What are the losses through a Zf? I've heard everything from 15-25% Does anyone have dyno proof?
Will
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This has been kicked around many times before.
Bear in mind engine flywheel dyno readings vs in-car readings have more than ZF factors:
ZF
exhaust system
u joints
bearings
tires/wheels
IIRC, an 'agreed' figure for loss was about 23%
To figure HP using this figure:
FWHP times .77 = RWHP
RWHP times 1.3 = FWHP
Larry
Bear in mind engine flywheel dyno readings vs in-car readings have more than ZF factors:
ZF
exhaust system
u joints
bearings
tires/wheels
IIRC, an 'agreed' figure for loss was about 23%
To figure HP using this figure:
FWHP times .77 = RWHP
RWHP times 1.3 = FWHP
Larry
19% was the figure I was quoted from various sources when I asked this question, which seems logical being slightly less than a conventionl front engined cars transmission loss.
This has been kicked around before, but shouldn't the loss be an absolute number and not a relative number? Ie, 75 hp is absorbed through the ZF. If it were a percentage, then a high HP engine would lose more power through the same ZF than a stock engine. The ZF doesn't change, so the loss should remain constant.
RIght?
RIght?
I would imagine the higher the torque of the motor the more the pressure is on the gear tooth face. This would cause more friction therefore a greater power loss. A percentage makes sense in this case.
Blaine
Blaine
From memory, some pretty high powered engines were showing some pretty modest RWHP on the dyno days.
If that is an accurate observation, a percentage loss would tend to explain it. However, the logic of that assumption is not apparent to me. I would assume that the hp drain was constant depending on the fluid temp and thickness and perhaps the tightness of the ZF bearings in a particular engine.
I do know that in the past there have been transmission specialty shops that would "blueprint" your box. Amongst other things, they would go through with a hand grinder polishing the edges of the gears.
I guess, thinking about it that it is possible to pick up a few hp by reducing parasitic losses in the drive train just about anywhere.
OK, I'll bet on a constant hp loss. Now we have to gather about a dozen Pantera and go dyno them. Anybody in?
If that is an accurate observation, a percentage loss would tend to explain it. However, the logic of that assumption is not apparent to me. I would assume that the hp drain was constant depending on the fluid temp and thickness and perhaps the tightness of the ZF bearings in a particular engine.
I do know that in the past there have been transmission specialty shops that would "blueprint" your box. Amongst other things, they would go through with a hand grinder polishing the edges of the gears.
I guess, thinking about it that it is possible to pick up a few hp by reducing parasitic losses in the drive train just about anywhere.
OK, I'll bet on a constant hp loss. Now we have to gather about a dozen Pantera and go dyno them. Anybody in?
quote:Originally posted by Charlie McCall:
This has been kicked around before, but shouldn't the loss be an absolute number and not a relative number? Ie, 75 hp is absorbed through the ZF. If it were a percentage, then a high HP engine would lose more power through the same ZF than a stock engine. The ZF doesn't change, so the loss should remain constant.
RIght?
While I can't cite any engineering sources, logically the loss has to be a function of the input (and that is always the way that I've seen flywheel HP calculated from rear wheel horsepower).
Think about it like this... using your hypothetical 75 HP constant loss, if someone put a 50 HP engine in their Pantera and put it on a rear wheel dyno, could the 50 HP engine turn the rear wheels through the ZF? If so, (and I think it would) then the loss couldn't possibly be a constant 75 HP.
The actual function almost certainly isn't a straight percentage either. It's just close to linear in the normal power range of most passenger car engines - so losses are estimated using a percentage. More than likely, the percentage loss increases at the power increases. (this might be a poor analogy, but sort of like wind drag, where the drag force increases as a function of the square of the velocity)
The number is a percentage, not a fixed value, this is due to the fact that inertia, friction, heat increase as the power being transmitted increases, the speed of the moving components increases, etc.
On an engine dyno a motor generally inhales cooler air, has no air filter assembly, and exhausts through an un-muffled exhaust system. The air filter, warmer air of the engine compartment, the muffled exhaust system all contribute to reducing the motors output when it is installed in a car. Once on the chassis dyno, the weight & friction of the rotating parts also contribute to the reduction in bhp.
There is no one fixed percentage per se, because each Pantera is a little different, and things like the varying weight of the rear wheel & tire assembly from one car to the next will influence the numbers. However, time and time again, when I have seen both numbers for the same vehicle, the difference always falls in the 19% range. I normally round that off to 20%. You can bet your numbers will not be far off if you calculate using 20%.
cowboy from hell
On an engine dyno a motor generally inhales cooler air, has no air filter assembly, and exhausts through an un-muffled exhaust system. The air filter, warmer air of the engine compartment, the muffled exhaust system all contribute to reducing the motors output when it is installed in a car. Once on the chassis dyno, the weight & friction of the rotating parts also contribute to the reduction in bhp.
There is no one fixed percentage per se, because each Pantera is a little different, and things like the varying weight of the rear wheel & tire assembly from one car to the next will influence the numbers. However, time and time again, when I have seen both numbers for the same vehicle, the difference always falls in the 19% range. I normally round that off to 20%. You can bet your numbers will not be far off if you calculate using 20%.
cowboy from hell
Heres a wrench in the systsem.
How much power do you lose a 2000 ft elevation compared to sea level? also, what about correctve numbers? I had my pontiac dynoed at sea level (629 HP @ 5500RPMS ) engine dyno.
then dynoed at 2000ft, chassis dyno with no corrective numbers simply it is what it is to the ground and made 392 rwhp.
