Please explain physics to me

[0-120]

Originally Posted by pyroguy

And Force = mass x acceleration but that doesn’t help here.



Interesting read, but unless I missed it did it spell out that the drivetrain loss is a linear percentage as horsepower increases?

Nor is it possible to apply a fixed drivetrain loss figure to your car (say 60 whp from my RevUp G35 example), because as you modify the engine and increase its output its ability to generate thrust force and angular acceleration also increases (though not in a linear fashion).

[christianchevell]

https://www.automobilemag.com/news/c...ic-drag-racer/

[Moto-Mojo]

Originally Posted by christianchevell

https://www.automobilemag.com/news/c...ic-drag-racer/

Pay close attention to this post. Notice it’s an “e” COPO. I was just chatting with the boys in the Chevy booth at SEMA this week. That car was there. I’ll post a pic.

[Moto-Mojo]

[Moto-Mojo]

Ok, sorry, I inserted the gasser AND the eCopo...

[christianchevell]

Yup look at all them wires they have to discombobulate the quantum physic regulator to get it up to 66.76 giggawatts at 76 MPH to go to SEMA Scientific Energy Manufacturers Association....theres got to be a MR. Nuke coffee maker in the console that converts energy also.....

Wish I could make it there....

[NEstyle]

The more power you make, the more loss there is. That's just how it works. Think about pushing your hand through water in a pool slowly. There is barely any resistance (loss) felt by you. Now push it through as fast as you can. You will now feel a lot of resistance (loss), even though you're pushing the same hand through the same medium. The only difference is that you have added power. It's a similar concept.

[Norm Peterson]

Originally Posted by 2015ghost

So if I have a car and for arguments sake it makes 400 hp. People say figure 20% for drivetrain loss. Ok so 80hp. But if my car makes 800hp and you still say 20% drivetrain loss then the drivetrain now takes 160hp to turn! How does this translate? What in the flux compasitor is this witchcraft?

The losses probably are not 'linear' with respect to the power being transmitted, but the absolute amounts of power lost to friction aren't going to remain constant either.

If you're doubling the torque (IOW, doubling the HP at any given rpm), you're doubling the gear tooth contact pressures, so in absolute terms the losses would be at least somewhat greater at the higher torque.


Norm

[Moto-Mojo]

They were pretty proud of that thing. It was a race car in every sense - from a how quick it went down the track, point. Me and the old-school sales dude just stood there and talked about our dinosaur, SS Camaros and speculated how many millennials would embrace the e car after we’re all gone.

[Royal Tiger]

Originally Posted by NEstyle

The more power you make, the more loss there is. That's just how it works. Think about pushing your hand through water in a pool slowly. There is barely any resistance (loss) felt by you. Now push it through as fast as you can. You will now feel a lot of resistance (loss), even though you're pushing the same hand through the same medium. The only difference is that you have added power. It's a similar concept.

That’s a neat way to explain it. Cool. Thanks.

[CHRISCAM]

I'm having a difficult time understanding how a given transmission and rear end assembly doesn't use the same amount of torque/ hp to turn, regardless of what engine is mounted to the front of the bellhousing.

Imagine a dyno shop that has a given transmission (let's say it's a T6060) mounted in their test stand, along with a 48" drive shaft, backed up by a full Gen. 5 Camaro IRS rear end and rear wheels (essentially, a full Gen 5 Camaro, minus the body and interior, just the frame with a drivetrain).

The shop installs an electric test motor to determine the amount of torque/ hp required to turn the existing driveline at a given rpm. It's determined by this test procedure that 50 hp/ lb. ft of torque are required to turn the assembly at 3000 rpm.

Now, if the shop then installs a 430 hp LS3 ahead of that T6060, wouldn't that driveline still consume 50 hp/ tq at 3000 rpm? That would leave 380 net hp at the rear wheels (which figure I have intentionally used in this example because 380 hp is pretty close to average rwhp on these Gen 5 LS3s, it seems).

Now, if we were to add a cam that is dyno-tested to add 20 hp to that same LS3 and T6060/ ISR rear (thus now making a dyno-tested 450 at the crank)-- wouldn't the rwhp now register at 400?

Or, per the OP's question-- 800 crank hp should have around 750 at the rear wheels with this same driveline?

Someone explain in the flaw in my logic. I'm not being sarcastic here, by the way.

[Dragonman]

I don't if this will help you Chriscam but, as I understand that water analogy above, the quicker (higher torque values) you try to turn that same driveline, the more resistance there is to turning it. It's just not a 1:1 ratio as you up the torque.

[mikeinmobile]

Originally Posted by Moto-Mojo

The good news is, a study in 2010 (coincidence with introduction of the 5th gen?) suggests that the laws of physics as we know them, may not be the same in all parts of the universe. I’m thinking the HP gains in another sector could be huge! Or really parasitic.

So I'm probably making around 1000 hp on the moon? YES!!!!

[mikeinmobile]

Of course I'm probably only making around 2 hp on Jupiter. (if that)