Rogan

Turbochargers while simple in design, can get very complex in theory. From deciding what compressor trim is desired, to what turbine housing to be used is confusing to most who enter the field of forced induction. This article will hopefully take away all the confusion in turbocharger selection. Terminology Before I can begin writing an article explaining turbochargers, terminology must be learned. Here is a list of the more commonly used turbocharger terms: Compressor - Essentially a fan that spins and compresses air within an enclosed area (compressor housing). In order to allow the air to compress and build pressure within the housing, the fan must be spun at certain rpm levels. Compressor Housing - Housing that encloses the compressor. Pictured is a compressor housing. Compressor Map – A map that allows the ability to plot compressor pressure ratio vs. engine airflow. An “island” shape is created on the plot showing where the compressor is efficient. Compressor Efficiency – Compressors efficiency is the ability to produce lowest possible temperature from the compressor air. When air is compressed heat is generated, at certain range of speeds of the compressor rotation the heat can be keep to a minimum. This is what is known as “being in the efficiency range of a compressor” Higher efficiency, lower outlet temperatures. Highest possible efficiency of compressors are 78~82%. Lower outlet temperature=lower intake air temperature. Lower intake air temperature=more dense mixture of air=more oxygen available in the combustion to burn. The greater amount of oxygen present with fuel provides more energy. More energy=more heat=more torque=more power. Compressor Trim – The trim of the compressor refers to the squared ratio of the smaller diameter divided by the larger diameter multiplied by 100 of the compressor wheel. The smaller diameter of the wheel is known as the inducer, and the larger diameter of the wheel is known as the exducer. Compressor Families – Beyond compressor trim levels there is compressor family of wheels. In the Garret turbo line of older technology compressors there is T22, T25, T3, T350, T04b, T04e, T04s and T04r families. In each family there is trim levels to the family. Although there is a 60 trim in both the t3 and to4e family wheels, the main difference is the inducer diameter of the wheels. The trim is only a ratio of the exducer/inducer, so while the inducer size of the compressor wheels are vastly different, the ratio between the exducer/inducer remains constant since it’s the comparison between the exducer to inducer size. Turbine – A fan that uses exhaust energy to rotate. The rotation of the turbine is transmitted through a shaft that is connected to the compressor. Faster the turbine spins, faster the compressor spins. Compressor uses the rpm translation through the turbine/compressor-connecting shaft to compress air at the rpm level that dictates compressor flow. Turbine Housing – Housing that encloses the turbine wheel. Turbine housing size affects the ability of the turbine to transmit rpm to the compressor wheel. Smaller turbine housing, quicker spool up due to quicker translation of rpm’s to compressor. Trade-off is increased low-end response for less high-end response from turbocharger. Picture below is a turbine housing. Turbine Trim – The trim of the turbine refers to the squared ratio of the smaller diameter divided by the larger diameter multiplied by 100 of the compressor wheel. The smaller diameter of the wheel is known as the inducer, and the larger diameter of the wheel is known as the exducer. Turbine Families – As with compressor families, there is turbine families. The most common evidence of the turbine families is the t31, t350 and t04 wheel used in the t3, t3/t4 turbos sold on the market. Precision offers the t31, aka stage 3 blade in their smaller line of sport compact series turbochargers. The t31 comes in two different trim levels the 69 and 76 trim. The t350, aka stage 5 blade comes in two different trim levels as well, 69 and 76 trim. The t31 will spool faster than the t350 due to the physical size differences (t31 being smaller). The smaller the trim level the quicker spool, but less top end. Essentially you are changing the turbine pressure ratio when you are selecting the family and trim level of the turbine wheel you are using. The larger family and trim level you choose the more power the turbocharger will produce at the expense of lag. As with the compressor trim levels both the t31 and t350 have the 69 and 76 trim levels, which are not the same. The turbine trim is the ratio of the exducer compared to the inducer size of the turbine wheel, since its only a ratio the size of the inducer/exducers are completely different. Turbine Map – A map that allows the ability to plot turbine expansion ratio vs. engine airflow. An “island” shape is created on the plot showing where the turbine is efficient. A/R – Ratio of the area of the compressor/turbine housing to the radius of the compressor/turbine wheel. In order to find out the A/R of the compressor or turbine housing select a point where the compressor/turbine housing begins and measure the cross-sectional area at that point. Cross sectional area is A=P*(Radius)2. Next step is to measure the distance between the center of the area and the center of the compressor/turbine wheel, this is the radius measurement. If you choose a different point on the compressor/turbine housing and remeasure the area and radius, you’ll find that it stays constant. This is due to the housing getting constantly smaller in diameter as it gets closer and closer to the compressor/turbine wheel. When you upgrade from a .48 to a .63, or .63 to a .82 A/R you are essentially increasing the area of the housing. Increasing the area increases the amount airflow to the turbine wheel. The smaller area of the smaller turbine housing builds pressure quickly and transmits this pressure to the turbine. The pressure gives the turbine enough rpm’s to allow the compressor to compress air at lower engine speeds (less engine speed, less airflow from engine). The trade off is that pressure builds up quickly in the housing to obtain quick spool up, but the pressure quickly becomes to great and backpressure builds up. The backpressure is the restriction that limits shaft speed of the compressor, and as the rpm increase (engine airflow increases) the torque curve begins to drop off due to the volumetric efficiency of the engine decreasing. Think of the turbine housing sizing as increasing/decreasing inlet pressure to the housing in order to gain low end, midrange or top-end response from the turbocharger. The smaller turbine housing wont carry the torque curve to a high rpm, limiting the amount of peak whp. Excellent low-end and midrange gains are felt through smaller housings. Compressor/Turbine Mismatch – When “matching” a compressor and a turbine you are seeking to balance the turbine characteristics to the compressor characteristics. When you increase the size of the turbine wheel you are decreasing the pressure ratio of the turbine, essentially decreasing the shaft speed connecting the compressor/turbine. When pairing a larger turbine wheel to a small compressor wheel, the smaller the compressor wheel the higher the rpm the wheel has to be spun at to compress the air. This becomes a problem in that the smaller compressor cannot generate adequate shaft speed to compress air. The same can hold true when pairing a huge compressor to a small turbine wheel. The larger compressor needs less shaft speed to compressor airflow, but the smaller turbine wheel will spin at a much higher rpm level that is what is necessary. The result is crossing over the choke or surge line on the compressor map (this will be explained in part 2 of this article). Note the two different compressor maps, one of a 60 trim t3 compressor wheel, the other a t64 compressor wheel. Part 2 will explain compressor and turbine maps, and all the terminology that goes along with topics.

