Talk:Air–fuel ratio

lambda
I just noticed that the German page and the English page don't align in the definition of the lambda value. the German definition is mass of air devided by stöchiometric air mass while the english definition talks about air fuel ratio devided by stoichiometric (different spelling but i guess its english) air fuel ratio.

after writing this realized that the fuel just devides out and you are left with the german version, but i still note it here because of the inconsistency.

[Untitled]
Thanks, but surely the lower fuel consumed would generate more CO2 if it is burning that much better? I can understand that one can reduce CO2 emission from changes in vehicle weight, drive system etc. but improving combustion just means increasing co2 from a given amount of fuel burnt. Or is it that vehicles do not have close to 100% combustion efficiency but maybe closer to 97-98% combustion efficiency and so need to burn extra fuel to compensate and the unburnt and partially burnt portions of the fuel including HCx, aldehydes, particulates, smoke etc. is released in to the atmosphere eventually to form Co2 and so any combustion efficiency improvement would increase CO2 per litre of fuel burnt but would also reduce the total amount of fuel burnt hence the Co2 reduction claim. The increase and decrease not being linear. Would appreciate your thoughts on this. —Preceding unsigned comment added by 81.153.23.27 (talk) 11:24, 30 March 2009 (UTC)

Can someone explain how CO2 is being reduced with 'cleaner' hydrocarbon fuels. I have come across claims that these new fuels with detergent and dispersant additives reduce fuel consumption by improving combustion. This I can understand i.e. cleaner injectors better atomisation etc. but don't understand how this reduces CO2 emissions. Surely, as combustion improves then the emission of CO2 increases at the expense of CO and HCx. Can someone give a more technical explanation of CO2 emission with changes in the fuel/air balance. —Preceding unsigned comment added by 81.157.25.184 (talk) 11:36, 27 March 2009 (UTC)


 * It is probably just that lower fuel consumption causes lower CO2 productions. For example, if fuel consumption is reduced by 1% then clearly CO2 production will be reduced by 1%, O2consumption will be reduced by 1%, and so on.  If so, this is an obvious outcome and does not really justify fuel refiners boasting about reducing CO2 emissions.  Dolphin51 (talk) 06:25, 28 March 2009 (UTC)

Can someone put the air fuel ratio into laymans terms? If the ratio is 14.7 to 1 and air weighs .0807 lbs per cubic foot, and gasoline weighs 46.24 lbs per cubic foot, then what is the % of fuel to air, presuming the fuel is gasoline? I think is should be about 5% fuel to air. Hhilliard (talk) 14:18, 12 May 2008 (UTC)

According to "Automotive Technology, A systems approach. 6th edition" by Jack Erjavec and Rob Thompson "the ideal air-fuel or stoichiometric ratio for a gasoline engine is approximately 14.7 pounds of air mixed with 1 pound of gas. This provides a ratio of 14.7:1. Different fuels have different stoichiometric ratios. Because air is so much lighter than gasoline, it takes nearly 10,000 gallons of air mixed with 1 gallon of gasoline to achieve this air-fuel ratio. Lean ratios of 15 to 16:1 provide the best fuel economy. Rich mixtures have a ratio below 14.7:1 and provide more power from the engine but greater fuel consumption." P.808 Chapter 28 Gasoline, Diesel, and other fuels. Djcookiii (talk) 20:43, 10 May 2016 (UTC) c

AFR
AFR is measured in volume not weight. For every cubic feet of gasoline that enters into the combustion chamber as vapor or mist (as near as possible)approxamatly 14.7 cubic feet of air is needed to obtain a Stiochimetric mixture. This is measured. in CFM (cubic feet per minute). —Preceding unsigned comment added by 134.134.136.3 (talk) 17:51, 5 August 2008 (UTC)


