Hammond drives the GT500

[SVT666]

And my point all along is that an increase in RWHP from lighter drivetrain components is not possible.

If a dyno shows a power gain, how can it be wrong?

[Nethead]

Sometime before GT500s were shipped to dealerships, the Society of Automotive Engineers were brought in to certify the power output of the 5.4 engine in the OEM hardware & software specifications.  On the dynamometer used by the SAE, the GT500 OEM specs engine produced 503 horsepower.  After that certification, Ford announced that the GT500 would be available with 500 HP engines. 

Googling correctly will probably find the article about the certification, since it dates only back to around the late summer of  2005.  The title was something like "GT500 to offer 500 horsepower" and may have been an Edmunds or AutoWeek news item.

[GoCougs]

If a dyno shows a power gain, how can it be wrong?

In short, the test procedure is flawed. To maintain proper test procedure, there should be a way to alter the change in system inertia. (SdV, MX793 and myself explained it on the previous page why the observation is incorrect.)

To state that a lighter flywheel or driveshaft increases RWHP is the same thing as stating that removing 500lbs of weight from the car increases RWHP.

[JYODER240]

To state that a lighter flywheel or driveshaft increases RWHP is the same thing as stating that removing 500lbs of weight from the car increases RWHP.

Wouldn't a lighter driveshaft=less rotating mass=less drivetrain loss= increased whp?

[Eye of the Tiger]

A lighter drivetrain has less inertia, therefore takes less power to accelerate, thereby increasing power to the wheels, henceforth making your car faster, and in effect, making you have fun.

[GoCougs]

Wouldn't a lighter driveshaft=less rotating mass=less drivetrain loss= increased whp?

Actually, no. Less rotating mass does not equal less drivetrain loss.

A lighter drivetrain has less inertia, therefore takes less power to accelerate, thereby increasing power to the wheels, henceforth making your car faster, and in effect, making you have fun.

Actually, no. Power does not accelerate.

[MX793]

That is all I understood from your post.

Well, let me give you an example.  Let's say you've got a 32 lb chunk of steel that slides on a frictionless surface.  You want to accelerate that block from rest to 50 ft/s (about 35 mph).  The work energy required will be equal to the desired change in kinetic energy, the change in KE being the final KE minus the starting KE.  KE for linear motion like this is 0.5 x mass x velocity^2.  If you accelerate the block to 50 ft/s, the final KE is 1250 lb-ft and the starting KE is 0 (it's at rest).  The work required is 1250 lb-ft of energy.

Now suppose you want that acceleration to happen in 2 seconds.  Divide the change in KE by the time and you get the power.  625 lb-ft/s, or roughly 1.1 hp, would be the power needed to accelerate this piece of steel from 0 to 50 ft/s in 2 seconds.

Now, instead of a piece of steel sliding on a frictionless surface, let's say we've got a 32 lb disk made of steel that is rolling along the surface with no slip (like a wheel).  Let's say this disk is 2 ft in diameter.  Again, the work required to accelerate this disk is equal to the desired change in kinetic energy.  For this disk, the total KE is 0.5mV^2 + 0.5Iw^2, where I is the polar moment of inertia and w is the rotation speed.  At 50 ft/s, the disk will be rotating 8 times per second.  I in this simple case (a solid disk) is 0.5 x mass x radius^2, which is 0.5 here.  Put it all together and the total kinetic energy at 50 ft/s would be 1882 lb-ft, so you would require that much work to accelerate the rolling disk.  That's 632 more than the non-rotating block of the same mass that is simply being propelled (roughly 50% higher).

If you want this rolling disk to accelerate from rest to 50 ft/s in 2 seconds, you'd need 1.7 hp.  Over half a horsepower more than the non-rotating block of the same mass (roughly 50% more power).

For a rolling wheel, V and w are directly related.  2 x velocity / diameter = w.  Since our diameter here is 2 ft, V = w.  Substitute that into the total KE formula and you find

KE = 0.5mV^2 + 0.5IV^2 = 0.5(m+I)V^2

The last expression is of the same form as the linear kinetic energy equation, but with (m+I) in the place of just m.  You can call (m+I) the equivalent mass.  The energy and power requirements for accelerating a 32 lb, 2 ft diameter rolling disk are the same as those for accelerating a 48 lb non-rotating mass.

[JYODER240]

It takes less power to rotate the driveshaft. So are you saying that if my driveshaft weighs 300lbs i'll have the same whp as with a 15lb one?

[omicron]

*Wind rushes over head*

[GoCougs]

It takes less power to rotate the driveshaft. So are you saying that if my driveshaft weighs 300lbs i'll have the same whp as with a 15lb one?

No, it does not take less power to rotate a heavier driveshaft.

