the dips are stutters when the frequency valve is shutting and opening the actuators... and something is wrong. This tells the tuner that the ECU is doing something funny... and should be corrected.
however TQ , ideally should keep going up til it crosses the HP (meets it) and then HP should take over and keep pulling up( meanwhile the TQ slowly drops).

 

The bigger question is why it is commonly assumed that driveline loss is a fixed % and not a constant horsepower value. That is, I don't see why doubling the power output of an engine would double the associated driveline losses.

I'm sure there's got to be some power-related losses involved in the mix but it would seem logical that they would be a relatively small component. In the case above, the car picked up 134 horsepower at the wheels. If you add that to the estimated stock flywheel horspower, you get 554. It's a pretty conservative approach but I'd say it is closer than some of the other numbers thrown about. And besides, bigger claims dont make your car faster! They just make it harder to live up to expectations

-shiv

 

I have wondered this very same question and discussed it recently with a collegue:

why would a 1000 horsepower car have twice as much drive train loss as
a 500 horsepower car ??????????????????
This is another reason why I like Jean/TB/Lat, etc. 60-130 mph comparison using AX-22!!
Marty

 

Because you have resistance. As the drive becomes higher (HP) the friction becomes greater and therefore less efficient. Friction creates heat, heat creates loss. Take your hands and rub them together, stay at a constant rate. Your hands will become hotter and hotter even though your rate has remained the same. You rub faster and the heat not only is achieved faster, but hotter. You can not expect the same level of friction with different levels of energy.

This is simple laws of physics. Anything has loss that increases. Blowing through a straw, moving fluid through a pipe. The loss does not stay constant.

You have what is called Kinetic energy. I am no expert so bear with me. Kinetic energy is the energy of motion. As you increase the Kinetic energy you also increase resistance. For example, a car that is going 30MPH and increase to 33MPH has about a 20% increase in Kinetic energy. What creates resistance and the increase of inertia? Temperature, vibration motion and force. So the inertia and resistance can not remain the same. All these factors play a part in increased inertia.

This will probably create more confusion, but it is as simplistic as I can explain it. The number achieved as loss can not be the same for a car creating 200HP and one creating 1000HP.

I agree, the times are a good way to judge performance. The best way in fact. HP gains do not always equal like performance gains. You have lots of issues to deal with. I have run lots of cars with more power and done well against them. There is a point where the driver can’t shift fast enough, the tires spin to much and you basically have little gain for added HP. It is a fine line and doesn’t always add up in the end. So many factors involved.

.

 

I don't think anyone is saying that the losses stay 100% constant regardless of output. A 1000hp 996tt certainly gererates more tire deflection and heat than a 500hp one. So, yes, it will be more lossy.

... but not twice as lossy. That was my point. If I had to bet my dollar, I'd say that true driveline losses are closer to fixed hp loss than % hp loss (but certainly somewhere in the middle). Assuming a fixed ramp-up rate (which holds the inertial losses constant). The bhp estimates wont be as glamorous as assuming fix % losses. But they will be far more reasonable.

Unfortunately, there is nothing simple about quantifying every single component of driveline loss (frictional losses, tire deflection on the ground, component interna, mechanical inefficiency, gear oil resistance, etc,.) and mapping out how each individual component reacts to increasing output.

shiv

 

No way. Not twice as much. I have debated this before with guys that believe that a constant number should be used. Not a percentage but say 75HP for all cars. This makes about as much sense as a screen door on a submarine.

The loss percentages will remain the same. Sure, it will move around some as the dyno heats, the joints get hot, sidewall give and so on. The general point is if that percentage is a given value, that value will dictate the actual loss number. We know what the answer to the question is, we just need to know the question. This is 11th grade physics.

We can look at the data calculated aside from laws of Newton 2nd or 3rd laws of motion and know that the loss number (not percentage) will increase. An example is data we have from motors on an engine dyno then transferred to the car and on a chassis dyno. It is all very trackable.

