A good read on horsepower and drivetrain loss.
04-06-2006, 08:55 PM
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A good read on horsepower and drivetrain loss.
I did not write this. It came to me in an e-mail so I do not know the original source, but it'* good reading!
Understanding Horsepower loss through drive train
This article has taken on a life of its own. I originally created it (doing the research) to verify my assumptions and to document as many facts as I possibly could. After posting I have been deluged with people emailing me to either ask a question, tell me how right I was, or to argue and tell me how wrong I was. I really do appreciate any and all comments including those from people who didn't agree. It was the people who argued the point that helped prove to myself that the data contained in this article was accurate. After all, if it can't stand the scrutiny of people with strong opinions in the opposite belief then it wasn't worth the time of doing!
Index:
Why did I write this article?
Explanation of Horsepower and Torque
Background on proofs
What does a dyno measure?
Proof #1
Proof #2
Real world example
Real world example #2
Why does an Automatic take more power than a manual?
Footnotes
Why did I write this article?
Horsepower ratings for any vehicle are not a complete representation of how they perform on the track or on the street. In fact a horsepower rating (in my opinion) is only a way of measuring the success of a performance modification or to compare two dissimilar vehicles. How else can you compare a Mustang to a Camaro or even a Mustang to a Honda or Ferrari? Horsepower numbers offer the owner the ability to "bench race" with authority and offer bragging rights when no track is available to proof who is best.
Automobiles are rated by horsepower ratings from the factory. The number a manufacture uses is a measurement of how the engine performs without the drag of the drive train on an engine dyno. These numbers are bantered about in the same manor as performance enthusiasts use track results or chassis dyno numbers. The problem is an engine dyno is a very difficult thing to perform as the engine must be out of the vehicle to perform the test. This makes back-to-back testing of modifications difficult if not impossible. Of course a chassis dyno is an easier way to test your modifications - the problem? How do you compare the chassis dyno numbers to the original engine dyno numbers? The relation between engine dyno ratings and chassis dyno ratings was what I was trying to document. The problem is I couldn't find a lot of data but I did find a lot of "opinions". Here are the two primary opinions I found.
Opinion 1: Horsepower loss through drive train is a constant percentage based on the type of transmission you have. A manual transmission loses around 15%-17% of engine horsepower and an automatic transmission loses between 20%-25%.
Opinion 2: Horsepower loss through a given transmission is consistent. A typical manual transmission takes around 35 horsepower where an Automatic takes around 50.
So my reason for doing the article was to be able to relate my chassis dyno numbers to what an equivalent engine dyno would produce. Remember though - the purpose is to compare ENGINE DYNO results with CHASSIS DYNO results. In the real world (the street) many other factors will affect how a vehicle performs and how much horsepower is lost in drive train. This is why a car that has more horsepower (and the same vehicle weight) will not always win the race. ]
Explanation of Horsepower and Torque:
Previously I had explained horsepower and Torque in my own words but recently found an article by Marlan Davis in the January 2004 Hot Rod Magazine. His explanation is one of the best and most accurate I have ever seen. You can skip this section if you already have a strong understanding of the principles of Horsepower and Torque. If you understand completely the following 3 statements then skip to [Background on Proofs].
1.
Force is a measurement of pressure applied to an object. Force does not require the object to move. Torque is a measurement of force (ft-lb)
2.
Work is force in motion. For work to be performed the object must move. The required information to measure work is the weight of the object and the distance moved. Work is measured in pound-feet (lb-ft). A dyno measures torque as work.
3.
Power is work over time. To measure power you must already have an accurate measurement of work and the time required for the work to be performed.
Below is his explanation (used without permission):
Work Explained:
Force is a pushing or pulling action of one body against another. Depending on the resistance to the application of force, it may or may not result in movement. Say you try to push on a stalled car with 125 pounds of force, but it remains stuck in the mud. You've exerted force, but no movement has occurred because the car (being heavier than you) has too much resistance.
If force is applied and movement does occur, you have performed work, or the movement of an object from one position to another. For example, if you use a hoist to lift a 600-pound engine 6 feet in the air the work done would be 6 feet x 600 pounds or 3,600 foot-pounds (ft-lb).
Torque Explained:
By definition, work is calculated as a vector force, exerted in a straight line. But engines (as well as nuts and blots when they are tightened or loosened) rotate around an axis. The expression of this rotational or twisting force is called "torque", which is measured in units of force times distance from the axis of rotation. If you have a 1-foot-long wrench and you exert a force of 10 pounds on the end of it you then apply a torque of 10 pound-feet (10 lb-ft). If the wrench were 2 feet long, the same force would apply a torque of 20 lb-ft. When an engine is said to make "200 lb-ft of torque", it means that 200 pounds of force on a 1-foot lever is needed to stop its motion.
To avoid confusion in the U.*. measurement system, the unit of measure for torque is the pound-foot (lb-ft), while for work it is to foot-pound (lb-ft). Remember, work and torque aren't exactly the same. Movement must occur for work to be done, but that doesn't necessarily hold true for torque: Exerting 10 lb-ft of torque on a bolt that'* already been tightened to 50 lb-ft won't produce any movement.
If torque does produce movement-as is the case with your engine unless it'* locked up- any "distance" traveled as the crank rotates is equal to the circumference of a circle, not a straight line ... so 1 lb ft of torque produced during one revolution actually is about equal to 6.28 ft-lb of work or mechanical energy. Huh? Just recall your geometry: Fin a circle'* circumference by multiplying its radius (r) by 2pi. With a 1-foot lever
2*pi*r * 1 lb-ft of torque =
2x3.1416 x 1-foot lever x 1 lb-ft =
6.2832 ft-lb of work
Power Explained:
Torque and work measurements tell us how much has been accomplished, but provides no clue how fast a given amount of work (or torque) is done. That'* the job of power, an expression of the rate or speed at which work is performed. The more power that is generated, the more work is done in a given time-period.
Suppose it takes a constant 100lb-ft of torque to spin a nut onto a blot one complete revolution. Your girlfriend takes 10 seconds to do this, You, being a real stud (pun intended), take only 5 Seconds to perform the same task. you would be twice as powerful, because you performed the same work in half the time.
In the U.*. system, power is expressed as "horsepower" (HP). One hp is the amount of power it takes to perform 33,000 ft-lb of work in 1 minutes, as based on 18th century engineer James Watt'* observations of the work performed by a strong horse as it operated a gear driven mine pump by pulling a lever connected to the pump (sounds like a dyno, doesn't it?).
Because 1 hp represents the production of 33,000 ft-lb of mechanical energy per minute, horsepower equals ft-lb/minute divided by 33,000, as expressed by:
ft-lb per minute D x F
hn = ------------------ = ----------
33,000 33,000 x t
Here, D is the distance in feet the weight is to be moved; F, the force in pounds required to move the weight; and t, the time in minutes required to move the weight (F) through the distance (D).
An engine dynamometer measures torque (lb-ft), not mechanical energy (ft-lb); and an engine'* "time" is expressed in revolutions per minute (RPM). Since 6.28 ft-lb of work is about equal to 1 lb-ft of torque, you can substitute:
torque x 6.28 x rpm
hp = ----------------------
33,000
torque x (6.28/6.2
x RPM
= ----------------------------
(33,000 / 6.2
torque x rpm
= ---------------
5,252
The result is the classic equation with which we are all familiar.
What the Math Tells Us:
Looking at this horsepower equation, several relationships are apparent. First, knowing that dividing the same number by itself cancels it out (because the product is "1"), it'* obvious that at 5,252 rpm, the horsepower and torque values are always equal. That'* why an engine'* torque and horsepower curves always cross at 5,252.
Second, assuming engine displacement is fixed, raising the engine'* operating rpm range is the most effective way to make big naturally aspirated horsepower numbers. Suppose an engine makes 400 lb-ft at 3,000 rpm. The equation tells us that at 3,000 rpm it would produce 228hp. But if the engine made 400 lb-ft at 6,000 RPM, it would produce 457 hp. It'* true that doubling the torque output to 800 lb-ft at 3,000 rpm would also yield 457 hp, but that large an increase on a given displacement engine is unlikely in the real world without forced induction.
(the article continued but for the purpose of this page I am not including the remainder of the article. I encourage you to find the magazine and read the remainder)
Background on Proofs:
Horsepower loss through drive train is caused by a few different things. Some have greater affect on resultant horsepower loss than others. If you can't agree with or understand these next few statements then the rest of the article will not help.