 * No. The stoichiometric mixture depends on MASS. You can measure air flow in volume and work back to mass if you wish but the air fuel ratio is very definitely dependant upon the ratio in mass. As density of air varies significantly with changes in pressure and temperature it is not possible to define the mixture by reference to volume alone. At the top of Pike's Peak for instance, a cubic feet of air contains only 2/3 as much oxygen by mass as it would at Daytona, but it's still a cubic foot of air. Beck daross (talk) 11:35, 10 February 2012 (UTC)

Stoichiometric Ratio
Someone might want to look into stoichiometric AF ratio a little more. I understand it to be 14.6 and not 14.7, as per Heywood

marine engine air fuel ratio on large container vessels - probable design faults attributable to stochiometric air requirement caused extensive damages like scuffiing, sudden severe wear and assosciated commercial losses to due to its effect on ship's schedule is becoming more and more common.

According to "Automotive Technology, A systems approach. 6th edition" by Jack Erjavec and Rob Thompson "the ideal air-fuel or stoichiometric ratio for a gasoline engine is approximately 14.7 pounds of air mixed with 1 pound of gas. This provides a ratio of 14.7:1. Different fuels have different stoichiometric ratios. Because air is so much lighter than gasoline, it takes nearly 10,000 gallons of air mixed with 1 gallon of gasoline to achieve this air-fuel ratio. Lean ratios of 15 to 16:1 provide the best fuel economy. Rich mixtures have a ratio below 14.7:1 and provide more power from the engine but greater fuel consumption." P.808 Chapter 28 Gasoline, Diesel, and other fuels. Djcookiii (talk) 20:44, 10 May 2016 (UTC)

[Untitled]
Indeed, I came to this subject expecting to see the 14.7 number proven as stoichiometric for some example, perhaps octane. After hours of calculations and searching I have become convinced that a 14.7 AFR for octane does indeed not produce stoichiometricity. Having a look at the aricles provided by the SAE I am suspicious that this number may have been derived using instrumentation to make measurements of the ehxaust gas and intake values of standard engine and fuel combination.

Having the calculations available in this section would seem to me to be critical to the understanding.

Bill S

Proposed Merger
It’s been added to Proposed mergers to gain a little more attention and help it merge faster. AeroKnight 18:18, 24 September 2006 (UTC)

[Untitled]
"air/fuel mixture is approximately 14.7 times the mass of air to fuel"

Sentence is ambiguouly structured. After reading this part of the sentence it is not clear to me what that means. The mixture has more mass than air to fuel? Is it 14.7 : 1 or 1 : 14.7?

Stoich Redirect?
How prevelant is the "stoich" abbreviation for Air-fuel ratio? I've always known "stoich" as a shortened versio of stoichiometry. Google seems to agree, if you search for "stoich", it brings up "See results for: stoichiometry". If no one has any reasonable objections, I will change the redirect in a few days. Brien Clark 07:41, 3 January 2007 (UTC)

Equivalence Ratio - Atomic mass of O = 8?
In the calculation of the equivalence ratio, we ca read that the mass of O2 is 2 times 8... BUT the atomic mass of O is 16g/mol so it would be 2*16 instead of 2*8. Morever, it is repeated several times while for the other masses (H and C) it's the good value.

Thanks for explaining me why it is not 16 and if it was a mistake, would you correct it?

Regards, Yannick


 * You are correct. It should be 16 and not 8. I will fix it. Myth ( Talk) 18:33, 26 January 2007

Please Comment on this
Why does a lean mixture run hot?? If fuel is added to improve cooling, the release of unburned hydrocarbons (soot) into the air would be very large. It would overwhelm the catalytic converter.

I do not buy the argument that fuel vaporization contributes to engine cooling, if there is not a corresponding increase in hydrocarbon emission.

This is a common question, and one to which I have not yet found a satisfactory answer. Could you add a few sentences on this subject to your entry?

