Omega Doom

iTrader: (2)

another thread got me thinking and I'm bored but anyway I know theres like a 20% frictional loss thru the auto trannys but is it more if you have a big stall?

so if my car dynode 414rwhp 390rwtq what would the crank HP and TQ #s be and if I have a 4K yank stall along with 4.10 gears if that matters? remember I know the track #s are what matters but I'm just curious

AnotherWs6

iTrader: (17)

I always use 15% for parasitic power loss. You're in the 475-500 range.

Omega Doom

iTrader: (2)

yea I was always led to believe 15% was for the manual guys

G Atsma

15% is a nice round number; it can vary widely depending on the transmission and rear axle. All have varying levels of internal friction, hence the variance.

RealQuick

iTrader: (14)

Using a percentage is a quick and easy way to calculate... but not accurate at all. Manual drivetrain usually lose 40-60hp depending on the entire drivetrain, and autos are usually in the 80-100hp range.

Darth_V8r

iTrader: (4)

I've wondered about this myself. How does the drivetrain know the motor makes more power? And then decide it's going to consume more power?

Seems like it takes a certain amount of work to turn the drivetrain and that's it

G Atsma

But all drivetrains are different; different levels of friction with different axles and transmissions, and every combination thereof. The possibilities are endless.
You would have to know the drag of every component in your drivetrain, then add them up. Every combo will have a different number

RealQuick

iTrader: (14)

exactly right..

example, horsepower TV did a cutlass awhile back and the 455ci made 400fwhp on engine dyno. With the th400 and big rearend it made 320rwhp on their chassis dyno. People will say "well there ya go, 20% drivetrain loss". The reality is if the added a nice cam and heads it could have made 500fwhp, so is it only going to make 400whp (20% loss)? No, the drivetrain wouldn't go from robbing 80hp to 100hp just because the engine output went up.

as mentioned, every combo of drivetrain components vary, using a percentage as a calculator can be somewhat close at lower power levels, but starting making a lot of power and the percentage calculation over inflates the drivetrain loss.

another example, I made 461rwhp (if I use 12% for a manual I would be making 524fwhp). I then made 703rwhp with dame drivetrain, so my 12% now means I make 799fwhp). I went from losing 63hp to 96hp with no changes to the drivetrain. Doesn't make sense.

Darth_V8r

iTrader: (4)

That was almost the exact type of thing I was thinking. My "325hp" stock TA made 285 rwhp. Being realistic a 99 LS1 made 350 flywheel not 325, so that's 65 HP or its 24% if you use the rwhp number, 18% if you use flywheel.

Now I'm 519 rwhp. Now I use 24%, I'm at 638 flywheel??? Or I add 65 im at 584? Seems much more realistic to me.

JakeFusion

iTrader: (7)

It's a set number. It's not a percentage. I think Tony Mamo proved that every time he took an engine from the Westech engine dyno and stuck it in his Vette. It was always the same HP loss.

Now, spinning the motor higher where you increase heat/drag can affect power output from the same motor/drivetrain setup. Otherwise, it comes down to differences in drivetrains. A TH400 takes more to spin than a 4L60.

The easiest way to think bout this is a turbo car that spins to 6500. It's not spinning any higher than a stock LS1. But you can see 1000rwhp out of a turbo motor. Does that mean it's making 1250CHP? When before the turbo was installed it was making 400rwhp with a low compression 370 with a cam and heads through a 4L80 and S60? Did it take 250hp to drive those same components on the 400rwhp low compression 370? Why did it go from 70-80hp to 250 with just the addition of boost? Is there more friction within the 4L60 at 5k rpm because there's more torque? I imagine there may be a little, but is there any way those friction losses equate to 3X as much power being consumed?

What?

kingtal0n

iTrader: (1)

I think one problem(with comparisons) is whenever you change the power output of the engine, and try comparing it to any number of previous outputs, whatever it was you did to change the output in the first place could affect the overall loss or not, depending on what it was.

Next is, some parts probably do work on a percentage basis, like the torque converter. In an unlocked state, I know there is some kind of fluid coupling multiplication effect (I've never taken fluid dynamics for mechanical devices, only biological systems like the heart/capillaries) and multiplication is essentially division (i.e. there is no such thing as division, its really just a form of multiplication) which works on a percentage base (1.0 times 0.8 or 80% of 1, for example). While other parts may indeed always take X horsepower to spin at Y rpm, like a tire rolling on it's own down the street perhaps.

JoeNova

Its not a set number. Its not an exact % either.
If it was a set number and it took 100 HP to turn the drivetrain, you wouldn't be able to spin any of it by hand. If it was only 15%, you would be able to spin it almost effortlessly by hand.

