It's Official! 2020 GT500 Makes 760HP

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DEARBORN, Mich., June 19, 2019 – Venomous strike: The all-new 2020 Mustang Shelby GT500 will produce 760 horsepower and 625 lb.-ft. of torque, making it the most powerful street-legal Ford ever – with the most power- and torque-dense supercharged production V8 engine in the world.

Enough said.
 
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No chance. PDK is at least 12 but more likely above 15%. 991.1 GT3 PDK at 400whp is a 16% loss.


You think crank is going to closer to 800hp? 700rwhp at 12% is 795hp. Kaneda would pull throw his toys away...
View attachment 1591889 View attachment 1591890

I don’t buy any of the % loss stuff.

I know it takes torque to turn items, and that thermal loss is mute at 150-300°F in a metal or carbon fiber gear set or driveshaft. You don’t magically see the metal or carbon fiber melt and eat up torque much like gum at 300°F and no transmission or differential I know on a factory car is very comfy at 300°f. Most set them to operate in and at 1500-250°.

So if the metals aren’t molten in any way, there is at most a 1% thermal loss that could be assumed only because of surface gumming. And that’s a stretch imo.

So from there let’s say linearly torque is eaten up to rotate the transmission, differentials, half shafts, hubs and wheels to 7500rpm, totaling about 50tq.

Considering that 50tq and 80hp is enough to motivate a 1800lbs 1980’s vw big to 100mph, I think it’s safe to assume it only takes about 50tq to move the rotating assembly on the gt500 to 7500rpm in a 1:1 gear on a 28” tire.

The main reason I think the trinity gt500’s ate up so much torque to the wheels is because of the 3.31 rear. The 3.73 in the gt500 is much more mechanically advantageous, as are the much lighter wheels. The heavier brakes are something I can’t readily figure into the mix but seeing they also get centrifugal affects much like the wheel/tire, I’m not as worried. There isn’t a gear being turned to specifically spin the brake rotators separate of the drivetrain assembly so I don’t even factory it in honestly.

I’m not a mechanic, I just play one on tv. Mechanics and technicians I do talk to agree with this general assessment, but don’t know the values so they say they just use the % as it’s easier.
 
It’s just that when on an engine dyno they put out so much and the when on a chassis dyno they register 14/15% less with manual and 18/20% less with auto(about 5 years ago with amg auto trans.

The vette lost less power as the transmission was out back in front of drive wheels.
 
It’s just that when on an engine dyno they put out so much and the when on a chassis dyno they register 14/15% less with manual and 18/20% less with auto(about 5 years ago with amg auto trans.

The vette lost less power as the transmission was out back in front of drive wheels.

Look at dodge cars. Almost same driveline in the 375hp cars, 485hp scatpack cars, and hellcats and demon.

They all roughly lose 40-50hp at 6200rpm fuel cuts.

It takes less torque to spin 6k rpm than 9k rpm. This is precisely why high rpm vehicles lose a lot more to the ground.

Exhibit a: Yamaha R1’s and virtually all liter bikes lose a ton of power to the wheel. Harly’s Don’t.

15% of 200hp r1’s is 170whp which is very common ‘17+ R1’s.. In theory, look, you are right! 15%!

Buuuuut, Harly 107ci’s 2018+ dyno 101-107 wtq and are rated at 111tq... so what gives?

Rpm matters. Component weight and torque required to spin them matters.

In the dodge world, the hellcats act very much like scatpacks. Hellcats dyno from 600-675whp with the a8, and are rated at 707hp.

Scatpacks dyno 410-460whp and are rated at 485hp.

Dodge for whatever reason has very wide power variance with their engines. Most hellcats dyno 640-660whp. Most scatpacks dyno 420-440whp. In both cases that is ~40-50hp lost. Redline 6200
 
I’m not speaking of horsepower torque curves. I’m speaking of engine dyno vs chassis dyno. There is a 15% loss + between those numbers.
 
I’m not speaking of horsepower torque curves. I’m speaking of engine dyno vs chassis dyno. There is a 15% loss + between those numbers.

No. I know what you’re speaking of. The theory is disproven. Component weight, rpm, final drive ratio, gearing, wheel/tire weight and height all factor into the loss.

Why does a v8 m3 make 320-340whp from 414hp. 8500rpm... 15-20% drivetrain loss.

Now, look at this

https://www.motortrend.com/news/mustang-power-we-dyno-the-2011-ford-mustang-50-v8-and-37-v6-8060

Notice how the v6 mustang with the same transmission and almost the same drivetrain as the gt gets assumed to make 308hp vs 305hp with the “15%” loss, but the GT gets revised to an assumed 435hp?

That gt also made substantially more power than bmw’s e92 v8 m3’s at the wheels. The gt350 loses more power to the wheels than the mustang gt, while also revving 1000rpm higher. I rest my case.
 
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Prove it to me

Mustang Power! We Dyno the 2011 Ford Mustang 5.0 V-8 and 3.7 V-6 - MotorTrend

I updated that response too.