Everyone seems to grasp that a larger exhaust on a turbocharger vehicle will gain faster spool up, increased power and thus faster acceleration. Here is the theory behind the phenomenon: The turbine housing is merely a volume for which exhaust gas (energy) is transmitted from the engine, to the turbine blade, then dispelled into the atmosphere. In order to better understand how this works we need to take a look at pressure. Turbines in general work off of a pressure differential. A pressure differential in layman's terms is the ratio of pressure before the turbine blade, and after the turbine blade. The greater the pre-turbine pressure compared to the post-turbine pressure, the greater the amount of work can be transmitted through the turbine/compressor shaft. This is where some engineering comes into play. Work is defined as the integration of force and displacement, keeping force constant. Again in layman's terms this merely means work is the force exerted on an object while taking into account the change in displacement, or position of the object. In relation to turbines, the greater the pressure differential, the greater the amount of work is created. The greater the amount of work created, the greater the amount of energy transmitted through the turbine, into the compressor through the connecting shaft. To break this down into another small explanation, compressors work off of rpm. The amount of air the compressor is able to “flow”, or merely put the lbs/minute the compressor can flow is dictated by the amount compressor blades, angles of the blades, etc. What the compressor uses to “compress” the air through the inlet is based off of how fast the compressor wheel is spun. The rpm at which the compressor blade has to be spun to “compress” the air varies from turbo to turbo from the different compressor blade characteristics. Another critical aspect of the compressor is the physical size and weight of the blade. The larger the blade, more amount of energy must be transmitted to allow the shaft to spin to the rpm at which the compressor can compress the air. In engineering terms rotating mass is called inertia, so smaller turbos have smaller inertia demands, larger compressors have larger inertia demands. To bring the focus back on the turbine side of things, the increased inertia demands more energy to be supplied from the turbine. The greater the pressure differential discussed above, the greater the amount of work that be be supplied to overcome the inertia effects of the compressor. Now looking at the exhaust, or post turbine the larger the exhaust, the larger the pressure differential can become. The increase in area of the exhaust, gives the exhaust gas much more room to expand. Hot gas has only one goal, to expand as quickly as possible. The goal pre-turbine is to focus the energy into the turbine to carry as much energy as possible. As the exhaust expands the energy dissipates, so the goal post turbine (i.e exhaust) is to have the largest area possible for the gas to expand. Looking at the immediate exit of the turbine housing, the downpipe, the exhaust gas is traveling at a very high rate of speed. The exhaust gas is expanding rapidly, and is in a very turbulence state from being flung from the turbine. At this point having a 3″ down-pipe becomes critical since the exhaust gas is both in a turbulent state, and is expanding. In turbulence, the expansion of an area the turbulence is forced to become more laminar (although this doesn't happen very quickly). Also the increased area allows the gas to expand rapidly, allowing the energy in the exhaust gas to dissipate quickly and letting the pressure created by the exit from the turbine housing to drop. Essentially you are creating a greater pressure drop. Looking at the turbine housing the sizing becomes a critical part in how the pressure differential is created. Take for instance the .48 A/R housing. Changing turbine A/R has many effects. By going to a larger turbine A/R, the turbo comes up on boost at a higher engine speed, the flow capacity of the turbine is increased and less flow is wastegated, there is less engine back pressure, and engine volumetric efficiency is increased resulting in more overall power. The .48 A/R is able to create the pressure differential at a much lower engine rpm, giving the compressor ability to make its maximum rpm speed sooner. As the engine rpm climbs, the pressure differential is lowered due to the physical volume of the housing size becoming a restriction on the post turbine side. As the housing size is increased, it take greater engine rpm speed (greater exhaust energy) to spool up the turbine, but the pressure differential is less effected by the physical volume of the housing. If you are after maximum midrange gains smaller housings are essentially, if top end gains are essential larger housing are essential. Selecting the power band of the engine is essentially dictated through the housing size, and the turbine physical characteristics.

