Post by Luis on Aug 12, 2005 22:25:09 GMT -5
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's 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 I disagreeumption 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.
* 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's 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's 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's 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's 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's a topic for another time. Interestingly, a 35hp loss for stick-shifted drivetrains is strikingly similar to the difference between Ford's horsepower ratings and the rear-wheel numbers we've observed on dynos across the nation. Hmmm...
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's 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 I disagreeumption 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.
* 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's 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's 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's 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's 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's a topic for another time. Interestingly, a 35hp loss for stick-shifted drivetrains is strikingly similar to the difference between Ford's horsepower ratings and the rear-wheel numbers we've observed on dynos across the nation. Hmmm...
