As I've mentioned previously I do some drafting of trucks to increase my gas mileage. I did some estimations and calculations, described in that post, that indicate that, done with extreme caution, it could be safe and that it should be effective. But how practical is it? It's turned out that I'm only able to find a truck to draft between 10% and 15% of the time I'm at freeway speed. Trucks are frequently going too fast for my fuel saving regime, thus leading to the question of what the break even point would be for the fuel used in speeding up to draft a fast-moving truck versus maintaining a leisurely 55 m.p.h. with no truck in front of me. More on this later.
But the fact is, there is rarely a suitable truck around when I need one. I look for trucks with the following characteristics, just based on intuition: low trailer; as square a back as possible (preferably not a milk or cement tanker); not hauling rock, dirt, etc. (no dump trucks); maintains a reasonably steady speed; doesn't do a lot of lane switching, I guess that pretty much covers the candidates. But at 55 m.p.h. I'm not doing much passing of them, so I have to wait for a truck that has the listed attributes to pass me. It is rarer than I would have guessed, though if I see a likely prospect in the rear view mirror, I can slow down to let him catch me.
So when does it pay to speed up to draft? There are two aspects to this - the fuel used in accelerating to a new speed, and the balance between the reduction in drag from being behind the truck (this would be in the density term in the drag equation) and the increase in the speed. Additionally, road loads would increase by a small amount, but I assume this to be linear, and thus the increase in road load force is compensated by the increase in distance covered.
There will be a number at which my fuel savings from reduced density (the low pressure zone behind the truck) will be overcome by the increase from the speed term, since it's squared in the drag equation. For the purposes of this post, I'll ignore the fuel used to get up to a higher speed - this fuel is used to increase the kinetic energy of the vehicle, I'll assume I can recapture this energy (though of course I can't, at least not with 100% efficiency).
I have to use the figures in my previous post on drafting to calculate the reduction in drag to attribute to the truck's wake, and calculate from there. Using my best estimate of the increase in gas mileage while drafting, from 21.5 m.p.g. to about 25 m.p.g., I can calculate that air density is decreased by about 14.0% (from about 1.16 kg/m^3 to about 0.998 kg/m^3). Again, unless I receive a huge outcry demanding the details of the mathematics involved, I'll only outline the process and give the results.
Plugging these density results into the drag equation and realizing that I want to minimize fuel/distance (gallons per mile) = energy/distance = work/distance = force * distance/distance = force, I merely need to determine when drag using the decreased density behind the truck but an increased speed exceeds drag at normal density and 55 m.p.h. As it turns out, that speed is a little over 59 m.p.h. So if I have to go faster than 59 m.p.h. to draft a truck, I will lose fuel efficiency compared to driving 55 m.p.h. in the clear.
Thus, the battle becomes one of finding a truck with all the characteristics listed above AND that is not going faster than 59 m.p.h. This has turned out to be extremely difficult. As I refine my data, I'll revisit these figures.
Friday, June 29, 2007
Tuesday, June 26, 2007
Turning resources to refuse
I read somewhere, I don't remember where, something to the effect that "an economy is a system for turning natural resources into refuse." Obviously, that's hyperbole, but it's not completely without basis. A couple of posts back I looked at national and global energy consumption, comparing time frames and comparing countries and the world. I thought that I could pursue that a little further and see what the energy requirements are to "turn natural resources into refuse."
As I stated in the post linked above, in 2006 the United States consumed 100.41*10^15 btu, or 1.059*10^20 joules of energy. In that time, we produced 245 million tons, or 222*10^9 kilograms of "MSW," municipal solid waste. I can't find an authoritative source for total air pollution in 2006, but in 2005 the total was 141 million tons. Now, that number is on a strong downward trend, it was 188 million in 1995 and 160 million in 2000. Bearing this in mind, I'll use 136 million tons or 123*10^9 kilograms in 2006. I can't find good numbers for water pollution and liquid waste, so I'll blithely assume it's negligible. With that in mind, I'll say we produced 345*10^9 kilograms of waste. So for a start, we used 307 million joules of energy per kilogram of waste. Now, 307 million joules is the amount of energy in about 6.5 kilograms or 2.5 gallons of gasoline.
So our economy used the energy available in 6.5 kilograms of gasoline to produce a kilogram of waste. Let's be optimistic and assume that the combined efficiency of all the energy conversion processes we use is 40%. That would mean that we really needed 16.25 kilograms of gasoline or equivalent to produce a kilogram of waste. Now what does that mean? Good question. From one point of view, if our energy to waste ratio were high, it could mean we didn't waste much, that that energy went into useful things. On the other hand, if our energy to waste ratio were low, it could mean we were efficient, since it wouldn't have taken much energy to produce our byproducts. But I suspect a big number is what we'd like to see.
The complexity of the situation comes from the fact that, taken independently, we'd like to produce small amounts of waste, and we'd like to consume small amounts of energy. I think the way to get a handle on this would be to find out how much energy in joules is contained in an "average" kilogram of manufactured product. In the end, what we want is as little of our consumed energy as possible to go into things that are discarded.
As I stated in the post linked above, in 2006 the United States consumed 100.41*10^15 btu, or 1.059*10^20 joules of energy. In that time, we produced 245 million tons, or 222*10^9 kilograms of "MSW," municipal solid waste. I can't find an authoritative source for total air pollution in 2006, but in 2005 the total was 141 million tons. Now, that number is on a strong downward trend, it was 188 million in 1995 and 160 million in 2000. Bearing this in mind, I'll use 136 million tons or 123*10^9 kilograms in 2006. I can't find good numbers for water pollution and liquid waste, so I'll blithely assume it's negligible. With that in mind, I'll say we produced 345*10^9 kilograms of waste. So for a start, we used 307 million joules of energy per kilogram of waste. Now, 307 million joules is the amount of energy in about 6.5 kilograms or 2.5 gallons of gasoline.
So our economy used the energy available in 6.5 kilograms of gasoline to produce a kilogram of waste. Let's be optimistic and assume that the combined efficiency of all the energy conversion processes we use is 40%. That would mean that we really needed 16.25 kilograms of gasoline or equivalent to produce a kilogram of waste. Now what does that mean? Good question. From one point of view, if our energy to waste ratio were high, it could mean we didn't waste much, that that energy went into useful things. On the other hand, if our energy to waste ratio were low, it could mean we were efficient, since it wouldn't have taken much energy to produce our byproducts. But I suspect a big number is what we'd like to see.