Thanks,

Dave 70.231.147.222 09:26, 27 April 2007 (UTC)

You can not find a satisfactory answer because that statement is false. The hottest combustion temperatures occur at approximately the stoichiometric ratio, and drop off as you go rich OR lean from there.

http://www.n66ap.alexap.com/Mixture_article.htm

Vessbot (talk) 09:37, 27 September 2008 (UTC)


 * Mixtures a little lean of stoichiometric cause the engine to run hotter than stoichiometric. (Very lean mixtures run cool, eventually running so cool that combustion is no longer supported and the engine stops.)  It is my understanding that lean mixtures burn more slowly than richer mixtures so the heat is generated later in the power stroke, causing the engine to run hotter even though power output is less than at stoichiometric.  Very lean mixtures burn so much slower that eventually the fuel-air mixture is not yet fully combusted when the exhaust valve opens, possibly leading to back-firing.


 * The reason rich mixtures burn cooler than stoichiometric is explained in the following sentences in Air-fuel ratio: Rich mixtures produce cooler combustion gases than does a stoichiometric mixture, primarily due to the excessive amount of carbon which oxidises to form carbon monoxide, rather than carbon dioxide. The chemical reaction oxidizing carbon to form carbon monoxide releases significantly less heat than the similar reaction to form carbon dioxide. (Carbon monoxide retains significant potential chemical energy. It is itself a fuel whereas carbon dioxide is not.)  Dolphin51 (talk) 12:26, 28 September 2008 (UTC)

non-rocket scientist, D in chemistry viewpoint
198.176.208.75 (talk) 17:49, 7 October 2008 (UTC) I would like the article to explain practical internal combustion engine air-fuel ratio better in terms of flame temperature. In blasting class for mining engineering, evaluating explosive mixtures for maximum flame temperature at "oxygen balance" (=ideal or stoichiometric conditions, I guess?) in order to impart the most energy to the rock being blasted was the goal. I know that in gasoline engines, going from a rich air-fuel mixture toward oxygen balance, by leaning, increases engine temperature and reduces CO and unburned hydrocarbons in the exhaust (obviously there are many other factors). I am also pretty sure that real internal combustion engines cannot run for very long at oxygen balance before they overheat and fail, hence part of the need for all the secondary means of reducing CO and other pollutants. Conversely, diesel engines run hotter by increasing the amount fuel or decreasing air-fuel ratio, in fact leaning a diesel engine results in cooler temperatures until it stops running. Aslo, diesel engines can be used underground without killing people because they produce almost no CO, compared to gasoline engines. Does all this mean that gasoline engines generally run on the fuel rich side of oxygen balance, and diesel engines generally run on the fuel poor side of oxygen balance - when they are running correctly anyway? Please expand the article in this practical direction or re-educate me as necessary here, if possible. Thanks.

Proposed changes
I plan of making some changes to the article. Before I do that I would like to know your suggestions on them.

Equivalence ratio section
Update the equivalence ratio section by making the oxidizer to be more specific, in this case air. The article should be talking about air and fuel mixtures, but equivalence ratio is a more broader concept, since it does not restrict the oxidizer definition to air or oxygen. It would be better to have a separate article for it. -- Myth (Talk) 20:01, 4 July 2007 (UTC)

Fuels
Most of the article assumes that the fuel is a hydrocarbon. While this maybe true for most applications, I believe the definition does not have anything to do with it. It would be better if the article also reflects this where ever possible be more general in the definition of the fuel. -- Myth (Talk) 20:01, 4 July 2007 (UTC)

Other terms used section
Should AFR be again mentioned here? -- Myth (Talk) 20:01, 4 July 2007 (UTC)

Effect on temperature for rich/lean mixtures
Someone recently added the explanation as to why rich / lean mixtures have lower temperature than stoichiometric mixtures. While the explanation is correct, it does not belong here. A more appropriate section for such explanation will be the adiabatic flame temperature article. As such this explanation should be removed from this article and moved to the flame temperature article. -- Myth (Talk) 20:01, 4 July 2007 (UTC)

