Video & Photo Gallery: 2018 Lexus LS 500 in Autumn Shimmer

mmcartalk

With FWD, HP is not a much of an issue as torque. Torque, especially with unequal-length half-shafts, is what actually tugs at the steering wheel on hard acceleration, not HP. More HP generally means more torque, but not always...several different factors in the engine come into play.

That is simply untrue. Torque does not matter at all once power is transmitted through the transmission; it is one of the most misunderstood fact about automobile on the Internet. A high rpm, low torque engine will have exactly the same effect on torque steer as a low rpm, high torque engine under hard acceleration, as long as they have the same exact peak power and transmission profile. Torque steer is caused by uneven traction, not engine torque. And it's not torque that accelerates the car; it's power, period. There are lots of professionals who interchange the two terms, because they gave up on lecturng the general public and just use whichever is more convenient instead.

The correct form of your statement is that turbocharged engines are likely more prone to torque steer than naturally aspirated engines, because they have more low end torque, hence more power at normal driving rpms. This is also why turbocharged engines "pull harder" than NA engines in daily driving.

mmcartalk

With FWD, HP is not a much of an issue as torque. Torque, especially with unequal-length half-shafts, is what actually tugs at the steering wheel on hard acceleration, not HP. More HP generally means more torque, but not always...several different factors in the engine come into play.

That is simply untrue. Torque does not matter at all once power is transmitted through the transmission; it is one of the most misunderstood fact about automobile on the Internet. A high rpm, low torque engine will have exactly the same effect on torque steer as a low rpm, high torque engine under hard acceleration, as long as they have the same exact peak power and transmission profile. Torque steer is caused by uneven traction, not engine torque. And it's not torque that accelerates the car; it's power, period. There are lots of professionals who interchange the two terms, because they gave up on lecturng the general public and just use whichever is more convenient instead.

The correct form of your statement is that turbocharged engines are likely more prone to torque steer than naturally aspirated engines, because they have more low end torque, hence more power at normal driving rpms. This is also why turbocharged engines "pull harder" than NA engines in daily driving.

ssun30

And it's not torque that accelerates the car; it's power, period.

Hm, sorry my friend, but you are wrong here.
F=m*a, remember?
So, acceleration of the car is proportional to the force it gets from the torque at the driving wheels.
Now, torque at the wheels is related to the engine/transmission gear combo rather than the engine torque alone, but still, it is the torque that does the job, not power.

ssun30

And it's not torque that accelerates the car; it's power, period.

Hm, sorry my friend, but you are wrong here.
F=m*a, remember?
So, acceleration of the car is proportional to the force it gets from the torque at the driving wheels.
Now, torque at the wheels is related to the engine/transmission gear combo rather than the engine torque alone, but still, it is the torque that does the job, not power.

ssun30

And it's not torque that accelerates the car; it's power, period.

Hm, sorry my friend, but you are wrong here.
F=m*a, remember?
So, acceleration of the car is proportional to the force it gets from the torque at the driving wheels.
Now, torque at the wheels is related to the engine/transmission gear combo rather than the engine torque alone, but still, it is the torque that does the job, not power.

ssun30

And it's not torque that accelerates the car; it's power, period.

Hm, sorry my friend, but you are wrong here.
F=m*a, remember?
So, acceleration of the car is proportional to the force it gets from the torque at the driving wheels.
Now, torque at the wheels is related to the engine/transmission gear combo rather than the engine torque alone, but still, it is the torque that does the job, not power.

And here the catalog of LS 500 & LS 500h, surprisingly there are another colors for the interior other than what we know so far, the most obvious is the Red for F SPORT trim :

And here the catalog of LS 500 & LS 500h, surprisingly there are another colors for the interior other than what we know so far, the most obvious is the Red for F SPORT trim :

And here the catalog of LS 500 & LS 500h, surprisingly there are another colors for the interior other than what we know so far, the most obvious is the Red for F SPORT trim :

And here the catalog of LS 500 & LS 500h, surprisingly there are another colors for the interior other than what we know so far, the most obvious is the Red for F SPORT trim :

Madi

And here the catalog of LS 500 & LS 500h, surprisingly there are another colors for the interior other than what we know so far, the most obvious is the Red for F SPORT trim :

Yes, Rioja red... and there is one in the states with it... Good catch!