The complexity of the situation comes from the fact that, taken independently, we'd like to produce small amounts of waste, and we'd like to consume small amounts of energy. I think the way to get a handle on this would be to find out how much energy in joules is contained in an "average" kilogram of manufactured product. In the end, what we want is as little of our consumed energy as possible to go into things that are discarded.
Taming the LR3
As I've mentioned repeatedly, I have had very little success in utilizing driving techniques to make a large impact on the fuel economy of the LR3. This lack of success led me down a road of trying to understand the aerodynamics of the vehicle, the specifics of the engine and transmission, and the effects of driveline friction. It's been educational, but not satisfying with respect to enabling me to goose my gas mileage.
As my brother said, "it kinda takes the fun out of it." Hence, for several months, starting around the beginning of 2007, I stopped using fuel economy maximizing techniques. I didn't go "hog wild" and floor it from the light, drive as fast as traffic would allow, etc., but I would accelerate with the traffic, drive at typical speeds, eschew coasting in neutral and turning off the engine, etc.
The last three weeks or so, however, I have renewed my efforts. I've managed to bring my three tank moving average of gas mileage from a little under 16 m.p.g. to a little under 19 m.p.g. My most recent fill-up indicated 19.37 m.p.h. The improvement is probably a result of 55 m.p.h. or slower, drafting where possible, stoplight shutdowns, and coasting with the engine off. I did most of these before, but I tried being ultra aggressive in doing them.
So fuel can be saved. The LR3 is rated by the EPA at 18 highway, 14 city. By my best estimate of the relative amounts of highway and city driving in my regime (about 60/40) the EPA thinks I should see overall mileage of 16.4 m.p.g. (0.6*18 + 0.4*14). That means that the 18.85 m.p.g. in my most recent three tank moving average is 14.9% over the weighted EPA estimate.
As a point of comparison, in the Jeep Grand Cherokee Limited in which I started this experiment, I was able to achieve about 30.6% better mileage than the weighted EPA estimate. I don't know the reason that the LR3 has only allowed me to exceed the EPA estimate by about 1/2 as great a percentage as the Grand Cherokee. It may be that the LR3, about six years newer, incorporates engine management techniques that make the vehicle more efficient when driving normally. If this is the case, I will have to assume that all manufacturers utilize these management techniques and hence reduce my estimate of the impact on national fuel consumption of universal adoption of extreme mileage enhancing driving techniques. I'll save that for another post.
I will say that I am pleased that I can exceed the 19 m.p.g. barrier, and 20 m.p.g. is in my sights.
As my brother said, "it kinda takes the fun out of it." Hence, for several months, starting around the beginning of 2007, I stopped using fuel economy maximizing techniques. I didn't go "hog wild" and floor it from the light, drive as fast as traffic would allow, etc., but I would accelerate with the traffic, drive at typical speeds, eschew coasting in neutral and turning off the engine, etc.
The last three weeks or so, however, I have renewed my efforts. I've managed to bring my three tank moving average of gas mileage from a little under 16 m.p.g. to a little under 19 m.p.g. My most recent fill-up indicated 19.37 m.p.h. The improvement is probably a result of 55 m.p.h. or slower, drafting where possible, stoplight shutdowns, and coasting with the engine off. I did most of these before, but I tried being ultra aggressive in doing them.
So fuel can be saved. The LR3 is rated by the EPA at 18 highway, 14 city. By my best estimate of the relative amounts of highway and city driving in my regime (about 60/40) the EPA thinks I should see overall mileage of 16.4 m.p.g. (0.6*18 + 0.4*14). That means that the 18.85 m.p.g. in my most recent three tank moving average is 14.9% over the weighted EPA estimate.
As a point of comparison, in the Jeep Grand Cherokee Limited in which I started this experiment, I was able to achieve about 30.6% better mileage than the weighted EPA estimate. I don't know the reason that the LR3 has only allowed me to exceed the EPA estimate by about 1/2 as great a percentage as the Grand Cherokee. It may be that the LR3, about six years newer, incorporates engine management techniques that make the vehicle more efficient when driving normally. If this is the case, I will have to assume that all manufacturers utilize these management techniques and hence reduce my estimate of the impact on national fuel consumption of universal adoption of extreme mileage enhancing driving techniques. I'll save that for another post.
I will say that I am pleased that I can exceed the 19 m.p.g. barrier, and 20 m.p.g. is in my sights.
Cuantos caballos?
In my last post I tried to calculate the top speed of my Land Rover LR3 HSE. In doing so, I used the horsepower found in the vehicle's specifications, i.e., 300 horsepower. Earlier in my musings about the LR3, I was marginally successful in some calculations relating to the car's (truck's?) fuel consumption. Now I'd like to calculate the vehicle's horsepower using known facts. These facts are the engine displacement and red line r.p.m. (6200)and the r.p.m. at which it was rated as specified (5500).
I'll use the method that was employed before, i.e., I'll determine how much fluid volume (air is a fluid) is going through the engine at those r.p.m.'s and how much fuel would be in this fluid. From there, I'll use the energetic content of the fuel and some thermodynamic calculations of efficiency to see what's available at the flywheel. Here we go. One advantage I'll have is that I can estimate the efficiency by figuring out how much heat energy per second comes from putting air/fuel mixture through the engine at 5500 r.p.m. and utilizing the rated brake horsepower of 300 at that r.p.m. This should give me an approximation of how many joules per second go to the flywheel versus how many are discarded to the environment.
A straight ratio of (6200/5500)*300 tells me that I should be able to produce 338 horsepower at 6200 r.p.m. As mentioned in the earlier post linked above, Car and Driver stated that the LR3 is governor limited to a top speed of 121 m.p.h. Let's see what kind of overall efficiency is indicated if 5500 r.p.m. produces 300 horsepower. To do so, I need an estimate of manifold pressure - for starters I'll assume wide open throttle and maybe something like 0.2 p.s.i. losses for an absolute pressure of about 14.3 p.s.i.