Statement 1: Frictional force increases the power required to move an object. This is the most major contributing factor to horsepower loss in drive train. Here is an example.
If you tie a board to a rope and drag it across a linoleum floor and then drag it across a carpeted floor it will be harder to pull on the carpet. If you put a stack of bricks on the board it will be even harder to pull. If you were to drag the same board for the same distance at the same speed and took a temperature measurement on the bottom of the board it would get hotter as you added weight. The heat generated would increase linearly as compared to the amount of weight you put on the board. The force required to move the board would also increase linearly.
For this same reason if you apply 10 ft-lb of torque on a gear it will produce a given amount of friction. If you change the force from 10 ft-lb to 100 ft-lb more friction will be generated by the turning gears. In a transmission or rear end (drive train) this increased friction is seen as heat generated. This is why when you add more power to a vehicle you need to install better coolers to keep everything cool.
Statement 2: It takes more power to accelerate any given object more quickly. This is probably the most obvious statement but is also very critical to understand. Second to frictional losses is acceleration power loss.
If you take a given car that can at best run the quarter mile in 14 seconds the only way to increase this acceleration (other than changing its weigh or drag - next two statements) is to increase the power. The more power you add the faster the vehicle can perform the work of traveling down the quarter mile. This is why we all build more powerful engines! If it wasn't true then we would all be driving Yugo'* at the track.
Statement 3: It requires more power to accelerate a heavier object. This is one of the explanations for the difference in horsepower loss between automatics and manual transmissions and why a Turbo 400 takes more power than a Turbo 350. Here is an example:
Grab a 10lb bag of potatoes and see how long it takes you to get to a full running speed. Now grab a 100lb bag. I can think of millions of more examples but....
Now picture the rotational mass of your transmission. If it weighs more it will take more power to make it accelerate to the same speed.
Statement 4: The faster you want to move an item the more power it takes (drag). Here is an example:
If you take a stock Mustang six cylinder and try and obtain a top speed rating it will probably be around 110 mph. Now get in a V8 Mustang and it will go faster. For an average vehicle to break 200 mph it takes around 500-600 horsepower. As an item accelerates it becomes harder and harder to push it because of the drag induced by the wind. See footnotes for more detailed information.
For the same reason it will take more power to spin a transmission at 7,000 RPM than it will at 6,000 RPM. The gears moving through the fluid in the transmission will require more power at higher RPM'*.
What does a dyno measure?
If you don't have a firm understanding of what horsepower and torque are please read that section before continuing with this proof - this is actually very important. On a dyno sheet you see both torque and horsepower readings, to understand what a dyno measures requires a background in what each is a measurement of.
An engine dyno actually measures engine output torque. It also measures engine speed (RPM). It can then calculate HP from the HP = T*RPM/5252 equation.
A dyno jet type chassis dyno (no load on the drum) measures the instantaneous drum speed, and time during a pull. From both, it can calculate drum acceleration. Knowing the rotational inertia of the drum, it can calculate the applied torque. Then finally, from the calculated applied torque, and the speed of the drum, it can calculate HP applied to the drum using the above formula
Proof #1
An internal combustion engine produces power by �combusting� fuel and air. This combustion process burns fuel in the cylinder at the end of the compression stroke and causes the atmosphere inside the cylinder to expand. This expansion (caused by the heat energy of the combustion process) pushes the piston back the other direction. In an average engine, around 33% of the heat (power is heat) is spent in pushing the piston. Another 33% is lost out the exhaust (the exhaust gas continues to expand for around 18�, this is why turbos work) and another 33% is lost through the radiator (heat transferred through the block).
Now lets state a fact of physics (you might remember this from High School). Energy can not be destroyed, only transformed into another state. What this means is, to optimize power in an internal combustion engine you try to optimize the amount of �heat� (that is, the power created by internal combustion) to be transferred into the �push� process of the piston. You can optimize this in many methods. Ignition timing is the most obvious. By changing timing you can optimize the amount of time the heat has to expand and push the cylinder. The whole point of this is to understand that power is created by heat and lost through the dissipation of heat. Heat expands the atmosphere inside the cylinder and that is how you make power.
So, if heat is power, then you can measure the amount of power by the dissipation of heat throughout a system. Have you ever wondered why you need a bigger radiator when your engine makes more power? Part of the reason is when you produce more horsepower you lose more energy through heat loss into radiator. How about why you start to need things like transmission coolers? Now we are starting to get to the root of my argument on why power loss is based on percentages �vs- a static value.
As you make more power you apply more force on transmission gears and (in automatic transmissions) fluid. As you make more and more power you need to run bigger and bigger transmission coolers and (in extreme cases) rear-end coolers. Why do you need to do this? Because you are generating more heat! That is obvious, right?
Now let�* look at the laws of physics. Energy can not be destroyed, only transformed. So, if your transmission and rear-end are generating more heat, where do you think this heat is coming from? The magic heat fairy? The heat is a direct result of power being transformed (lost) in the form of heat. The more power you produce (in your engine) the more power (heat) you lose in your drive train.
Short version: Increase in heat generated by your transmission or rearend is a sign that more power is being consumed (converted to heat) and therefore lost.
Proof #2: (more complicated and involves some math)
If you don't have a firm understanding of what horsepower and torque are please read that section before continuing with this proof:
To calculate torque, A dyno jet type chassis dyno (no load on the drum) measures the instantaneous drum speed, and time during a pull. From both, it can calculate drum acceleration. Knowing the rotational inertia of the drum, it can calculate the applied torque. This work is represented as lb-ft of torque. A "Dynojet" brand dynamometer uses a "fixed inertia" load, whereas a "Mustang" brand dynamometer uses "loaded inertia" that can simulate driving stresses the vehicle would encounter in actual driving conditions. Typically a "loaded" dynamometer will spin up slower than a fixed resistance dynamometer (this is important later).
Now lets take a look at drag inside the transmission and rearend (the drag of the gears pressing against each other and traveling through the fluid inside the transmission). Land speed racers know that the drag coefficient of a moving object is exponential based on speed (for more definition on drag components see footnotes). What this means is if wind resistance (drag) takes 10 horsepower to overcome in a vehicle moving at 25 mph it will take 40 horsepower at 50 mph (2x speed = (2�) or 4x power requirement). To move the vehicle at 100mph you need 160 horsepower (4x speed = (4�) or 16 x horsepower (10) = 160). Take that to 200 mph and you lose 640 horsepower to drag (8x speed = (8�) or 64 x horsepower (10) = 640). So, the same vehicle that took only lost 10 horsepower due to drag coefficient at 25mph loses 640 horsepower to drag at 200mph.
Short version: The faster an object moves that is subject to drag (air, liquid, or pressure) the more horsepower it takes to move it.
Take two identically equipped vehicles (same transmission and rearend) with one vehicle producing 400 horsepower and the other producing 800 horsepower. The vehicle producing 800 horsepower will accelerate faster than the vehicle producing 400 horsepower. This goes back to our definition - horsepower = work over time. The same vehicle will cause the fluids, pressures, and gears inside the transmission to spin up more quickly (as evident by the increased acceleration). Because this same work is performed in less time, the net result is that more horsepower was required (horsepower = work over time).
Short version: To accelerate an object faster requires more power (horsepower). If this were not true then why would we build high horsepower cars? This is also true of spinning gears inside a transmission or rearend.
Now factor in the drag coefficient of the spinning gears inside the transmission and rearend. Since the force required to spin the gears grows with speed, and since increased horsepower is causing the rate of acceleration to increase (your engine revs faster on the dynamometer) you will find that the actual amount of work required to spin the gears greater than with lower horsepower. Therefore as horsepower goes up the percentage of loss power loss through drivetrain actually increases. This is probably a minimal effect on total drivetrain loss but is still a factor..
Short version: The higher the RPM of the vehicle the more parasitic drag that exists inside the transmission and rearend, regardless of acceleration rate. Therefore more power is spent to spin the drivetrain at high RPM.
Another thing to note: The reported HP (flywheel -or- rear wheel), is highly dependent on acceleration rate. The higher the acceleration rate, the lower the reported HP. So on a dynojet-type dynamometer (fixed inertia), higher horsepower will accelerate the drum quicker, which will increase drivetrain inertial losses, thereby reporting even lower RWHP numbers. This means that the percentage of horsepower drivetrain loss through a Dyno-Jet type dynamometer will actually grow as horsepower increases at a rate larger than a "loaded inertia" dynamometer.