For the last century, horsepower has been used to describe the power output of the internal combustion engine. The horsepower unit was created by James Watt in the 18th century. Its origin is based from how much power a horse could lift in foot pounds, 33,000 ft-lbs to be exact in one minute. The unit is derived from torque, which is the true measurement of the engine physical power production. What is strange about the units of horsepower is that it has no physical meaning. Its an arbitrary unit that has no real significance in describing the characteristic of the engine. For those that are curious to calculate horsepower: horsepower = (rpm/5252) * torque From this equation you can see that horsepower is nothing more than a contrived unit that is based purely from torque and rpm. You’ll notice the number 5252 in the equation, this represents the point at which every dyno graph must intersect horsepower and torque. Its a mathematical relationship, both strange and interesting since horsepower is a function of torque and rpm. There has been much confusion and rumors across the internet about gaining more horsepower. In essence, gaining more horsepower is gaining torque. If you are after “peak” horsepower, you are interesting in carrying the torque curve as high in the rpm range without falling as possible. You can see from the equation that as the rpm’s increase, and the torque remains the same you get a higher horsepower number. What physically is happening is that the engine is able to produce enough torque to overcome frictional forces through the air, tires, etc. As you are able to keep the torque from falling off on the top end, you are able to maintain a steady torque curve that will “pull” the car through the mph you are trying to reach. So people who are after “peak” horsepower really want to extend their torque curves as far towards redline as possible, without letting the torque fall off. Check out some dyno graphs and see what I mean. Horsepower doesn't describe the true nature of how the engine performs, its the torque curve. From a tuners perspective, I don't tune off of the horsepower curves. The physical relevance towards the engine performance is arbitrary, since the torque is truly what is effected by the fuel, timing, breathing, etc of the engine. The horsepower is merely a concocted unit of measure, showing no true characteristics of the engine power output. A good tuner will only make changes from the torque curves, see what increase/decrease the curves show from the changes. So next time you are thinking horsepower, think “what would I want my torque curve to be”?

What is an A/R ratio and how is it calculated?: The A/R in a relationship that is obtained when dividing the interior area of the turbine where the inner walls are found, through the turbine housing radio from the center to the tongue as the illustration indicates. A/R values are expressed as .35, .47, .68, .84, 1.00, 1.15, etc. A small A/R indicates a small interior volume in the small turbine and a large A/R indicates a greater volume. At a minimum A/R the motor's response is produced at small revolutions per minute but at high revolutions we will not achieve an adequate caudal. We should always find a compromise between achieving the lowest response possible and have enough caudal at high revolutions. The picture below is for reference: As stated, the turbine housing A/R is the cross sectional area of the turbine housing divided by the distance from the center of that cross section to the center of the wheel. This makes sense if you look at the graphic. If you take for example, the area in A1 and divide it by R1, you will have found the A/R for this turbine housing. Each cross section and radius have the same proportions so the A/R will be found by using any cross section/radius. What is the Trim of a turbo and how is it calculated? Each turbine wheel y compressor wheel model generally have the same turbine diameter (highest diameter), but different steps (lowest diameter). Each type of step (trim), has different blowing characteristics. [*]TRIM values are expressed as 45, 50, 55, etc... and can only go from 0 to 100. A value of 100 means Dp = Dg [*]A large TRIM indicates a large turbine diameter. [*]A TRIM of 55, gives 10% more caudal than a TRIM of 50. [*]TRIM is used in the same way for turbine wheels as for compressor wheels. [*]TRIM is calculated through the following formula. TRIM = ( Dp / Dg )² x 100 Si Dg = 50 mm y Dp = 35 mm TRIM = ( 35/50 )² x 100 = 49 What are the different flanges and what are the sizes? [*]All most all of your turbo head units come with the flanges described below. The T3 housing is the smallest and flows the least, with the T6/Thumper flange being the biggest and flowing the most. The flange plays a role in spool up, backpressure...etc. The rule of thumb here is use the largest flange you can possibly fit. Of course this will be limited by what headers you use, since most are pre-fabbed and come with a flange already, and under hood space will also be a limitation. [*]Basic T3 [*]Basic T4 [*]Basic T6