So at 5500 r.p.m., the engine (as per my policy, I'll spare readers many of the actual calculations) will pump about 0.0147 kilograms of fuel air mixture through the engine each second. This takes into account manifold pressure of 14.3 p.s.i. as mentioned above. Oxidation of this mass of gasoline will release about 705,000 joules of energy each second. Since joules/second are watts, a measure of power, we can convert to horsepower. Doing so, if the heat of oxidation of that amount of gasoline could be converted to mechanical energy with 100% efficiency, we would develop 946 horsepower. The rated horsepower at 5500 r.p.m. is 300, implying an efficiency of about 32%. This seems very reasonable.
OK, so what's available at 6200 r.p.m.? Well, since all these calculations are based on mass flow of fuel air mixture through the engine, the simple calculation above gets close, at 338 horsepower. However, I assume that there's a slight improvement in volumetric efficiency, in other words, a slightly higher manifold pressure than at 5500 r.p.m. Making this assumption and calculating as above, it seems possible that, if full throttle at redline can be achieved, an absolute maximum of 358 horsepower could be measured at the flywheel.
I'll use the method that was employed before, i.e., I'll determine how much fluid volume (air is a fluid) is going through the engine at those r.p.m.'s and how much fuel would be in this fluid. From there, I'll use the energetic content of the fuel and some thermodynamic calculations of efficiency to see what's available at the flywheel. Here we go. One advantage I'll have is that I can estimate the efficiency by figuring out how much heat energy per second comes from putting air/fuel mixture through the engine at 5500 r.p.m. and utilizing the rated brake horsepower of 300 at that r.p.m. This should give me an approximation of how many joules per second go to the flywheel versus how many are discarded to the environment.
A straight ratio of (6200/5500)*300 tells me that I should be able to produce 338 horsepower at 6200 r.p.m. As mentioned in the earlier post linked above, Car and Driver stated that the LR3 is governor limited to a top speed of 121 m.p.h. Let's see what kind of overall efficiency is indicated if 5500 r.p.m. produces 300 horsepower. To do so, I need an estimate of manifold pressure - for starters I'll assume wide open throttle and maybe something like 0.2 p.s.i. losses for an absolute pressure of about 14.3 p.s.i.
So at 5500 r.p.m., the engine (as per my policy, I'll spare readers many of the actual calculations) will pump about 0.0147 kilograms of fuel air mixture through the engine each second. This takes into account manifold pressure of 14.3 p.s.i. as mentioned above. Oxidation of this mass of gasoline will release about 705,000 joules of energy each second. Since joules/second are watts, a measure of power, we can convert to horsepower. Doing so, if the heat of oxidation of that amount of gasoline could be converted to mechanical energy with 100% efficiency, we would develop 946 horsepower. The rated horsepower at 5500 r.p.m. is 300, implying an efficiency of about 32%. This seems very reasonable.
OK, so what's available at 6200 r.p.m.? Well, since all these calculations are based on mass flow of fuel air mixture through the engine, the simple calculation above gets close, at 338 horsepower. However, I assume that there's a slight improvement in volumetric efficiency, in other words, a slightly higher manifold pressure than at 5500 r.p.m. Making this assumption and calculating as above, it seems possible that, if full throttle at redline can be achieved, an absolute maximum of 358 horsepower could be measured at the flywheel.
Sunday, June 24, 2007
Top speed
Those who may have stumbled upon my little blog have probably noticed that I enjoy putting numbers to things. The Land Rover LR3 HSE is certainly an example, though the figures that I've been able to derive have borne only a passing resemblance to measured values. So I think I'll derive one that I won't be able to disprove myself - the maximum speed of the vehicle.
As we've previously noted, in unaccelerated travel the sum of the forces acting on the vehicle must be zero. The rated power of the 4.4 liter V8 engine in the LR3 is 300 horsepower. We'll convert that to 223710 watts using the Google Calculator. Now I'm going to estimate, based on various sites I've visited, that the efficiency of transmitting the engine's power at the flywheel to the road is 78%. This may seem a little low to some, typical figures are often in the 80% to 85% range. But this is a full time four wheel drive vehicle, so losses will be higher.
That leaves 174794 watts to the road. Now, power is force times speed, so if I add the external forces and multiply them by the speed, I'll have the power being utilized to overcome forces. The aerodynamic drag reduces to 0.7491*s^2 and the road load is estimated to be 14.65*s, where s is the speed of the vehicle. If anyone leaves a comment that they would like to know where these figures came from, I'll be happy to oblige.
In any case, since those are forces and speed times force equals power, we'll multiply by speed (s) and equate it to 174794 Thus: 174794=0.7491*s^3+14.65*s^2. I have various computer algebra systems, but the simplest is called Derive 6. Much to my regret, Texas Instruments will stop development and shipment of the program this week. I've had most versions, and where truly exotic math isn't required, I prefer it to Mathcad, Maple, and Mathematica, as capable as those programs are.
In any event, Derive easily solves this cubic equation and determines that s=55.65 meters per second, or 124 miles per hour. For purists, this is the solution in the real domain. I'm not sure how realistic this is - I haven't come anywhere close to topping out the speed in the LR3, nor do I intend to do so. This is the basis of my contention at the outset of this post that I won't be able to disprove the number. However, Car and Driver's site gives the top speed as 121 m.p.h. but refers to it as "governor limited." According to my calculations, you'd have to be going down a hill to exceed this by more than 2.5% so I'm not sure why a governor is needed. I still find this published number satisfying.
As we've previously noted, in unaccelerated travel the sum of the forces acting on the vehicle must be zero. The rated power of the 4.4 liter V8 engine in the LR3 is 300 horsepower. We'll convert that to 223710 watts using the Google Calculator. Now I'm going to estimate, based on various sites I've visited, that the efficiency of transmitting the engine's power at the flywheel to the road is 78%. This may seem a little low to some, typical figures are often in the 80% to 85% range. But this is a full time four wheel drive vehicle, so losses will be higher.
That leaves 174794 watts to the road. Now, power is force times speed, so if I add the external forces and multiply them by the speed, I'll have the power being utilized to overcome forces. The aerodynamic drag reduces to 0.7491*s^2 and the road load is estimated to be 14.65*s, where s is the speed of the vehicle. If anyone leaves a comment that they would like to know where these figures came from, I'll be happy to oblige.