I disagree strongly that the information as to why rich and lean mixtures have lower combustion temperatures does not belong on this page. On 27 April, on this page, Dave wrote "Please comment on this." He asked how rich mixtures can run cool without depositing excessive soot throughout the exhaust system. He said "This is a common question." In the 2 months since then no-one has posted any information in response to Dave's questions. Clearly this is a common question, but one that doesn't have a common answer. Any discussion about stoichiometric mixture must be accompanied by some information about why this mixture burns hotter and produces more power than mixture strengths either side of stoichiometric. Dolphin51 12:56, 6 July 2007 (UTC)


 * First of all, rich mixture can achieve higher temperature than stoichiometric mixture (This is because CO2 has higher specific heat capacity than CO). So even though less energy is released in case of rich mixture, it is easier to raise temperature of the mixture. Furthermore, this largely depends on the fuel we are using. Here we assumed it is a hydrocarbon.
 * Secondly, a "common question" is a relative term. The reason why I proposed to move it the adiabatic flame temperature article is because it more appropriate for discussing temperature changes and would be easier to explain it in detail there.
 * Just because "Dave" asked a question here and no one answered it for two months, does not make it necessary to include that explanation here. It will just make the article too cluttered with bits and pieces of information.
 * btw I am also not totally convinced that we should move the explanation, but I would like to discuss it further, before I make up my mind. Thanks. -- Myth (Talk) 18:13, 6 July 2007 (UTC)

Hi Myth. Thanks for your prompt response. My view is that, seeing the article has such a strong focus on the concept of stoichiometric mixture the article should present the complete picture. Stoichiometric mixture is more than just an academic concept in chemistry. It has an intensely practical application in that carburettors and fuel injection systems are designed to deliver charge that is close to the stoichiometric mixture, rather than very rich or very lean. Until recently this article revealed only that fuel injection systems aim to achieve stoichiometric mixture; and that stoichiometric mixtures burn very hot (presumably hotter than other mixtures.) These pieces of information beg the question "Why?" Good question. A reader thinking intuitively might imagine that the highest temperature (and the highest power) would be achieved with maximum rich mixture. You and I know that isn't so. My experience indicates that this really is a paradox. I have read a statement by a very competent practitioner who wrote that with rich mixtures, the exhaust gas contains unburnt fuel!

My view is that in any document where stoichiometric mixture is mentioned the explanation must carry on far enough to give the reader an understanding of the two major processes that cause combustion temperature and power output to vary with mixture strength, and to display a maximum that is neither fully rich nor fully lean - those two major processes are dilution by excess air, and the contribution made by carbon monoxide. You say that it is easier to explain the process in detail in the context of adiabatic flame temperature. The reader who is new to the concept of fuel-air ratio is not ready to understand the process to the level of detail that you are able to provide. He or she is looking for some information of the same level of difficulty as "air-fuel ratio", and that isn't rocket science. I agree the article is cluttered at present. Let's get rid of the extraneous information about n-heptane, iso-octane, alkanes, detergents, oxygenators, MTBE and ethanol. These things are not essential to an introductory article about air-fuel ratio (certainly not in the 3rd paragraph.) And on the matter of fuels, let's assume we are talking exclusively about hydrocarbons. So far the article is focussing on internal combustion engines and I'm not aware of any such engine that burns something other than hydrocarbons. Best regards Dolphin51 12:23, 7 July 2007 (UTC)


 * Well then it would be better to move the discussion about temperature to a separate section in the article. We can also include small details like why lean mixtures are preferred in IC engines especially because of better performance and lower NOx emissions.
 * Also it would be better not exclusively talk about hydrocarbons, reason being fuels like hydrogen are likely to be used extensively in future. Also the temperature variation with air-fuel ratio is valid for most of the fuels, so it is not necessary to restrict the article to hydrocarbons. -- Myth (Talk) 22:34, 7 July 2007 (UTC)

AFR for solid fuels
I suggest adding info or a link to info about the AFR for a typical grade of coal, or perhaps if there are significant differences, various grades of coal, coke, etc. GilesW (talk) 13:41, 10 March 2008 (UTC)

stock ECU information incorrect
This statement "Vehicle manufacturers do not provide a means of altering predefined fuel maps making such alterations impossible without replacing the stock ECU with a customizable system." is incorrect. I'm just deleting it for now.

