EV Dynamics Physics Experiment

[larryh]

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.

Edited May 11, 2015 at 12:14 AM by larryh

[flyingcheesehead]

Good stuff Larry!

 

Was that motor map provided by Toyota, or is it something that was put together by a Prius enthusiast? Is there any chance you could use your data to put such a map together for the Energi?

 

I also frequently wonder about acceleration vs. road slope. On a level road, lower acceleration should take less energy, in theory. Somewhere there is a practical limit to how low that can be. When accelerating from a stop and then going up a hill while still accelerating, though - Something I do 7 times during my daily round trip commute - I believe it would be advantageous to accelerate a little faster. Even though you'd be using even more energy during the acceleration in that scenario, you'd be helping to minimize the amount of time with a negative acceleration component, that is, gravity. I'm really curious how much faster that is for a given slope.

 

This is where automated cars could really excel in efficiency - They know where they're going and they could know the slope of the roads at each point. Given the appropriate data, they could automatically accelerate in the most efficient manner.

[larryh]

Good stuff Larry!

 

Was that motor map provided by Toyota, or is it something that was put together by a Prius enthusiast? Is there any chance you could use your data to put such a map together for the Energi?

 

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. 

Edited May 12, 2015 at 08:30 PM by larryh

[Hybridbear]

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.

FORScan seems to update multiple times per second for me in the Focus Electric. I imagine that this is because it's only reading one module at a time instead of many like Torque Pro. Would you be able to get more accurate measurements with FORScan? I think it has a logging feature.

[Hybridbear]

Why Fast Acceleration is Inefficient

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.

What would you say qualifies as "fast" acceleration in the Energi? How many kW of power? How many bars on the Empower screen?

 

What is "slow" acceleration? How many kW? How many bars on the Empower screen?

[larryh]

FORScan seems to update multiple times per second for me in the Focus Electric. I imagine that this is because it's only reading one module at a time instead of many like Torque Pro. Would you be able to get more accurate measurements with FORScan? I think it has a logging feature.

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.

[larryh]

What would you say qualifies as "fast" acceleration in the Energi? How many kW of power? How many bars on the Empower screen?

 

What is "slow" acceleration? How many kW? How many bars on the Empower screen?

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).

Edited May 13, 2015 at 10:59 PM by larryh

[larryh]

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.

 

Edited May 14, 2015 at 11:59 PM by larryh

[larryh]

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. 

Edited May 15, 2015 at 03:12 AM by larryh

[larryh]

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%.

 

Edited May 16, 2015 at 06:58 PM by larryh

[larryh]

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.

 

Edited May 16, 2015 at 06:49 PM by larryh

[larryh]

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.

[larryh]

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.

 

Edited May 16, 2015 at 07:48 PM by larryh

[larryh]

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:

1. Car’s mass is 1895 kg, transmission efficiency is 97%, and friction is 50% of normal friction on a level road.
2. Car’s mass is 1895 kg, transmission efficiency is 93%, and friction is 10% of the normal friction on a level road.
3. Car’s mass is 1770 kg, transmission efficiency is 97%, and friction is 100% of the normal friction on a level road

Case 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).

 

Edited May 17, 2015 at 05:13 PM by larryh

[larryh]

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.

 

Edited May 17, 2015 at 02:35 PM by larryh

[larryh]

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.

[larryh]

I also frequently wonder about acceleration vs. road slope. On a level road, lower acceleration should take less energy, in theory. Somewhere there is a practical limit to how low that can be. When accelerating from a stop and then going up a hill while still accelerating, though - Something I do 7 times during my daily round trip commute - I believe it would be advantageous to accelerate a little faster. Even though you'd be using even more energy during the acceleration in that scenario, you'd be helping to minimize the amount of time with a negative acceleration component, that is, gravity. I'm really curious how much faster that is for a given slope.

 

 

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. 

[larryh]

Energy Consumption Climbing a Hill

 

The analysis presented in the previous posts doesn’t quite work when climbing hills.  Something strange is going on.

...

We are forced to conclude that friction must be reduced when going uphill for some unknown reason.  Either there is less friction when climbing a hill or the transmission is more efficient when climbing a hill. 

 

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%.

Edited May 17, 2015 at 11:37 PM by larryh

[flyingcheesehead]

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. 

 

Am I wrong in my thinking? 