The author of this article is a turbo engineer for Garrett. I'm submitting it because it is an informative read and therefore is beneficial to the community. Enjoy. Compressor Efficiency and More by Khiem Dinh Compressor efficiency is a term thrown around whenever people mention forced induction, but what does it really mean? How does it affect an engine's performance? And what role does air temperature play in all of this? The cool thing about thermodynamics is that we can explain effects with equations. Using basic compressor equations, we'll put some numbers to the affects of air temperature and compressor efficiency on compressor power requirements and air temperature increase. Some fundamental compressor equations are below. I like working in SI units because doing calculations with English units sucks! So, mass flow rate is in kg/sec, the average constant specific heat value for air I used is 1.007 kJ/kg*k, and k for air is 1.4. The specific heat of air actually varies with temperature, but the change is basically nothing within the range of temperatures we're using, so I'm assuming a constant value. After calculating all the values in SI units, it is a simple conversion to English units. The table and chart I generated assumes: air inlet temperature of 298K/24.85C/76.7F, and 100% compressor efficiency (isentropic compression). Looking at the chart and table above, it becomes very obvious that increases in mass flow rate and pressure ratio require more power. Also, the graph shows lines of constant power. For a given compressor power, you can get a lot of flow and little pressure ratio, a lot of pressure ratio and little flow, or somewhere in the middle. To get a feel for what the numbers mean in the real world, we'll use 2.0L 4-cylinder engine as an example. Automotive engines of this displacement and cylinder count will make roughly 500hp with a mass flow rate of 50lbs/min and a pressure ratio of 2.75, or about 25psi of boost. Looking at the table, a 100% efficient compressor would require 51.1hp! Looking at the compressor map for a GT3076, it shows a compressor efficiency of ~72% at this point. So the actual power requirement becomes 71hp. A GTX3582 has a compressor efficiency of 77% resulting in a power requirement of 66.4hp. That extra hp required over a 100% efficient compressor ends up as extra heat in the air. More efficient is better! The other variable in the equation that's very important, but many people seem to neglect in the practice of building turbo cars, is the temperature of the air going into the compressor. Many people have the misconception that performance is unaffected because the intercooler will cool the air enough regardless of the air temp going into the turbo. What they are neglecting is the fact that compressor performance improves with cooler air. Said another way, it gets worse with taking in hotter air. Using values of 50lbs/min and a PR of 2.75, the table below shows compressor power required, change in temperature of the air (Delta T), and the exit temperature of the air based on the inlet temperature of the air. Hotter air in equals mo' hotta air out! Takes more power too. Translation? Laggy turbo. Notice that the hotter the air going into the compressor, the more power is required to compress it. Also, the increase in the air temperature is greater. Of course, this results in the final temperature being even hotter. I think comparing air inlet temps of 77F and 122F is reasonable; 77F being ambient air temp and 122F being the temp if you ingest air from the engine bay. So by sucking up the hotter engine bay air instead of cooler air from the front of the car, the compressor power required increases by 4.3hp, or about 8%. The difference in the temperature coming out of the compressor is a toasty 60.1F making the intercooler work that much harder. Of course, there's no such thing as a perfect compressor, so let's see what happens with a 60% efficient compressor; this is often where the tuner industry operates as they try to squeeze as much power as possible out of turbos. Looking again at 77F and 122F inlet temps, the power and temperature differences are now 7.1hp and 70.2F! Crappy compressor efficiency means supa dupa hotta air! You could boil an egg on it. Mo' laggy too. So now we know how crappy compressor efficiency and sucking in hot air increase the power required of a compressor to move and squeeze air. But where does that power come from? A supercharger gets it off the crankshaft of the engine and a turbocharger uses a turbine wheel in the exhaust. If you go by old school nomenclature, what we call a turbocharger was referred to as a turbo supercharger. So basically a special type of supercharger with a turbine wheel to get work out of the exhaust. Anyways, we like turbos because they use otherwise wasted energy. Going back to the very first chart, a perfect compressor doing 100lbs/min at a pressure ratio of 4 would need almost 150hp. Given the option of taking that from the engine crankshaft or the exhaust energy, we'll take the exhaust energy. That's the reason why pretty much every diesel engine and the majority of gasoline engines use turbos instead of superchargers. So does the turbo engine make 150hp more than the supercharged engine? The answer is no because the turbine wheel and turbine housing create back pressure in the engine reducing its volumetric efficiency. But a turbo engine will still make significantly more power than a supercharged engine given the turbo is properly sized. Sizing of the turbine wheel is important so as to get the maximum efficiency from it. Our worst case scenario of the 2.0L engine, 60% efficient compressor, 50lbs/min mass flow rate, 2.75 PR, and 122F air inlet temperature requires 92.4 hp of shaft power from the turbine to drive the compressor wheel. A 100% efficient turbine wheel would need to get 92.4 hp worth of energy out of the exhaust. However, like compressor wheels, there’s no such thing as a 100% efficient turbine wheel. Throw in some moderate turbine efficiency and you end up needing a lot of exhaust power to spin a compressor. Plugging in a value of 60% for turbine efficiency, we can see how much energy needs to be pulled from the exhaust based on the temperature of the air going into the turbo. The difference in power required going from 77F and 122F is a lag inducing 11.9 hp, or about 8%. So what have we learned? Maximizing compressor and turbine efficiencies reduces the exhaust energy required to get a turbo going. Sucking in colder air instead of hot air also makes compressors happy. The final conclusion to all of this is that math is cool, stay in school!

LOL Stroke ain't killed me, yet..

I've been googling and found that A LOT of people have complained about this. Seems no one has found a 'fix' for it. $60+/gal fluid is a common theme, so I tried it.. Absolutely no real change. Maybe a decibel or two in reduction, but nothing all "OMG this is so much better" that such a costly fluid should bring to the table.. The removing of any slack (by pressing the shifter a little) is understandable, as it quietens it to nill, but that's not a fix.. By the way, @Mopar1973Man, you still , as I am?

