My first Dyno....disappointed

[JD Cross]

I agree JD,
Strictly speaking though: torque (lb ft) IS a measure of work (force applied over a distance), and power (hp) is a measure of power (work per unit time)
The whole 5252 rpm thing is neat but is only true when working with power in hp and torque in lb ft. Working with kW for power and kg m for torque, the magic rpm is 973.
As mentioned by dablackgoku the key to all this is that the gearing/final drive/wheel diameter/dyno diameter all must be known such that there is a 1:1 ratio from crank to dyno measure. Else the numbers on the screen are scaled up or down biasing the torque while power will always be true regardless of gearing.
This may be what rope was trying to get at.
I ask the dyno gods to feel free to correct me. I'm doing napkin stuff here.
Lucas

I stand corrected.
I should have said torque is a force and is not dependent on rotational speed.
Torque (force) measured with rotational speed is indeed a measurement of work and is the normal number stated by engine manufacturers on an at the flywheel basis.
I think that's what rope meant.

And yes hp is a measure of work done over time.

I know power is supposed to be the same regardless of gear ratio
What puzzles me is that a dyno calculates power from the measured torque.
If torque was measured on a dyno in say second gear and not corrected for the multiplication of torque in second would the power rating of that engine be a larger number?

Love to hear from dyno/hp gurus on this.

 

[I1O101I]

^that is just gross, lol.

lol

 

[Rope]

I stand corrected.
I should have said torque is a force and is not dependent on rotational speed.
Torque (force) measured with rotational speed is indeed a measurement of work and is the normal number stated by engine manufacturers on an at the flywheel basis.
I think that's what rope meant.

And yes hp is a measure of work done over time.

I know power is supposed to be the same regardless of gear ratio
What puzzles me is that a dyno calculates power from the measured torque.
If torque was measured on a dyno in say second gear and not corrected for the multiplication of torque in second would the power rating of that engine be a larger number?

Love to hear from dyno/hp gurus on this.

yes

 

[Supra TT]

Call lovefab..

Wait, no. Call SOS

 

[jaja]

What puzzles me is that a dyno calculates power from the measured torque.
If torque was measured on a dyno in say second gear and not corrected for the multiplication of torque in second would the power rating of that engine be a larger number?

Yeah, it begins to get confusing real fast.
The dyno will simply feel said force over its diameter, so yes it can register a torque, and knowing its own rotational velocity, it can calculate the power.

That power will be the same power that the transmission, rear dif, wheels, and flywheel would all feel if drivetrain losses didn't exist.

So essentially, the measured power is independent of gear ratios, dyno size, wheel size etc.

The dyno would then have to back-calculate from the power to determine the torque outputted based on the gear ratios.

So all said and done, if you fly into a dyno, and put the pedal to the metal, you should get a realistic power number regardless of what gear. But you will have no idea about you "wheel torque" unless you know your ratios.

Another thing to make things a little more confusing is how the dyno actually works, some are inertial, some use resisting electric motors, its kinda cool.

Lucas

 

[JD Cross]

Lucas
I need more help here.
My reading on dynos suggest they all measure torque and then multiply torque by rpm to calculate power.
Most dynos in North America state the measured torque in Imperial units, so torque in lb. ft., multiply that torque by rpm, then divide by 5252 to get hp.
I believe the dyno, whether inertial or otherwise, uses a load and determines engine torque from the engine's measured force against that load.

Is it true that the greater the torque output of an engine the less time it should take to spin up a given dyno?
If it takes less time then is the engine making more power?
If this is true then an acceleration run in second gear would measure more torque in a faster time and so the dyno would calculate more hp?

This does not mean the engine is making more power, but that the dyno would calculate it making more power?
You help is appreciated as this is a deep mystery to me.

 

[jaja]

I have prepared something that I hope clears a few things up.

The summary of the math I have down there is:
We proved that power is conserved from gear to gear. Therefore, the dyno power (Pdyno) is the same as the crank power (Pcrank).

If the dyno can measure its power (wheel torque x the angular velocity of the dyno), it is measuring the crank power.
From that, we know the crank power is just the crank torque x the angular velocity of the crank (tachometer).

So knowing the tachometer, the dyno can simply calculate for the crank torque (Tcrank).

Reality Check: Drivetrain losses will result in Pdyno < Pcrank, aka "Wheel power". From this and the tachometer, the "wheel torque" is calculated.



Lucas

 

[JD Cross]

Lucas

I think your math is beyond me but I'm guessing using velocity is similar to using rpm.
Here's an example of my confusion:

Let's assume 1 ft. lb. of torque at a flywheel (no rotation just force) and how the torque changes going through the driveline (ignoring parasitic driveline losses).
1 ft.lb. x (2nd gear ratio) 1.727 x (final drive) 4.062 x (wheel assuming 24.85" diameter) .9654 = 6.77 lb. ft. of torque at the wheel.