This deserves more explanation. Typically you will find a car that produces 200 horsepower on a Dyno-jet (fixed inertia) will only produce 180 horsepower on a Mustang (loaded inertia) dynamometer - 10% less. However as horsepower increases the numbers (as a percentage) will become closer. So a vehicle making 800 horsepower on a Dynojet may only make 740 horsepower (7.5% less) on a Mustang dynamometer. The percentage has become closer. This is the result of time being reduced (with the dynojet dynamometer) in the equation of work over time.
Real World Example:
(from Muscle Mustang & Fast Ford Magazine, Febuary 2003, Article titled "Mass VS. Myth" by David Vizard)
Sometimes you just luck out. I was trying to figure out how I could put some "proof" in the form of a chart or real world Dyno results. I was just reading the latest issue of Muscle Mustang & Fast Ford magazine and ran across the article "Mass VS. Myth". The article is discussing the pro'* and con'* of using a light weight flywheel. The cool part is the same data they used is also a proof for horsepower loss through drivetrain. Let me explain.
In the article they dyno'd a car in both 1st gear and 4th gear (this was with a manual transmission). Normally all dyno runs are done in whatever gear provides a 1:1 ratio through the transmission (3rd in Automatics, 4th in Manuals typically). What was amazing was how much less rear wheel horsepower was generated when the car was in first gear. Take a look at the chart and the authors notes (quote: "Because of the rapid rate the engine internals and components back to the wheels are accelerated the power absorbed is greater. As can be seen the difference in rear wheel output between first and fourth gear is an amazing 85 hp and a staggering 140 lbs.-ft of torque.")
So, why would a dyno show less power in 1st gear than in 4th gear? The engine is producing the exact same amount of power, the only change is the gear the transmission is in. The answer: In first gear the engine can accelerate the dyno wheel more quickly because of the mechanical advantage it has (lower transmission gear). To accelerate the dynamometer wheel requires the same amount of work in 1st gear and in 4th gear. Because it happens in less time in first gear it must take more horsepower. horsepower = work / time. If you reduce the amount of time and if work remains the same the horsepower must be higher (this is realized as greater horsepower "loss" as it was horsepower "required" to do the "work" of spinning the dynamometer wheel).
Update 12/22/03: I had a person ask me to clarify the reason why horsepower and torque would drop when the car was dyno'd in first gear so I spent a little more time thinking through what is happening. After my first attempt at explaining it my friend Ed Hohenberg helped me refine this explanation. Here is what we came up with.
The reason for the large HP difference between 1st and 4th gear is due to 2 factors:
1. In first gear, the power flow through the transmission involves more gear meshes (high gear is straight through, 1st gear uses the countershaft, with gear meshes at both ends). The additional gear meshes reduce the transmitted power by typically 2% per mesh (for a helical cut gear, less for straight cut), for a combined loss of 4% or so.
2. In first gear, because of the additional torque multiplication produced by the gearing, the applied torque to the dyno drum increases proportionally to the gear ratio (less the additional losses from the gear meshes), which causes the dyno drum to accelerate much faster. Now you might think this would indicate higher HP, but the drum speed is similarly proportionally reduced by the added trans ratio, so neglecting friction and inertia effects, the power would remain the same. But the power measured at the drum DOES decrease, because all the drivetrain components (engine rotating and reciprocating masses, flywheel, trans gears and shafts, driveshaft, rear gears, axles, wheels and tires) all accelerate faster now (but in first gear, the starting and finishing speeds are much less compared to high gear). Since it takes power to accelerate mass, the higher driveline component accelerations use up more power in first gear than high gear, and thus the available power at the dyno drum is decreased by the HP "robbed" in the vehicle drivetrain due to the higher driveline component accelerations.
[EDIT 8/31/04] The following section has caused a lot of confusion lately. As a point I have made a few changes to the section below and will re-state that torque is measured two different ways:
1) Torque measured as force requires no movement and is measured as foot-pounds (ft/lb). It is simply a measurement of how hard something is being pushed on. Imagine a lever with a fulcrum. No picture 1 foot on each side of the fulcrum. Imagine one end is under a 10,000 lb rock. If you pushed with one pound of force you would be applying 1 foot pound of torque against the rock - but it would not move. If you made the lever 10,000 foot long on the side away from the rock and pushed with the same 1 pound of force you would lift the rock. At 9,999 feet the rock would still not move but you would be applying 9,999 ft-lbs of torque. Using first gear in a car will apply more force than in fourth gear however less work is done (vehicle speed is slower).
2) Torque as measured as work requires movement and is measured as pound-feet (lb/ft). Since work requires movement torque measured as work requires the item to move. 1 lb/ft of torque = amount of force to move 1 pound 1 foot. Picture a rope with a 1 pound rock tied to the end. If you pull the rope up 1 foot you have done 1 lb/ft of work. If you move it two feet you have done 2 lb/ft of work. If you put 1,000 lbs on the rope and moved it 1 foot you have done 1000 lb/ft of work.
A dyno measures torque as work as measured by knowing the mass of the dyno drum, the speed of the dyno wheel at the time the measurement starts, and the speed of the dyno wheel at the time the dyno measurment stops. For a dyno each measurement of force is measured per engine revolution.
HP is down, but torque (measured as force) is certainly up in first gear, as the dyno sees it. The reason HP is down is because of the increases losses, of course. If you ignore all losses (frictional, parasitic and inertial) the HP in any gear would be the same, because while the torque (measured as force, not work) is increased proportionally, the speed is decreased proportionally, so the amount of work performed would otherwise stay the same.
But the torque at the dyno and torque at the engine are 2 different things, and I think this may be where some confusion is. The dyno measures the applied torque at the drum, which is higher in first gear - but the speed at the drum is lower. Ignoring the losses the HP *should be* the same. But then the dyno is also measuring the engine speed (RPM), so it takes the dyno measured power, then goes back and calculates torque from the engine speed and dyno HP. The "torque" reported by a chassis dyno is really meaningless (as a measurement of force), since it'* neither the torque measured at the engine (how could it be?), nor the actually torque measured at the dyno drum (that would include the gear multiplications, so unless you have a 1:1 rear axle ratio in addition to a 1:1 trans ratio, the rwtq reported is certainly NOT the actual force at the rear wheels). It is instead a measurement of the amount of work that was performed at the rear wheels and therefore a measurement of torque at the rear wheels as a unit of work.
Further in the article the comparison was done between a lightweight clutch/flywheel and a stock/heavy combination. You will see that the exact same engine produced slightly (14.5 hp) more horsepower with the lightweight setup than the stock/heavy setup. The reason for this was the heavy flywheel requires more "work" to spin. Anytime greater work is required the result will be a reduction in power.
The author in this article included a lot of good information regarding inertia and resultant horsepower recovery. If you have the opportunity I would recommend picking up this magazine for this article. It does make me wonder what would happen if you were to dyno your car in overdrive. Unfortunatly my AOD will not take the abuse of a dyno run in overdrive so I won't be the one to figure this out.
Real World Example #2: (added 9/29/2003)
The November 2003 issue of Car Craft has a great article titled "The Brutal Truth" by Jeff Smith (page 40). In the article they place two engines on an engine dyno and then dyno the engine again once it is installed in the vehicle. One engine is a 357 cubic inch Ford Windsor engine and the other is a 455 Buick. The Ford 357 was installed in a 63 Comet using an AOD transmission and Ford 9" rearend with 3:50 gears (exact combination of drivetrain in my Mustang except my AOD is a non-lockup which means it is even less efficient or should lose even more horsepower). The 455 was installed in a 70 Buick GS with a Muncie 4-speed and a 12 bolt rearend with 2.73 gears. The point of the article was to show how things like a belt driven cooling fan or poor vehicle exhaust could affect the engine output in the vehicle but was equally as valid for showing drivetrain induced power loss.
Without reading any further it would be my assumption that the Ford combination would lose a larger percentage of power through the drivetrain. Not only is the AOD an Automatic it is also a 4 speed Automatic that has substantial weight and rotating mass. The 9" rearend is also larger and heavier than the 12 bolt Chevrolet.