Thanks, Mike. A lot has changed here, that's for sure... Well, the 06 would've had a NV5600, correct? The G56 is a different creature, but the principles are the same. I've read where 'gear rollover' noise increase is common in the G56, but this is nuts. Everything works fine, shifts fine, engagement is fine.. Just this horrible rollover. And like I said, slightly adding pressure to a shift-gate stops all of it. I know the G56 has brass or bronze fork pads, rather than Delrin or the likes.. I'r afraid I may end up going inside the G56 before too long..

Good morning, all, and hello old friends 2008 2500 with G56 trans. Changed to a SMF and DD clutch. After doing so, I get horrid "gear rollover" in neutral, as well as in gears. In neutral, no clutch depressed, it's quite noisy. If I slightly push the shift lever towards a gear, it immediately goes quiet. Does this at all gear positions. If I let the shifter rest, it's quite noisy again.. Anyone experienced this? I mean, I understand there'll be 'gear rollover' when going from DMFW to SMFW, as the vibration transmitted is increased a little, but this is nuts.. Thanks.

Ultra Low Sulfur Diesel 101 (ULSD Diesel) Emissions Standards Continue to Tighten You may have heard that The U.S. Environmental Protection Agency has adopted a tough set of diesel emissions standards aimed at drastically reducing the sulfur content of diesel fuel to improve air quality. A good thing indeed. To meet those standards, petroleum refiners are producing Ultra Low Sulfur Diesel (ULSD) or S15, a cleaner diesel fuel that has a maximum sulfur content of 15 parts per million (ppm). When the full retail phase-in is complete, this fuel will be a direct replacement for Low Sulfur Diesel or S500 (which has a sulfur content of 500 ppm) culminating in more than a 95% sulfur reduction. In a move similar to eliminating lead from gasoline to protect catalytic converters, sulfur reduction in diesel fuel is necessary to preserve the proper function of the advanced emissions control systems on new diesel engines. The combination of this cleaner burning fuel and the sophisticated emissions devices on the new engines will result in more than a 90% reduction in soot and oxides of nitrogen. Look for the Label Beware 2007 vehicle owners: Not all retail outlets (excluding California) are required to offer ULSD fuel until December 1, 2010. For a period of time, through this transition period, some fueling stations may not have ultra low sulfur diesel (ULSD) available. The pumps must have a label stating that they dispense ULSD, so if the label isn't on a pump, don't put the fuel in your 2007 vehicle—severe damage to the emissions control systems could result. Find a station that has ULSD to fill your 2007 vehicle. Please note, however, that all diesel fuel in California was transitioned to ULSD by September 1, 2006, and no labeling is required. But Can I Use ULSD in My Older Diesel vehicle? All 2007 and later diesel vehicles sold in the U.S. for on highway use will be required to use ULSD. No problem here, these engines are designed to run on this fuel. But what about older engines—engines that were designed to run on Low Sulfur Diesel (S500) or its very early brother, S5000? The argument here certainly isn’t in defense of higher sulfur fuel—its many drawbacks are quite evident. But sulfur in the fuel enhances lubricity, and older engines depend on it to protect their pumps and injectors from premature wear. To combat the loss of this lubrication, packages of additives that increase lubricity will be blended with the fuel prior to distribution. Biodiesel to the Rescue While synthetic additives will initially fill the role of lubricant, there is plenty of room for development of biodiesel blends that can easily handle the task. Because of its vegetable oil base, biodiesel contains no sulfur and has a very high lubricity factor. These two qualities make it an ideal candidate for blending with ULSD to solve the problem of lubricity while keeping the sulfur ratio intact. Stay tuned. Biodiesel started out as a niche fuel touted mainly by environmentalists and the agriculture industry, but it is proving to be quite useful as an additive as well as transitional fuel while cleaner alternatives continue to be developed. For it pure versatility—as a straight fuel, and as a multifaceted additive—biodiesel has certainly earned its stripes. Clearly the advantages are compelling, but as with any transition from one technology to another, there are hurdles to be cleared. Mainly it boils down to logistics. It will take time to develop the blends that perform well under various conditions (cold flow characteristics, adequate lubricity, fuel efficiency) and to get the blends into the infrastructure pipeline. Cold Weather Considerations The refining process used to attain the sulfur ratio of ULSD affects the naturally occurring paraffins (wax) inherent in diesel fuel in such a way that can cause the fuel to gel more readily in cold temperatures. Though most retail fuel should be properly winterized for your local climate, some testing and experimenting with additional treatment processes will probably be in order. Higher doses of anti-gel additives may be necessary for extremely low temperatures—and not all additives that have traditionally been used for treating conventional low sulfur diesel will be compatible with—or effective on—ULSD. Keep the Following Points In Mind: First look for packaging labels that indicate compatibility with ULSD. Secondly, be aware that mixing conventional low sulfur kerosene with ULSD as a flow enhancer will increase the sulfur ratio and could damage emissions control equipment. Finally, even in cases where it is available—cutting ULSD with ultra low sulfur kerosene may be ineffective as a flow enhancer.