In any case, since those are forces and speed times force equals power, we'll multiply by speed (s) and equate it to 174794 Thus: 174794=0.7491*s^3+14.65*s^2. I have various computer algebra systems, but the simplest is called Derive 6. Much to my regret, Texas Instruments will stop development and shipment of the program this week. I've had most versions, and where truly exotic math isn't required, I prefer it to Mathcad, Maple, and Mathematica, as capable as those programs are.
In any event, Derive easily solves this cubic equation and determines that s=55.65 meters per second, or 124 miles per hour. For purists, this is the solution in the real domain. I'm not sure how realistic this is - I haven't come anywhere close to topping out the speed in the LR3, nor do I intend to do so. This is the basis of my contention at the outset of this post that I won't be able to disprove the number. However, Car and Driver's site gives the top speed as 121 m.p.h. but refers to it as "governor limited." According to my calculations, you'd have to be going down a hill to exceed this by more than 2.5% so I'm not sure why a governor is needed. I still find this published number satisfying.
Saturday, June 23, 2007
The Almanac
One of my favorite books is the annual World Almanac and Book of Facts. Some of the book covers topics which don't interest me in the slightest, e.g., entertainment facts and much (but not all I hasten to admit) sports information. But it is chock full of nuggets to please a fact and number junkie such as myself.
I thought I'd look into per capita energy consumption in the United States and see how it's increased in the range of time covered by the Almanac for such information. In 1960 the U.S. consumed 45.09 "quads." A quad is a quadrillion, or 10^15, btu's (british thermal units). This is total consumption of every kind - industrial, agricultural, commercial, transportation, etc., and from all sources. In 2006 the consumption was 100.41 quads.
To get these figures to a unit with which we are familiar, energy per time (quads/year) is power, so the figure for 1960 converts to 1.51*10^12 watts. In 2006 the figure is 3.36*10^12 watts. The population in 1960 was 179,323,175 for a per capita power usage (that is, rate of energy consumption per unit of time) of 8421 watts per person. In 2006 with a population of 298,444,215 we consumed energy at the rate of 11,248 watts per person, a 33.6% increase in consumption rate. Frankly, this is a smaller increase than I would have guessed.
Just to put this into perspective, this is as if, in 2006, each of us had 112 100 watt light bulbs lit 24 hours per day, seven days per week. Or, since 11,248 watts is 15.1 horsepower, each of us had about 15 lawn mowers following us around 24/7.
Let's take a look at a comparison of the U.S. with China. In 2006, the U.S. consumed energy at a rate of 11,248 watts per capita. China, with a total consumption of 59.57*10^15 quads and a population of 1,313,973,713 consumed energy at a rate of 1,514 watts per capita, or about 13.4% of the U.S. rate of consumption. I'd be curious to know how much of China's figure relates to production of consumer goods for shipment to the United States.
Worldwide, in 2005 humanity consumed 446*10^15 quads, or used energy at the rate of 1.49*10^13 watts. With a population of about 6.4 billion, we consumed energy at the rate of about 2330 watts per capita, about 21% of the U.S. rate. This is very scary stuff. Standard of living is very strongly correlated with rate of energy consumption, so to bring the world to "our" standard of living, we'd have to approximately quintuple our rate of energy consumption on a worldwide basis. This seems ludicrous. Never mind climate change, there is no chance of converting (all human energy use is conversion, never creation) energy at such prodigious rates.
How about looking at it from the other direction? How much energy consumption could we, as a society, forgo? I think I personally could struggle by on 1/2 the personal energy consumption. Remember, though, that this would include reducing the energy content of my consumer purchases, my food, etc., since the numbers above are all-inclusive. As I look around, listen to the television on downstairs, listen to the waterfall in my pool as water is pumped through the filtration system, listen to my wife in the shower as the heater provides hot water, etc., I know there's a long way to go.
I thought I'd look into per capita energy consumption in the United States and see how it's increased in the range of time covered by the Almanac for such information. In 1960 the U.S. consumed 45.09 "quads." A quad is a quadrillion, or 10^15, btu's (british thermal units). This is total consumption of every kind - industrial, agricultural, commercial, transportation, etc., and from all sources. In 2006 the consumption was 100.41 quads.
To get these figures to a unit with which we are familiar, energy per time (quads/year) is power, so the figure for 1960 converts to 1.51*10^12 watts. In 2006 the figure is 3.36*10^12 watts. The population in 1960 was 179,323,175 for a per capita power usage (that is, rate of energy consumption per unit of time) of 8421 watts per person. In 2006 with a population of 298,444,215 we consumed energy at the rate of 11,248 watts per person, a 33.6% increase in consumption rate. Frankly, this is a smaller increase than I would have guessed.
Just to put this into perspective, this is as if, in 2006, each of us had 112 100 watt light bulbs lit 24 hours per day, seven days per week. Or, since 11,248 watts is 15.1 horsepower, each of us had about 15 lawn mowers following us around 24/7.
Let's take a look at a comparison of the U.S. with China. In 2006, the U.S. consumed energy at a rate of 11,248 watts per capita. China, with a total consumption of 59.57*10^15 quads and a population of 1,313,973,713 consumed energy at a rate of 1,514 watts per capita, or about 13.4% of the U.S. rate of consumption. I'd be curious to know how much of China's figure relates to production of consumer goods for shipment to the United States.
Worldwide, in 2005 humanity consumed 446*10^15 quads, or used energy at the rate of 1.49*10^13 watts. With a population of about 6.4 billion, we consumed energy at the rate of about 2330 watts per capita, about 21% of the U.S. rate. This is very scary stuff. Standard of living is very strongly correlated with rate of energy consumption, so to bring the world to "our" standard of living, we'd have to approximately quintuple our rate of energy consumption on a worldwide basis. This seems ludicrous. Never mind climate change, there is no chance of converting (all human energy use is conversion, never creation) energy at such prodigious rates.
