ZR1 vs GTR vid, ZR1 a disappointment

[ndubz]

Ok guys watch this vid. http://www.youtube.com/watch?v=mC-PQca6FJU

Yes i realize that edmunds declares the ZR1 the winner, but lets look a little closer... First off, on the dyno the ZR1 makes 505whp that from 638bhp, thats a 21% drivetrain loss!!! from a rear drive vette? thats more than my awd sti loses!!! the gtr made 404awhp from 483bhp, thats a normal 17% drivetrain loss for an awd car. The vette is losing double what most rwd cars lose at the wheels. So if u got a GTR, send it to COBB tuning, they will give it similar hp to the stock ZR1, and you will actually make more whp too. with an awd car!!! not to mention, the GTR even after being sent to COBB, is still a few G $tack$ less than the ZR1. Also interesting how they had to compare the GTR to the ZR1 over the more appropriately priced Z06...hmm. Sorry chevy, i think Nissan has u beat either way. Not that either of these cars are bad, just saying that the ZR1 disappointed a bit from my expectations. And that the GTR has risen even higher in my respect than before.

Also, Edmunds sucks A$$ at getting good test results. 0-60 figures for both cars should be in the low 3s and they got 3.8. The vette i understand, but in the gtr, all u have to do is set up launch control and floor the gas, its retardedly easy so edmunds must be retarded. LOL

[Scapegoat]

oh here we go again...

whoooooooo caaaaaares

people buying the ZR1 are buying a car like that for a specific reason, and those buying the GTR have different reasons... we all know that japanese engineering > american...

but forever it will hold true... american muscle > japanese finesse...

[Oakes]

That was the most boring commentator they could have possibly put for those two awesome cars. That is a serious power loss though on the ZR1.

[GHost240SX]

Just a note Nissan as well as GM put out factory freaks. And you have to take into account the different drivetrains and what is attached to them.

[SovXietday]

The transmission and rear in the ZR1 actually will hold up to the power though. We already know the GT-R has issues with it's transmission not being able to hold the power, and Nissan themselves say they've only gotten a handful of launches out of the thing before it axed itself.

That ZR1 drivetrain is damn near bulletproof, if I remember correctly they are using very similar if not the same thing in the ACR Viper, which makes an UNGODLY amount of torque (over 700 if I remember correctly). When talking drivetrain, you only need talk about torque, HP doesn't matter as it is derived from torque.

Take into account that the ZR1 is very torquey, RWD, lightweight, and has very little driver assists. You will need to be behind the wheel of that car for a long ass time before you start to get the full potential out of it. The GT-R is an easy to learn car and most drivers are going to pick it up real quick.

[Oakes]

While I agree with your point, I still feel like a 133 hp loss is kind of weird.

[OBEEWON]

The Laplace transform replaces one function F(t) of t by another f(s) of the new variable s by the rule: f(s) = ∫(0,∞) e-stF(t)dt. f(s) is called the transform of F(t): f(s) = L[F(t)]. An easy integration by parts gives the transform of the derivative of F(t): L[F'(t)] = sf(s) - F(0). If we replace s by p, and F(0) = 0, we recover Heaviside's expression. The transform of U(t) is 1/s, as can be found by doing the easy integral. As far as F(t) is concerned, the transform uses only the values for t > 0, just as in the operational calculus.

Returning to our transient problem, we transform the equation L(dI/dt) + RI = EU(t) to find Lsi(s) + Ri(s) = E/s. This gives i(s) = E/[s(Ls + R)]. To find I(t), we must find the inverse transform of the function of s on the right-hand side. In the elementary use of the Laplace transform, this is done as follows. First, we note that 1/[s(s + k)] = (1/k)[1/s - 1/(s + k)]. This is a partial fraction expansion (used also by Heaviside). We already recognize one term: 1/s = L[U(t)]. As for the other term, it is the transform of e-ktU(t), which can be found by direct integration. If L-1[1/s] = 1 and L-1[1/(s + k)] = e-kt, then L-1[1/s(s+k)] = (1 - ekt)/k, and we find that I(t) = (E/R)(1 - ekt). It is usual to drop the U(t) when using the Laplace transform. Comparing the Heaviside and Laplace methods shows how very similar they are in practice, if not in theory.