Why is the gt assumed up to 435hp, but the v6 assumed the same as rated by Ford? Because 15% doesn’t matter when you take virtually the same drivetrain and add 100hp.

Think of it logically. You have a power source. You create a cog and wheel system to transfer that force into motion. You add more power to the motor. The drive system is still easily capable of handling this load.

Will it
A. Transfer the additional power in full
B. Magically lose some percentage of the new power because the original power lost 15% when transferred through the drive system
C. Become a very long winded tt335ci03cobra opinion thread
D. It’s time for beer.
 
Once you effectively turn all the drive components, and spin roughly the same rpm, there is no magic 15% being sapped out of additional power. Thermal loss is less than 1% under 300° with the present metals and fluids in an engine, and they operate at 150-250°.
 
Once you effectively turn all the drive components, and spin roughly the same rpm, there is no magic 15% being sapped out of additional power. Thermal loss is less than 1% under 300° with the present metals and fluids in an engine, and they operate at 150-250°.
I've never thought of it that way, seems to make sense though
 
IF the rest of the drivetrain is exactly the same then you are correct. The percentage system is commonly accepted because an engine with more horsepower and torque will usually warrant stronger, heavier, components in the transmission or rear end than the same car with a lower HP output.
 
Mustang Power! We Dyno the 2011 Ford Mustang 5.0 V-8 and 3.7 V-6 - MotorTrend

I updated that response too.

Why is the gt assumed up to 435hp, but the v6 assumed the same as rated by Ford? Because 15% doesn’t matter when you take virtually the same drivetrain and add 100hp.

Think of it logically. You have a power source. You create a cog and wheel system to transfer that force into motion. You add more power to the motor. The drive system is still easily capable of handling this load.

Will it
A. Transfer the additional power in full
B. Magically lose some percentage of the new power because the original power lost 15% when transferred through the drive system
C. Become a very long winded tt335ci03cobra opinion thread
D. It’s time for beer.
They don’t don’t pull the motor and strap it to an engine dyno though. Until you test the same motor to both an engine and chassis dyno you don’t have the loss of power.
 
The amount of power sent through the drive train will dictate the amount of power lost in noise/heat/slippage/etc.

Cruise along at 40 mph in a truck for 20 minutes and get out and check the temp of the transmission and rear end. Probably somewhat warm, not too bad.

Now hook up to a big trailer that requires 4x the amount of power to drive at the same 40 mph for 20 minutes. The temps will be a lot higher, which are losses.

The drive train doesn't have a magic 50 horsepower to get it to move. Can you imagine needing 50 hp to pull forwards in a drive through? The rough number from what I can find for a car to maintain speed on level ground is about 10-20 horsepower.

The same idea works at RPM too, click a coyote into 3rd at 70 mph and just breathe on the throttle, now do the same thing with a pair of chutes hanging off the back of it and try to maintain 70 mph. Which situation is going to generate more heat/noise (losses) in the transmission and rear end?
 
Mustang Power! We Dyno the 2011 Ford Mustang 5.0 V-8 and 3.7 V-6 - MotorTrend

I updated that response too.

Why is the gt assumed up to 435hp, but the v6 assumed the same as rated by Ford? Because 15% doesn’t matter when you take virtually the same drivetrain and add 100hp.

Think of it logically. You have a power source. You create a cog and wheel system to transfer that force into motion. You add more power to the motor. The drive system is still easily capable of handling this load.

Will it
A. Transfer the additional power in full
B. Magically lose some percentage of the new power because the original power lost 15% when transferred through the drive system
C. Become a very long winded tt335ci03cobra opinion thread
D. It’s time for beer.
Comparing Chassis and Engine Dynos - Popular Hot Rodding Magazine

Engine dyno 675hp crank.
Chassis dyno 564hp at the wheels. Same engine.

This is a drivetrain loss of 16.5%

If the GT500 has a similar efficiency loss the new GT500 would be dyno around 640rwhp.
 
The amount of power sent through the drive train will dictate the amount of power lost in noise/heat/slippage/etc.

Cruise along at 40 mph in a truck for 20 minutes and get out and check the temp of the transmission and rear end. Probably somewhat warm, not too bad.

Now hook up to a big trailer that requires 4x the amount of power to drive at the same 40 mph for 20 minutes. The temps will be a lot higher, which are losses.

The drive train doesn't have a magic 50 horsepower to get it to move. Can you imagine needing 50 hp to pull forwards in a drive through? The rough number from what I can find for a car to maintain speed on level ground is about 10-20 horsepower.

The same idea works at RPM too, click a coyote into 3rd at 70 mph and just breathe on the throttle, now do the same thing with a pair of chutes hanging off the back of it and try to maintain 70 mph. Which situation is going to generate more heat/noise (losses) in the transmission and rear end?

Yes but also indeterminable.