I love the new carbon fiber clock detail on the F-Sport

View attachment 2490

Madi

And here the catalog of LS 500 & LS 500h, surprisingly there are another colors for the interior other than what we know so far, the most obvious is the Red for F SPORT trim :

Yes, Rioja red... and there is one in the states with it... Good catch!

I love the new carbon fiber clock detail on the F-Sport

View attachment 2490

Madi

And here the catalog of LS 500 & LS 500h, surprisingly there are another colors for the interior other than what we know so far, the most obvious is the Red for F SPORT trim :

Yes, Rioja red... and there is one in the states with it... Good catch!

I love the new carbon fiber clock detail on the F-Sport

View attachment 2490

Madi

And here the catalog of LS 500 & LS 500h, surprisingly there are another colors for the interior other than what we know so far, the most obvious is the Red for F SPORT trim :

Yes, Rioja red... and there is one in the states with it... Good catch!

I love the new carbon fiber clock detail on the F-Sport

View attachment 2490

View attachment 2491

View attachment 2491

View attachment 2491

View attachment 2491

mikeavelli

Yes, Rioja red... and there is one in the states with it... Good catch!

I love the new carbon fiber clock detail on the F-Sport

View attachment 2490

I think that looks more like the GSF's circuit red interior than the Rioja Red found in the LC500.

mikeavelli

Yes, Rioja red... and there is one in the states with it... Good catch!

I love the new carbon fiber clock detail on the F-Sport

View attachment 2490

I think that looks more like the GSF's circuit red interior than the Rioja Red found in the LC500.

mikeavelli

Yes, Rioja red... and there is one in the states with it... Good catch!

I love the new carbon fiber clock detail on the F-Sport

View attachment 2490

I think that looks more like the GSF's circuit red interior than the Rioja Red found in the LC500.

mikeavelli

Yes, Rioja red... and there is one in the states with it... Good catch!

I love the new carbon fiber clock detail on the F-Sport

View attachment 2490

I think that looks more like the GSF's circuit red interior than the Rioja Red found in the LC500.

DFGeneer

Hm, sorry my friend, but you are wrong here.
F=m*a, remember?
So, acceleration of the car is proportional to the force it gets from the torque at the driving wheels.
Now, torque at the wheels is related to the engine/transmission gear combo rather than the engine torque alone, but still, it is the torque that does the job, not power.

As you've correctly pointed out the "F" in the "F=m*a" you just quoted is the traction at the driving wheels. And you've also correctly pointed out that the torque at the wheels is not the same as engine torque, because engine torque does not matter once power is transmitted through the transmission.

A quick thought experiment: let's say we have a big engine that has 400 Nm of torque at 3000rpm and a small engine that has 200 Nm of torque at 6000rpm, so both of them have a peak power of 125 kW. Now we gear the transmission such that both cars (also assuming same weight) reach 60mph at their peak torque output. Such a setup will ensure both cars accelerate to 60mph at the same time. Why? Because remember KE=1/2*m*v^2, and t=KE/P. Both cars will hit the same kinetic energy, while having the same amount of power, so it will take the same amount of time for the two engines to impart the kinetic energy onto the cars. You can apply the same simulation to a tank engine with 2000 Nm of torque at 600rpm or a go-kart engine with 100 Nm of torque at 12000rpm. The acceleration figure will not change, because the power is the same.

Now in real world we can't assume all engines have similar torque curve. The aforementioned "big engine" will generally perform better at normal driving rpms (1500-2500) so a driver gets that power very easily, while the "small engine" will require regular downshifting to, say, pass a car on the highway. That's why people say they "feel the torque" when they press down the gas pedal on a powerful car. But it's not because their engine has more torque, but because their engine has more power at the rpm they are requesting. Modern turbocharged cars have high low-end torque so they can be more efficient; it is a good solution to have both efficiency and power. The similar can be done through continuous variable valve lift on NA engines.

But our conversation was about driving the car near its peak power output, i.e. full throttle acceleration. That's when torque steer really becomes apparent. You wouldn't feel torque steer at all during normal driving.

