larryh

Looking more closely at the data I have collected on a 0.5 mile section of a hill at 8% grade and constant speed, I observe the following conversion efficiencies from the motor mechanical energy or the HVB electrical energy to potential energy: Speed Direction Mechanical Electrical (mph) Efficiency Efficiency 20 Uphill 87% 74% 30 Uphill 86% 72% 20 Downhill 76% 75% 30 Downhill 72% 72% So when going up the hill at 20 mph, 87% of the mechanical energy output from the motor is converted to potential energy which propels the car up the hill. The remaining 13% is lost to friction (aerodynamic drag, tire rolling resistance, etc.). The motor is 85% efficient converting electrical to mechanical energy, so 87% * 85% = 74% of the electrical energy consumed from the HVB is converted to potential energy. When going down hill at 20 mph, 76% of the potential energy from descending the hill is converted to mechanical energy via the motor for regen, the remaining 24% is lost due to friction. The motor is 99% efficient during regen, so 76% * 99% = 75% of the potential energy is applied to the HVB. This seems very strange. Conversion efficiency is asymmetric with respect to mechanical energy. but symmetric with respect to electrical energy. At 30 mph, the conversion efficiency from mechanical to potential energy is 86%. The reverse conversion from potential to mechanical energy (during regen) is much lower at 72%. However, the conversion in either direction between electrical and potential energy is the same at 72%. The same holds true for 20 mph. For kinetic energy, the symmetries are reversed. The conversion efficiency between kinetic and motor mechanical energy is the about the same in either direction, but the conversion efficiency from electrical energy to kinetic energy is much lower than the conversion efficiency from kinetic energy to electric energy via regen. Motor efficiency is about 85% when accelerating and is greater than 95% during regen. Note that when stopping quickly, the conversion efficiency from kinetic to electrical energy is at most 80%, i.e. 80% of the kinetic energy is converted to electrical energy by the motor. This is comparable to the 75% conversion efficiency from potential to electrical energy observed going down the hill. Since the conversion between electrical and potential energy is around 75% efficent and the conversion between potential and kinetic energy is 100% efficient, this suggests that you want to slow down going up the hill, converting some of the kinetic energy to potential energy, and let the car speed up going back down the hill, converting some of the potential energy to kinetic energy.

Note that I maintained the same constant speed with each of the different methods. If you go down the hill faster, you will loose slightly more energy from friction and it will not be a fair comparison between the methods to determine which yields the best regen. All the hills I use have stop signs at the bottom.

I don't have a Scan Gauge. The Scan Gauge is not useful for the analyses I have been doing. I have the OBDLink MX scanner. It makes no difference whether you use the brakes, grade assist, or Low and Cruise Control to maintain constant speed going down the hill. The car does the same thing and you will get identical results. They all use the motor for regen and hold back the car in exactly the same way. I have tried it. The only thing that could make a difference is how much you allow the car to slow down when going up the hill and how much you allow it to speed up when going back down. I don't have any good hills without stop signs, too much traffic, or too many curves to try that on at the moment. I did try going down a relatively steep hill at 20 mph vs. 30 mph. Regen was about the same. For 20 mph, it was 0.491 kWh / mile. For 30 mph, it was 0.476 kWh / mile. There is a little more energy lost due to friction at the higher speed.

The following plot shows Energy Loss vs. Distance on a hilly road. The road is about 1.4 miles long. The tops three curves show the energy loss (presumably due to friction) at 20, 29, and 33 mph. The bottom two curves show the kinetic energy of the car (for the trip at 20 mph) and change in potential energy (due to the hills). For the top three curves, I have plotted the mechanical energy output of the electric motor, subtracting out the kinetic and potential energy that must be supplied by the motor to get the car up to speed or to slow down, and to climb or descend hills. The curves are not what I would expect--something is missing. I would have expected them to be fairly straight lines. There should be no bumps in the curves. I am plotting energy loss. There is no way to reclaim lost energy. However, at about 1.2 miles, the energy loss decreases. That is impossible. The car has just reached the top of a hill and is now beginning to descend and gain speed. There has to be some other mechanism that is storing and releasing energy other than kinetic energy and potential energy. That mechanism seems to be related to the rate of change of speed of the car. It probably has something to do with changes in the moment of inertia of the various rotating components inside the car, such as the wheels, planetary gear system, motor, and generator. But I have no idea what it is. Without understanding this mechanism, I don't have a good way of analyzing efficient driving techniques for hills.

