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Can somebody explain to me why people think drivetrain loss is a set percentage based on layout? Such as FWD, RWD, AWD.
Example, owner of AWD car "A" tells me that drivetrain loss for AWD 5spd cars is 20%.
Soo... if the car makes 200hp to the crank that means it makes 160whp.
400hp to the crank = 320whp.
800hp to the crank = 640whp.
This makes no sense to me.
TLAR calculations.
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What's worse is that I have not seen data explaining any of it. The closest that I have seen is a Hot Rod/Car Craft test of rear end types. Each of the ones tested took between 10-15 hp if I remember correctly.
Go dyno your car. That will give you an answer.
- N. Sperlo -:-:- "Never take life seriously. No one ever gets out alive anyway." ~ ~ A strong tail wind can't hurt either...~~ K0HOF
I thought driveline loss was when the engine and transmission fell out of the car?
N Sperlo wrote: Go dyno your car. That will give you an answer.
It would... if i had time and access to both an engine and a chassis dyno.
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I agree that it sounds very sloppy, and I'd like to know what the "real" best guesses from an engineering standpoint would be.
That being said, the losses are frictional in nature (except on a pure inertial dyno, in which case driveline mass will affect numbers? are any of them pure inertial with no load?), so it makes sense to me that as the amount of power you put through increases, so do the losses.
Boost torque 10%, and the forces pressing gear teeth together go up 10%. Rev higher, all the touchy-spinny bits spin faster.
EDIT: distinguishing cars for ballparking by FWD, RWD, AWD makes some sense to me. For instance, only RWD and AWD need to muck about with hypoid gears and longitudinal/horizontal transitions, right? The 5-speed thing I think is a red herring, and you'll more variation between transmission models for reasons other than number of gears.
It seems plausible to me that it would be fairly linear. Though I would also think that a figure like 20% is understood to be very ballpark.
"In theory, there is no difference between theory and practice. But, in practice, there is." - Jan L.A. van de Snepscheut
Why does loss make no sense to you?
It takes energy to turn a rotating mass and the more of it you have between the engine and tires, the more loss you have.
The amount isn't set - but a front engined rear drive car can only mitigate so much. It still has a transmission, flywheel, driveshaft, differential and two axles to waste energy on and since those components are designed for loads similar to each other - they have a similar combined mass and similar losses across vehicles so you can ballpark it as some percentage and be accurate to a few percent.
Lord of drivel and harbinger of Floundering
What doesn't make sense to me is that the same drivetrain when attached to an 800hp motor magically takes 4x as much power to turn as it did when attached to a 200hp motor. It's the same drivetrain. Same gears, same force required to turn said gears.
I'm not convinced that a static percentage is anywhere near accurate.
The friction needed to be overcome would vary by RPM, not by whatever amount of power happens to be attached to the other side.
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92CelicaHalfTrac wrote: The friction needed to be overcome would vary by RPM, not by whatever amount of power happens to be attached to the other side.
I think this is where you've gone awry.
Friction is a combination of the friction coefficient of the contact (how much force it takes to move it per unit of force pressing the contact point together) multiplied by that force pressing the contact together. To give a simple but obvious example, you can slide a tire along the ground just fine, but press down on it with 500 pounds of the corner of a car, and then, not so much.
A contact point with a friction coefficient of 1 would require 100 pounds to move it for every 100 pounds pressing on the contact point. Mercifully, gears do much better than 1...
An increase in power is the result of either more torque or more RPM. Torque in this case increases the loading of the contact point, RPM increases how much that contact point moves for a given period of time. Increasing either one increases the losses.
"In theory, there is no difference between theory and practice. But, in practice, there is." - Jan L.A. van de Snepscheut
Why is it so much of the stuff I like best gets files under titles like "one of THOSE discussions..."?
"In theory, there is no difference between theory and practice. But, in practice, there is." - Jan L.A. van de Snepscheut
ransom wrote:92CelicaHalfTrac wrote: The friction needed to be overcome would vary by RPM, not by whatever amount of power happens to be attached to the other side.I think this is where you've gone awry.
Friction is a combination of the friction coefficient of the contact (how much force it takes to move it per unit of force pressing the contact point together) multiplied by that force pressing the contact together. To give a simple but obvious example, you can slide a tire along the ground just fine, but press down on it with 500 pounds of the corner of a car, and then, not so much.
A contact point with a friction coefficient of 1 would require 100 pounds to move it for every 100 pounds pressing on the contact point. Mercifully, gears do much better than 1...
An increase in power is the result of either more torque or more RPM. Torque in this case increases the loading of the contact point, RPM increases how much that contact point moves for a given period of time. Increasing either one increases the losses.
Why would i press down on the tire? Wouldn't i just push it from behind to go faster? (Make more power)
I think what i'm not getting my brain around is that when a gear is presented with more force... it's somehow harder to turn?
The only things i can think of that are like that would be those old exercise bikes with the fans in place of the wheels, but i'm not able to draw that parallel to this.
So basically, you would agree with the power figures in the first post?