In order to calculate power we multiply torque by rpm and divide by 5252 correct?
So at 5000 rpm we have (6.77 ft.lb. of torque x 5000 rpm)/5252 = 6.45 hp

The same example in 1st gear is (12.04 ft. lb. of torque x 5000 rpm) = 11.46 hp

It appears that the greater the torque multiplication by the driveline the greater the hp at the wheels.

Where am I going wrong?

Jim

 

[jaja]

Let's assume 1 ft. lb. of torque at a flywheel (no rotation just force) and how the torque changes going through the driveline (ignoring parasitic driveline losses).
1 ft.lb. x (2nd gear ratio) 1.727 x (final drive) 4.062 x (wheel assuming 24.85" diameter) .9654 = 6.77 lb. ft. of torque at the wheel.

In order to calculate power we multiply torque by rpm and divide by 5252 correct?
So at 5000 rpm we have (6.77 ft.lb. of torque x 5000 rpm)/5252 = 6.45 hp

The same example in 1st gear is (12.04 ft. lb. of torque x 5000 rpm) = 11.46 hp

It appears that the greater the torque multiplication by the driveline the greater the hp at the wheels.

Where am I going wrong?

Jim,

So it looks like the confusion is you are multiplying apples with oranges.

Lets take a look back at your exmaple:
At the crank, you are providing 1 lb ft at 5000 rpm. With P=T*w/5252, we get the crank power to be P=0.952 hp.

For the same example, lets look at the power at the wheels:
At the wheels, in 2nd gear, 1 lbft --> 6.77 lbft at the wheels. NOW, the confusion is what is the RPM of the wheels?
Using the same gear ratios provided, the wheels should be spinning at: 5000 rpm/1.727/4.062/0.9654 = only 738 rpm.
So, the Power at the wheels = Torque at the wheels *wheel RPM
In our example, P= (6.77 * 738)/5252 = 0.952 hp!! Just like before when we took crank torque and crank rpm.

So there in-lies the confusion. As shown, the power is said to be conserved from crank to wheels, where the torque is directly related to the gearing. However, since the speed of the wheel is inversely related to the gearing, power is unaffected by the gearing.

So while more torque gets delivered to the road at lower gears, the wheels are spinning very slow, as such, a dyno pull in a lower gear won't give any power gains.

I hope that helps :redface:.

Lucas

 

[JD Cross]

Jim,

So it looks like the confusion is you are multiplying apples with oranges.

Lets take a look back at your exmaple:
At the crank, you are providing 1 lb ft at 5000 rpm. With P=T*w/5252, we get the crank power to be P=0.952 hp.

For the same example, lets look at the power at the wheels:
At the wheels, in 2nd gear, 1 lbft --> 6.77 lbft at the wheels. NOW, the confusion is what is the RPM of the wheels?
Using the same gear ratios provided, the wheels should be spinning at: 5000 rpm/1.727/4.062/0.9654 = only 738 rpm.
So, the Power at the wheels = Torque at the wheels *wheel RPM
In our example, P= (6.77 * 738)/5252 = 0.952 hp!! Just like before when we took crank torque and crank rpm.

So there in-lies the confusion. As shown, the power is said to be conserved from crank to wheels, where the torque is directly related to the gearing. However, since the speed of the wheel is inversely related to the gearing, power is unaffected by the gearing.

So while more torque gets delivered to the road at lower gears, the wheels are spinning very slow, as such, a dyno pull in a lower gear won't give any power gains.

I hope that helps :redface:.

Lucas

Lucas
The light has gone on!
Thank you.
Jim

 

[tof]

Setting aside the technical discussion (which is interesting and valuable) here's the important part...

Chassis dynos are great for comparing different cars or for gauging the effectiveness of tuning on performance...but only when all the measurements are taken from the same dyno at the same location (if the dyno is portable) set up the same way, under nearly identical environmental conditions. Altitude, humidity, ambient temperature, air circulation, barometric pressure, and operator practices ALL affect chassis dyno readings. So in terms of determining absolute hp and torque at the wheels, any chassis dyno result is just a good guess.

 

[JDM EM1]

I've got a 92' manual bone stock Nsx that dyno'd on a dyno jet and made 251whp/191wtq and then got a one off made custom exhaust weighing in at 11lbs all together and used same dyno jet making 261whp/198wtq afterwards.

Hoping to get an intake and headers to net me around 270-275whp.

I guess mines on crack bc with and without bolt ons it produces some good numbers and is very healthy even on the street against much higher horsepower cars.

 

[stuntman]

FYI: E46 M3s are rated at 333hp and put down ~270 to the tire.

Also Dynojet 248s and 224xLCs can read over 10hp different from each other as well so it also depends on what dynojet you're on. My E36 M3 read the same on a DJ224 as it did on a Mustang, so those two seem fairly comparable while I think 248s read higher.

Your correction factor could also shift the # ~10hp so there are tons of variables.