How about looking at it from the other direction? How much energy consumption could we, as a society, forgo? I think I personally could struggle by on 1/2 the personal energy consumption. Remember, though, that this would include reducing the energy content of my consumer purchases, my food, etc., since the numbers above are all-inclusive. As I look around, listen to the television on downstairs, listen to the waterfall in my pool as water is pumped through the filtration system, listen to my wife in the shower as the heater provides hot water, etc., I know there's a long way to go.
Sunday, June 17, 2007
Turning off the engine
I have linked a blog called "Daily Fuel Economy Tip in the right column of this blog. I enjoy reading it, and have left quite a few comments. Brian Carr's (the host of the blog) most recent post is a follow up to a previous post regarding turning off the engine at stop lights. I also do this, and I have done a fair amount of googling on the topic.
As anyone reading this will likely be aware, the web is awash in sites discussing fuel saving methods. A lot of these discuss avoidance of idling, and there are large differences in the "break even" times for turning off the engine and avoiding burning fuel while idling versus extra fuel used to start the engine. Brian estimates an increase in m.p.g. of about 5.8% from this technique alone. On his site, I left a comment contemplating whether that was a plausible number, I'll copy the comment here:
"I also do this, however, I wonder if a significant portion of your increased mileage in the second period comes from this. Let’s see if it’s plausible:
If you had driven the 1559.9 miles in the second period with the miles per gallon (31.25) from the first, you would have used 49.9 gallons. Instead you used 47.2 gallons, a savings of 2.7 gallons. Now, I’ve had a Jeep Grand Cherokee Limited with a 4.7 liter engine and now have a Land Rover LR3 with a 4.4 liter engine. The Grand Cherokee used 0.38 gallons per hour at idle, the LR3 about 0.5 gallons per hour. Your car probably uses less than either of these at idle but let’s assume 0.4 gallons per hour. It would have taken you 6.75 hours of idling at stoplights to burn 2.7 gallons. In a 30 day month, that would be about 13.5 minutes of sitting at stoplights each day, every day. Assuming that a stop is an average of 40 seconds (don’t have data, just a guess), that would mean that, every single day, you were stopped at a little over 20 stoplights.
Now, if you have a much smaller engine (I forgot what you drive), that would mean you’d have to be spending even more time at stoplights to have turning the vehicle off at stoplights be primarily responsible for your savings. Do you think that this is the case?
Also note that these calculations assume absolutely NO extra fuel use on startup. I’ve searched the web and can find no figures for extra fuel used on startup, but the assumptions above are clearly the most favorable for fuel savings by engine shutoff."
It would seem that his savings must be due to some other factors as well, yet the numbers above aren't completely out of the question. As mentioned there, I know well how much fuel I'm burning as I sit at a light (or coast to it) but it's been very difficult finding any information on extra fuel used to start a spark ignition internal combustion engine. Various sites claim six seconds, 10 seconds, 30 seconds, and one minute as the break even point. No one cites any data to back up their number, and they vary by a factor of 10. This is not especially helpful.
However, after replying to Brian Carr's post, I tried changing my google search string to "idling versus shutting down engine". This led me to this site. It's the closest I've found to giving me the answer I need to determine whether shutting off the engine at stop lights is a fuel saver. It says the following: "Our research showed that a V6 restart takes about the same fuel as 5 seconds of idling. We expect a V8 to save more and a 4-cylinder less." It doesn't state the nature of the research, whether it was actual measurement by the authors, by associates of the authors, or was located in a literature search. It ain't much but it's all I've found.
As I've mentioned before, I haven't done experiments by varying a single parameter and keeping all others as constant as possible to determine the effect of that parameter. Brian Carr says he has done so and the article from the ASME Florida Section provides some verification. So I suspect that, while Brian's entire 5.8% increase in fuel economy isn't due to this procedure, a significant portion of it is.
I am going to run an experiment to determine this once and for all. I don't have a way to directly measure the fuel consumption on startup, but I'll find a vacant stretch of road forming a closed course a couple of miles long. I'll run it for an hour or so stopping and idling for a measured period every mile and find the total fuel consumed. Then I'll drive the same distance, turning the car off and back on every mile - it won't matter how long the shut off period is. I'll be concentrating very hard on duplicating the acceleration, cruising, and deceleration regime of the first series. I'll start both with a full fuel tank.
At the end of this process, I should have enough data to approximate the break even time and from there, the excess fuel used on startup. No doubt this number won't reflect what's happening before the car warms up, but it should provide reasonably definitive data on the question of turning the car off at lights. I can't take credit for formulating this methodology - I read it on someone's web site. I'd credit the site, but I can't find it now.
As anyone reading this will likely be aware, the web is awash in sites discussing fuel saving methods. A lot of these discuss avoidance of idling, and there are large differences in the "break even" times for turning off the engine and avoiding burning fuel while idling versus extra fuel used to start the engine. Brian estimates an increase in m.p.g. of about 5.8% from this technique alone. On his site, I left a comment contemplating whether that was a plausible number, I'll copy the comment here:
"I also do this, however, I wonder if a significant portion of your increased mileage in the second period comes from this. Let’s see if it’s plausible:
If you had driven the 1559.9 miles in the second period with the miles per gallon (31.25) from the first, you would have used 49.9 gallons. Instead you used 47.2 gallons, a savings of 2.7 gallons. Now, I’ve had a Jeep Grand Cherokee Limited with a 4.7 liter engine and now have a Land Rover LR3 with a 4.4 liter engine. The Grand Cherokee used 0.38 gallons per hour at idle, the LR3 about 0.5 gallons per hour. Your car probably uses less than either of these at idle but let’s assume 0.4 gallons per hour. It would have taken you 6.75 hours of idling at stoplights to burn 2.7 gallons. In a 30 day month, that would be about 13.5 minutes of sitting at stoplights each day, every day. Assuming that a stop is an average of 40 seconds (don’t have data, just a guess), that would mean that, every single day, you were stopped at a little over 20 stoplights.
Now, if you have a much smaller engine (I forgot what you drive), that would mean you’d have to be spending even more time at stoplights to have turning the vehicle off at stoplights be primarily responsible for your savings. Do you think that this is the case?
Also note that these calculations assume absolutely NO extra fuel use on startup. I’ve searched the web and can find no figures for extra fuel used on startup, but the assumptions above are clearly the most favorable for fuel savings by engine shutoff."