The Ford 357 produced 371 horsepower on the engine dyno at 5,000 RPM. On a 1990 Mustang that came stock with an AOD and a 3.27 8" rearend (more efficient than the 9") the rear wheel horsepower is typically 180 hp. That represents a loss of 45 horsepower given the rated 225 flywheel horsepower on that vehicle. Using this 45 horsepower and even giving it another 5 for the 9" rearend the 357 would have produced 321 peak horsepower on the chassis dyno. Well, it didn't! Even after removing all the factors that could have contributed to extra power loss in the vehicle (removing the belt powered cooling fan) the chassis dyno only showed 283 hp. In fact over the entire power curve the difference between the engine dyno and the chassis dyno was 24%. This provides more evidence that the power loss through common drivetrain remains a percentage even as power is increased rather than remaining a static loss value.
The Buick 455 produced 329hp and made 280 through the drivetrain at 4,500 RPM. The average drivetrain horsepower loss in this vehicle was 18.3%. This can be accounted for by the fact that the 4 speed Muncie is more efficient (require less power to accelerate) as well as the 12 bolt rearend being more efficient than the 9".
Why does an Automatic take more power than a Manual transmission?
So, what about the 15%-manual and 20% automatic? Well, they are good places to start. Some transmissions are more efficient than others and some rear ends are more efficient than others. In the end, none of this matters as wheel horsepower is what is actually used. You will find the heavier the transmission parts (gears, shafts, etc.) the more power they will take. I have seen engine dyno comparisons to rear wheel and you typically see 15-17% for manual transmissions and 20-25% for automatics. How much does your transmission take? Take the basic values 15 and 20 percent and consider the following.
HP loss in auto vs stick is mostly related to the converter slip (there is always some slip in a normal converter). However, even with a mechanically locked up converter, in the planetary gear system used in autos, there are more gear meshes occurring, which increases HP losses since each gear mesh results in a HP loss (relates back to statement 1). And don't forgot, in an auto, you have direct pumping losses from the oil pump (you don't have this in a manual trans). And the higher the line pressure, or fluid flow rate, the greater the pumping losses.
In an Automatic transmission you will find several factors that determine power requirements.
Weight: a comparable C4 take less power than a C6 - primarily because of the weight difference of the moving parts. A turboglide would take even less and an AOD would require around the same as a C6 (just using weight)
Lockup/non-lockup: Many modern Automatics use a lockup feature in drive and overdrive. This will increase horsepower available at the rear wheels because you don't have any slip in the torque converter. The slip in the torque converter generates heat which is an indicator that power is being lost. Even at over the stall point of a torque converter where you have complete hydraulic lock up you will experience some power loss through the hydraulic coupling process. Having said this I still prefer non-lockup converters as they are easier on the drivetrain (less harsh shifts) and allow the engine to run in the proper RPM power range.
Stall speed: The purpose of installing a high stall torque converter is to allow the engine to spin up to a desired RPM range quickly. Once this stall speed is reached you have a hydraulic lockup and the transmission is locked to the same RPM as the engine. Measuring horsepower below the stall speed of a torque converter is completely meaningless. The slip will give bad readings. Normally you can see the "flash" or hydraulic lockup point on the dyno sheet (I will post an example later). Having said all of this a high speed stall converter will normally take less power than a stock type converter. The reason is a high speed stall converter is typically smaller and lighter - that is the only reason.
Torque converter design: I do not know enough about this to go into detail however some designs are more efficient than others. With torque converters expect to spend between $500-$1000 for a good quality converter - if you try to go cheap here you will only hurt yourself later. In my AOD I run a Lentech 9" 2,800 RPM converter that flashes (hydraulic lockup) at 3,400 RPM (higher torque output will push your flash point further in the RPM band).
Valve Body: Again, I don't know enough to tell you exactly why but some are better than others. Some valve bodies will route hydraulic fluid more efficiently and activate clutch packs more effectively. For my AOD I use a Lentech Strip Terminator valve body and I love it.
With a manual transmission your primary factor in power loss is weight. The heavier the gears, clutch, flywheel, etc. the more power required to turn it.
Thank you for your time! If you have any questions or comments please email me at garnerjohn@earthlink.net
* Footnotes:
Explanation for the value of 5,252 RPM being used to calculate horsepower provided by Ed Hohenberg (thank you Ed).
Work is force through a distance, as you correctly stated. For linear motion it'* simply Force X distance. For angular motion, the distance the force moves is again used, only now that distance will be the circumference of the circle that the force moves through, or 2 * PI * radius. If you assume an imaginary 1 foot long radius for a moment arm, the distance is simply 2*PI. So then for angular WORK, the Force X distance = Torque X 2*PI. If we're turning at 1 RPM, then the 2*PI distance is covered every 1 minute (1 revolution per minute). At 2 RPM for example, we're now doing the same work, but in half the time (1 revolution in 1/2 minute), and so on for higher speeds. So our angular Power, Work/time = T X 2*PI X RPM, which gives us power in units of lb-ft/minute. To get that into lb-ft/sec divide by 60, so Power = T X 2*PI X RPM/60, or T X RPM X 2*PI/60. Lastly, "Horse Power", as defined by Mr. Watt, is 550 lb-ft/second, so to get our angular power from lb-ft/second into HP, we divide by 550, so finally:
Power = T X RPM X 2*PI/60/550, or T X RPM X .0001904 or T X RPM/5252.11
Explanation of Drag coefficient provided Ed Hohenberg (thanks again!)
HP losses for a moving vehicle have 3 basic components: inertial, rolling, and aerodynamic. The inertial HP losses will be proportional to mass, acceleration and velocity. The rolling resistance losses will be proportional to mass, rolling/friction coefficients and velocity. The aerodynamic HP losses will be proportional to drag coefficient, frontal area, and velocity squared.
The magnitude of each component can vary significantly, depending on the input variables. For constant speed (no acceleration) the inertial component is zero, so only rolling resistance and aerodynamic losses are present. For a typical car, the rolling resistance component will have the greatest magnitude up to about 50 mph. For a very sleek car like a new Corvette, the aerodynamic component probably won't exceed the rolling HP losses until 65 mph.
Article from 5.0 Mustang & Super Ford magazine that caused me to write this page.
(February 2002, page 54, sub content titled "What'* the loss" from the article "Heads or Tails", page 47, by Eric English)
(Note: I really enjoyed the article in whole and only have an issue with the authors opinion on horsepower loss through drivetrain. I have heard this authors opinion expressed by many others and after a cursory review of his logic it does appear to make sense. When you dig deeper and analyze each of his assumptions you realize they are flawed. I realize the Internet and Magazines are a great way to spread both information and dis-information. I only hope that those who read this page will take the time to understand the math and theories expressed and know how to answer the next person who expresses the opinion expressed in the article below)
What'* the Loss?
When we began to contemplate flywheel horsepower figures for our different combinations, we bantered about some concepts that can be deceiving. Most enthusiasts have been exposed to the idea that flywheel and rear-wheel horsepower can be equated by factoring in a given percentage for drivetrain loss - the drag that occurs from all the items between the flywheel and the rear tires. You may have seen factors such as 15 percent for stick-shift cars and 25 percent for automatics, applied by dividing rear-wheel horsepower by either 0.85 for stick-shifts or 0.75 for automatics.
Now take a time-out and consider the following. Our original baseline indicated 195 hp at the rear wheels, which when divided by 0.85 equates to 229 flywheel horsepower, and implies that the drivetrain is absorbing some 34 horsepower. On the other hand, our combination of blower and traditional bolt-ons netted nearly 340 hp on the Blood Enterprises dyno, which when divided by 0.85, equates to 400 flywheel horses, and implication that the drivetrain is now absorbing 60 hp.
Nothing has changed between the flywheel and the rear wheels on our '93 LX, so does it make sense to figure the drivetrain is now absorbing nearly twice as much power? Such a concept just doesn't jibe in our little brains, so we asked a couple of people in the biz what they thought. Lee Bender of C&L Performance and Paul Svinicki of Paul'* High Performance are both well versed in evaluating Mustangs on the dyno, and they both agreed that extrapolating drivertrain horsepower loss via percentages is flawed. Lee believes that the stick Mustangs experience roughly a 35hp loss through the drivetrain, whether they make 200 hp or 400 hp. He did explain that ultra-high-powered vehicles - typically race cars - can be and exception to this rule, but that'* a topic for another time. Interestingly, a 35hp loss for stick-shifted drivetrains is strikingly similar to the difference between Ford'* horsepower ratings and the rear-wheel numbers we've observed on dynos across the nation. Hmmm...