Does anyone know of an article where they took identical cars with the same drivetrain, one with more power than the other, and did both an engine and chassis dyno to measure the loss?
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92CelicaHalfTrac wrote:N Sperlo wrote: Go dyno your car. That will give you an answer.It would... if i had time and access to both an engine and a chassis dyno.
Not to be too much of a smart ass, but maybe a good estimate of horsepower minus 10-15% would be a safe guess. Go find someone with a similar setup (knowing you, that doesn't exist) and see what kinds of ponies are popping out of their hood.
- N. Sperlo -:-:- "Never take life seriously. No one ever gets out alive anyway." ~ ~ A strong tail wind can't hurt either...~~ K0HOF
I think what i'm not getting my brain around is that when a gear is presented with more force... it's somehow harder to turn?
Yes. The more force applied the more friction that is generated to resist it right up until something gives.
Lord of drivel and harbinger of Floundering
N Sperlo wrote:92CelicaHalfTrac wrote:N Sperlo wrote: Go dyno your car. That will give you an answer.It would... if i had time and access to both an engine and a chassis dyno.
Not to be too much of a smart ass, but maybe a good estimate of horsepower minus 10-15% would be a safe guess. Go find someone with a similar setup (knowing you, that doesn't exist) and see what kinds of ponies are popping out of their hood.
Ohh... i think you're missing my confusion.
I'm trying to figure out why driveline loss isn't as static as i think it should be.
So i would need my car, someone else's stock car. An engine dyno, and a chassis dyno.
Measure my engine on engine dyno, then my car on chassis dyno. See how much was lost.
Then do the same for the stock car. Compare losses. I would bet that they would be extremely close, representing a change in percentage, really.
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I've chosen my metaphors sloppily...
In my tire example, I was simply pointing out that any interface of two materials gets harder to slide the more you press the materials together. (EDIT: specifically, I was talking about how you can slide a tire by hand with only the weight of the tire pressing it down, but not when there's a car pressing it down)
Now imagine that you have two gears. One of them is fixed so it cannot rotate (welded to the shaft, which is welded to, let's say a workbench). The other gear which is meshed to the first has a handle on it. If you press on that handle, you are applying torque. You'd agree that the harder you press on the handle, the harder the gear teeth are pressed together, right?
That's what happens with increased torque.
Gear meshing involves contact between the teeth, and a certain amount of sliding. Increasing torque causes that sliding to require more force. Increasing torque increases how hard the teeth are pressed together, resulting in a proportional increase in friction.
The alternative of NOT increasing torque, but increasing RPM, allows the amount of force to slide to remain the same, but you have to increase the distance slid by the same proportion you increase RPM.
I would not necessarily agree with the numbers in the first post because I have no baseline numbers. I'm merely pointing out that the increase in losses with increase in power makes sense. Whether 20% is anywhere near correct isn't something I know.
"In theory, there is no difference between theory and practice. But, in practice, there is." - Jan L.A. van de Snepscheut
Giant Purple Snorklewacker wrote:I think what i'm not getting my brain around is that when a gear is presented with more force... it's somehow harder to turn?Yes. The more force applied the more friction that is generated to resist it right up until something gives.
So the amount of force necessary to turn a gear changes, even if it's the same gear, merely because the power the gear is attached to changes?
This is why i'm not an engineer... i still see friction, or at least the force necessary to overcome it, as a static number at a certain RPM.
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ransom wrote: I've chosen my metaphors sloppily... In my tire example, I was simply pointing out that any interface of two materials gets harder to slide the more you press the materials together. (EDIT: specifically, I was talking about how you can slide a tire by hand with only the weight of the tire pressing it down, but not when there's a car pressing it down) Now imagine that you have two gears. One of them is fixed so it cannot rotate (welded to the shaft, which is welded to, let's say a workbench). The other gear which is meshed to the first has a handle on it. If you press on that handle, you are applying torque. You'd agree that the harder you press on the handle, the harder the gear teeth are pressed together, right? That's what happens with increased torque. Gear meshing involves contact between the teeth, and a certain amount of sliding. Increasing torque causes that sliding to require more force. Increasing torque increases how hard the teeth are pressed together, resulting in a proportional increase in friction. The alternative of NOT increasing torque, but increasing RPM, allows the amount of force to slide to remain the same, but you have to increase the distance slid by the same proportion you increase RPM. I would not necessarily agree with the numbers in the first post because I have no baseline numbers. I'm merely pointing out that the increase in losses with increase in power makes sense. Whether 20% is anywhere near correct isn't something I know.
Ok, i think i'm with you now....
One more question, i think: Aren't the gears "stationary" in automotive usage, though? I know they turn, but there's no handle keeping them down...? Or at least i think? I'm not seeing how power has an effect on teeth clearance.
But i think i'm starting to get it. Would you agree with me that keeping a fixed percentage isn't exactly accurate, and serves more as a very rough baseline, though? (Not speaking specifically to 20%, just in terms of using an absolute percentage in general)
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In reply to 92CelicaHalfTrac:
So this is a question in theory. Is this what when you're bored?
