Post by Luis on Aug 12, 2005 22:23:40 GMT -5
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 I disagreeumptions 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
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].
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)
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.
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.S. 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's 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's 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's 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's 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.S. 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's 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's "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 (6.28/6.28) x RPM
= ----------------------------
(33,000 / 6.28)
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's obvious that at 5,252 rpm, the horsepower and torque values are always equal. That's why an engine's torque and horsepower curves always cross at 5,252.
Second, assuming engine displacement is fixed, raising the engine's 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's 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's 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's.
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’s 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 = (2²) or 4x power requirement). To move the vehicle at 100mph you need 160 horsepower (4x speed = (4²) or 16 x horsepower (10) = 160). Take that to 200 mph and you lose 640 horsepower to drag (8x speed = (8²) 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's and con's 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.
CONTINUE NEXT MESSAGE
This article has taken on a life of its own. I originally created it (doing the research) to verify I disagreeumptions 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
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].
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)
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.
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.S. 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's 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's 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's 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's 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.S. 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's 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's "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 (6.28/6.28) x RPM
= ----------------------------
(33,000 / 6.28)
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's obvious that at 5,252 rpm, the horsepower and torque values are always equal. That's why an engine's torque and horsepower curves always cross at 5,252.
Second, assuming engine displacement is fixed, raising the engine's 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's 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's 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's.
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’s 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 = (2²) or 4x power requirement). To move the vehicle at 100mph you need 160 horsepower (4x speed = (4²) or 16 x horsepower (10) = 160). Take that to 200 mph and you lose 640 horsepower to drag (8x speed = (8²) 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's and con's 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.
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