Now I need to do something a little simpler like time these cams in a 996TT  Before my head explodes. Cobwebs

 

So a 1000bhp 996tt will have twice the driveline loss of a 500bhp 996tt? Just wanted to make sure I'm understanding you correctly.

-shiv

 

The acceleration produced by a net force on an object is directly proportional to the magnitude of the net force. F(force)=M(mass)A(acceleration), or F=dp/dt = d(mv)/dt which states that net force is directly proportional to time rate change of momentum.

There are several ways it can be broken down. This is the basic theory that is openly shared in the world of physics. If you disagree with these theories then I guess Newton was just a crazy old man

 

I still don't quite understand what you are saying. Are you arguing that the driveline loss (which you believe to be a percentage) remains reasonably constant regardless of power output? If that is the case, a 1000bhp 996tt will have twice the driveline loss as a 500bhp 996tt. Is this what you are saying? If not, just a simply statement will suffice. English isn't my first language so go easy on me

-shiv

 

Hi Shiv,

I can't find the link so you might want to search drivetrain loss. The answer is yes, the 1000bhp car will have twice the driveline loss. There is a considerably LARGER amount of friction in the gears when you apply 1000 than when you only apply 500. The friction causes heat. The more heat, the more loss. Like I said, rather than arguing, you could do a search...

 

Your percentage of loss does not change. The loss percentage stays constant. Your subtraction of numbers will not. I am not sure how I can explain this differently. Are you suggesting that a number be chosen like 75HP as a standard given across the board? What physics principles is that based on?

Everything revolves around resistance and the increase in loss as it moves forward. In electronics R = V/I , again the same now with the length of wire included R = ρL / A. Add temperature resistance R = R0 + αR0( T - T0 ). Do the math. This is very simple.

Look at fluid dynamics and how resistance applies here. All the rules remain the same. The only one I can think of at the moment is Navier's equation. I am sure someone on here can site some FD equations.

 

Well I be darned I knew there was some sense in there somewhere... Anyone want to borrow a copy of:

 

Thanks for the suggestions guys. Here's my take:

There are many components that, when added together, give us total driveline loss. Among these components are:

Mechanical losses (in transmission and differentials)
Bearing losses
Transmission material deflection/deformation
Tire deflection losses
Windage losses within the gearbox
Windage losses within the spokes of the wheels
Inertial losses induced by having to accelerate lots of mass (gears, shafts, wheels, tires, rotors, etc,.)
And so on...

And not all of them increase proportionally with power output.

But before we go any further though, let's try to establish the conditions of this experiment (albeit theoretical). First, let's assume that we are using a proper load bearing dyno. This is useful because they allow us to hold many variables constant. One of them being dyno run duration. That is, let's assume that all the dyno pulls will be held at a fixed duration and with a fixed ramp-up rate (change in vehicle speed with respect to time). Obviously, we're going to hold gear selection constant as well since different gears will generate different ramp-up rates and tractive efforts which will influence some of the components mentioned above.

With these testing conditions fixed, we can reasonably state that the only thing that is variable is going to be power output. And our goal is to determine how driveline losses react in light of changes to power output. Now let's look at the aforementioned list of driveline loss components, one by one, and see how power output influences them. We don't really need to rely on inappropriate physics equations either since common sense should suffice just fine (although I may be giving myself too much credit here )

1) Mechanical Losses (trans, diffs, etc,.)- This can be defined by how much power is needed to spin all the gears, shafts, clutch plates, etc,. in the drivetrain. Whether a car makes 500hp or 1000hp, the amount of power to spin these components from the beginning of a dyno pull to the end of the dyno pull is going to be constant. This job is only going to take a certain amount of work/hp. Any additional hp applied to this job is going to be passed along elsewhere and put to good use.

2) Bearing losses- Unlike #1 and #2, it is conceivable that more torque applied to a bear will generate more bearing heat induce more associated losses. Is it a direct 1:1 relationship? Who knows? But even assuming that it is, the bearing loss component in the whole scheme of things should be reasonably minor. That is if they are doing their job, of course.