Tech: Tires, Gears, MPH and RPM A common 4x4 application is an upgrade to larger-than-stock tires. Once completed, this change immediately alters vehicle speed at a given rpm, rpm at a given speed, and effective gear ratio, which in turn affect both acceleration and fuel economy. Tire size, gear ratio, mph and rpm weave an intricate pattern of performance. Change one and all four are affected.; knowing any of the three, the fourth can be easily determined. The following four formulas illustrate the point: Tire diameter = ((MPH x Gear Ratio) x 336) / RPM Gear ratio = ( rpm x tire diameter) / (Mph x 336) Mph = (rpm x tire diameter) / (Gear ratio x 336) Rpm = ( (mph x gear ratio) x 336) / Tire diameter If you are contemplating a tire size upgrade and know your rear-end gear ratio, you can measure your tire size and observe rpm and mph, you can calculate what gears are in your axles. How To Calculate Actual Speed: With the change to taller tires, your speedometer will real "slower" than the actual vehicle speed. To determine the percentage of speedometer error, the formula is a simple relationship between old and new tire diameters. Actual Speed = (new tire diameter x indicated speed) / Old tire diameter Example: You’ve replaced your 30-inch OEM rubber with a new set of 35-inch all-terrains and you want to know your actual speed when the speedo reads 60 mph: (35 x 60) / 30 = 70 mph Speedometer ratio adjustment calculation Formula used (New Tire Diameter / Old Tire Diameter) * Speedometer MPH = Actual MPH Another way of looking at this relationship would be to figure what the indicated speed would be if you were actually going 60 mph. In this case, the tire diameter relationship is flip-flopped to: Indicated Speed = old tire diameter x actual speed / New tire diameter Using the previous example, your speedometer reading at an actual 60 mph is: 30 x 60 / 35 = 51 mph Gearing up: Using the above tire change as an example, lets say that your vehicle is currently running a 3.40:1 final-drive gear set. Now that you have changed to a taller tire, you want to determine the actual, or effective, final ratio. This can be figured by dividing the old tire diameter by the new, and multiplying by the current gear ratio (:1): 30 x 3.40 / 35 = 2.91:1 Dropping from a 3.40:1 to a 2.72:1 ratio will reduce off-the-line responsiveness and severely affect slow-speed trail capabilities. If your new 35-inch rubber is just what you want, but you now need to restore your vehicle’s low-end, the following formula will allow you to determine what gear set (equivalent) ratio should be installed to compensate: Equivalent ratio = new tire diameter x original ratio Old tire diameter Or, in this example: 35 x 3.40 / 30 = 3:85:1 By installing a gear set in the range of 4.25:1, you will not only restore your vehicle’s low-end responsiveness, you will likewise restore your speedometer’s accuracy. Figuring gear ratio: Knowing what gears are in a given axle is a must when considering that axle for a swap. The actual ratio or reference code, will normally be found on either a tag attached to a bolt, or will be stamped into the axle housing. If it cannot be found, there is a simple method for manually (and mathematically) determining the ratio for any axle installed on a vehicle. Raise both wheels of the axle, with the transmission in Neutral. (Make sure you support the vehicle with safety stands and block the front tires.) Make a reference mark on the driveshaft and on the differential housing. Next, without rotating them, make a mark on both tires and their respective fender wells. With a friend watching the driveshaft, carefully rotate both tires at the same time exactly one revolution. The number of turns the driveshaft makes will indicate the ratio. If the driveshaft rotates 4 ½ turns, for instance, the axle ratio is roughly 4.5:1. METRIC TIRE TO DIAMETER (INCHES) CALCULATION Formula used ((Section Width x Aspect Ratio x 2) / 25.4) + Rim Diameter = Tire Diameter Width in inches = section width / 25.4 Section Height in inches = Width in inches X Aspect Ratio (%) GEAR RATIO CALCULATION Calculating Gear Ratio For two standard round gears, the gear ratio is calculated by counting the number of teeth on each gear and dividing the number of teeth on the driver gear by the number of teeth on the driven gear. For example, a gear with 25 teeth drives a gear with 75 teeth. Dividing 25 by 75 gives you a ratio of 3/1, meaning that for every three rotations the driver gear makes, the larger gear turns once. Formulas used (RPM x Tire diameter) / (Diff Ratio x Trans Ratio x TCase Ratio x 336) = MPH and (Trans Ratio x TCase Ratio x Axle Ratio) = Crawl Ratio Definition of Gearing Ratios In the automotive industry, gear ratio refers to the difference between the input and output speed of a differential or trans-axle. The purpose of the gear ratio is to give the best combination of performance and economy for a given engine size and vehicle weight. Some gear ratios are designed to operate outside of this condition, such as those used by hot rods in drag racing. Function 1. In the simplest terms, a gear ratio is the difference in size between two gears that have the same tooth spacing. One common ratio is the 4.10 gear ratio used in heavy-duty pickup trucks and drag-racing cars. This ratio is considered "low-geared"; for every revolution the larger gear (attached to the axles and wheels) makes, the small gear (attached to the drive shaft) makes 4.10 turns. This gear reduction gives increased power at the expense of top speed and fuel economy. An example of a "high-geared" ratio would be 2.62, where the large gear that drives the wheels turns once every time the drive shaft gear turns 2.62 times. Effects 2. The effect of the gear ratio is to convert rotational energy--in this case coming from the drive shaft--to a desired level of energy at the wheels. The gear ratio does not increase