Understanding Horsepower loss through drive train
This article has taken on a life of its own. I originally created it (doing the research) to verify my assumptions and to document as many facts as I possibly could. After posting I have been deluged with people emailing me to either ask a question, tell me how right I was, or to argue and tell me how wrong I was. I really do appreciate any and all comments including those from people who didn't agree. It was the people who argued the point that helped prove to myself that the data contained in this article was accurate. After all, if it can't stand the scrutiny of people with strong opinions in the opposite belief then it wasn't worth the time of doing!
Index:
Why did I write this article?
Explanation of Horsepower and Torque
Background on proofs
What does a dyno measure?
Proof #1
Proof #2
Real world example
Real world example #2
Why does an Automatic take more power than a manual?
Footnotes
Why did I write this article?
Horsepower ratings for any vehicle are not a complete representation of how they perform on the track or on the street. In fact a horsepower rating (in my opinion) is only a way of measuring the success of a performance modification or to compare two dissimilar vehicles. How else can you compare a Mustang to a Camaro or even a Mustang to a Honda or Ferrari? Horsepower numbers offer the owner the ability to "bench race" with authority and offer bragging rights when no track is available to proof who is best.
Automobiles are rated by horsepower ratings from the factory. The number a manufacture uses is a measurement of how the engine performs without the drag of the drive train on an engine dyno. These numbers are bantered about in the same manor as performance enthusiasts use track results or chassis dyno numbers. The problem is an engine dyno is a very difficult thing to perform as the engine must be out of the vehicle to perform the test. This makes back-to-back testing of modifications difficult if not impossible. Of course a chassis dyno is an easier way to test your modifications - the problem? How do you compare the chassis dyno numbers to the original engine dyno numbers? The relation between engine dyno ratings and chassis dyno ratings was what I was trying to document. The problem is I couldn't find a lot of data but I did find a lot of "opinions". Here are the two primary opinions I found.
Opinion 1: Horsepower loss through drive train is a constant percentage based on the type of transmission you have. A manual transmission loses around 15%-17% of engine horsepower and an automatic transmission loses between 20%-25%.
Opinion 2: Horsepower loss through a given transmission is consistent. A typical manual transmission takes around 35 horsepower where an Automatic takes around 50.
So my reason for doing the article was to be able to relate my chassis dyno numbers to what an equivalent engine dyno would produce. Remember though - the purpose is to compare ENGINE DYNO results with CHASSIS DYNO results. In the real world (the street) many other factors will affect how a vehicle performs and how much horsepower is lost in drive train. This is why a car that has more horsepower (and the same vehicle weight) will not always win the race. ]
Explanation of Horsepower and Torque:
Previously I had explained horsepower and Torque in my own words but recently found an article by Marlan Davis in the January 2004 Hot Rod Magazine. His explanation is one of the best and most accurate I have ever seen. You can skip this section if you already have a strong understanding of the principles of Horsepower and Torque. If you understand completely the following 3 statements then skip to [Background on Proofs].
1.
Force is a measurement of pressure applied to an object. Force does not require the object to move. Torque is a measurement of force (ft-lb)
2.
Work is force in motion. For work to be performed the object must move. The required information to measure work is the weight of the object and the distance moved. Work is measured in pound-feet (lb-ft). A dyno measures torque as work.
3.
Power is work over time. To measure power you must already have an accurate measurement of work and the time required for the work to be performed.
Below is his explanation (used without permission):
Work Explained:
Force is a pushing or pulling action of one body against another. Depending on the resistance to the application of force, it may or may not result in movement. Say you try to push on a stalled car with 125 pounds of force, but it remains stuck in the mud. You've exerted force, but no movement has occurred because the car (being heavier than you) has too much resistance.
If force is applied and movement does occur, you have performed work, or the movement of an object from one position to another. For example, if you use a hoist to lift a 600-pound engine 6 feet in the air the work done would be 6 feet x 600 pounds or 3,600 foot-pounds (ft-lb).
Torque Explained:
By definition, work is calculated as a vector force, exerted in a straight line. But engines (as well as nuts and blots when they are tightened or loosened) rotate around an axis. The expression of this rotational or twisting force is called "torque", which is measured in units of force times distance from the axis of rotation. If you have a 1-foot-long wrench and you exert a force of 10 pounds on the end of it you then apply a torque of 10 pound-feet (10 lb-ft). If the wrench were 2 feet long, the same force would apply a torque of 20 lb-ft. When an engine is said to make "200 lb-ft of torque", it means that 200 pounds of force on a 1-foot lever is needed to stop its motion.
To avoid confusion in the U.*. measurement system, the unit of measure for torque is the pound-foot (lb-ft), while for work it is to foot-pound (lb-ft). Remember, work and torque aren't exactly the same. Movement must occur for work to be done, but that doesn't necessarily hold true for torque: Exerting 10 lb-ft of torque on a bolt that'* already been tightened to 50 lb-ft won't produce any movement.
If torque does produce movement-as is the case with your engine unless it'* locked up- any "distance" traveled as the crank rotates is equal to the circumference of a circle, not a straight line ... so 1 lb ft of torque produced during one revolution actually is about equal to 6.28 ft-lb of work or mechanical energy. Huh? Just recall your geometry: Fin a circle'* circumference by multiplying its radius (r) by 2pi. With a 1-foot lever
2*pi*r * 1 lb-ft of torque =
2x3.1416 x 1-foot lever x 1 lb-ft =
6.2832 ft-lb of work
Power Explained:
Torque and work measurements tell us how much has been accomplished, but provides no clue how fast a given amount of work (or torque) is done. That'* the job of power, an expression of the rate or speed at which work is performed. The more power that is generated, the more work is done in a given time-period.
Suppose it takes a constant 100lb-ft of torque to spin a nut onto a blot one complete revolution. Your girlfriend takes 10 seconds to do this, You, being a real stud (pun intended), take only 5 Seconds to perform the same task. you would be twice as powerful, because you performed the same work in half the time.
In the U.*. system, power is expressed as "horsepower" (HP). One hp is the amount of power it takes to perform 33,000 ft-lb of work in 1 minutes, as based on 18th century engineer James Watt'* observations of the work performed by a strong horse as it operated a gear driven mine pump by pulling a lever connected to the pump (sounds like a dyno, doesn't it?).
Because 1 hp represents the production of 33,000 ft-lb of mechanical energy per minute, horsepower equals ft-lb/minute divided by 33,000, as expressed by:
ft-lb per minute D x F
hn = ------------------ = ----------
33,000 33,000 x t
Here, D is the distance in feet the weight is to be moved; F, the force in pounds required to move the weight; and t, the time in minutes required to move the weight (F) through the distance (D).
An engine dynamometer measures torque (lb-ft), not mechanical energy (ft-lb); and an engine'* "time" is expressed in revolutions per minute (RPM). Since 6.28 ft-lb of work is about equal to 1 lb-ft of torque, you can substitute:
torque x 6.28 x rpm
hp = ----------------------
33,000
torque x (6.28/6.2
x RPM= ----------------------------
(33,000 / 6.2
torque x rpm
= ---------------
5,252
The result is the classic equation with which we are all familiar.
What the Math Tells Us:
Looking at this horsepower equation, several relationships are apparent. First, knowing that dividing the same number by itself cancels it out (because the product is "1"), it'* obvious that at 5,252 rpm, the horsepower and torque values are always equal. That'* why an engine'* torque and horsepower curves always cross at 5,252.
Second, assuming engine displacement is fixed, raising the engine'* operating rpm range is the most effective way to make big naturally aspirated horsepower numbers. Suppose an engine makes 400 lb-ft at 3,000 rpm. The equation tells us that at 3,000 rpm it would produce 228hp. But if the engine made 400 lb-ft at 6,000 RPM, it would produce 457 hp. It'* true that doubling the torque output to 800 lb-ft at 3,000 rpm would also yield 457 hp, but that large an increase on a given displacement engine is unlikely in the real world without forced induction.
(the article continued but for the purpose of this page I am not including the remainder of the article. I encourage you to find the magazine and read the remainder)
Background on Proofs:
Horsepower loss through drive train is caused by a few different things. Some have greater affect on resultant horsepower loss than others. If you can't agree with or understand these next few statements then the rest of the article will not help.
Statement 1: Frictional force increases the power required to move an object. This is the most major contributing factor to horsepower loss in drive train. Here is an example.