Vehicles at least back into the 80's (Ford and GM included) have ECU's that can be remapped with a chip. Many more modern vehicles can be flashed through the OBD-II port or by other means. --167.102.224.45 (talk) 13:08, 18 March 2008 (UTC)

AFR measurments
AFR is measured in volume not weight. For every cubic feet of gasoline that enters into the combustion chamber as vapor or mist (as near as possible)approxamatly 14.7 cubic feet of air is needed to obtain a Stiochimetric mixture. This is measured. in CFM (cubic feet per minute). —Preceding unsigned comment added by 134.134.136.36 (talk) 18:03, 5 August 2008 (UTC)


 * This statement suggests an automobile gasoline tank of 100 liters (26 US gallons or 3.53 cubic feet) would consume only 51.9 cubic feet of air during combustion! If the automobile achieves 10 liters per 100 kilometres, it could travel 1000 km on its 100 liter tank of gas, and in traveling 1000 km it would require only 51.9 cubic feet of air.  This could not possibly be correct!


 * Chemical equations are always based on mass rather than volume, because volume changes so significantly when pressure and temperature change. The correct statement is that the air-fuel ratio (eg 14.7) is a ratio of the masses of the air and the fuel, not a ratio of their volumes.  Dolphin51 (talk) 02:38, 27 January 2009 (UTC)

Lambda?
The article describes a parameter called lambda, after the Greek letter $$\lambda$$. I don't doubt that there is a parameter that matches the description given here, and I'm sure the parameter is abbreviated $$\lambda$$, but what is the name of the parameter? Lambda is simply the Greek letter used as an abbreviation. Citing a source for this information would be a good move. Dolphin51 (talk) 11:59, 18 January 2009 (UTC)


 * I now believe lambda, $$\lambda\;$$, is the stoichiometric ratio. This is not to be confused with the stoichiometric air-fuel ratio (or chemically-correct air-fuel ratio) which is about 14.7:1 for gasoline.  When an air-fuel mixture is chemically correct it is said to be at a stoichiometric ratio, $$\lambda\;$$, of unity (1.0).  Unfortunately, I don't yet have a citation for this.  Dolphin51 (talk) 01:24, 27 March 2009 (UTC)

Air Press v Stoich
Please explain why air pressure at sea level and stoichiometric are the same value? —Preceding unsigned comment added by 71.102.8.187 (talk) 17:04, 23 March 2009 (UTC)


 * Purely coincidence! Standard air pressure at sea level is 14.696 pounds per square inch, commonly truncated to 14.7 pounds per square inch.  The air-fuel ratio of a stoichiometric mixture of air and gasoline is about 14.7.  This is often called a fuel-air ratio of 0.067.


 * If air pressure is given SI units of 1013 hPa, and the stoichiometric mixture of gasoline air is described as having a fuel-air ratio of 0.067 there is no similarity. Dolphin51 (talk) 21:43, 23 March 2009 (UTC)
 * I presume the stoich AFR 14.7:1 was based on certain "classical" condition:
 * 1. Fuel RON of at least 90
 * 2. The ratio was obtained at NTP (not STP condition).
 * 3. The engine is normally aspirated that it does not observe significant temperature increase/less denser as in pressure induced engine (super/turbo charged), on moving vehicle, or static engine with far enough air intake source.
 * 4. On pressure charged engine the must keep the AFR stochio, as power increased, more fuel is burned too, unlike diesel cycle which does not mix air and fuel in compression stroke.
 * 5. To increase the air density, the temperature of compressed air must be run through intercooler after turbine before entering AFR chamber.
 * 6. Some high performance rally car use water injection to obtain fuel ron of 100+ to avoid detonation in extreme ambient termperature & extreme charging boost rate.
 * 7. Modern ECU injection car has oxygen, temp, & other sensors as parameter when dynamically adjusting the combustion telemetry (ignition, variable valve, fuel, throttle body, etc).
 * I am not an automative engineer, just an automative enthusiast from 1976 carb to some of modern engine, excluding EV/motor. Supeskrim (talk) 18:34, 20 August 2023 (UTC)

