Pack Performance and Launch Mode Limits

[St Charles]

What I'm describing is how an open differential works. It always delivers the same amount of torque to both wheels. If one wheel is in the sand, it controls the amount of torque that goes to both wheels. That's why traction control works. By applying the brake to the wheel in the sand, the torque it can generate goes up thereby increasing the torque to the wheel on the road. You don't have to bring the wheel to a stop. You just have to create enough drag with the brake so that the wheel on the street can move the car out of the sand.

Of course you can always limit the performance of the P cars. But I'm talking about fiksegts' contention that Tesla is intentionally decreasing the performance of the non-P Model X. Looking at the power curve he included in his video, the ratio of power to speed up to max power is the same. This indicates that they are applying max torque the entire time.

You just described my point. See we agree. Without traction control one wheel breaks free and this is what happens. Without limited slip and/or traction control you cannot 100% prevent this. TC will cut power/apply brakes to force torque to the other wheel. Since the Model X has traction control, wheel slip is prevented.

The reality is this. The drive units output more enough torque to make a wheel break traction. You even did the math to prove it. Unless you glue the tires to the ground, there is nothing mechanically preventing this condition. This is 100% my point. Traction control is always watching, always measuring, to ensure that the wheels don't slip. This is especially true at launch when the most torque can be applied to the wheels.

Certainly you agree that if the model x had a 5 horsepower motor that only generated 50 ft-lbs of torque at the wheels that it wouldn't spin the tires. So there is a certain amount of torque that will cause the wheels to spin. That is when the tangential force at the road/wheel interface is enough to overcome the friction of the tire. In this case it's about 1250 lbs, which the small motors just can't provide.

Hyperbole is not an argument. The drive units can provide almost double the torque needed to break loose one wheel in optimal conditions. To say nothing about less than optimal traction. You did the math, now apply it to the real world.

Unless someone knows how to measure the behavior of the Tesla Traction Control system, we'll never actually know what's going on. We are just reading incomplete data and extrapolating. That said, I somehow doubt that Tesla is allowing even the small motors output 100% of their capability. That is ultimately what I took away from the video, there is more on the table.

 

[TIppy]

You just described my point. See we agree. Without traction control one wheel breaks free and this is what happens. Without limited slip and/or traction control you cannot 100% prevent this. TC will cut power/apply brakes to force torque to the other wheel. Since the Model X has traction control, wheel slip is prevented.

The reality is this. The drive units output more enough torque to make a wheel break traction. You even did the math to prove it. Unless you glue the tires to the ground, there is nothing mechanically preventing this condition. This is 100% my point. Traction control is always watching, always measuring, to ensure that the wheels don't slip. This is especially true at launch when the most torque can be applied to the wheels.



Hyperbole is not an argument. The drive units can provide almost double the torque needed to break loose one wheel in optimal conditions. To say nothing about less than optimal traction. You did the math, now apply it to the real world.

Unless someone knows how to measure the behavior of the Tesla Traction Control system, we'll never actually know what's going on. We are just reading incomplete data and extrapolating. That said, I somehow doubt that Tesla is allowing even the small motors output 100% of their capability. That is ultimately what I took away from the video, there is more on the table.

No. We disagree. The differential will always divide the torque in half and apply an equal amount to each wheel. All of the torque will never be applied to a single wheel. No more than 1054 lbs of force will be applied to one wheel. It could be less if the road conditions are bad. 1054 lbs of force is not enough force to break the friction force of the tire.

With decent road conditions the traction control won't be needed for straight line acceleration.

 

[TIppy]

That is ultimately what I took away from the video, there is more on the table.

There isn't more unless you're saying the motor has more max hp and max torque than is being used. fiksegts said he felt the initial acceleration was less than what he felt later. But the power curve doesn't support this. The torque is constant up to the max power.

 

[yak-55]

There isn't more unless you're saying the motor has more max hp and max torque than is being used. fiksegts said he felt the initial acceleration was less than what he felt later. But the power curve doesn't support this. The torque is constant up to the max power.