horsepower; it merely changes how it is used. A lower ratio offers greater power while sacrificing top speed, whereas a higher ratio offers greater top speed with less power. Gear ratios also affect the vehicle's fuel economy. A higher gear ratio offers greater fuel economy on the highway, and a lower ratio will be less efficient. This is because the higher a gear ratio is, the fewer revolutions the engine has to make to achieve a desired speed. Since one of the things that affect fuel economy is engine RPM, fewer revolutions per minute use less fuel. History 3. Gear ratios have existed since the first wooden gears were created. They were in use in applications such as windmills, where the weak rotational energy from the blades of the windmill was made usable by through slower speeds that turned a large grinding stone. Some uses predate even this. One of the simplest ancient gear ratios involves the well-bucket winch. A wooden cylinder with a handle was used to wind a rope and pull the bucket full of water up to the operator. This is an example of gear reduction, where ease of use was preferred to speed. Considerations 4. Gear ratio changes to a vehicle can change seemingly-unrelated aspects of vehicle operation. For example, changing the gear ratio of most vehicles will cause the speedometer to register inaccurately. The gear that drives the speedometer cable is sized according to the gear ratio. This gear must be replaced to ensure accuracy of the speedometer. One of the most frequently overlooked gear ratio changes occurs with changes to tire size. While a tire is not a gear, changes to its diameter have the same effect. Putting larger tires on a car effectively increases its gear ratio, costing power and gaining fuel economy, whereas smaller tires increase power but reduce fuel economy. Tire size changes also affect speedometer accuracy. Benefits 5. Changing the existing gear ratio can be highly beneficial if done to meet specific needs. For example, a large farm truck, or a truck used to pull a boat or trailer, can benefit from a lower gear ratio. The increased performance when stopping and starting, or going up hills, may easily outweigh the loss of fuel economy. The same is true for sports cars used in drag racing. A lower gear ratio gives more power to the wheels, at the cost of top speed; occasionally drag racers will benefit from a higher ratio, trading excessive wheel power for rotational speed. This depends on the weight of the car and the horsepower available. Many times the ideal ratio for a particular race car is not the ratio that the differential came with. As a 'general' rule of thumb, choose green to stay close to factory gear ratios, yellow for better economy and highway usage and red for better off road capabilities. What's in a Ratio? An automobile uses gear ratios in both the transmission and the drive axle to multiply power. The two ratios multiplied together equal the final drive ratio. Spend a few minutes in any bench-racing session and soon you'll hear rear axle gear ratios discussed. For many performance cars, 3.73s and 4.10s are common gear choices. The rearend gear ratio refers to the relationship between the ring gear and the pinion gear. By simply dividing the ring gear tooth count by the pinion gear tooth count, the ratio is determined. For example, if we divide a ring gear with 41 teeth by a pinion gear with 10 teeth we find that the gear ratio is 4.10:1 (41/10 = 4.10). Tire diameter will also have an effect on a vehicle's final drive ratio. As tire diameter changes, so will engine rpm at a given speed. We can demonstrate this with the simplified formula: rpm = (mph x final gear ratio x 336*) / tire diameter (*see "Formulas for Success" sidebar). For example, given 65 mph, a tire diameter of 30 inches, and a final gear ratio of 4.10, the engine speed will be approximately 2,984 rpm--(65 mph x 4.10 final gear ratio x 336) / 30-inch diameter tire. If we reduce the tire diameter to 25 inches, the engine speed increases to 3,581 rpm. By installing shorter tires, the vehicle will accelerate as though it has a 4.73 (higher numerically) gear without the expense of gear swapping. Because transmissions are comprised of several gear choices, the transmission allows the vehicle to accelerate quickly with lower gears and to maintain a cruising rpm using higher gears. In the '60s and '70s, most transmissions offered three or four gears with a 1:1 high gear. Using a TH400 as an example, First gear is 2.48:1, Second gear is 1.48:1, and Third gear is 1:1. Multiplying the 2.48 First gear by the 4.10 rear axle results in a final drive ratio of 10.16:1 (2.48 x 4.10 = 10.16). For most street performance applications, a 10:1 final First gear ratio is usually considered optimal. The disadvantage of operating a 4.10:1 axle ratio on the street with a 1:1 high gear is excessive freeway engine speed. Fortunately, today's transmissions frequently utilize Overdrive high gears in the neighborhood of 0.70:1, which allow reduced engine speeds. Combine these overdrive transmissions with a 4.10 axle ratio and you have a fuel-friendly final drive ratio of 2.87:1 (4.10 x 0.70 = 2.87) in high gear. A TH200-4R overdrive automatic utilizes a First gear of 2.74, a Second of 1.57, a Third of 1.00, and a 0.67 Overdrive. With this transmission's First gear ratio of 2.74 combined with a 3.73 axle ratio, the final drive ratio >> yields a 10.22 (2.74 x 3.73 = 10.22). In overdrive, the final drive ratio equates to a Bonneville-ready 2.49:1. Making Torque Multiply Acceleration is all about torque. One way to accelerate more quickly is to multiply the torque at low speeds to help move the vehicle forward. That's what a torque converter does. The torque converter features a component called a stator. The stator changes the direction of oil flow to the pump impeller's rotating direction and also incorporates a one-way clutch assembly. This redirection