My calculator says 0.068 regardless of the number of digits, whether I set it to 3 or floating = 3 or 11 digits. And whether I use 14.7 or 14.696 as a divisor. — Preceding unsigned comment added by 96.58.168.144 (talk) 00:46, 12 December 2013 (UTC)

Definition of Stoichiometric Wrong in Article
In the article, it says:

"For gasoline fuel, the stoichiometric air–fuel mixture is approximately 14.7; i.e. for every one molecule of fuel, 14.7 molecules of O2 are required"

This is incorrect as the ratio is weight of air to to fuel, not the mole ratio.

The article then goes on to correctly discuss the combustion of octane with 12.5 moles of O2.

If gasoline were composed of octane, then the stoichiometric ratio would be about 15.1 based on mass of air consumed per unit mass of octane. (Air is about 21% oxygen by volume/mole fraction, 78% nitrogen (N2), 0.93% argon, and 0.039%CO2)

It would require a fuel of composition of about C8H15.3 to have an ideal 14.7 ratio. Apparently the ideal ratio is based upon empirical testing, since gasoline is a complex mixture of many hydrocarbons.

Lambda (the Greek letter) is used in the automotive industry to represent the ratio of the actual air/fuel ratio to the stoichiometric air/fuel ratio.

Article needs to be corrected, but not sure how to handle the octane combustion example with the 14.7 "standard" stoichiometric ratio, and I can find no references.

`Wwwaller (talk) —Preceding undated comment added 21:27, 7 July 2012 (UTC)
 * I agree. I have changed "molecule" to "gram". I also changed "oxygen" to "air" because air-fuel ratio is all about the mass of air, not of oxygen. Dolphin  ( t ) 07:55, 8 July 2012 (UTC)

Criticism
I am a Professor of Engineering. This entire page is a bunch of mumbo jumbo. Due to the improper use of Greek symbols, meaningless formulas and ratios, as a Professor of Engineering I'd give you a failing grade. I could start at the top of this page and find fault after fault but the more I read it, the more stupidity I find in it. It's not worth any more of my time. - — Preceding unsigned comment added by Std1234 (talk • contribs) 02:09, 26 August 2012‎ (Transferred from article - see the (diff)

Can't agree with this one
"Because the composition of common fuels varies seasonally, and because many modern vehicles can handle different fuels, when tuning, it makes more sense to talk about lambda values rather than AFR." This statement contradicts itself.

Because no one actualy knows the ACTUAL stoich for most fules i would have to dissagree with the above. Since lambda is calculated from AFR using a constant that must be accuratly known for the fule. Slightly different blends of gas will not have the same stoich ratio (summer/winter)and therefore the lambda value will be in-accurate. AFR is directly measured and the optimum value can be found for that fule. — Preceding unsigned comment added by 65.51.205.194 (talk) 12:59, 5 September 2012 (UTC)