The following chart shows Regen during a stop. At distance 0 miles, the car is going 40 mph. The stop sign is at distance 0.32 miles. The plot shows three different stops. For the red curve, I coast for a long time in Drive starting at distance 0.04 miles and then use the brakes starting at distance 0.3 miles to come to a complete stop at the stop sign. For the orange curve, I continue at 40 mph for 0.14 miles, coast in Drive until distance 0.26 miles, and then apply the brakes to complete the stop. For the purple curve, I maintain 40 mph until distance 0.22 miles and then shift into Low. I apply the brakes at distance 0.32 right before the stop sign to complete the stop. Each curve shows the energy consumed from the HVB vs. distance. The car consumes energy from the HVB to maintain 40 mph until coasting begins or I shift into Low, after which regen commences. The curves now start falling as the motor generates electricity from the kinetic energy of the car which is supplied to the HVB. For the purple curve, 0.04 kWh of energy is consumed to maintain a constant speed of 40 mph until I shift into Low (you can see the the value of the purple curve at 0.22 miles is 0.04 kWh). While in L, regen supplies about 0.067 kWh of energy to the HVB and the curve falls to -0.027 kWh at distance 0.32 miles. The kinetic energy of the car at 40 mph is approximately 0.08 kWh. So regen captures approximately 84% of the kinetic energy. For the red curve, 0.008 kWh of energy is consumed to maintain constant speed until coasting begins. Regen supplies about 0.045 kWh of energy to the HVB and the curve falls to -0.037 kWh at the stop sign. This time regen is only 56% efficient. The orange curve is in between the other two. We end up with the most energy in the HVB with the long coast (the red curve) at -0.037 kWh. We end up with the least amount of energy in the HVB when using Low to stop (the purple curve) at -0.027 kWh. The difference is 0.01 kWh. While there is more regen when using Low to stop, 0.067 kWh vs. 0.045 kWh of energy, the extra energy from the HVB required to maintain constant speed of 40 mph until closer to the stop sign more than offsets the additional 0.022 kWh gained from regen.

I do not use the ICE to commute to work. 97% of my trips are EV only. For longer commutes on the weekends, I have to use EV mode on the freeway in order to use up all the energy in the HVB. The first 10 miles of freeway are generally down hill and require less power, so I want to stay in EV mode for this portion of the trip. I do not want to start the ICE before I get on the freeway. After the downhill section of freeway, I then switch on EV later mode while I am still on the Freeway and beginning to go up hill. For the return trip, the first 15 miles are slower speed 55 mph highway driving, so I want to stay in EV mode during this portion of the commute to use up the HVB. The highway turns into a freeway. At this point I switch to EV later mode for higher speeds.

There are more techniques to take into consideration. Since energy conversion between potential and kinetic energy is 100% efficient, it would be interesting to measure the impact of slowing down going up the hill and then allowing gravity to assist speeding the car back up again going down the hill.

Without actually trying both ways of starting the ICE, I don't know which is better. You will have to use a trip odometer to measure how much gas is consumed during warmup for both methods and how much HVB energy is used before the on-ramp and while getting up to speed for both methods.

Prior to this week, the average SOC of 12 V battery reported by the BCM was about 85%. The highest SOC reported was 94%. For some unknown reason, this week my car has been reporting the SOC to be between 96% and 97%. This seems very strange.

I don't think the ICE is simply idling at a stop light--it should be charging the HVB. Having the car charge the HVB on the freeway is inefficient. Assuming about 75% of the mechanical power from the motor can be stored in the HVB and later output by the electric motor, and x kWh of energy from the HVB are consumed during the warmup while getting up to speed, the ICE will have to output x/0.75 = 1.33x kWh of additional energy to recharge the HVB. Suppose it takes 0.3 kWh of energy from the HVB to get up to speed. We can either start the ICE early and use it to accelerate up to speed. Or we can start the ICE on the on ramp and use 0.3 kWh of energy from the HVB. If it comes from the HVB, the ICE will have to output an additional 1.33*0.3 = 0.4 kWh of energy. If it comes directly from the ICE, the ICE will have to output only 0.3 kWh of energy instead. Going through the math, starting the ICE on the on ramp will use 0.008 more gallons of gas (assuming the ICE is 36% efficient). I'm not sure how much gas will be consumed during warmup when the ICE is started before the on ramp vs. when it is started on the on ramp. That may or may not overwhelm the 0.008 gas savings from starting the ICE before the on ramp. However, note that the ICE will have assisted the electric motor and charged the HVB when started before the on-ramp. So there will be more energy in the HVB which further reduces gas consumption. So when starting the ICE early, there is actually more than the 0.008 gallons of gas savings mentioned above.