- N. Sperlo -:-:- "Never take life seriously. No one ever gets out alive anyway." ~ ~ A strong tail wind can't hurt either...~~ K0HOF
If you're trying to sell things, use a percentage. It lets you artificially jack up your power numbers nicely. I've seen one shop use a combination of fictitious correction factors to actually double the measured horsepower to come up with a flywheel number.
And it is partially a percentange, partly due to inertial effects. The faster you accelerate a drivetrain (ie, when you're increasing power and using an inertial dyno) the more power it will suck up. It's equivalent to measuring the difference in flywheel weight. I've done tests on our dyno where all I changed was the timespan of the test sweep and seen a surprising variation in the measured power output.
But it's also a fixed amount, and that's going to depend on drivetrain layout as well as how good the engineers for Brand X are. So really, there is no nice clean correction factor. I tend to use a fixed amount when someone really really really wants a flywheel number (26 hp on a Miata seems to be pretty consistent from 1990-06 stock cars) because I'd rather be a bit conservative.
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Ah, here's that dyno run. Same car, same dyno - the only difference is that one run went through the sweep in 15s while the other took 25s.
http://www.flyinmiata.com/tech/dyno_runs/NC_sweep_times.pdf
One nice thing about our dyno is that we can control the rate of acceleration, which does keep inertial losses as a constant.
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Keith wrote: If you're trying to sell things, use a percentage. It lets you artificially jack up your power numbers nicely. I've seen one shop use a combination of fictitious correction factors to actually double the measured horsepower to come up with a flywheel number. And it is partially a percentange, partly due to inertial effects. The faster you accelerate a drivetrain (ie, when you're increasing power and using an inertial dyno) the more power it will suck up. It's equivalent to measuring the difference in flywheel weight. I've done tests on our dyno where all I changed was the timespan of the test sweep and seen a surprising variation in the measured power output. But it's also a fixed amount, and that's going to depend on drivetrain layout as well as how good the engineers for Brand X are. So really, there is no nice clean correction factor. I tend to use a fixed amount when someone really really really wants a flywheel number (26 hp on a Miata seems to be pretty consistent from 1990-06 stock cars) because I'd rather be a bit conservative.
Ok, this makes sense.
So you give 26hp loss on a stock 1.8 Miata. If you had to hazard a rough guess (no, i won't hold you to it ) as to how much power is being lost by Igor, what would you say?
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N Sperlo wrote: In reply to 92CelicaHalfTrac: So this is a question in theory. Is this what when you're bored?
Yeah, that, or scrounging around my garage wondering what other horrible things i can do to my car.
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92CelicaHalfTrac wrote: One more question, i think: Aren't the gears "stationary" in automotive usage, though? I know they turn, but there's no handle keeping them down...? Or at least i think? I'm not seeing how power has an effect on teeth clearance.
I'm not sure what you mean by "keeping them down"...
It's not a matter of clearance. They're touching (save for a film of oil, hopefully?)... Again I've chosen my examples a bit sloppily. In my example with the handle, we're not seeing the whole process. There's no motion, so there's no sliding. I was just trying to point out that as torque increases, the teeth are pressed together harder.
But torque is torque, whether or not there is motion. It's a force. If I apply a one-pound force to the end of a lever, that's one pound-foot of torque, whether it's stationary or turning 1 RPM or 10,000 RPM. The amount of power changes with RPM if the torque remains the same.
Wait, I think I've got it... I think you might think I'm talking about torque pressing the gears against each other harder, as in trying to move the centers toward one another. That's not it. What I'm talking about is the teeth themselves. if you look at two teeth right in the middle of the meshing section, one from each gear, they are pressed together. That's how the force is transferred. The harder you try to turn gear A, the harder gear A's teeth have to push gear B's teeth. As they mesh and unmesh, there's a sliding motion, and the harder the teeth are pushing one another, the more friction there is at that point.
I get the feeling I'm doing a consistently horrible job trying to give a clear explanation of what should be a pretty simple principle...
But i think i'm starting to get it. Would you agree with me that keeping a fixed percentage isn't exactly accurate, and serves more as a very rough baseline, though? (Not speaking specifically to 20%, just in terms of using an absolute percentage in general)
I'm not 100% certain, but I would expect frictional losses to be fairly linear for both increases in torque and RPM. So I could be wrong, but I don't see a big problem with using a percentage.
Again, saying something like "20% for drivetrain loss" is such a ballpark figure that I would expect the generalization to have at least as much error as the changes that occur with relation to power.
I'm pretty confident in the underlying concepts here, but I don't have any empirical data, which would trump the ever-living crap out of what I've got.
"In theory, there is no difference between theory and practice. But, in practice, there is." - Jan L.A. van de Snepscheut
EDIT: The more torque you make, the harder the flanks of the "pinion" teeth are pressed against the flanks of the "gear" teeth.
EDIT EDIT: The sliding motion we've been referring to is visible as you watch the contact between the gears happen along that red line.
"In theory, there is no difference between theory and practice. But, in practice, there is." - Jan L.A. van de Snepscheut
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