I have been told and believe correct that you lose a point of compression per 2000ft elevation.
wheel-stock rims/255-60/15
TKO600 5spd trans
stock 12 bolt w/3:73
same fuel both times.
I will do the same testing with the GT5.
MME final #680HP, I will install it and then post the RWHP numbers, again with no corrected numbers.
Daniel
How much power do you lose a 2000 ft elevation compared to sea level? also, what about correctve numbers? I had my pontiac dynoed at sea level (629 HP @ 5500RPMS ) engine dyno.
then dynoed at 2000ft, chassis dyno with no corrective numbers simply it is what it is to the ground and made 392 rwhp.
I have been told and believe correct that you lose a point of compression per 2000ft elevation.
wheel-stock rims/255-60/15
TKO600 5spd trans
stock 12 bolt w/3:73
same fuel both times.
I will do the same testing with the GT5.
MME final #680HP, I will install it and then post the RWHP numbers, again with no corrected numbers.
Daniel
I had always been told to allow 3% reduction in HP for every 1,000 ft in elevation above sea level. On the local Mallory dyno in Carson City, that's an 18% loss! I definitely noticed the difference when I went to Monterey for the Concurso. I also know at least one local Pantera guy running 12.5:1 CR on pump gas at this elevation and puttung down around 510 RWHP.
quote:Originally posted by Perry H:quote:Originally posted by Charlie McCall:
This has been kicked around before, but shouldn't the loss be an absolute number and not a relative number? Ie, 75 hp is absorbed through the ZF. If it were a percentage, then a high HP engine would lose more power through the same ZF than a stock engine. The ZF doesn't change, so the loss should remain constant.
RIght?
While I can't cite any engineering sources, logically the loss has to be a function of the input (and that is always the way that I've seen flywheel HP calculated from rear wheel horsepower).
Think about it like this... using your hypothetical 75 HP constant loss, if someone put a 50 HP engine in their Pantera and put it on a rear wheel dyno, could the 50 HP engine turn the rear wheels through the ZF? If so, (and I think it would) then the loss couldn't possibly be a constant 75 HP.
The actual function almost certainly isn't a straight percentage either. It's just close to linear in the normal power range of most passenger car engines - so losses are estimated using a percentage. More than likely, the percentage loss increases at the power increases. (this might be a poor analogy, but sort of like wind drag, where the drag force increases as a function of the square of the velocity)
OK, the same question phrased differently. How does the ZF know if it is connected to a 75hp engine or a 750hp engine? Shouldn't the internal resistance be the same? A higher powered motor will be able to overcome that resistance more easily, but the resistance will be the same, right?
(btw - general consensus seems to be that it is a percentage.. just trying to understand the logic behind that)
quote:OK, the same question phrased differently. How does the ZF know if it is connected to a 75hp engine or a 750hp engine? Shouldn't the internal resistance be the same? A higher powered motor will be able to overcome that resistance more easily, but the resistance will be the same, right?
(btw - general consensus seems to be that it is a percentage.. just trying to understand the logic behind that)
It knows it by the bearing surfaces being pushed on harder. It is like when you lift weights. The heavier the weights, the harder you sweat and the more load it puts on your joints. The extra friction causes extera heat, which makes things expand, which cause extra friction, and so on clear until she freezes right up and breaks something. That is why they say you shouldn't hook a 1200 horsepower engine up to a stock ZF.
This question has come up a bunch of times and as has been mentioned, a figure in the low 20%s range is the typical answer. Most results I have seen correlate this figure to cars in 450-550 bhp (at flywheel) range of power.
Drive train losses tend to be related to friction, viscous behavior of lubes, and other mechanical losses in gears, bearings, & rotating joints. These losses add through every component in series fashion (the percentage loss through each component multiplying to the next). Friction is not constant but a function of load. More torque = more friction. For drive train, it’s also a function of speed at steady state because of viscous behavior of lubricants. These types of losses can become strong functions of velocity. During acceleration, you also have to throw in inertial losses for the energy to spin the drive train mass up to speed. –You don’t get that for nothing.
Most of what you don’t get to the ground is lost to heat, but vibration, inertial, and slip creep in there too. These can become strong functions of velocity. In a Pantera, you can typically gain back a couple percentage points (some claim more) with CV joints as opposed to the u-joints. U-joints actually exhibit angular acceleration and deceleration multiple times per revolution depending upon suspension/axle geometry. This causes a parasitic loss and can also affect tire grip.
IMO, low ~20%s drive train loss is not bad for a manual transmission street car.
My favorite form of energy loss is the tire slip that routinely occurs through first, second, and well into third gear.
Kelly
Drive train losses tend to be related to friction, viscous behavior of lubes, and other mechanical losses in gears, bearings, & rotating joints. These losses add through every component in series fashion (the percentage loss through each component multiplying to the next). Friction is not constant but a function of load. More torque = more friction. For drive train, it’s also a function of speed at steady state because of viscous behavior of lubricants. These types of losses can become strong functions of velocity. During acceleration, you also have to throw in inertial losses for the energy to spin the drive train mass up to speed. –You don’t get that for nothing.
Most of what you don’t get to the ground is lost to heat, but vibration, inertial, and slip creep in there too. These can become strong functions of velocity. In a Pantera, you can typically gain back a couple percentage points (some claim more) with CV joints as opposed to the u-joints. U-joints actually exhibit angular acceleration and deceleration multiple times per revolution depending upon suspension/axle geometry. This causes a parasitic loss and can also affect tire grip.
IMO, low ~20%s drive train loss is not bad for a manual transmission street car.
My favorite form of energy loss is the tire slip that routinely occurs through first, second, and well into third gear.
Kelly