If you tie a board to a rope and drag it across a linoleum floor and then drag it across a carpeted floor it will be harder to pull on the carpet. If you put a stack of bricks on the board it will be even harder to pull. If you were to drag the same board for the same distance at the same speed and took a temperature measurement on the bottom of the board it would get hotter as you added weight. The heat generated would increase linearly as compared to the amount of weight you put on the board. The force required to move the board would also increase linearly.
For this same reason if you apply 10 ft-lb of torque on a gear it will produce a given amount of friction. If you change the force from 10 ft-lb to 100 ft-lb more friction will be generated by the turning gears. In a transmission or rear end (drive train) this increased friction is seen as heat generated. This is why when you add more power to a vehicle you need to install better coolers to keep everything cool.
Statement 2: It takes more power to accelerate any given object more quickly. This is probably the most obvious statement but is also very critical to understand. Second to frictional losses is acceleration power loss.
If you take a given car that can at best run the quarter mile in 14 seconds the only way to increase this acceleration (other than changing its weigh or drag - next two statements) is to increase the power. The more power you add the faster the vehicle can perform the work of traveling down the quarter mile. This is why we all build more powerful engines! If it wasn't true then we would all be driving Yugo'* at the track.
Statement 3: It requires more power to accelerate a heavier object. This is one of the explanations for the difference in horsepower loss between automatics and manual transmissions and why a Turbo 400 takes more power than a Turbo 350. Here is an example:
Grab a 10lb bag of potatoes and see how long it takes you to get to a full running speed. Now grab a 100lb bag. I can think of millions of more examples but....
Now picture the rotational mass of your transmission. If it weighs more it will take more power to make it accelerate to the same speed.
Statement 4: The faster you want to move an item the more power it takes (drag). Here is an example:
If you take a stock Mustang six cylinder and try and obtain a top speed rating it will probably be around 110 mph. Now get in a V8 Mustang and it will go faster. For an average vehicle to break 200 mph it takes around 500-600 horsepower. As an item accelerates it becomes harder and harder to push it because of the drag induced by the wind. See footnotes for more detailed information.
For the same reason it will take more power to spin a transmission at 7,000 RPM than it will at 6,000 RPM. The gears moving through the fluid in the transmission will require more power at higher RPM'*.
What does a dyno measure?
If you don't have a firm understanding of what horsepower and torque are please read that section before continuing with this proof - this is actually very important. On a dyno sheet you see both torque and horsepower readings, to understand what a dyno measures requires a background in what each is a measurement of.
An engine dyno actually measures engine output torque. It also measures engine speed (RPM). It can then calculate HP from the HP = T*RPM/5252 equation.
A dyno jet type chassis dyno (no load on the drum) measures the instantaneous drum speed, and time during a pull. From both, it can calculate drum acceleration. Knowing the rotational inertia of the drum, it can calculate the applied torque. Then finally, from the calculated applied torque, and the speed of the drum, it can calculate HP applied to the drum using the above formula
Proof #1
An internal combustion engine produces power by �combusting� fuel and air. This combustion process burns fuel in the cylinder at the end of the compression stroke and causes the atmosphere inside the cylinder to expand. This expansion (caused by the heat energy of the combustion process) pushes the piston back the other direction. In an average engine, around 33% of the heat (power is heat) is spent in pushing the piston. Another 33% is lost out the exhaust (the exhaust gas continues to expand for around 18�, this is why turbos work) and another 33% is lost through the radiator (heat transferred through the block).
Now lets state a fact of physics (you might remember this from High School). Energy can not be destroyed, only transformed into another state. What this means is, to optimize power in an internal combustion engine you try to optimize the amount of �heat� (that is, the power created by internal combustion) to be transferred into the �push� process of the piston. You can optimize this in many methods. Ignition timing is the most obvious. By changing timing you can optimize the amount of time the heat has to expand and push the cylinder. The whole point of this is to understand that power is created by heat and lost through the dissipation of heat. Heat expands the atmosphere inside the cylinder and that is how you make power.
So, if heat is power, then you can measure the amount of power by the dissipation of heat throughout a system. Have you ever wondered why you need a bigger radiator when your engine makes more power? Part of the reason is when you produce more horsepower you lose more energy through heat loss into radiator. How about why you start to need things like transmission coolers? Now we are starting to get to the root of my argument on why power loss is based on percentages �vs- a static value.
As you make more power you apply more force on transmission gears and (in automatic transmissions) fluid. As you make more and more power you need to run bigger and bigger transmission coolers and (in extreme cases) rear-end coolers. Why do you need to do this? Because you are generating more heat! That is obvious, right?
Now let�* look at the laws of physics. Energy can not be destroyed, only transformed. So, if your transmission and rear-end are generating more heat, where do you think this heat is coming from? The magic heat fairy? The heat is a direct result of power being transformed (lost) in the form of heat. The more power you produce (in your engine) the more power (heat) you lose in your drive train.
Short version: Increase in heat generated by your transmission or rearend is a sign that more power is being consumed (converted to heat) and therefore lost.
Proof #2: (more complicated and involves some math)
If you don't have a firm understanding of what horsepower and torque are please read that section before continuing with this proof:
To calculate torque, A dyno jet type chassis dyno (no load on the drum) measures the instantaneous drum speed, and time during a pull. From both, it can calculate drum acceleration. Knowing the rotational inertia of the drum, it can calculate the applied torque. This work is represented as lb-ft of torque. A "Dynojet" brand dynamometer uses a "fixed inertia" load, whereas a "Mustang" brand dynamometer uses "loaded inertia" that can simulate driving stresses the vehicle would encounter in actual driving conditions. Typically a "loaded" dynamometer will spin up slower than a fixed resistance dynamometer (this is important later).
Now lets take a look at drag inside the transmission and rearend (the drag of the gears pressing against each other and traveling through the fluid inside the transmission). Land speed racers know that the drag coefficient of a moving object is exponential based on speed (for more definition on drag components see footnotes). What this means is if wind resistance (drag) takes 10 horsepower to overcome in a vehicle moving at 25 mph it will take 40 horsepower at 50 mph (2x speed = (2�) or 4x power requirement). To move the vehicle at 100mph you need 160 horsepower (4x speed = (4�) or 16 x horsepower (10) = 160). Take that to 200 mph and you lose 640 horsepower to drag (8x speed = (8�) or 64 x horsepower (10) = 640). So, the same vehicle that took only lost 10 horsepower due to drag coefficient at 25mph loses 640 horsepower to drag at 200mph.
Short version: The faster an object moves that is subject to drag (air, liquid, or pressure) the more horsepower it takes to move it.
Take two identically equipped vehicles (same transmission and rearend) with one vehicle producing 400 horsepower and the other producing 800 horsepower. The vehicle producing 800 horsepower will accelerate faster than the vehicle producing 400 horsepower. This goes back to our definition - horsepower = work over time. The same vehicle will cause the fluids, pressures, and gears inside the transmission to spin up more quickly (as evident by the increased acceleration). Because this same work is performed in less time, the net result is that more horsepower was required (horsepower = work over time).
Short version: To accelerate an object faster requires more power (horsepower). If this were not true then why would we build high horsepower cars? This is also true of spinning gears inside a transmission or rearend.
Now factor in the drag coefficient of the spinning gears inside the transmission and rearend. Since the force required to spin the gears grows with speed, and since increased horsepower is causing the rate of acceleration to increase (your engine revs faster on the dynamometer) you will find that the actual amount of work required to spin the gears greater than with lower horsepower. Therefore as horsepower goes up the percentage of loss power loss through drivetrain actually increases. This is probably a minimal effect on total drivetrain loss but is still a factor..
Short version: The higher the RPM of the vehicle the more parasitic drag that exists inside the transmission and rearend, regardless of acceleration rate. Therefore more power is spent to spin the drivetrain at high RPM.
Another thing to note: The reported HP (flywheel -or- rear wheel), is highly dependent on acceleration rate. The higher the acceleration rate, the lower the reported HP. So on a dynojet-type dynamometer (fixed inertia), higher horsepower will accelerate the drum quicker, which will increase drivetrain inertial losses, thereby reporting even lower RWHP numbers. This means that the percentage of horsepower drivetrain loss through a Dyno-Jet type dynamometer will actually grow as horsepower increases at a rate larger than a "loaded inertia" dynamometer.