If you are making significant elevation changes, as in your tests, then the elevation difference between the origin and destination is what is going to dominate your results. Speed and the elevation profile will only be minor factors affecting the results since most of the energy is spent providing the necessary potential energy to get up the hill vs. overcoming friction. You did not observe much difference in MPGe for the various speeds. The same is true when you went down the hill. If there is some alternative path that avoids the hill completely, you would want to compare the MPGe for this route vs. one the goes up and down the hill to see just how expensive hills are. Rather than considering just one steep hill, another interesting case would be how to tackle a road with many hills to provide the best MPGe for a given average speed.

I tried Low on a very steep hill (70% max grade according to Google Earth) going less than 15 mph--I didn't dare go much faster. I had to use the brakes to slow the car down with brake score in the lower 80s. The ICE did not come on. The max regen was only about 21 kW.

Use the trip odometers and observe MPGe. Record the results for each method. You might want to reset the trip odometer at the beginning of each method. But I think the differences may be small and hard to detect. You need to make sure you execute each method very precisely.

I would not expect the ICE to be used to slow the car down unless regen exceeds 35 kW or the HVB is full.

That site only has elevation data for Minnesota. I don't know if any other states have LiDAR data available. The elevation data provided by Google Earth is not very accurate. I do not have enough experience with hills yet to answer those questions.

Make sure you are taking elevation into account when comparing the outbound vs. return trip energy consumption. Go to this web site to determine the origin and destination elevations using the elevation tools: http://arcgis.dnr.state.mn.us/maps/mntopo/. The site provides elevation accurate to better than 0.1 meters. You can use it to generate an elevation profile for a trip. If the destination elevation is greater than the origin elevation, then an additional m*g*h Joules of energy is required, where m = mass of car in kg (I use 1871 kg), g = 9.81 (Earth’s gravitational constant), and h is the elevation change in meters. To convert Joules to kWh, divide by 3600 seconds in a hour and by 1000 watts in a kilowatt.

I do not have adaptive cruise control. I manually use the steering wheel controls to adjust cruise control and maintain distance to the cars in front of me.

You need to set cruise control when driving in L down a hill. That will maintain constant speed down the hill similar to grade assist. You will average slightly slower speed down the hill (and get slightly more regen) using L and cruise control than with grade assist.

I would expect grade assist and low to work about the same going down hill. But I can try it out and see if there is a difference.

I am a software engineer, but I do have undergraduate degrees in math and physics so I should be able to figure figure out basic mechanics. I only use low when I intentionally want to slow down quickly and recover as much regen as possible. Yes there is a lot to digest, but it should help improve driving efficiency. You should try to achieve the highest driving score possible via the driving coach. It is giving correct advice on how to improve driving efficiency. In city driving, you should be able to achieve scores in the mid 90's. For my commute to work on city streets with a speed limit of 55 mph, my score is usually 94 - 97. If you have highway driving, you are going to be penalized for high speeds and your score will not make it to the 90's.

During the winter, it requires about 2.5 kWh of electricity for my 8 mile commute to work in subzero temperatures. During the summer, it takes 1.6 kWh. So it takes 56% more electricity in the cold. I think much of that additional energy is due to the additional drivetrain frictions when cold. Only a fraction of it is from higher aerodynamic drag and increased rolling resistance with colder temperatures. I suspect that if I had a heated garage, it would take much less energy. I tried using the Engine Block Heater (EBH), but that did not warm up the transmission fluid very much. You need a heated garage in the winter. If your commute is longer than the EV range, a good strategy might be to start the ICE at the beginning of the commute to warm up the car and reduce the internal drivetrain frictions.