Yes, you will have the same RWHP whether your driveshaft weighs 300 lbs or 15 lbs. (Note that this is in theory. In reality, that much weight would probably bind up the bearings on the tranny tailshaft or differential pinion.)

I think you are confusing acceleration, power and torque. A lighter driveshaft will yield slighter greater acceleration but that is not the same thing as saying a lighter driveshaft will yield slightly higher RWHP.

[JYODER240]

No, it does not take less power to rotate a heavier driveshaft.

Yes, you will have the same RWHP whether your driveshaft weighs 300 lbs or 15 lbs. (Note that this is in theory. In reality, that much weight would probably bind up the bearings on the tranny tailshaft or differential pinion.)

I think you are confusing acceleration, power and torque. A lighter driveshaft will yield slighter greater acceleration but that is not the same thing as saying a lighter driveshaft will yield slightly higher RWHP.

That makes no sense. Then why do high-performance cars use carbon fiber and other lightweight materials on their driveshafts? How can you say that it takes no more effort to rotate something that weighs 15lbs vs. 300lbs?

[Raza]

It also puts down 20 more hp at the wheels. There's no way a Mustang GT has that little drivetrain loss. Most 911's have about a 15% loss, I don't see how a FR layout Mustang could have even less than that.

Perhaps the VQ is overrated...

[JYODER240]

Perhaps the VQ is overrated...

Its SAE certified at 297.

[SVT666]

I'm done.  All I know is it would be easier for me to spin a 15 lbs flywheel or driveshaft with my hands then it would a 300 lbs flywheel or driveshaft.  To me that equals more power gets to the end component...being the wheels.  If a chassis dyno shows a gain in whp, then I can't say it's wrong.  If it is wrong, then no dyno can be trusted.

[JYODER240]

I'm done.? All I know is it would be easier for me to spin a 15 lbs flywheel or driveshaft with my hands then it would a 300 lbs flywheel or driveshaft.? To me that equals more power gets to the end component...being the wheels.? If a chassis dyno shows a gain in whp, then I can't say it's wrong.? If it is wrong, then no dyno can be trusted.

I agree completely.

[MX793]

If a dyno shows a power gain, how can it be wrong?

It's a false reading.  A lot of dynos out there are inertial dynos.  How they work is they measure the rate at which a drum is accelerated and the drums speed.  The drum has a known mass and moment of inertia, so once you know the rate of acceleration over a range of speeds, the power can be calculated.  Power is moment of inertia x angular acceleration x angular velocity.

The reason typical chassis dynos will show that there's more power when you switch to lighter driveline components is because the dyno computer will read more acceleration, but nobody informed it that there was a reduction in mass (or moment of inertia) when it goes to calculate the power.  It's not that there's more power, it's that there's less mass in the system.

To give a parallel to this, have you seen those in-car G meters that can estimate your horsepower based on the measured acceleration of the vehicle?  How they work is they measure the acceleration, and then you provide the weight of the car and the gear ratio and tire size and the computer spits out the horsepower.  Let's say your car weighs 3200 lbs and you do a run down the drag strip with one of these G meters.  Given all of the info about your car, it'll give you an estimated horsepower.  Now you remove 200 lbs worth of extra weight from the car and try again, but you neglect to change the weight input in the G meter's computer to reflect that you've removed 200 lbs.  Your acceleration will be better because your car is lighter and the meter will detect this change in acceleration.  Because the meter is still working with the old 3200 lb weight, it will tell you that you're making more power.  It's a false reading.

[JYODER240]

It's a false reading.? A lot of dynos out there are inertial dynos.? How they work is they measure the rate at which a drum is accelerated and the drums speed.? The drum has a known mass and moment of inertia, so once you know the rate of acceleration over a range of speeds, the power can be calculated.? Power is moment of inertia x angular acceleration x angular velocity.

The reason typical chassis dynos will show that there's more power when you switch to lighter driveline components is because the dyno computer will read more acceleration, but nobody informed it that there was a reduction in mass (or moment of inertia) when it goes to calculate the power.? It's not that there's more power, it's that there's less mass in the system.

To give a parallel to this, have you seen those in-car G meters that can estimate your horsepower based on the measured acceleration of the vehicle?? How they work is they measure the acceleration, and then you provide the weight of the car and the gear ratio and tire size and the computer spits out the horsepower.? Let's say your car weighs 3200 lbs and you do a run down the drag strip with one of these G meters.? Given all of the info about your car, it'll give you an estimated horsepower.? Now you remove 200 lbs worth of extra weight from the car and try again, but you neglect to change the weight input in the G meter's computer to reflect that you've removed 200 lbs.? Your acceleration will be better because your car is lighter and the meter will detect this change in acceleration.? Because the meter is still working with the old 3200 lb weight, it will tell you that you're making more power.? It's a false reading.