Combustion stoichiometry
Perhaps I can clear up some of the confusion here about the relationship to the highest possible adiabatic flame temperature and the air/fuel ratio. First, it is common to calculate an adiabatic flame temperature based on the combustion products reaching the equilibrium position. However, in an internal combustion engine, or even in an industrial boiler, the time to reach equilibrium is longer than the residence time of the flame in the combustion chamber. Also, there is some heat loss to the surroundings as the fuel-air mixture is burning, so the situation is not truly adiabatic. So given a short residence time, the maximum attainable combustion temperature depends on the reaction mechanism and the kinetics. In some cases, the maximum attainable combustion temperature could be on the slightly excess air side of perfect stoichiometry, while in other conditions or situations, Tmax could be reached with a slight excess of fuel. Second, for combustion temperatures above about 1800 K, and near stoichiometric, there are small (but significant) amounts of minor species present at the equilibrium position, such as O, OH, N, H, and of course NO if there is a slight excess of air. These species have a heat of formation and contain both O and fuel elements, and so they affect the heat of combustion. Third, the system pressure affects the equilibrium position of the combustion products. So combustion of a fuel at a specified air/fuel ratio in a constant volume device will yield a different temperature than in a constant pressure device, and the composition of the combustion gas will also be different, even for the same air/fuel ratio. For these and other reasons, the calculated adiabatic flame temperature obtained by assumption of equilibrium conditions will not be met in practice. The AFT, or some version of it, must be obtained by appropriate measurements at the conditions of fuel consumption.

Finally, the article is strongly oriented towards the internal combustion engine, with some added reference to industrial furnace combustion. I suggest either calling the article "Air-Fuel ratio for internal combustion engines"; if so, why not move this article to the article on the internal combustion engine. Or, if this is to be a general article on air-fuel ratio for any kind of combustion process, make it more general, and include coverage of industrial furnaces, where the air/fuel ratio is controlled for a different purpose than for the internal combustion engine. Thermbal (talk) 05:07, 30 June 2013 (UTC)

Recent edits have made the article more general to non-internal-combustion engine applications. Thermbal (talk) 21:32, 7 July 2013 (UTC)

Question regarding this Text vs. Air-Fuel Gauges.
This appeared on the article page. I'm shifting it here instead.  Stepho  talk 04:09, 12 December 2013 (UTC)

HELP - I am not an engineer. Just a 74 year old jack-of-many-trades building his first street rod, {a 55 Chevy, 2nd series, 3100 Pickup, which is naturally aspirated with 3-2 barrel carbs on progressive linkage.}.

Air-Fuel Gauges and Ratios have been driving me batty. I understand as the ratio number gets smaller, less than 14.7 (or 15) the mixture is Richer. I understand this is because the number denotes the amount of AIR in the mixture and considers (for calculation and/or discussion purposes) Fuel to be a constant. Actually, in the newer engines, the computer changes the amount of fuel to "correct" the ratio. BUT, mine has NO computer, no engine management system, just me to tweak the carburteors.

So, regarding the GAUGES, IF the smaller number is a richer (less air) mixture and a larger number (more air) is a leaner mixture (as denoted above), why do the Analog and Digital Air-Fuel gauges show the larger (Red) numbers for a Rich running (mixture) engine and smaller (Amber) numbers for a Lean running (mixture) engine? And why is the Stoichiometric are on these meters (Green) so wide (covers several numbers)? Most easily seen on the Auto Meter website = http://www.autometer.com/cat_gaugeseries.aspx - then go their their ADVANCE SEARCH (top left) and scroll way down to choose the Digital Wide Band Air Fuel gauges in 2.0625" size..

Further, if running TOO Lean is potentially engine damaging, why is Lean shown as Amber instead of Red and Rich is Red vice Amber (too rich can also be bad, but not as bad - ?). Or am I more confused than I imagine? In lay language please, just as the other fella asked. User talk:Surfmurf

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Other types of engines
I don't think this paragraph makes any sense. Could someone look at it, please?


 * ''In the typical air to natural gas combustion burner, a double cross limit strategy is employed to ensure ratio control. (This method was used in World War II).[citation needed] The strategy involves adding the opposite flow feedback into the limiting control of the respective gas (air or fuel). This assures ratio control within an acceptable margin.

Longinus876 (talk) 12:56, 4 November 2019 (UTC)

Air-to-fuel ratio