Pulse and Glide (P&G) requires that you shift into neutral during the glide. If you don't shift into neutral, you will get very poor results--see post #18. You can also see the posts I have made regarding P&G starting at: http://www.fordfusionenergiforum.com/topic/1683-obd-ii-data-for-hvb/?p=15551. P&G relies on the fact that the motor operates more efficiently during acceleration than cruising at constant speed. Motor efficiency might be around 84% during acceleration and 72% for constant speed. So the motor is 84% / 74% = 1.14 times more efficient. Unfortunately, as seen in post #18, their is a severe penalty for accelerating. You may gain 14% in motor efficiency, but you also lose about 10% of the energy used to accelerate in the drivetrain components during acceleration. The net improvement might be around 5% if you are lucky. It takes skill and experience to use P&G effectively. Quite likely you will end up with worse mileage than maintaining constant speed. Yes, you will end up using more energy from the HVB by waiting to brake. You may recover 80% of the kinetic energy with fast braking vs. 45% with slow braking. However, maintaining speed until closer to the stop sign requires more energy than you will have gained through fast braking. When going downhill, close to 100% of the mechanical energy applied to the motor is converted to electrical energy. I measured it at 101%--obviously there is some measurement error. The problem is that a significant fraction of the potential energy difference from descending the hill is not making it to the motor for some reason. I don't know why. So far I have only discussed EV driving. I'm not sure what the best way to accelerate is when using the ICE.

My statement was not well worded. I was trying to be concise, but ended up being too concise. In this case, I am measuring the amount of energy loss due to friction starting from the time the brakes are pressed until the car has stopped. I am not considering the additional energy required to maintain speed longer while waiting to begin the stop. So a fast stop might require 20 seconds to come to a complete stop and slow stop might take 60 seconds. At all times, after pressing the brake, you will be going slower for the fast stop than the slow stop. Hence, there will be less energy lost due to friction for the fast stop, which is then available for regen. After 20 seconds, you will have come to a complete stop with the fast stop and will no longer be losing energy due to friction. After 20 seconds for the slow stop, you will have only slowed down slightly and will still be losing a significant amount of energy due to friction. The longer it takes to stop, the more energy loss due to friction during the stop, which is not available for regen. For the fast stop, you might recover up to 80% of the kinetic energy during regen. For the stop stop, you might only recover 45%. The difference is due to the greater energy loss from friction during the slow stop.

For some reason, regen going downhill seems to be significantly less efficient than regen on level surfaces. I have no idea why. I have repeated the measurements many times on different hills and always come up with the same results. Going uphill, energy losses are in line with what I expect--the same as driving on a level road. But going downhill, I observe more than double the expected losses. The hill I am considering in this post has an elevation change of 264 feet and an average grade of 8-9%. I am using 1 meter resolution LiDAR Data to determine the elevation of the road with an elevation accuracy of better than 0.1 meters. The Potential Energy errors in my measurements should be very small. I use cruise control to go up the hill and grade assist to go downhill, both at 30 mph. The motor mechanical power loss when going uphill is 3.07 kW (about 13%), which is what I would expect for 30 mph on a level road at 45 F due to friction. The power loss when going down hill is 6.79 kW (about 32%), more than twice as much. Considering the middle portion of the hill where speed was constant for both the down and up hill trips, the change in Potential Energy was 0.28 kWh. During the uphill, the motor output 0.31 kWh of energy. Thus there was a loss of 0.03 kWh. During the downhill, the motor was supplied with 0.19 kWh of energy. Thus there was a loss of 0.09 kWh of energy. The large loss of regen during downhills is going to make driving on a hilly road significantly less efficient than a level road if you maintain constant speed. In this instance, it requires 35% more energy to go up and down the hill vs. driving on a level road.

At 60 mph, it takes about 0.3 - 0.4 kW more power from the HVB to propel the car with the windows open. The AC takes a minimum of about 0.6 kW of power. So at 60 mph, you are better off opening the windows. You would probably have to exceed 70 mph before operating the AC would be more efficient. But if you are going that fast, you are not driving efficiently to begin with. Initially, the AC requires much more than 0.6 kW of power--up to 5 kW of power (that's more power than the central A/C for a large home consumes). It will take quite some time for the power required by A/C to reduce to 0.6 kW. For my normal 8 mile city commute, using the AC uses far more energy than opening the windows--there is no contest.