Dyno % lost, Engine vs Chassis dyno

[JustenGT8]

Ah yes... the drivetrain "loss" argument.

I had quite the exchange a couple years ago on the Corvette Forum with an engineer who did power transfer systems on huge ship (cruise liners, supertankers) systems. He blew off my arguments as "proven" false.

I find the "engineering" arguments USUALLY highly valid except when the engineering "impossible" is somehow made possible with a couple examples being the "impossible" gear cutting angles on the original torsen LSDs and the theoretical ET and MPH limits in 1/4 mile racing of 8 seconds and 200mph.:laugh2:

The "constant" loss theory is dead. If a drivetrain where to take say 50hp as a constant then you would need 50hp just to move the vehicle. Since you can move most vehicles with less than ONE horsepower obviously the losses are dynamic.

The QUICKER you try to apply the force the higher the loss. If you take a 3500 pound car, on level ground, in neutral, with the brakes off... just leaning on the vehicle will slowly start to move it with almost no "loss". If instead you punch or kick the vehicle as hard as you can the loss will be nearly 100% uch: as the vehicle will probably not move.

The loss that I argued with the very qualified engineer was what I call the "spring" loss. Springs have extensive engineering definitions but somehow these engineering rules do not describe what I have observed.

I would love to hear some of this forums members thoughts on my "spring theory".

My "springs" in a vehicle include the obvious suspension springs but also the less obvious things including: engine transmission and differential mounts; torsional deflection of metal parts including driveshaft, gear shafts, gear teeth, axle shafts and even the wheels themselves; tire deflection, body twisting and flexing and on and on... if you think about it there are a HUGE number of spring like deflections going on within a vehicle as it is accelerated.

Spring compression takes time. Power = force x time. Any addition to the time by the springs has to reduce measured power. Engineering seems to argue that once the compression has occured there is no more loss. I argue that force MUST be used to maintain the compression. Picture a vehicle doing a wheelstand. Force must be used to lift the weight off the ground. Even though the vehicle is moving SOME of the force available for forward motion must instead be used to lift the weight against gravity. This is actually a VERY complicated thought. Every little compressible "spring" has the same characteristics of the vehicle wheelstanding. Some force MUST be used to hold the compression and that detracts from the force available to accelerate the vehicle.

The stronger the force the higher the degree of deflection and therefore the higher the loss.

Sounds kinda reasonable...my point was only that it's not a constant "%" loss.

there is a 'fixed' loss as well, just that it is probably quite small.

 

[XR8tt]

% may not be the right method ??
I'm not too much into maths..
It seems what it takes to turn transmission etc..
to turn it at the speed of acceleration of engine..
There would be some inertia also ??
High rpm converters play havoc with power numbers,
This where a 450 h.p or low-sh power car can can good times
over the 1/4 esp in light well set up car...
My son's car has 3000 converter and reads 80 RWKW lower on new tune..
We knew we made more power due to bigger injectors, fuel pumps..
Run .8 second quicker 1/4's.. Mid 12's to low 11's..

 

[JBrady]

Fixed loss?

Let's lift the car off the ground, transmission in gear, torque wrench on dampner bolt. How much torque to rotate assembly? Let's say for argument it is 10 lb/ft at say 10 rpm. Math says < 0.02 horsepower. Let's now take this up to 7000rpm and we have > 13 horsepower. Intuition tells me the torque required to maintain 7000rpm would be higher than 10rpm so this too is variable.

Unless we are studying fixed rpm applications like stationary pumps or generators we must include acceleration in the equation. This brings inertia into the mix. The inertia variables include rotational mass and rate of acceleration. Highly variable. More power, faster acceleration, higher inertial loss.

Friction? Except for the tire to road contact all drivetrain parts are lubricated. As the lube is squeezed between contacts heat is created. If the lube fails and metal to metal contact occurs component failure becomes imminent. The amount of friction is based on load which includes mass, power, gearing, aerodynamics and inertia. All but mass being variables.

Spring effect? As I describe above and includes hundreds of potential points of "spring" deflection. Think about a transmission; the case flexes with torque and the gears try to push away from each other stretching the case away from the gearshaft mounting points, all the gearshafts twist torsionally, the splines all flex slightly... literally everything and anything that power touches will "spring" distort. highly variable.

So, is the sum total of all variables describable in a percentage? Yes and no. If you measure on a engine dyno and then on a chassis dyno you can calculate the percentage loss for THAT application at THAT peak power and rpm level on that day/time temp humidity pressure... ect. Now, if you create a table or graph for the full RPM range comparing both dynos... you can see if the percentage is the same of different at different rpm points. My guess is no.