That doesn't mean that a lighter driveshaft is robbing less hp does it?

[The Pirate]

That makes no sense. Then why do high-performance cars use carbon fiber and other lightweight materials on their driveshafts? How can you say that it takes no more effort to rotate something that weighs 15lbs vs. 300lbs?

I could be wrong, and one of the ME's will correct me if this is the case, but the effort to rotate said piece at a constant speed doesn't require more effort.  It's accelerating that rate of rotation where you will see improvement with the lighter mechanism.

[JYODER240]

I could be wrong, and one of the ME's will correct me if this is the case, but the effort to rotate said piece at a constant speed doesn't require more effort.? It's accelerating that rate of rotation where you will see improvement with the lighter mechanism.

but wouldn't you see an increase in whp during the accelerating?

[Nethead]

You have no clue what you're talking about and you completely f-ed up the facts here, as usual, comparing apples to oranges.

On a like for like basis, and expressed in US$, essentially ALL cars are significantly more expensive in Australia than in the US (just as ALL cars are significantly more expensive in the UK than they are in the US).? And it's not because they add or remove x thousand dollars of "good stuff" when they ship them overseas, it's the market dynamics.? A non-z06 corvette can cost the equivalent of $100k in the UK, for example, and they don't add anything to it (other than different spec brake lights and stuff like that).? You can NOT take an Australian price and compare it to a US price when talking about value.? The US version of the GTO is essentially identical to the monaro, which essentially matched the GT500 around their track, which is a fact.?

So no, you can't take the Australian price of one car, the US price of another car, and then try to imply what's what they both cost in the UK.? The GT500, for example, isn't $40,900 in the UK - hell he even said in the clip it was '30 grand' (that's $60k US), had you been paying attention maybe you would have picked up on that.

Try as hard as you want to bungle the facts by comparing prices across countries, the bottom line is the monaro went around the track as fast as the GT500, the monaro is essentially the same as the GTO in the US, and in the US the GTO was a hell of a lot cheaper than the morbidly obese GT500.

And I never said any car was riding on 37 year old tires, I mis-read the article and thought the car set that lap time 37 years ago, you intentionally extracted that and tried to make it sound like I said something that I didn't, stop putting words into other peoples' mouths.

Leblowski: LebDude!? 'Bout time we heard from you!? Sure, the Nethead wasn't being serious about the prices--Monaros even with whatever are just their basic V8s are $50,000 US IN AUSTRALIA WHERE THEY ARE BUILT .? God knows why, but there it is???? GM paid to ship those Monaros to the US, where they were MSRP'd for $34,000 US after some interior trim was added and "GTO" badges were applied.? GO FIGURE????

Now the Nethead here knows very little was left off those Monaros--"Monaro" and "Holden" badges, for sure--and maybe the interior pieces that would be given "Pontiac" or "GTO" monikers once they got off the boat.? ?I was having a little fun with dudes who were also having a little fun comparing an export version of a Monaro with a US-legal version of the GT500 (which would be $208,482,965,376.62 in Australia, and they won't take any country smaller than France as a trade-in for one at a dealership).? You can save a bundle by buying the GT500 in the US and driving it to Australia yourself.? *** NO!? I'M KIDDING, LEBDUDE, I'M KIDDING!!!? REALLY, I AM!!! DON'T--REPEAT--DON'T DRIVE THE GT500 TO AUSTRALIA FROM THE US!!! ***

But it's true nonetheless that a US-legal version of the Monaro, impersonating a GTO, got whupped bad by the nearly 500 pounds heavier, automatic-transmissioned, four-door SRT8 Charger in the slalom, the figure-eight, and the 60--0 braking test.? Don't take my word for it--read the Motor Trend article yourself!? And read it six or seven times so you don't miss much...You might also notice how the GTO's performance figures compared to those of the GT500 on your fourth or fifth reading...

I know, I know--the SRT8 and the GT500 had better tires.? ?But that's how DCX and Ford build these cars--I don't know about the SRT8 but the GT500 doesn't even come with lesser tires that coulda been used for the comparo.? Maybe they coulda put some 37-year-old racing tires on the GT500 and run through that battery of performance tests again...

So lighten up some, LebDude--this here's a forum and we're tongue-in-cheek at least half the time.? OK, OK--more than half the time...Like HEMI666 suggested--that I make up half of this stuff as I type.? Nah, I WOULD make up half this stuff as I type, but you guys provide such great raw material that there just isn't any room nor need for the Nethead here to wax creative to point out the humor in people taking themselves too seriously, too often!?

Your Honor, I move that we adjourn for lunch--all this bringing LebDude up to speed has made us hungry.

[The Pirate]

but wouldn't you see an increase in whp during the accelerating?