This deserves more explanation. Typically you will find a car that produces 200 horsepower on a Dyno-jet (fixed inertia) will only produce 180 horsepower on a Mustang (loaded inertia) dynamometer - 10% less. However as horsepower increases the numbers (as a percentage) will become closer. So a vehicle making 800 horsepower on a Dynojet may only make 740 horsepower (7.5% less) on a Mustang dynamometer. The percentage has become closer. This is the result of time being reduced (with the dynojet dynamometer) in the equation of work over time.
Real World Example:
(from Muscle Mustang & Fast Ford Magazine, Febuary 2003, Article titled "Mass VS. Myth" by David Vizard)
Sometimes you just luck out. I was trying to figure out how I could put some "proof" in the form of a chart or real world Dyno results. I was just reading the latest issue of Muscle Mustang & Fast Ford magazine and ran across the article "Mass VS. Myth". The article is discussing the pro'* and con'* of using a light weight flywheel. The cool part is the same data they used is also a proof for horsepower loss through drivetrain. Let me explain.
In the article they dyno'd a car in both 1st gear and 4th gear (this was with a manual transmission). Normally all dyno runs are done in whatever gear provides a 1:1 ratio through the transmission (3rd in Automatics, 4th in Manuals typically). What was amazing was how much less rear wheel horsepower was generated when the car was in first gear. Take a look at the chart and the authors notes (quote: "Because of the rapid rate the engine internals and components back to the wheels are accelerated the power absorbed is greater. As can be seen the difference in rear wheel output between first and fourth gear is an amazing 85 hp and a staggering 140 lbs.-ft of torque.")
So, why would a dyno show less power in 1st gear than in 4th gear? The engine is producing the exact same amount of power, the only change is the gear the transmission is in. The answer: In first gear the engine can accelerate the dyno wheel more quickly because of the mechanical advantage it has (lower transmission gear). To accelerate the dynamometer wheel requires the same amount of work in 1st gear and in 4th gear. Because it happens in less time in first gear it must take more horsepower. horsepower = work / time. If you reduce the amount of time and if work remains the same the horsepower must be higher (this is realized as greater horsepower "loss" as it was horsepower "required" to do the "work" of spinning the dynamometer wheel).
Update 12/22/03: I had a person ask me to clarify the reason why horsepower and torque would drop when the car was dyno'd in first gear so I spent a little more time thinking through what is happening. After my first attempt at explaining it my friend Ed Hohenberg helped me refine this explanation. Here is what we came up with.
The reason for the large HP difference between 1st and 4th gear is due to 2 factors:
1. In first gear, the power flow through the transmission involves more gear meshes (high gear is straight through, 1st gear uses the countershaft, with gear meshes at both ends). The additional gear meshes reduce the transmitted power by typically 2% per mesh (for a helical cut gear, less for straight cut), for a combined loss of 4% or so.
2. In first gear, because of the additional torque multiplication produced by the gearing, the applied torque to the dyno drum increases proportionally to the gear ratio (less the additional losses from the gear meshes), which causes the dyno drum to accelerate much faster. Now you might think this would indicate higher HP, but the drum speed is similarly proportionally reduced by the added trans ratio, so neglecting friction and inertia effects, the power would remain the same. But the power measured at the drum DOES decrease, because all the drivetrain components (engine rotating and reciprocating masses, flywheel, trans gears and shafts, driveshaft, rear gears, axles, wheels and tires) all accelerate faster now (but in first gear, the starting and finishing speeds are much less compared to high gear). Since it takes power to accelerate mass, the higher driveline component accelerations use up more power in first gear than high gear, and thus the available power at the dyno drum is decreased by the HP "robbed" in the vehicle drivetrain due to the higher driveline component accelerations.
[EDIT 8/31/04] The following section has caused a lot of confusion lately. As a point I have made a few changes to the section below and will re-state that torque is measured two different ways:
1) Torque measured as force requires no movement and is measured as foot-pounds (ft/lb). It is simply a measurement of how hard something is being pushed on. Imagine a lever with a fulcrum. No picture 1 foot on each side of the fulcrum. Imagine one end is under a 10,000 lb rock. If you pushed with one pound of force you would be applying 1 foot pound of torque against the rock - but it would not move. If you made the lever 10,000 foot long on the side away from the rock and pushed with the same 1 pound of force you would lift the rock. At 9,999 feet the rock would still not move but you would be applying 9,999 ft-lbs of torque. Using first gear in a car will apply more force than in fourth gear however less work is done (vehicle speed is slower).
2) Torque as measured as work requires movement and is measured as pound-feet (lb/ft). Since work requires movement torque measured as work requires the item to move. 1 lb/ft of torque = amount of force to move 1 pound 1 foot. Picture a rope with a 1 pound rock tied to the end. If you pull the rope up 1 foot you have done 1 lb/ft of work. If you move it two feet you have done 2 lb/ft of work. If you put 1,000 lbs on the rope and moved it 1 foot you have done 1000 lb/ft of work.
A dyno measures torque as work as measured by knowing the mass of the dyno drum, the speed of the dyno wheel at the time the measurement starts, and the speed of the dyno wheel at the time the dyno measurment stops. For a dyno each measurement of force is measured per engine revolution.
HP is down, but torque (measured as force) is certainly up in first gear, as the dyno sees it. The reason HP is down is because of the increases losses, of course. If you ignore all losses (frictional, parasitic and inertial) the HP in any gear would be the same, because while the torque (measured as force, not work) is increased proportionally, the speed is decreased proportionally, so the amount of work performed would otherwise stay the same.
But the torque at the dyno and torque at the engine are 2 different things, and I think this may be where some confusion is. The dyno measures the applied torque at the drum, which is higher in first gear - but the speed at the drum is lower. Ignoring the losses the HP *should be* the same. But then the dyno is also measuring the engine speed (RPM), so it takes the dyno measured power, then goes back and calculates torque from the engine speed and dyno HP. The "torque" reported by a chassis dyno is really meaningless (as a measurement of force), since it'* neither the torque measured at the engine (how could it be?), nor the actually torque measured at the dyno drum (that would include the gear multiplications, so unless you have a 1:1 rear axle ratio in addition to a 1:1 trans ratio, the rwtq reported is certainly NOT the actual force at the rear wheels). It is instead a measurement of the amount of work that was performed at the rear wheels and therefore a measurement of torque at the rear wheels as a unit of work.
Further in the article the comparison was done between a lightweight clutch/flywheel and a stock/heavy combination. You will see that the exact same engine produced slightly (14.5 hp) more horsepower with the lightweight setup than the stock/heavy setup. The reason for this was the heavy flywheel requires more "work" to spin. Anytime greater work is required the result will be a reduction in power.
The author in this article included a lot of good information regarding inertia and resultant horsepower recovery. If you have the opportunity I would recommend picking up this magazine for this article. It does make me wonder what would happen if you were to dyno your car in overdrive. Unfortunatly my AOD will not take the abuse of a dyno run in overdrive so I won't be the one to figure this out.
Real World Example #2: (added 9/29/2003)
The November 2003 issue of Car Craft has a great article titled "The Brutal Truth" by Jeff Smith (page 40). In the article they place two engines on an engine dyno and then dyno the engine again once it is installed in the vehicle. One engine is a 357 cubic inch Ford Windsor engine and the other is a 455 Buick. The Ford 357 was installed in a 63 Comet using an AOD transmission and Ford 9" rearend with 3:50 gears (exact combination of drivetrain in my Mustang except my AOD is a non-lockup which means it is even less efficient or should lose even more horsepower). The 455 was installed in a 70 Buick GS with a Muncie 4-speed and a 12 bolt rearend with 2.73 gears. The point of the article was to show how things like a belt driven cooling fan or poor vehicle exhaust could affect the engine output in the vehicle but was equally as valid for showing drivetrain induced power loss.
Without reading any further it would be my assumption that the Ford combination would lose a larger percentage of power through the drivetrain. Not only is the AOD an Automatic it is also a 4 speed Automatic that has substantial weight and rotating mass. The 9" rearend is also larger and heavier than the 12 bolt Chevrolet.
The Ford 357 produced 371 horsepower on the engine dyno at 5,000 RPM. On a 1990 Mustang that came stock with an AOD and a 3.27 8" rearend (more efficient than the 9") the rear wheel horsepower is typically 180 hp. That represents a loss of 45 horsepower given the rated 225 flywheel horsepower on that vehicle. Using this 45 horsepower and even giving it another 5 for the 9" rearend the 357 would have produced 321 peak horsepower on the chassis dyno. Well, it didn't! Even after removing all the factors that could have contributed to extra power loss in the vehicle (removing the belt powered cooling fan) the chassis dyno only showed 283 hp. In fact over the entire power curve the difference between the engine dyno and the chassis dyno was 24%. This provides more evidence that the power loss through common drivetrain remains a percentage even as power is increased rather than remaining a static loss value.