It would seem that his savings must be due to some other factors as well, yet the numbers above aren't completely out of the question. As mentioned there, I know well how much fuel I'm burning as I sit at a light (or coast to it) but it's been very difficult finding any information on extra fuel used to start a spark ignition internal combustion engine. Various sites claim six seconds, 10 seconds, 30 seconds, and one minute as the break even point. No one cites any data to back up their number, and they vary by a factor of 10. This is not especially helpful.
However, after replying to Brian Carr's post, I tried changing my google search string to "idling versus shutting down engine". This led me to this site. It's the closest I've found to giving me the answer I need to determine whether shutting off the engine at stop lights is a fuel saver. It says the following: "Our research showed that a V6 restart takes about the same fuel as 5 seconds of idling. We expect a V8 to save more and a 4-cylinder less." It doesn't state the nature of the research, whether it was actual measurement by the authors, by associates of the authors, or was located in a literature search. It ain't much but it's all I've found.
As I've mentioned before, I haven't done experiments by varying a single parameter and keeping all others as constant as possible to determine the effect of that parameter. Brian Carr says he has done so and the article from the ASME Florida Section provides some verification. So I suspect that, while Brian's entire 5.8% increase in fuel economy isn't due to this procedure, a significant portion of it is.
I am going to run an experiment to determine this once and for all. I don't have a way to directly measure the fuel consumption on startup, but I'll find a vacant stretch of road forming a closed course a couple of miles long. I'll run it for an hour or so stopping and idling for a measured period every mile and find the total fuel consumed. Then I'll drive the same distance, turning the car off and back on every mile - it won't matter how long the shut off period is. I'll be concentrating very hard on duplicating the acceleration, cruising, and deceleration regime of the first series. I'll start both with a full fuel tank.
At the end of this process, I should have enough data to approximate the break even time and from there, the excess fuel used on startup. No doubt this number won't reflect what's happening before the car warms up, but it should provide reasonably definitive data on the question of turning the car off at lights. I can't take credit for formulating this methodology - I read it on someone's web site. I'd credit the site, but I can't find it now.
Saturday, June 16, 2007
Drafting
In thinking about what I've done to attempt to minimize fuel consumption there are three things that stand out in my mind as potentially dangerous. I say "potentially dangerous" because for all three of them it is "common knowledge," stated all over the web, that they are. For the sites and blogs that I frequent (devoted to fuel saving measures and techniques) this is typically in the context of "while this may save fuel, it is very dangerous and no amount of fuel savings is worth a serious injury or your life."
So what are the three? The first is filling my tires (slightly) beyond the recommended maximum. The second is turning off the engine on long downhills. The last and most controversial is drafting trucks. It is this that is the subject of this post.
There are two obvious question. The first is "does it work?" The second is "is it dangerous?" As to its value in fuel savings, while there are some naysayers, most references agree that it is quite effective. Coincidentally, the Discovery Channel hit show Mythbusters covered this topic recently. Their results are summarized here. As is their common practice in such "myths," Kari, Tori, and Grant started with a scale model in a wind tunnel and achieved very encouraging results. Their full scale testing, while it didn't achieve quite the drag reduction of the wind tunnel tests, indicated significant fuel savings even at a following distance as large as 100 feet.
I have played with this idea off and on, both in the Grand Cherokee and in the LR3. However, because I have been so very unsuccessful with the LR3 in achieving the dramatic enhancements to gas mileage I was able to accomplish in the Jeep, I recently decided to pursue this avenue more aggressively. I know that this is controversial, many will call it selfish - if I crash, it will cause a huge backup for those behind me. I agree that this technique is extremely selfish if there is any significant chance that I will crash into a truck and tie up a freeway. I'll get back to that momentarily.
But does it work? According to the information from my Scan Gauge II, it does. I attempted to judge how far I was behind trucks by using my stopwatch to see what fraction of a second elapsed between the back of a truck crossing a highway mark and the front of my car crossing the same mark. Needless to say, tailgating a truck while looking at a Scan Gauge II reading and timing intervals with my stopwatch is sort of living on the edge, but I concluded that I was about 35 feet behind the truck. According to Mythbusters' results, I can look for an increase in miles per gallon of somewhere between 20% and 27%. As best I could tell, I saw an increase from about 21.5 m.p.g. to about 25 m.p.g., an increase of about 16%. These are pretty fuzzy figures - the readouts are constantly changing with changes in slope, etc., and the trucks typically don't maintain a particular speed quite as efficiently as the LR3 (due to the huge mass of the trucks no doubt). Nevertheless, every time I try it I get significant indications of increased fuel efficiency.
So apparently it works, how dangerous is it? Obviously, the worst case scenario is for the truck to apply maximum braking suddenly with no advance indication to me. I don't know if trucks have anti-lock braking systems, for the analysis to follow I won't assume they do. Judging from the condition of a lot of the trucks I see, this is a valid assumption. It should go without saying that, when following a truck at 35 feet, my eyes are focused on the truck's brake lights and my foot is ready to instantly hit the brake pedal. So the question is, when I see his (male pronouns are to be understood as gender-free) brake lights when he implements maximum braking, can I avoid hitting him?
I have to calculate the circumstances under which the distance between my front and his rear decreases to zero. The data necessary for this calculation is his maximum deceleration rate, my maximum deceleration rate, and my time to go from his brake lights on to my application of maximum braking. I will obviously limit my calculations to dry pavement in good condition - I'm not suicidal.
Truck braking, like everything else when looked at by academics, is ridiculously complex, see here if you are skeptical. But the nugget for my purposes in that paper is that the best that an empty semi-tractor trailer can do in deceleration is right at 20 ft/sec^2, or 6.10 m/s^2. For my LR3, it's (maybe, subject to correction) about 8.0 m/s^2. My reaction time, when paying close attention (as I do when tailgating a large truck) was determined using this online reaction timer to be 0.303 seconds in an average of 10 trials. With a lot of practice and knowing what to expect, I was able to bring the average way down, to around 0.20 seconds, but I'll leave it at 0.303 to be conservative.