The only correct way to determine a car's acceleration is to give the customer the full torque curve and the transmission ratios, as well as the dynamic weight distribution and tire grip. But it's unnecessarily complex. If I need a rough estimate, I only need to know the Power-to-Weight ratio. Do yourself a favor and calculate the power-to-weight ratio values of Lexus/BMW/Mercedes/Audi vehicles and plot them against the 0-60 figures, you will see a good correlation there. But try doing that with the torque-to-weight ratio and nothing will come out of it.

DFGeneer

Hm, sorry my friend, but you are wrong here.
F=m*a, remember?
So, acceleration of the car is proportional to the force it gets from the torque at the driving wheels.
Now, torque at the wheels is related to the engine/transmission gear combo rather than the engine torque alone, but still, it is the torque that does the job, not power.

As you've correctly pointed out the "F" in the "F=m*a" you just quoted is the traction at the driving wheels. And you've also correctly pointed out that the torque at the wheels is not the same as engine torque, because engine torque does not matter once power is transmitted through the transmission.

A quick thought experiment: let's say we have a big engine that has 400 Nm of torque at 3000rpm and a small engine that has 200 Nm of torque at 6000rpm, so both of them have a peak power of 125 kW. Now we gear the transmission such that both cars (also assuming same weight) reach 60mph at their peak torque output. Such a setup will ensure both cars accelerate to 60mph at the same time. Why? Because remember KE=1/2*m*v^2, and t=KE/P. Both cars will hit the same kinetic energy, while having the same amount of power, so it will take the same amount of time for the two engines to impart the kinetic energy onto the cars. You can apply the same simulation to a tank engine with 2000 Nm of torque at 600rpm or a go-kart engine with 100 Nm of torque at 12000rpm. The acceleration figure will not change, because the power is the same.

Now in real world we can't assume all engines have similar torque curve. The aforementioned "big engine" will generally perform better at normal driving rpms (1500-2500) so a driver gets that power very easily, while the "small engine" will require regular downshifting to, say, pass a car on the highway. That's why people say they "feel the torque" when they press down the gas pedal on a powerful car. But it's not because their engine has more torque, but because their engine has more power at the rpm they are requesting. Modern turbocharged cars have high low-end torque so they can be more efficient; it is a good solution to have both efficiency and power. The similar can be done through continuous variable valve lift on NA engines.

But our conversation was about driving the car near its peak power output, i.e. full throttle acceleration. That's when torque steer really becomes apparent. You wouldn't feel torque steer at all during normal driving.

The only correct way to determine a car's acceleration is to give the customer the full torque curve and the transmission ratios, as well as the dynamic weight distribution and tire grip. But it's unnecessarily complex. If I need a rough estimate, I only need to know the Power-to-Weight ratio. Do yourself a favor and calculate the power-to-weight ratio values of Lexus/BMW/Mercedes/Audi vehicles and plot them against the 0-60 figures, you will see a good correlation there. But try doing that with the torque-to-weight ratio and nothing will come out of it.

DFGeneer

Hm, sorry my friend, but you are wrong here.
F=m*a, remember?
So, acceleration of the car is proportional to the force it gets from the torque at the driving wheels.
Now, torque at the wheels is related to the engine/transmission gear combo rather than the engine torque alone, but still, it is the torque that does the job, not power.

As you've correctly pointed out the "F" in the "F=m*a" you just quoted is the traction at the driving wheels. And you've also correctly pointed out that the torque at the wheels is not the same as engine torque, because engine torque does not matter once power is transmitted through the transmission.

A quick thought experiment: let's say we have a big engine that has 400 Nm of torque at 3000rpm and a small engine that has 200 Nm of torque at 6000rpm, so both of them have a peak power of 125 kW. Now we gear the transmission such that both cars (also assuming same weight) reach 60mph at their peak torque output. Such a setup will ensure both cars accelerate to 60mph at the same time. Why? Because remember KE=1/2*m*v^2, and t=KE/P. Both cars will hit the same kinetic energy, while having the same amount of power, so it will take the same amount of time for the two engines to impart the kinetic energy onto the cars. You can apply the same simulation to a tank engine with 2000 Nm of torque at 600rpm or a go-kart engine with 100 Nm of torque at 12000rpm. The acceleration figure will not change, because the power is the same.