The Laplace method, however, has much more powerful mathematics behind it. This includes the convolution theorem, that states that L-1 [f(s)g(s)] = F(t)◊G(t) = ∫(0,t) F(t - τ)G(τ)dτ, the convolution of the two functions F(t) and G(t). The symbol ◊ is used here because the browser does not support the usual symbol, a cross in a circle. It also shows that Laplace transforms are closely related to Fourier transforms, so that similar properties exist (such as convolution). The Fourier transform has an inverse transformation, and the transformation between time and frequency domains is equally convenient both ways, involving a simple integral along the real axes. The Laplace transformation is quite troublesome to invert. There is a formula involving real integrals, and another with repeated differentiation (see Churchill and Widder), but these are of little use because the integrals are too difficult to perform in practical cases. The only practical inversion integral uses analytic functions of a complex variable, and will be discussed below.

Most problems in elementary applications of the Laplace transform can be solved by reference to tables of transform pairs. This was also the usual method for the Heaviside calculus as well. Churchill gives a table with 122 transforms that is extensive enough for most purposes. It is curious that the Laguerre and Hermite polynomials, as well as Bessel functions, have relatively simple transforms. Students like Laplace transform because it seems to be a mechanical method that can be used to find answers without thinking. This is an erroneous impression.

[Oakes]

< head just assssplodeddd

[SovXietday]

Originally Posted by OBEEWON

E=MC^2

Is it bad that I kinda understood your post?

[TGilb2007]

Originally Posted by Oakes

< head just assssplodeddd

Mine too......

I now have a headache.

[OBEEWON]

Originally Posted by SovXietday

Is it bad that I kinda understood your post?

LOL

Yesss, stop paying attention in differential equations!!

[DSMTJ]

i have not seen a zr1 dyno sheet for a stock car, but you put a set of headers on one and it will make over 600whp. good luck getting that out a gtr with $2k invested.

anyway, these are totally different cars, transmissions, engines, and drivetrains. totally different. they were built for a different kind of person.

[ndubz]

Originally Posted by DSMTJ

i have not seen a zr1 dyno sheet for a stock car, but you put a set of headers on one and it will make over 600whp. good luck getting that out a gtr with $2k invested.

anyway, these are totally different cars, transmissions, engines, and drivetrains. totally different. they were built for a different kind of person.

tru, but ur forgetting that the ZR1 is waaay more expensive to begin with, so with that price difference, they are very competitive cuz u can put alot more than 2k into the gtr and still be cheaper. But tru, they are very different.

[ndubz]

Originally Posted by SovXietday

The transmission and rear in the ZR1 actually will hold up to the power though. We already know the GT-R has issues with it's transmission not being able to hold the power, and Nissan themselves say they've only gotten a handful of launches out of the thing before it axed itself.

That ZR1 drivetrain is damn near bulletproof, if I remember correctly they are using very similar if not the same thing in the ACR Viper, which makes an UNGODLY amount of torque (over 700 if I remember correctly). When talking drivetrain, you only need talk about torque, HP doesn't matter as it is derived from torque.

Take into account that the ZR1 is very torquey, RWD, lightweight, and has very little driver assists. You will need to be behind the wheel of that car for a long ass time before you start to get the full potential out of it. The GT-R is an easy to learn car and most drivers are going to pick it up real quick.

Very true. Bulletproof they are. I really hope nissan fixes their tranny issues asap cuz the gtr will then have NO flaws!!!!

[The Captain]

Worst comparison video i've ever seen. Obvious bias since the beginning, not even taking into account the fact that the two cars are COMPLETELY different.

[ndubz]

Originally Posted by OBEEWON

The Laplace transform replaces one function F(t) of t by another f(s) of the new variable s by the rule: f(s) = ∫(0,∞) e-stF(t)dt. f(s) is called the transform of F(t): f(s) = L[F(t)]. An easy integration by parts gives the transform of the derivative of F(t): L[F'(t)] = sf(s) - F(0). If we replace s by p, and F(0) = 0, we recover Heaviside's expression. The transform of U(t) is 1/s, as can be found by doing the easy integral. As far as F(t) is concerned, the transform uses only the values for t > 0, just as in the operational calculus.