3) Transmission material deflection/deformation- Let's all hope that there is no gear material that is actually being deformed as power levels increase. That would make a pretty crummy gear. So this component should be reasonably indifferent to power output.

4) Tire deflection losses- By all accounts, tire deflection (where the tread deforms and starts to roll over upon itself) will become more of an issue as power increases. How much of an issue depends a lot of the stiffness of the sidewall and diameter of the dyno roller. But yes, these losses will increase as power increases. Will it increase proportionally? Maybe yes, maybe no. But it’s safe to say that this loss is not going to be a fixed figure independent of power output..

5) Windage losses within the gearbox- This loss is most definitely independent of power output in this given test. Although one could argue that the power-related loss components mentioned above could raise the temperature of the fluids which would have some effects on overall windage results. But trying to isolate, let alone quantify, such secondary sources of losses would be tough (and most likely very very minor).

6) Windage losses within the spokes of the wheels- This loss is also independent of power output. The speeds and ramp-up rate (dv/dt) of the wheels aren't changing with increases in power output.

7) Inertial losses induced by having to accelerate lots of mass (gears, shafts, wheels, tires, rotors, etc,.)- Inertial losses can play a big part in loss characteristics-- especially when rate of acceleration changes. The higher the rate, the greater the inertial losses. But, in this case, we are holding rate of acceleration equal. After all, if we are to compare chassis dyno numbers to engine dyno numbers (the difference being driveline loss), we need to keep testing procedures constant across the board. So, in this test, these losses are not going to be influenced by changes in power output.

Looking at these components, and the effect that changes in power output has on them, would tell us that cumulative driveline loss is neither a fixed hp number nor a fixed % of power output. Instead, it's going to be in the middle somewhere. Most likely towards one end than the other. And where in that range will depend on testing conditions (ramp-up rate, max vehicle speeds, etc,.)

And for every 10 people that insist that driveline losses are proportional to power output, we have at least one person who says something like this (nearly 10 years ago on the Miata forum):

http://realbig.com/miata/1996-11/2053.html

Corky Bell doesn't address or acknowledge some of the power-output influenced components above but he does point hit the major points of this argument pertaining to the losses that aren't necessarily influenced by power output increases.

Shiv
www.vishnutuning.com

 

As inappropriate these equations may be to the unenlightened in this area you just answered the same answer (more or less) to what is being said. The value remains fixed to a degree as I stated or as you would interpret it farther on one end of the scale than another. Like any standard equation you can take just the formula for loss in a simple form or you can then add all the variables which were our points.

On your dyno you have the ability to do a roll back calculation and loss configuration. When we looked at buying a DD they had this feature in the software. So I know this test is possible.

We do as well in ours, but require that I go through a few hoops to get those numbers. Like rolling the car then dropping off and running it again and literally remove the rear wheels from the rollers and allow roll back. The parasitic are then calculated for that vehicle and those numbers can be extracted to a percentage of loss. Given that information you can determine the over all rate of loss on several a 6 axis point based on Force, TQ and HP losses. These are independent from the dyno loss factors which are known and apply it to the chassis wheel numbers. This is one chart. It is a little hard to cut and past the print screen of a dozen pages so I will spare the masses of that.

 

After spending 15 years using a Maha 4WD dyno I would like to throw in my $0.02 !
From all the data that I collected on loss numbers I think that it is safe to say that tire losses are the largest factor. Although not a linear factor , tire losses are heavily influenced by road speed . The one piece of information that I dont see ( missed ? ) on the Mustang results is max road speed / gear used.

If a car is measured in 4th gear and then 6th gear, the rwhp will be very different as will be the losses.
This can makes the calculation of losses a difficult part of determining crank hp.
Loss measurement is obviously better .
I guess I would add to " the only number that counts is rwhp " is " if we all measure in the same gear ".

Thanks

Geoff