All right then, we have what we need. The initial conditions are that the truck and my LR3 are each going 60 m.p.h. or 26.8 m/s. The truck hits his brakes maximally and decelerates at 6.10 m/s^2. 0.303 seconds later, I hit my maximum brakes and decelerate at 8.0 m/s^2. How far back must I be to avoid hitting the truck? I suspect that detailed expositions of mathematics bore those who read here (though I'd certainly appreciate any feedback on this). So I'll just state that the answer is that I must be a minimum of 9.67 m or 31.7 feet behind the truck. Now, this assumes I can meet my reaction time and instantly apply maximum braking. I should add, say, a 50% safety factor for a total of 47.6 feet. Call it 50 feet. For those that would say that that's not enough of a safety factor, remember that I assumed that the truck applied perfect braking as well.
While I won't proselytize that drafting trucks is safe and should be done by one and all, I do think that with extreme caution and maximum alertness it can be safely done. I wouldn't want to try to get closer than 50 feet, and I will adjust my procedures accordingly, but I'm keeping this weapon in my arsenal.
So what are the three? The first is filling my tires (slightly) beyond the recommended maximum. The second is turning off the engine on long downhills. The last and most controversial is drafting trucks. It is this that is the subject of this post.
There are two obvious question. The first is "does it work?" The second is "is it dangerous?" As to its value in fuel savings, while there are some naysayers, most references agree that it is quite effective. Coincidentally, the Discovery Channel hit show Mythbusters covered this topic recently. Their results are summarized here. As is their common practice in such "myths," Kari, Tori, and Grant started with a scale model in a wind tunnel and achieved very encouraging results. Their full scale testing, while it didn't achieve quite the drag reduction of the wind tunnel tests, indicated significant fuel savings even at a following distance as large as 100 feet.
I have played with this idea off and on, both in the Grand Cherokee and in the LR3. However, because I have been so very unsuccessful with the LR3 in achieving the dramatic enhancements to gas mileage I was able to accomplish in the Jeep, I recently decided to pursue this avenue more aggressively. I know that this is controversial, many will call it selfish - if I crash, it will cause a huge backup for those behind me. I agree that this technique is extremely selfish if there is any significant chance that I will crash into a truck and tie up a freeway. I'll get back to that momentarily.
But does it work? According to the information from my Scan Gauge II, it does. I attempted to judge how far I was behind trucks by using my stopwatch to see what fraction of a second elapsed between the back of a truck crossing a highway mark and the front of my car crossing the same mark. Needless to say, tailgating a truck while looking at a Scan Gauge II reading and timing intervals with my stopwatch is sort of living on the edge, but I concluded that I was about 35 feet behind the truck. According to Mythbusters' results, I can look for an increase in miles per gallon of somewhere between 20% and 27%. As best I could tell, I saw an increase from about 21.5 m.p.g. to about 25 m.p.g., an increase of about 16%. These are pretty fuzzy figures - the readouts are constantly changing with changes in slope, etc., and the trucks typically don't maintain a particular speed quite as efficiently as the LR3 (due to the huge mass of the trucks no doubt). Nevertheless, every time I try it I get significant indications of increased fuel efficiency.
So apparently it works, how dangerous is it? Obviously, the worst case scenario is for the truck to apply maximum braking suddenly with no advance indication to me. I don't know if trucks have anti-lock braking systems, for the analysis to follow I won't assume they do. Judging from the condition of a lot of the trucks I see, this is a valid assumption. It should go without saying that, when following a truck at 35 feet, my eyes are focused on the truck's brake lights and my foot is ready to instantly hit the brake pedal. So the question is, when I see his (male pronouns are to be understood as gender-free) brake lights when he implements maximum braking, can I avoid hitting him?
I have to calculate the circumstances under which the distance between my front and his rear decreases to zero. The data necessary for this calculation is his maximum deceleration rate, my maximum deceleration rate, and my time to go from his brake lights on to my application of maximum braking. I will obviously limit my calculations to dry pavement in good condition - I'm not suicidal.
Truck braking, like everything else when looked at by academics, is ridiculously complex, see here if you are skeptical. But the nugget for my purposes in that paper is that the best that an empty semi-tractor trailer can do in deceleration is right at 20 ft/sec^2, or 6.10 m/s^2. For my LR3, it's (maybe, subject to correction) about 8.0 m/s^2. My reaction time, when paying close attention (as I do when tailgating a large truck) was determined using this online reaction timer to be 0.303 seconds in an average of 10 trials. With a lot of practice and knowing what to expect, I was able to bring the average way down, to around 0.20 seconds, but I'll leave it at 0.303 to be conservative.
All right then, we have what we need. The initial conditions are that the truck and my LR3 are each going 60 m.p.h. or 26.8 m/s. The truck hits his brakes maximally and decelerates at 6.10 m/s^2. 0.303 seconds later, I hit my maximum brakes and decelerate at 8.0 m/s^2. How far back must I be to avoid hitting the truck? I suspect that detailed expositions of mathematics bore those who read here (though I'd certainly appreciate any feedback on this). So I'll just state that the answer is that I must be a minimum of 9.67 m or 31.7 feet behind the truck. Now, this assumes I can meet my reaction time and instantly apply maximum braking. I should add, say, a 50% safety factor for a total of 47.6 feet. Call it 50 feet. For those that would say that that's not enough of a safety factor, remember that I assumed that the truck applied perfect braking as well.
While I won't proselytize that drafting trucks is safe and should be done by one and all, I do think that with extreme caution and maximum alertness it can be safely done. I wouldn't want to try to get closer than 50 feet, and I will adjust my procedures accordingly, but I'm keeping this weapon in my arsenal.
Tuesday, June 05, 2007
Another comparison
I'm still trying to understand the differences between the Jeep Grand Cherokee Limited I had until late November of 2006 and that was the topic of many of my posts, and the Land Rover LR3 HSE I have now. Right now, I'm thinking about the 31 m.p.g. the Jeep exhibits at 55 m.p.h. versus the 21.5 shown by the Land Rover. My last post dealt with the LR3 from the point of view of a big fuel burning air pump. I decided to compare what the two vehicles "should" require.