Returning to our transient problem, we transform the equation L(dI/dt) + RI = EU(t) to find Lsi(s) + Ri(s) = E/s. This gives i(s) = E/[s(Ls + R)]. To find I(t), we must find the inverse transform of the function of s on the right-hand side. In the elementary use of the Laplace transform, this is done as follows. First, we note that 1/[s(s + k)] = (1/k)[1/s - 1/(s + k)]. This is a partial fraction expansion (used also by Heaviside). We already recognize one term: 1/s = L[U(t)]. As for the other term, it is the transform of e-ktU(t), which can be found by direct integration. If L-1[1/s] = 1 and L-1[1/(s + k)] = e-kt, then L-1[1/s(s+k)] = (1 - ekt)/k, and we find that I(t) = (E/R)(1 - ekt). It is usual to drop the U(t) when using the Laplace transform. Comparing the Heaviside and Laplace methods shows how very similar they are in practice, if not in theory.

The Laplace method, however, has much more powerful mathematics behind it. This includes the convolution theorem, that states that L-1 [f(s)g(s)] = F(t)◊G(t) = ∫(0,t) F(t - τ)G(τ)dτ, the convolution of the two functions F(t) and G(t). The symbol ◊ is used here because the browser does not support the usual symbol, a cross in a circle. It also shows that Laplace transforms are closely related to Fourier transforms, so that similar properties exist (such as convolution). The Fourier transform has an inverse transformation, and the transformation between time and frequency domains is equally convenient both ways, involving a simple integral along the real axes. The Laplace transformation is quite troublesome to invert. There is a formula involving real integrals, and another with repeated differentiation (see Churchill and Widder), but these are of little use because the integrals are too difficult to perform in practical cases. The only practical inversion integral uses analytic functions of a complex variable, and will be discussed below.

Most problems in elementary applications of the Laplace transform can be solved by reference to tables of transform pairs. This was also the usual method for the Heaviside calculus as well. Churchill gives a table with 122 transforms that is extensive enough for most purposes. It is curious that the Laguerre and Hermite polynomials, as well as Bessel functions, have relatively simple transforms. Students like Laplace transform because it seems to be a mechanical method that can be used to find answers without thinking. This is an erroneous impression.

Ok i sort of get this, but i dont take this course for 2 more years. But damn bro did u just know this, or did google help u out. Be honest its impressive either way.

[SovXietday]

^ That is copy and paste to the max.

[92sileighty]

i wanna know how he got the integration symbol on here lol i didn't know that existed

[The Captain]

Originally Posted by OBEEWON

Blah blah blah math stuff blah blah blah

Awe, now why you gotta go and get all smart on us, Donnie?

[CleanNeon98]

Originally Posted by ndubz

Ok guys watch this vid. http://www.youtube.com/watch?v=mC-PQca6FJU

Yes i realize that edmunds declares the ZR1 the winner, but lets look a little closer... First off, on the dyno the ZR1 makes 505whp that from 638bhp, thats a 21% drivetrain loss!!! from a rear drive vette? thats more than my awd sti loses!!! the gtr made 404awhp from 483bhp, thats a normal 17% drivetrain loss for an awd car. The vette is losing double what most rwd cars lose at the wheels. So if u got a GTR, send it to COBB tuning, they will give it similar hp to the stock ZR1, and you will actually make more whp too. with an awd car!!! not to mention, the GTR even after being sent to COBB, is still a few G $tack$ less than the ZR1. Also interesting how they had to compare the GTR to the ZR1 over the more appropriately priced Z06...hmm. Sorry chevy, i think Nissan has u beat either way. Not that either of these cars are bad, just saying that the ZR1 disappointed a bit from my expectations. And that the GTR has risen even higher in my respect than before.

Also, Edmunds sucks A$$ at getting good test results. 0-60 figures for both cars should be in the low 3s and they got 3.8. The vette i understand, but in the gtr, all u have to do is set up launch control and floor the gas, its retardedly easy so edmunds must be retarded. LOL

But that would break it