. The purple line shows the predicted energy consumption. The predicted power consumed from the HVB is given by the model of the motor described in post 8: Po = e * Pi – Pl(rpm). I assume the efficiency e is 88%. The Power Loss, Pl(rpm), is as shown in post 6. The green and purple lines also coincide rather closely. The light blue line shows the error. The actual energy consumption from the motor is about 0.002 kWh more than predicted. The entire discrepancy occurs within the first 2.5 seconds. The model fails to take something into account during the initial 2.5 seconds of acceleration from rest.

Motor Torque and Rpms During Acceleration The first graph below shows the motor torque and power vs. rpms during fast acceleration. The x axis is rpms. 7000 rpms corresponds to about 50 mph. The blue line shows torque in newton-meter. The motor torque quickly ramps up to a maximum of about 230 N-m. Then torque falls off gradually after 2000 rpms. The red line shows power. Power ramps up linearly to a maximum of about 50 kW at 2000 rpms and then remains constant up to 7000 rpms (I didn’t accelerate past 50 mph). Torque, power, and rpms are related by the equation Power = 2*pi*torque*rpms/60. From this equation, we can deduce that acceleration is limited by torque below 2000 rpms (about 14 mph), and by power above 2000 rpms. Below 2000 rpms, torque is constant, so power is proportional to rpms. Above 2000 rpms, power is constant, so torque is proportional to the reciprocal of rpms. The motor output power is the power output of the HVB divided by the motor efficiency. In this instance, the HVB was supplying 56.5 kW of power and the motor was outputting 47.3 kW of power, i.e. efficiency was 84%.

Transmission Efficiency I collected some more data for fast accelerations and updated the transmission efficiency chart. The graph shows the percentage of energy output from the motor that is converted into kinetic energy to accelerate the car, i.e. transmission plus tire slippage efficiency. It looks like efficiency is 96% - 97%. I am unable to obtain accurate enough data to determine if transmission efficiency varies with motor output power. But if it does, it is not significant.

Determining Transmission Efficiency The following plot illustrates how I am computing transmission/tire slippage efficiency. It shows energy consumption vs. time during acceleration. The purple line is Potential Energy, i.e. energy required to climb the slight incline of the road. The light blue line is the energy required to overcome friction, i.e. aerodynamic drag, tire rolling resistance, and internal frictions. The green line is the Kinetic Energy of the car, which increases as the car speeds up. The blue line is the sum of the Kinetic Energy, Friction Losses, and Potential Energy. This is the total energy required to accelerate the car. The red line is the actual mechanical energy output of the motor. If the drivetrain were 100% efficient, the red and blue lines would coincide. Due to losses in the drivetrain, the motor outputs more energy than is required to propel the car. The excess energy is lost mostly to heat. Power is the derivative with respect to time of energy, i.e. the slope of the lines in the graph. The average power output of the motor is 28.61 kW, i.e. the slope of the red line. The average amount of power required to propel the car is 27.64 kW, i.e. the slope of the blue line. The efficiency of the transmission is thus approximately 27.64 / 28.61 = 97%.