That makes sense to me as well, but I don't think it's totally correct.  From a purely mathematically formulated approach, the result would be a quicker acceleration, but no power gains would be realized.

Again, my knowledge of this stuff is rudimentary at best.

[SVT666]

That makes sense to me as well, but I don't think it's totally correct.? From a purely mathematically formulated approach, the result would be a quicker acceleration, but no power gains would be realized.

Does not make any sense on a chassis dyno where it doesn't matter how much the car weighs because it's stationary.  If a car runs the quarter mile faster because of lighter body panels, that's one thing.  But if a car shows higher whp numbers on a chassis dyno where the weight of the car has no effect, then I can't believe that the dyno readings are wrong.

[SVT666]

To give a parallel to this, have you seen those in-car G meters that can estimate your horsepower based on the measured acceleration of the vehicle?? How they work is they measure the acceleration, and then you provide the weight of the car and the gear ratio and tire size and the computer spits out the horsepower.? Let's say your car weighs 3200 lbs and you do a run down the drag strip with one of these G meters.? Given all of the info about your car, it'll give you an estimated horsepower.? Now you remove 200 lbs worth of extra weight from the car and try again, but you neglect to change the weight input in the G meter's computer to reflect that you've removed 200 lbs.? Your acceleration will be better because your car is lighter and the meter will detect this change in acceleration.? Because the meter is still working with the old 3200 lb weight, it will tell you that you're making more power.? It's a false reading.

However the weight of the car has no effect on the reading that spits out of a chassis dyno.

[The Pirate]

Yeah, I'm out of my league here.

[GoCougs]

I could be wrong, and one of the ME's will correct me if this is the case, but the effort to rotate said piece at a constant speed doesn't require more effort.? It's accelerating that rate of rotation where you will see improvement with the lighter mechanism.

That is correct. The "improvement" though is not because of an increase of RWHP. It's the result of less inertia, which means greater acceleration. This greater acceleration does not occur because of an increase in RWHP though.

[SVT666]

That is correct. The "improvement" though is not because of an increase of RWHP. It's the result of less inertia, which means greater acceleration. This greater acceleration does not occur because of an increase in RWHP though.

What?

I give up.  There's something I'm obviously not understanding here.

[MX793]

That doesn't mean that a lighter driveshaft is robbing less hp does it?

The only thing that really "robs" power in a drivetrain is friction.  Ignoring friction or other losses (what are sometimes called "irreversibilities") acceleration is a function of power and mass.  Increase the power and you'll increase the acceleration.  Decrease the mass and you'll also increase the acceleration.  If you add 500 lbs to your car, it's going to accelerate slower.  Does that mean it's making less power at the wheels?

Friction actually robs power by creating a "negative power" that subtracts from the actual power.

[GoCougs]

I'm done.? All I know is it would be easier for me to spin a 15 lbs flywheel or driveshaft with my hands then it would a 300 lbs flywheel or driveshaft.? To me that equals more power gets to the end component...being the wheels.? If a chassis dyno shows a gain in whp, then I can't say it's wrong.? If it is wrong, then no dyno can be trusted.

Sounds like you're getting a bit snippy about it.

I don't intend to sound arrogant, but the three of us (SoupDeVille, MX793 and myself) that are in agreement that a lighter driveshaft and/or flywheel does not increase RWHP each have a mechanical engineering degree.

[MX793]

However the weight of the car has no effect on the reading that spits out of a chassis dyno.

While most of the car isn't moving at all when strapped to a chassis dyno, parts of it are undergoing an acceleration, and thus a change in kinetic energy.  Which parts?  All of rotating bits in the drivetrain (wheels, driveshaft, flywheel, clutch, etc).  Inertial chassis dynos work by measuring accelerations, then multiply by moment of inertia (which is a mass or weight term) to figure out the power.  If you reduce the moment of inertia of the system by reducing the weight of the rotating pieces, you'll get more rotational acceleration in the rotating bits for the same power, just as if you reduce the vehicle as a whole you'll get better acceleration at the drag strip.  But if you don't change the moment of inertia parameter in the dyno's computer to reflect this reduction in rotating mass, it will falsly read that you're generating more power because the dyno is measuring greater drum acceleration and, as far as it knows, the rotating mass in the system is the same as before.

[93JC]

Sounds like you're getting a bit snippy about it.

I don't intend to sound arrogant, but the three of us (SoupDeVille, MX793 and myself) that are in agreement that a lighter driveshaft and/or flywheel does not increase RWHP each have a mechanical engineering degree.

I have three quarters of a mechanical engineering degree, and I say a lighter driveshaft and/or flywheel can increase RWHP...

... fractionally, because friction losses would hypothetically be smaller. If the parts are lighter then, hypothetically, you won't lose as much energy to the bearings, etc.

But the difference is probably negligible at best.