Now in real world we can't assume all engines have similar torque curve. The aforementioned "big engine" will generally perform better at normal driving rpms (1500-2500) so a driver gets that power very easily, while the "small engine" will require regular downshifting to, say, pass a car on the highway. That's why people say they "feel the torque" when they press down the gas pedal on a powerful car. But it's not because their engine has more torque, but because their engine has more power at the rpm they are requesting. Modern turbocharged cars have high low-end torque so they can be more efficient; it is a good solution to have both efficiency and power. The similar can be done through continuous variable valve lift on NA engines.

But our conversation was about driving the car near its peak power output, i.e. full throttle acceleration. That's when torque steer really becomes apparent. You wouldn't feel torque steer at all during normal driving.

The only correct way to determine a car's acceleration is to give the customer the full torque curve and the transmission ratios, as well as the dynamic weight distribution and tire grip. But it's unnecessarily complex. If I need a rough estimate, I only need to know the Power-to-Weight ratio. Do yourself a favor and calculate the power-to-weight ratio values of Lexus/BMW/Mercedes/Audi vehicles and plot them against the 0-60 figures, you will see a good correlation there. But try doing that with the torque-to-weight ratio and nothing will come out of it.

DFGeneer

Hm, sorry my friend, but you are wrong here.
F=m*a, remember?
So, acceleration of the car is proportional to the force it gets from the torque at the driving wheels.
Now, torque at the wheels is related to the engine/transmission gear combo rather than the engine torque alone, but still, it is the torque that does the job, not power.

As you've correctly pointed out the "F" in the "F=m*a" you just quoted is the traction at the driving wheels. And you've also correctly pointed out that the torque at the wheels is not the same as engine torque, because engine torque does not matter once power is transmitted through the transmission.

A quick thought experiment: let's say we have a big engine that has 400 Nm of torque at 3000rpm and a small engine that has 200 Nm of torque at 6000rpm, so both of them have a peak power of 125 kW. Now we gear the transmission such that both cars (also assuming same weight) reach 60mph at their peak torque output. Such a setup will ensure both cars accelerate to 60mph at the same time. Why? Because remember KE=1/2*m*v^2, and t=KE/P. Both cars will hit the same kinetic energy, while having the same amount of power, so it will take the same amount of time for the two engines to impart the kinetic energy onto the cars. You can apply the same simulation to a tank engine with 2000 Nm of torque at 600rpm or a go-kart engine with 100 Nm of torque at 12000rpm. The acceleration figure will not change, because the power is the same.

Now in real world we can't assume all engines have similar torque curve. The aforementioned "big engine" will generally perform better at normal driving rpms (1500-2500) so a driver gets that power very easily, while the "small engine" will require regular downshifting to, say, pass a car on the highway. That's why people say they "feel the torque" when they press down the gas pedal on a powerful car. But it's not because their engine has more torque, but because their engine has more power at the rpm they are requesting. Modern turbocharged cars have high low-end torque so they can be more efficient; it is a good solution to have both efficiency and power. The similar can be done through continuous variable valve lift on NA engines.

But our conversation was about driving the car near its peak power output, i.e. full throttle acceleration. That's when torque steer really becomes apparent. You wouldn't feel torque steer at all during normal driving.

The only correct way to determine a car's acceleration is to give the customer the full torque curve and the transmission ratios, as well as the dynamic weight distribution and tire grip. But it's unnecessarily complex. If I need a rough estimate, I only need to know the Power-to-Weight ratio. Do yourself a favor and calculate the power-to-weight ratio values of Lexus/BMW/Mercedes/Audi vehicles and plot them against the 0-60 figures, you will see a good correlation there. But try doing that with the torque-to-weight ratio and nothing will come out of it.

krew

krew


View the original article post

hello...this is the news of LEXUS LSF...the link is here

http://www.caradvice.com.au/573008/2018-lexus-ls-f-spied/photos/

please insert in your site

krew

krew


View the original article post

hello...this is the news of LEXUS LSF...the link is here

http://www.caradvice.com.au/573008/2018-lexus-ls-f-spied/photos/

please insert in your site

krew

krew


View the original article post

hello...this is the news of LEXUS LSF...the link is here

http://www.caradvice.com.au/573008/2018-lexus-ls-f-spied/photos/

please insert in your site

krew

krew


View the original article post

hello...this is the news of LEXUS LSF...the link is here

http://www.caradvice.com.au/573008/2018-lexus-ls-f-spied/photos/

please insert in your site