 

My thought was that gravity introduces a negative acceleration component, thus when going up a hill, you could potentially use somewhat less energy, or at least make up for faster acceleration, by minimizing the time of exposure to the negative acceleration. To use the extreme example, if you were to get on an uphill grade and stop, using only the electric motor to keep you from rolling backwards, you could eat up the entire battery going nowhere. Your potential energy and kinetic energy won't change and you won't lose any energy to friction in this scenario, but it will clearly use up energy counteracting the negative acceleration due to gravity.

 

So, I would think that there is some benefit to a slightly increased acceleration when going uphill?

[larryh]

If you use the electric motor to keep you from rolling backwards, the motor does no work.  The motor outputs zero power and zero mechanical energy to keep the car from rolling backwards on a hill.  However, because the motor is not 100% efficient, some energy will be consumed from the HVB. 

 

In the absence of non-conservative forces, i.e. friction, it requires exactly the same amount of energy to climb the hill no matter how you do it, regardless of the speed/acceleration you choose.  It takes exactly mgh energy (referred to as potential energy), where m is the mass of the car, h is the height of the hill, and g is the earth's gravitational constant. 

 

However, because of friction, it requires more energy the faster you go up the hill.  Aerodynamic drag increases with the square of speed.  Thus to minimize the amount of energy lost to friction, you want to go slowly up the hill.  The total energy to go up the hill will be that lost due to friction plus the potential energy mgh.  You can't change the potential energy, but you can reduce the amount lost due to friction by driving slower.

 

Furthermore, I suspect the car is more efficient (less friction and/or the motor uses less electricity) when going uphill than on a level road.  If that is the case, you want to maximize the time going up the hill to keep the car operating more efficiently longer.  Which again suggests you want to go slowly up the hill.

[larryh]

The following chart demonstrates that you consume more energy the faster you climb a hill.  I drove on the same hill at three different speeds.  The faster the car traveled, the more energy consumed.  The main reason for more energy consumption at higher speeds is increased aerodynamic drag, tire rolling resistance, and internal friction--the same reason as when driving on a level road. 

 

Edited May 22, 2015 at 10:26 AM by larryh

[larryh]

Energy Consumption Climbing a Hill (Revised)

 

After doing several more coast down experiments and analyzing the data, I strongly suspect that the car does not report motor output power correctly when ascending or descending hills.  It is only seems to be accurate on level roads. 

 

Based on this assumption, I have revised the chart in Post 39.  The friction is the same traveling on a level road as it is climbing a hill.  The car is simply reporting motor output power incorrectly.   The difference between the red and dark blue lines shows the discrepancy.  It appears the motor is more efficient when producing higher output power required to climb hills vs. lower power output required for level roads.  For level roads, motor efficiency is about 88%.  When climbing this hill, motor efficiency is about 93%. 

 

Efficiency appears to be a function of motor output power.   The greater the power, the more efficient the motor.  This is the opposite from what we saw during acceleration.  However, with acceleration, motor rpm is rapidly increasing.  When climbing a hill, motor rpm remains constant. 

 

I suspect the car is computing the output power of the motor using the input power of the HVB and assuming efficiency is 88%.  That does not work when climbing hills when the motor outputs a lot more power and is more efficient.

 

Edited May 22, 2015 at 09:35 PM by larryh

[Hybridbear]

How could the car report accurate data on flat ground but then report inaccurate data when climbing a hill? That doesn't make sense to me...

[larryh]

Well it definitely does not report motor output power accurately when descending a hill.  If I were to believe what it reports, regen would be 100% efficient.  I suspect that motor torque (used to compute motor output power), is estimated based upon the power from the HVB.  The estimate doesn't take all the variables needed into account.  Motor efficiency is probably a very complex function of many variables.  That's another reason I can't generate a map of motor efficiency--I don't have accurate torque measurements.

Edited May 24, 2015 at 01:00 AM by larryh

[larryh]

The car also does not report motor output power accurately during regen.  The following chart shows regen with the car in Low decelerating from 40 mph to 17 mph.  The blue line is the predicted amount of energy available for regen (assuming the mass of the car and its contents if 1940 kg), i.e. kinetic energy minus frictional energy losses.  The red line is what the car reports as the mechanical energy input to the motor.  It should coincide with the blue line.  In order for them to coincide, the car's actual weight would have to be less than the curb weight given in Ford's specification. 

 

The green line is the regen supplied to the HVB, i.e the electrical energy output from the motor.  The red and green lines coincide.  That implies that the motor is 100% efficient during regen.  That is very unlikely.  Fords specifications state it is 95% efficient during regen. 

 

Motor power appears to be correct only when traveling on a level road at constant speed.