I gave some figures that should help out in this effort a few posts back. Let's run a few numbers. The LR3 has a frontal area of 3.15 m^2 and a coefficient of drag of 0.41. We have a dynamic pressure at 24.6 m/sec (approx. 55 mph) and air density of 1.16 kg.m^3 of 351 newtons/m^2. Thus, for this condition and drag coefficient, my aerodynamic drag is approximately 453 newtons (dynamic pressure times frontal area times drag coefficient).
Tire rolling resistance (as best I've been able to find) is about .015 times vehicle weight, or about 392 newtons. Total external forces to be overcome by the engine at 55 mph are therefore about 743 newtons or 167 pounds force. Now, force times speed is power, so the engine must provide 743 newtons at 24.6 m/s or about 18,280 watts. This equates to about 24.5 horsepower.
Obviously, much of the energy in the gasoline is lost to heat, engine and driveline friction, and pumping various fluids (refrigerant if the a.c. is on, water in the cooling system, air through the fan, oil, etc.). So enough gas must be burned per second to overcome all of these "dissipative" forces and still provide 24.5 horsepower.
For the Grand Cherokee, frontal area is 2.48 m^2, coefficient of drag is 0.44. Running through the same calculations, I get the external forces to be overcome by the Jeep at 55 mph are 383 newtons of aerodynamic drag and 289 newtons of rolling resistance for a total of 672 newtons or 151 pounds force. Enough energy must come from the fuel per second to overcome the dissipative forces and provide about 22.1 horsepower to maintain 55 mph against the external forces. So, if all else were equal, the LR3 should burn about 10.9% more fuel per mile at 55 mph. In fact, it burns about 44% more fuel. Or so the gauges say. So obviously, all else isn't equal.
These are the types of things I'm trying to understand.
I gave some figures that should help out in this effort a few posts back. Let's run a few numbers. The LR3 has a frontal area of 3.15 m^2 and a coefficient of drag of 0.41. We have a dynamic pressure at 24.6 m/sec (approx. 55 mph) and air density of 1.16 kg.m^3 of 351 newtons/m^2. Thus, for this condition and drag coefficient, my aerodynamic drag is approximately 453 newtons (dynamic pressure times frontal area times drag coefficient).
Tire rolling resistance (as best I've been able to find) is about .015 times vehicle weight, or about 392 newtons. Total external forces to be overcome by the engine at 55 mph are therefore about 743 newtons or 167 pounds force. Now, force times speed is power, so the engine must provide 743 newtons at 24.6 m/s or about 18,280 watts. This equates to about 24.5 horsepower.
Obviously, much of the energy in the gasoline is lost to heat, engine and driveline friction, and pumping various fluids (refrigerant if the a.c. is on, water in the cooling system, air through the fan, oil, etc.). So enough gas must be burned per second to overcome all of these "dissipative" forces and still provide 24.5 horsepower.
For the Grand Cherokee, frontal area is 2.48 m^2, coefficient of drag is 0.44. Running through the same calculations, I get the external forces to be overcome by the Jeep at 55 mph are 383 newtons of aerodynamic drag and 289 newtons of rolling resistance for a total of 672 newtons or 151 pounds force. Enough energy must come from the fuel per second to overcome the dissipative forces and provide about 22.1 horsepower to maintain 55 mph against the external forces. So, if all else were equal, the LR3 should burn about 10.9% more fuel per mile at 55 mph. In fact, it burns about 44% more fuel. Or so the gauges say. So obviously, all else isn't equal.
These are the types of things I'm trying to understand.
Saturday, June 02, 2007
First principles
This post is a continuation of the previous one in which I tried to reason the cause of my inability to coax better mileage out of the LR3. I was able to show through an analysis of manifold pressure, engine frequency (r.p.m.) and the chemical mixture requirements for combustion that the vehicle was burning fuel at a rate that equated to 13.3 m.p.g. at 55 m.p.h. on level ground. While this is in the ballpark, it's certainly not close to the base, so I tried reasoning from first principles, as the mathematicians say.
I started with the assumption that the fuel/air mixture in the manifold (and the cylinder during the intake stroke) is an ideal gas. I carried through an analysis on that basis (from the details of which I'll spare my patient readers) utilizing the absolute pressure and the manifold air temperature as reported by the Scan Gauge II I have attached to my engine.
I ran my analysis using the approximation that gasoline is normal heptane and that air is 22% O2 and 78% N2. From there, I proceeded to calculate from the ideal gas law. As Scotty used to say on Star Trek, "you canna change the laws of physics." I didn't expect to determine a fuel consumption number that matched the indicated m.p.g. from the Scan Gauge II to the nearest 0.1 m.p.g., but I did hope for something within, say, 10%.
Nope. The calculated result was worse than my previous calculation, coming in at 11.06 m.p.g. So what gives?? I have begun reading a treatise entitled "The Internal-Combustion Engine, Theory and Practice" by Taylor, since it's obvious that my level of understanding of the physics of internal combustion engines is inadequate to the problem at hand. It's a two volume tome, and exhaustive in scope and detail. When I learn enough to see where I've gone astray with my analyses, hopefully I'll also understand the mysteries of the LR3 vs. Grand Cherokee Limited engine comparison.
I started with the assumption that the fuel/air mixture in the manifold (and the cylinder during the intake stroke) is an ideal gas. I carried through an analysis on that basis (from the details of which I'll spare my patient readers) utilizing the absolute pressure and the manifold air temperature as reported by the Scan Gauge II I have attached to my engine.
I ran my analysis using the approximation that gasoline is normal heptane and that air is 22% O2 and 78% N2. From there, I proceeded to calculate from the ideal gas law. As Scotty used to say on Star Trek, "you canna change the laws of physics." I didn't expect to determine a fuel consumption number that matched the indicated m.p.g. from the Scan Gauge II to the nearest 0.1 m.p.g., but I did hope for something within, say, 10%.
Nope. The calculated result was worse than my previous calculation, coming in at 11.06 m.p.g. So what gives?? I have begun reading a treatise entitled "The Internal-Combustion Engine, Theory and Practice" by Taylor, since it's obvious that my level of understanding of the physics of internal combustion engines is inadequate to the problem at hand. It's a two volume tome, and exhaustive in scope and detail. When I learn enough to see where I've gone astray with my analyses, hopefully I'll also understand the mysteries of the LR3 vs. Grand Cherokee Limited engine comparison.
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