So, is there any use is quoting percentage loss estimations? Yes. Known/realistic loss ranges help in calculating fuel delivery and compressor matching. In both regards better to overcompensate than under. As far as performance goes using time to speed or distance gives a minimum power requirement. At the end of the day or race the only thing that matters is how much power is converted to acceleration.

 

[JustenGT8]

. At the end of the day or race the only thing that matters is how much power is converted to acceleration.

I think we all agree with this bit

 

[stevechumo]

With a higher stall speed converter for the auto tranny, will the loss be greater than a stock converter? If it is, then is it just a dyno number loss, or is it really more loss at the wheels? I was told a AWD car, such as the Subaru STI that has the low dyno number, in fact, it has more rwhp. Well I'd like to know. :boggled:

 

[ed_ma61]

Spring compression takes time. Power = force x time. Any addition to the time by the springs has to reduce measured power. Engineering seems to argue that once the compression has occured there is no more loss. I argue that force MUST be used to maintain the compression. Picture a vehicle doing a wheelstand. Force must be used to lift the weight off the ground. Even though the vehicle is moving SOME of the force available for forward motion must instead be used to lift the weight against gravity. This is actually a VERY complicated thought. Every little compressible "spring" has the same characteristics of the vehicle wheelstanding. Some force MUST be used to hold the compression and that detracts from the force available to accelerate the vehicle.

The stronger the force the higher the degree of deflection and therefore the higher the loss.

that makes absolutely zero sense

firstly: power = force x distance / time

once youve compressed a spring, theres no further change in distance (ie = 0) and it take no more power to remain compressed. to prove this in your head, think of the work done on the spring. work = power x time... if you keep the spring compressed, does it get hot? no! ie, no ongoing work being done to it, ie, not consuming any power over that time. youve stored a bit of energy in it, and when you relax, youll get that energy back (mostly)

if on the other hand you compress it, relax it, compress it, relax it, ie constantly take the spring through a distance, itll cosume power, have work done to it, and get hot.

a car on a dyno will compress all of your little springs along the driveline, then itll all equibrilate and will be null and void in a constate state calculation of the engines power output.

only the the things dynamically moving will consume power, ie gears/gear oil, bearings, uni joints, diff/diff oil, cv's, tyres on the rollers. these losses will increase with rpm. and the load on them. youd have to calculate it for each vehicle.

 

[JBrady]

that makes absolutely zero sense

The "rules" of springs apply to linear machines. I agree that once the spring in a linear machine is compressed it takes no more power. That said a car is FAR from a linear machine and is free to move in all directions and against gravity. This distinction is the basis for my points.

If I were to ask you to build a machine that uses an motor or engine to compress a spring could you? Of course. Let us add that your machine does not lock in some way once the spring is compressed so that that the spring would only stay compressed while the machine was turned on. Simple right? By observation wouldn't it make sense that it would take power to hold the spring in compression? Turn on... compressed. Power is being consumed by this "machine" for as long as you keep the spring compressed. What ever else you could use the engine for would have its total power available reduced by the act of holding the spring compressed.

The accelerating car is exactly this "machine". The total "spring" compression is powered by the engine and therefore reduces the amount of power available to accelerate the vehicle.

firstly: power = force x distance / time

While this is classically accurate the distance component is arguably not needed. Force ALWAYS causes motion. The distance may be minute such as the deflection the weight a grain of sand moves the earth or small even to the near quantum levels with the motion of sub atomic particles. Now measuring these distances can be very tough and for specific measurements a distance rule must be known. But to know if power is present all that is necessary to know is that force is present. To label it in specific units the specific distance must be defined.

once youve compressed a spring, theres no further change in distance (ie = 0) and it take no more power to remain compressed.

What then keeps the spring compressed? There is only one answer, power. Either a locking mechanism to keep the spring compressed or continues power are required. You will want to argue there is no "distance" to which I stand on the fact that there is distance for multiple reasons including the rotation of the engine holding the compression.

to prove this in your head, think of the work done on the spring. work = power x time... if you keep the spring compressed, does it get hot? no! ie, no ongoing work being done to it, ie, not consuming any power over that time. youve stored a bit of energy in it, and when you relax, youll get that energy back (mostly)

Again, something MUST hold that sping in compression. Either a locking mechanism or power.

The act of the spring being compressed of course takes power. I don't think we disagree on that point. In doing so the spring gives off heat but then applies the force to something else... such as lifting the vehicles front wheels off the ground. It is this "lifting" that provides some of the resistance between which the engine holds the spring in compression. Both power to compress and resistance to hold the spring are necessary to compress the spring.