Its determined by the amount of force required to accelerate the drivetrain.
Lets use completely random numbers because I'm on my cell phone and don't feel like doing real math lol.

Lets say the total rotating mass of your drivetrain is 100 kg.
Your engine makes 500 HP at the crank.
It is able to accelerate the 100 kg drivetrain by 1000 RPM/s
Lets say it takes 100 HP to accelerate the 100 kg drivetrain by 1000 RPM/s.
You now have 400 WHP and have a 20% drivetrain loss.

Then you do some upgrades.
Now you have 800 HP at the crank.
You are able to accelerate the 100 kg drivetrain by 1400 RPM/s, which takes 140 HP to do.
You now have 660 WHP and a 17.5% drivetrain loss.

Typically, drivetrain loss % goes down as power increases, even if the actual amount of power lost goes up.

SoFla01SSLookinstok

So the best way to figure out crank horsepower is yank the motor & put it on an engine dyno. Otherwise we're just trying to make good guesses.

Hey kingtal0n, who taught you math? You say "there is no such thing as division, its really just a form of multiplication"? Sorry but that's pretty stoopid. That's like " there is no such thing as subtraction, it's really just a form of addition". Try not to get stuck in the rabbit hole.

JoeNova

You could always try to figure out how much rotating mass your drivetrain has, figure out how fast you accelerate on the dyno at certain set points, find out how much power it takes to accelerate that mass by that speed, and find your flywheel power/torque at those determined set points.

Doesn't work if there is any slippage. Will need to be locked converter or clutch setup.

G Atsma

Speaking of set numbers, it IS a set number, BUT per component. And it's different for every tranny and rear end in the world. If you know the exact drag numbers for your tranny/rear, plug 'em in and calculate away! If everyone ran a T56/9 inch/4.11 gears, it WOULD be the "set number" mentioned above. But since that's not happening, NO "one size fits all" number.

AnotherWs6

iTrader: (17)

Don't bother reading let alone responding to kingtalon posts. They always make zero sense, are flat our wrong and prove that he is mentally fng -fill in the blank with whatever you'd like-

SoFla01SSLookinstok

True. You could always try those things. To me it just seems like a waste of time, or a lot of work, where you still would not be getting exact numbers. Just a well thought out guess. Even pulling the motor & putting it on an engine dyno, get numbers, then install & chassis dyno & see loss. This would still be using numbers from two different dynos. Say you get xhp loss from that test. Then do the same thing but use different dynos. You will probably get different numbers. So a different percent & loss? No, same loss just different dynos. Hey it's something to do lol.

SoFla01SSLookinstok

Lol, I do get some good laughs.

Darth_V8r

iTrader: (4)

That makes alot of sense. I'm going to try to translate to make sure I understand what you mean...

If the engine gains power and is capable of accelerating the drivetrain faster, then when it DOES accelerate the drivetrain faster, the raw power it consumes from the engine to do so increases.

However, the amount of additional power consumed due to faster drivetrain acceleration is less than the total power gained by the engine in the process of modding to make it more powerful.

So the raw power consumed by the drivetrain increased, but relative to the total engine power output, the percentage decreased. Is that about what I read there, Joe?

probably all academic anyway, since the only thing that matters is power to the ground.

spanks13

iTrader: (5)

There are going to be speed, force, and rate related losses.

Frictional losses increase with the amount of force applied. More torque = more loss. The differential ring and pinion very likely account for the most losses when in a 1:1 gear such as 4th gear. The ring and pinion very likely has a fixed efficiency percentage and losses will increase proportionally with power.

The transmission should be in a 1:1 gear. If you're in a lower or higher gear you would see a reduction in power at the tire due to efficiency loss through the gear multiplication. At a 1:1 gear ratio through the trans there will be some losses due to the work required to spin the mass of the gearset and overcome internal friction.

You will also see a reduction in power the FASTER you accelerate by the square of the acceleration rate. The inertial mass of your drivetrain (and dyno) rotating takes more work the faster you want to move it. This is why for the most accurate readings engine dynos do slow stepped pulls. If you rev them faster they will read lower.

There are going to be fixed minor losses such as windage. Windage is not going to increase with power level unless you're getting massive blow-by or you turn much higher rpm to achieve that power. You also have frictional forces that do not increase with power such as drag from seals and bearings (which aren't loaded up by the power flow).

Frictional forces do not increase with speed, but they do more "negative work" as stuff goes higher speed.

Long story short, losses will not be linear with RPM - higher rpm = more loss. A fixed loss for a given drivetrain is also incorrect as frictional forces increase linearly with torque, and the largest losses will be due to gear multiplication and redirecting the power flow in the differential.

More torque definitely = more loss.

For anyone wondering..the lost power turns into HEAT..our biggest enemy.