The Buick 455 produced 329hp and made 280 through the drivetrain at 4,500 RPM. The average drivetrain horsepower loss in this vehicle was 18.3%. This can be accounted for by the fact that the 4 speed Muncie is more efficient (require less power to accelerate) as well as the 12 bolt rearend being more efficient than the 9".
Why does an Automatic take more power than a Manual transmission?
So, what about the 15%-manual and 20% automatic? Well, they are good places to start. Some transmissions are more efficient than others and some rear ends are more efficient than others. In the end, none of this matters as wheel horsepower is what is actually used. You will find the heavier the transmission parts (gears, shafts, etc.) the more power they will take. I have seen engine dyno comparisons to rear wheel and you typically see 15-17% for manual transmissions and 20-25% for automatics. How much does your transmission take? Take the basic values 15 and 20 percent and consider the following.
HP loss in auto vs stick is mostly related to the converter slip (there is always some slip in a normal converter). However, even with a mechanically locked up converter, in the planetary gear system used in autos, there are more gear meshes occurring, which increases HP losses since each gear mesh results in a HP loss (relates back to statement 1). And don't forgot, in an auto, you have direct pumping losses from the oil pump (you don't have this in a manual trans). And the higher the line pressure, or fluid flow rate, the greater the pumping losses.
In an Automatic transmission you will find several factors that determine power requirements.
Weight: a comparable C4 take less power than a C6 - primarily because of the weight difference of the moving parts. A turboglide would take even less and an AOD would require around the same as a C6 (just using weight)
Lockup/non-lockup: Many modern Automatics use a lockup feature in drive and overdrive. This will increase horsepower available at the rear wheels because you don't have any slip in the torque converter. The slip in the torque converter generates heat which is an indicator that power is being lost. Even at over the stall point of a torque converter where you have complete hydraulic lock up you will experience some power loss through the hydraulic coupling process. Having said this I still prefer non-lockup converters as they are easier on the drivetrain (less harsh shifts) and allow the engine to run in the proper RPM power range.
Stall speed: The purpose of installing a high stall torque converter is to allow the engine to spin up to a desired RPM range quickly. Once this stall speed is reached you have a hydraulic lockup and the transmission is locked to the same RPM as the engine. Measuring horsepower below the stall speed of a torque converter is completely meaningless. The slip will give bad readings. Normally you can see the "flash" or hydraulic lockup point on the dyno sheet (I will post an example later). Having said all of this a high speed stall converter will normally take less power than a stock type converter. The reason is a high speed stall converter is typically smaller and lighter - that is the only reason.
Torque converter design: I do not know enough about this to go into detail however some designs are more efficient than others. With torque converters expect to spend between $500-$1000 for a good quality converter - if you try to go cheap here you will only hurt yourself later. In my AOD I run a Lentech 9" 2,800 RPM converter that flashes (hydraulic lockup) at 3,400 RPM (higher torque output will push your flash point further in the RPM band).
Valve Body: Again, I don't know enough to tell you exactly why but some are better than others. Some valve bodies will route hydraulic fluid more efficiently and activate clutch packs more effectively. For my AOD I use a Lentech Strip Terminator valve body and I love it.
With a manual transmission your primary factor in power loss is weight. The heavier the gears, clutch, flywheel, etc. the more power required to turn it.
Thank you for your time! If you have any questions or comments please email me at garnerjohn@earthlink.net
* Footnotes:
Explanation for the value of 5,252 RPM being used to calculate horsepower provided by Ed Hohenberg (thank you Ed).
Work is force through a distance, as you correctly stated. For linear motion it'* simply Force X distance. For angular motion, the distance the force moves is again used, only now that distance will be the circumference of the circle that the force moves through, or 2 * PI * radius. If you assume an imaginary 1 foot long radius for a moment arm, the distance is simply 2*PI. So then for angular WORK, the Force X distance = Torque X 2*PI. If we're turning at 1 RPM, then the 2*PI distance is covered every 1 minute (1 revolution per minute). At 2 RPM for example, we're now doing the same work, but in half the time (1 revolution in 1/2 minute), and so on for higher speeds. So our angular Power, Work/time = T X 2*PI X RPM, which gives us power in units of lb-ft/minute. To get that into lb-ft/sec divide by 60, so Power = T X 2*PI X RPM/60, or T X RPM X 2*PI/60. Lastly, "Horse Power", as defined by Mr. Watt, is 550 lb-ft/second, so to get our angular power from lb-ft/second into HP, we divide by 550, so finally:
Power = T X RPM X 2*PI/60/550, or T X RPM X .0001904 or T X RPM/5252.11
Explanation of Drag coefficient provided Ed Hohenberg (thanks again!)
HP losses for a moving vehicle have 3 basic components: inertial, rolling, and aerodynamic. The inertial HP losses will be proportional to mass, acceleration and velocity. The rolling resistance losses will be proportional to mass, rolling/friction coefficients and velocity. The aerodynamic HP losses will be proportional to drag coefficient, frontal area, and velocity squared.
The magnitude of each component can vary significantly, depending on the input variables. For constant speed (no acceleration) the inertial component is zero, so only rolling resistance and aerodynamic losses are present. For a typical car, the rolling resistance component will have the greatest magnitude up to about 50 mph. For a very sleek car like a new Corvette, the aerodynamic component probably won't exceed the rolling HP losses until 65 mph.
Article from 5.0 Mustang & Super Ford magazine that caused me to write this page.
(February 2002, page 54, sub content titled "What'* the loss" from the article "Heads or Tails", page 47, by Eric English)
(Note: I really enjoyed the article in whole and only have an issue with the authors opinion on horsepower loss through drivetrain. I have heard this authors opinion expressed by many others and after a cursory review of his logic it does appear to make sense. When you dig deeper and analyze each of his assumptions you realize they are flawed. I realize the Internet and Magazines are a great way to spread both information and dis-information. I only hope that those who read this page will take the time to understand the math and theories expressed and know how to answer the next person who expresses the opinion expressed in the article below)
What'* the Loss?
When we began to contemplate flywheel horsepower figures for our different combinations, we bantered about some concepts that can be deceiving. Most enthusiasts have been exposed to the idea that flywheel and rear-wheel horsepower can be equated by factoring in a given percentage for drivetrain loss - the drag that occurs from all the items between the flywheel and the rear tires. You may have seen factors such as 15 percent for stick-shift cars and 25 percent for automatics, applied by dividing rear-wheel horsepower by either 0.85 for stick-shifts or 0.75 for automatics.
Now take a time-out and consider the following. Our original baseline indicated 195 hp at the rear wheels, which when divided by 0.85 equates to 229 flywheel horsepower, and implies that the drivetrain is absorbing some 34 horsepower. On the other hand, our combination of blower and traditional bolt-ons netted nearly 340 hp on the Blood Enterprises dyno, which when divided by 0.85, equates to 400 flywheel horses, and implication that the drivetrain is now absorbing 60 hp.
Nothing has changed between the flywheel and the rear wheels on our '93 LX, so does it make sense to figure the drivetrain is now absorbing nearly twice as much power? Such a concept just doesn't jibe in our little brains, so we asked a couple of people in the biz what they thought. Lee Bender of C&L Performance and Paul Svinicki of Paul'* High Performance are both well versed in evaluating Mustangs on the dyno, and they both agreed that extrapolating drivertrain horsepower loss via percentages is flawed. Lee believes that the stick Mustangs experience roughly a 35hp loss through the drivetrain, whether they make 200 hp or 400 hp. He did explain that ultra-high-powered vehicles - typically race cars - can be and exception to this rule, but that'* a topic for another time. Interestingly, a 35hp loss for stick-shifted drivetrains is strikingly similar to the difference between Ford'* horsepower ratings and the rear-wheel numbers we've observed on dynos across the nation. Hmmm...
04-06-2006, 09:13 PM
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Junior Member
Posts like a Ricer Type-R
Been posted before. In fact, I have express permission from the Author to tailor the context of his article to our use in Techinfo. One of the many things on the 'list'.
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