Tracking with Forced Induction

[D K]

Are you guys saying a turbo car is MORE likely to tto than equally powerful NA car????

[cdrazic93]

By mathmatical reasoning the derivative of the power curve, if described as perfectly linear, should have its limit approaching zero. This mathmatical constant of approaching zero on a power curve ends up translating itself as a constant torque line.

In essance calling it a "curve" implies there is a change in the Y values.

If power was pefectly linear which it will never be then the torque will be as stated before, a flat horizontal line with a slope of zero.

[s2d4]

Originally Posted by cdrazic93

By mathmatical reasoning the derivative of the power curve, if described as perfectly linear, should have its limit approaching zero.

Sounds awesome but no idea what you are on about.
The derivative would be m, not zero.

[cdrazic93]

Slope is a constant in the basic form of linear power. The derivative of a constant is always zero. If the slope of a supercharged 86 is 43, then the derivative is zero, thus the torque of that 86 is a flat line; or as flat as it can be.

[s2d4]

Originally Posted by cdrazic93

Slope is a constant in the basic form of linear power. The derivative of a constant is always zero. If the slope of a supercharged 86 is 43, then the derivative is zero, thus the torque of that 86 is a flat line, or as flat as it can be.

You are talking about the torque curve now? That would be right for torque.
Like I said, what you said sounded nice but I would have failed you if you presented it as an explanation in intro calcs class.

[cdrazic93]

Originally Posted by s2d4

You are talking about the torque curve now? That would be right for torque.
Like I said, what you said sounded nice but I would have failed you if you presented it as an explanation in intro calcs class.

edit: my bad, I'm actually talking about acceleration, double derivative. d/dx of y=mx+b is m. y'' of y=mx+b is 0. Even so, the slope m, is still a constant. Acceleration would be even across the rpm band; QED torque "curve" is flat.

[s2d4]

Originally Posted by cdrazic93

edit: my bad, I'm actually talking about acceleration, double derivative. d/dx of y=mx+b is m. y'' of y=mx+b is 0. Even so, the slope m, is still a constant. Acceleration would be even across the rpm band; QED torque "curve" is flat.

All good man.

[cdrazic93]

Originally Posted by s2d4

All good man.

my professor had us spend much more time on the harder functions like d/dx of logs, trig and inverse trig. all the super fun ones.

[s2d4]

Originally Posted by cdrazic93

my professor had us spend much more time on the harder functions like d/dx of logs, trig and inverse trig. all the super fun ones.

My professor didn't spend much time at all on those. Fourier, laplace, Z transforms and maxwell equations were his happy places.
Didn't know professors teach in high school over there.
Edit: Inverse trigs are applicable to AC true, apparent and reactive power (complex plane) .

[clayrush]

Originally Posted by ptuning

Also, while a turbocharger will have a perceived transient lag to a supercharger, its powerband doesn't make it less than ideal for track use. The SC guys will say that the lack of transient boost lag will make it easier to drive fast on the track, I say that the lack of torque/power coming out of the corner makes it easier to drive in the hands of a less experienced driver. Once you've had a lot of seat time dialing in your car, you'll appreciate the much greater torque/horsepower output that a turbo powerband makes when you shift into the next gear.



- Toan

I don't follow F1 much but I thought I heard that they were having a harder time keeping their cars under control this season partly because of the way the turbo kicks in. On top of different aero and the ERS.
I think they have a bit of experience.

I also notice that many of the turbo Lemans cars like to fish tail coming out of hard turns.

[s2d4]

Originally Posted by clayrush

I also notice that many of the turbo Lemans cars like to fish tail coming out of hard turns.

Worn out tyres?

[clayrush]

Originally Posted by s2d4

Worn out tyres?

Could be.

We were at Laguna Seca for the tudor championship. It seemed like the first spot I really noticed it they were only about 10-15 laps in, on turn 3. I thought at first it was all cars and possibly sand at that spot, but after awhile of sitting I noticed that it seemed like cars with turbo power. It seemed the same at turn 2 and 11 that day as well.

I could be totally wrong, and I may have been wrong on what machines have turbos.It was just something I noticed.

[King Tut]

Originally Posted by clayrush

I don't follow F1 much but I thought I heard that they were having a harder time keeping their cars under control this season partly because of the way the turbo kicks in. On top of different aero and the ERS.
I think they have a bit of experience.

I also notice that many of the turbo Lemans cars like to fish tail coming out of hard turns.

It isn't the way the turbo kicks in, it is the larger amount of torque they have this season because of the turbo and K as Vettel likes to call it. You don't have much torque when you have a small displacement motor reving to 18,000 RPM like they did prior to this season.

[cdrazic93]

Originally Posted by s2d4

Didn't know professors teach in high school over there.
Edit: Inverse trigs are applicable to AC true, apparent and reactive power (complex plane) .

Im a sophomore in college haha hence the 'professor' and not 'teacher'. took me a while to pass pre-calc, I hated that class. Calc is so much easier!