When a vehicle accelerates some power is extended in moving it down the track/road. Some is used to compress/stretch the springs. Once compressed some of these springs will transmit power to move the car down the track and some to lift the car. The initial compression is a loss and the act of maintaining compression requires power for the duration of the run. The "return" of energy when the "springs" return happens after the run and that component is not measured for the race.

if on the other hand you compress it, relax it, compress it, relax it, ie constantly take the spring through a distance, itll cosume power, have work done to it, and get hot.

Agreed, heat is created for each act of compressing the spring. At each shift some cycling takes place. But my theory is more the power required to hold the compression.

a car on a dyno will compress all of your little springs along the driveline, then itll all equibrilate and will be null and void in a constate state calculation of the engines power output.

only the the things dynamically moving will consume power, ie gears/gear oil, bearings, uni joints, diff/diff oil, cv's, tyres on the rollers. these losses will increase with rpm. and the load on them. youd have to calculate it for each vehicle.

Again, what holds the compression? It is the engines power. The visual act of the car lifting against the dyno straps or lifting the front and or back and or sides of the vehicle against gravity are apparent in any high powered run.

Again, in a system that is linear once a spring is compressed it no longer reduces the LINEAR power consumed because it is added and equalized. But a vehichle is free to move in all directions and against gravity and therein lies the losses I am referring. Along this same line anything that reduces these non linear (in racing meaning straight down the track) spring compressions will add to the measured output of the engine vs measurement.

 

[Jake Breyck]

all this debate, just do a negitive pull on a dyno and know for sure.

 

[ed_ma61]

i think youre argument is suffering from a significant misuse of the term 'power'

 

[XR8tt]

AHhhh most cars have a type of four link which includes anti squat..
The springs don't compress down to bump stops..
Even the old elliptic spring set up doesn't pull suspension down too much..
I have seen a loose converter add power in upper rpm also..
But its like dumping the clutch on dyno, giving a spike on power graph..
Wheelspin can do this also, when it takes up..

 

[JBrady]

AHhhh most cars have a type of four link which includes anti squat..
The springs don't compress down to bump stops..
Even the old elliptic spring set up doesn't pull suspension down too much..
I have seen a loose converter add power in upper rpm also..
But its like dumping the clutch on dyno, giving a spike on power graph..
Wheelspin can do this also, when it takes up..

Suspension technology is a very important part of competitive racing. The more linear (in the direction of travel) you can focus all the forces the less "loss" is incurred.

A loose converter adding power in high RPMs is an interesting observation. Could be a number of things including allowing the engine to accelerate at a rate that is more efficient for its power production. Larry Widmer of Endyn calls this transient response. It could also be the converter multiplying torque as the engine accelerates more quickly giving more power for the measured tire speed.

 

[XR8tt]

Or converter finally taking up or if it has lock up ??
Which may shockload dyno slightly...

 

[stevechumo]

I found a very interesting article from a dyno shop. I might go there and check out. Here's the article: http://home.earthlink.net/~spchurch/churchautomotivetesting/id12.html

 

[JBrady]

I found a very interesting article from a dyno shop. I might go there and check out. Here's the article: http://home.earthlink.net/~spchurch/churchautomotivetesting/id12.html

Steve, good site for teaching readers that inertia is a calculation that needs to be included when considering the mysterious cloud called "drivetran loss"

They sum it up in their final paragraph...

"Please keep in mind that these are rough approximations based upon some loose physics and real world experiences. But they should provide a better insight into the difference between loaded and inertial dynos, and hub vs. roller dynos, along with assisting the user in trying to compare results across the different systems"

There is so much going on dynamically that trying to accurately calculate any single components effect is very complicated. It is useful to understand as many of the factors as possible so that changes can be made. In my example of springs if you modify the suspension dynamics to change some of the force from lifting the car (wheelie action) to forward energy (linear compression) you net a performance improvement that calculates out to a decrease in drivetrain loss.

Their mass x accerlation for calculating inertial mass effects on measured loss is more useful in pointing out that this is an important part of total loss. This load includes the weight of everything accelerated whether rotationally or linearly such as tire and wheel mass as well as driveshaft, axles, flywheel, clutch, torque converter and on and on. Within this also is moment of inertia. A lighter tire on a heavier wheel has the same net mass but has a different inertial effect.

Good mental calisthenics.

 

[stevechumo]

I like the term "mysterious cloud" on drivetrain loss. Assuming all dyno machines are manufacturered with the same specs. The dyno results are always different from cars to cars, even with the same car after dyno pulls.