of fluid increases torque by applying the energy remaining in the oil. By applying the basics of gear ratios and power leverage, you can easily improve acceleration without paying too steep a price in highway rpm. It's all in the ratios. The Numbers Game Choosing the proper gear ratio can give your car the performance you want. Remember, acceleration depends on torque. Engines that produce more torque generally require less gear for optimal acceleration. Tire diameter plays a big role in the final drive ratio and engine rpm. Shorter tires increase engine rpm and are sometimes an economical way to improve acceleration. Torque converters multiply torque at low speeds. If your engine produces low amounts of torque, a converter with a higher stall speed and greater torque ratio can improve acceleration. The trade-off is that a higher-stall converter slips more, which means a higher engine rpm at a given speed. Overdrive automatics like the 4xRE series offer a low First gear and a 0.7x Overdrive. This allows the use of performance gears during low-rpm freeway cruising. Rack-and-Pinion A rack-and-pinion gear system consists of a round gear known as the pinion and a flat, toothed component known as the rack. The principle is the same; however, rather than rotations, the ratio determines the linear distance traveled by the rack with each rotation of the pinion. Calculating Rack-and-Pinion Gear Ratio Instead of counting the number of teeth in each gear, the distance the rack moves is measured in inches. Measure the distance from the end of the rack to an arbitrary point, turn the pinion one full revolution and then measure the distance again. The difference is the gear ratio. Steering Gear Ratios Bear in mind that adding power assist does not quicken the steering; it only decreases the input effort. Remember, also, that the steering ratio required for your car is a function of the radius of the turns of the race track, and on dirt, the slide angle necessary to steer into. In general, converting from manual steering to power assisted steering will permit quickening the steering ratio by at least one step, and usually two. Typical applications of the various rack ratios appear in the chart (Figure 8). The "ratios" (1.57, 2.09, etc.) in the chart refer to rack and pinion gearing and are given as linear inches of rack shaft travel per turn of the pinion (or steering wheel). Since the rack steers the front wheels by means of levers (the steering arms out on the spindles), the actual overall steering ratio of the car depends as much on their length as on the diameter of the pinion gear driving the rack. How to measure the RACK ratio If there are no numbers stamped on the caps, or if you have reason to believe a different pinion has been installed, just measure the distance from one tie rod hole (or from the end of the rack shaft) out to some stationary object (a piece of flat stock clamped to the frame rail). Turn the steering wheel (or the pinion) one full revolution and remeasure; the difference is the linear travel of the rack, which is the "gear ratio" of the rack and pinion. For reference, each additional tooth on the pinion increases the linear travel in one turn by slightly more than 1/4 inch. How to calculate the OVERALL steering ratio The overall steering ratio (12:1, 14:1, etc.) is measurable using turntables under the front wheels. Beginning with the front wheels pointed straight ahead, rotate the steering wheel one turn (360 degrees) one way, and read the turning angle of the front wheels from the turntable scales. You will have to resolve the difference between the right and left due to the Ackerman or steering toe; the usual method is to read the angle of the inside wheel, which is the maximum value. In road racing some prefer to average the two. As an example, if your reading is 36 degrees, dividing this into 360 gives you a quotient of 10, and thus a 10:1 overall steering ratio (if it is not possible to get a full turn out of the steering wheel, use three quarters of a turn and divide into 270). With a six-inch steering arm, a result of 10:1 is roughly what you could expect with a 3.14 rack. This measurement becomes more approximate with quicker racks and shorter steering arms, and because of the prevalence today of rack and pinion steering in short track stock cars it is common now to refer simply to rack travel numbers. Considerations Using a lower gear ratio like 4.10 with normal city or highway driving results in poorer fuel economy. The vehicle isn't utilizing the power exerted by the drive shaft, but it is having to generate it instead. It is suggested to have a higher gear ratio for typical driving to save on gas. A good gearing/tire combo calculator, created by Roy Grimm, is available on the web, found here: http://www.grimmjeeper.com/gears.html He states: This form allows you to calculate final drive ratios as well as see a comparison of speeds and RPMs within operating ranges of the vehicle. This calculator is useful for planning your rig, allowing you to see what kind of performance to expect from different combinations. Mopar1973man Cummins Gearing Calc.gsheet

Mine was, yes. Although may not have been plastic -specific.. that, i don't remember.

Agreed, that this will work.. and will work longer than just buffing the lenses with a restoration kit. But, I've done the 'clear coat' paint method years ago, and it was yellowed WITH paint-checking within a year's time. To clean them up after that, it's much easier and time-saving to just buy new ones, as the paint will take hours to strip/sand off to get back to square one.

All three of my trucks (both manual and autos, 96, 97, 01) reset the cruise light to OFF when the ignition was cycled.

I wished it would snow here. This single digit temp with -10 to -20*F wind chills sucks more when I can still see the grass.. I mean, if it's gonna be this cold, then at least SNOW!!