You are getting to but never fully got to thermal loss being because as temperature rises, the metals and alloys become slightly tacky, much like a tire at 100° is slightly bendable just by hand. No, the metal is not sheering or breaking away, but yes, the metals and alloys are sticking to one another and starting to ever ever slightly melt together on surface contact. This is thermal loss. At 300°, it is so minimal that it doesn’t even equate to a 1% tranfer theft or thermal loss. The oils on the other hand have expanded their heat tolerances so general efficiency may be down more than 1%.

What you are describing is what I brought up in greater detail earlier. Keep in mind, cars are NOT dyno’d super duper hot and on the brink of sheering an axle or diff.

I’ve been to many dyno days, and temps don’t exceed 200° water, 240° trans etc, unless a car isn’t put together well. Factory cars easily handle dyno pulls and stay well below even the specified operating temp range.

All of this is based on time spent with big machinery, that said, I’m also speculating.

I don’t get what noise transfer would cause loss, vibration I get, but modern drive trains don’t vibrate if at all a thousandths.
 
Comparing Chassis and Engine Dynos - Popular Hot Rodding Magazine

Engine dyno 675hp crank.
Chassis dyno 564hp at the wheels. Same engine.

This is a drivetrain loss of 16.5%

If the GT500 has a similar efficiency loss the new GT500 would be dyno around 640rwhp.

I assume you realize you posted an article with a nova that has:


With a 9-inch rearend housing, 3.73 gears, a Powerglide trans, and a 3,500-stall converter spinning BFGoodrich drag radials set at 30 psi, the driveline was as common as anything else.”

I assume you don’t comprehend much of the article you posted. You should re read it because if more than validates everything I’ve claimed.

So let’s play with what you’ve presented. That drivetrain ensemble ate up ~125tq based on comparing 3 different sae protocols. The difference largely comes from:

I read the article. The eddy wheel hub dyno corrected down to 487whp because it was assuming there were no accessories being spun by the engine. Gross hp. Think 1970’s. Keep in mind the engine is using 35-40tq to spin it’s serpentine belt and accessories at 6000rpm. That means 45hp right there, and they admit the temps and humidity varied. It would have shown about 535whp. It was hotter, and we don’t know the temp the dyno corrects to with that rating. The torque is added back in because it’s being accounted out from the bottom up.

Next the 564whp number was corrected to 77°f meanwhile the 675hp engine rating was corrected to 60°f and GROSS HP. The torque must be reduced because the torque is being added in from the top down. Meaning it reads about 40tq higher in this case. So apart from over correcting because it was 17° colder in the settings, it also is adding in the gross values.

So that 675hp engine would be realistically a 625hp engine at 6000rpm. 40tq at 6k is 45hp. 675-45=630. I’m going to be nice and say 17° is only worth 5hp. It’s honestly about 8.5hp at this juncture.

Now on to the comparison, what do we find...

564whp from the Dynojet becoming 630hp is a 66hp (57tq) loss at 6000rpm. That is very fitting of this type of setup.

Somethings you also should consider...

The % suggestions state that:

FF and RR lose 8-10%
Mr and fr lose 10-15%
Awd loses about 15-18%
4wd loses about 18-25%.

All manuals.

Autos lose an addition 5% in the guessing game. So in this case, an auto fr car should lose 20%. Even using the craptastic numbers they presented of 564whp and 675hp, we see 16% loss, which doesn’t stand up to the % game.






Look, let me explain it to you like this, does an alternator, water pump, oil pump, and so on eat up gradually more tq just because the engine gets a pair of headers and a supercharger bolted on? Assume rpm doesn’t change, you might be spinning the oil pump a little harder, or the alternator a little harder to supply electricity for the fuel pump, but you don’t just assume you have an extra 25% more parasitic drag on the serpentine assembly because you add 25% more hp.



If you take a 20lbs drive shaft, and hook it up to a thrust dyno, then measure the force needed to spin it to 6,000rpm, its resistance to spin won’t read any differently if you then slowly accelerate it to 6000rpm to represent 300tq, or very quickly accelerate it to 6000rpm to represent 900tq.


Torque is being produced by the engine, and reaches the wheels after driving the drivetrain. The drivetrain is made out of metal and carbon fiber and alloys. None of those materials are spunge like globs of gum or honey that sap away torque.
 
My head hurts.
lol

% or logical assessment, that is the question.

Does a drivetrain magically eat more power just because it is being spun quicker to 6000rpm?

I am 100% confident that the speed to which you spin a drivetrain will eat more and more torque because inherently the opposite force of moving a system weighin let’s say 150lbs of components to 8000rpm is those components desire to be at rest. Resistance to spin is then represented as torque needed to spin to 8000rpm. It will take less torque to spin to 6000rpm or 4000rpm for obvious reason.

Go take a 15lbs bucket of water and spin it a circle with your hand at 30 revolutions per second, then again at 60 revolutions per second. How many calories will you burn at 30 and 60rpm?

Take the transfer from being more so mechanical and direct as in a dct or manual trans and now try to account for pushing fluid and gates and locking up a torque converter. The percentage game does make much more sense with a torque converter auto because of lock up.
 

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