I think we can all agree that the slope of power vs rpm is indeed torque. And what we do observe appears to be a constant slope initially. I suspect the disconnect occurs in assuming that this slope represents the maximum torque the drivetrain is capable of (with uncontrolled current into the inverter). While this slope is the maximum observed, it doesn't mean that active controls (via software) are not creating an artificial limit (for marketing, engineering, or other reasons). In this way, the observed data is not inconsistent with the car "being held back". So maybe everyone is "right" in their views.

I have no idea either way, just trying to facilitate the conversation.

 

[xborg]

I know we have distracted from the subject,
But could someone claims Tesla is not limiting power intentionally on non-P cars, could you please explain this:

Chevy Volt

• 0-30: 2.6 seconds
• 0-60: 8.4 seconds

Tesla S 85 Rwd

• 0-30: 2.7 seconds
• 0-60: 5.6 seconds

I'd expect a better 0-30 from 85. (or 90 or 100)

 

[St Charles]

The differential will always divide the torque in half and apply an equal amount to each wheel.

No. That is an untrue statement. An open differential will allow force applied by the motor will follow the path of least resistance. the 50/50 torque statement you are throwing around means that the torque transmitted through friction will always be the same. If you watch the video i linked above you'll see a chevy truck with an open differential doing a burnout. The torque applied by the wheels to the ground is equal, however one wheel is spinning while the other is in place. All extra torque supplied by the drivetrain is being used to rotate the spinning wheel. The video is displaying the exact situation i am describing. A clutch style limited slip would force whatever torque the clutch pack can hold to the stationary wheel. A gear style LSD would force the non spinning wheel to rotate at a fixed ratio in relation to the spinning wheel.

Lets play a logic exercise. If I immobilize one of the two wheels of the model X and disable traction control. is the drive unit capable of delivering enough torque to force the other wheel to break traction? [hint: yes]

The same effect happens if one wheel is on ice while the other is not. in an open differential one wheel will spin on ice and you would go nowhere....until traction control applies braking force to the spinning wheel.

Keep in mind that we are not even discussing things like weight transfer, suspension geometry, and adhesion. These factors and more can easily affect traction even in optimal situations. And then there is AWD which really does make it a lot less realistic that traction will be lost. (look at me all helping your argument n' stuff)

My point is that you keep saying it isn't possible to break traction. it is absolutely possible.

With decent road conditions the traction control won't be needed for straight line acceleration.

It's unlikely that traction control is going to get involved. It is unreasonable to assume it will never be needed.

 

[TIppy]

Did you mean bolt?
chevy bolt may be geared to get good low end acceleration, but quickly reaches the plateau of the power curve.

 

[St Charles]

There isn't more unless you're saying the motor has more max hp and max torque than is being used. fiksegts said he felt the initial acceleration was less than what he felt later. But the power curve doesn't support this. The torque is constant up to the max power.

This is what I am saying, on both counts. It is smart engineering to back off from the absolute maximum capability of any system relative to duty cycle. This whole thread started because the 90kW battery pack was clearly too close to it's maximum capability and some of us like to drive our cars to 11 more then Tesla expected.

Also, are you really thinking that the motors are going from 0 to maximum amperage in an instant? Clearly, they wouldn't do this. There has to be a ramp, even if it's a split second.

 

[TIppy]

Lets play a logic exercise. If I immobilize one of the two wheels of the model X and disable traction control. is the drive unit capable of delivering enough torque to force the other wheel to break traction? [hint: yes]

Correct answer is no.
Holding the wheel isn't any different than both wheels rolling under static friction. The differential is still going to divide the torque equally between two wheels.

The same effect happens if one wheel is on ice while the other is not. in an open differential one wheel will spin on ice and you would go nowhere....until traction control applies braking force to the spinning wheel.

Exactly. Both wheels get the torque that the ice wheel can generate.

 

[St Charles]

Correct answer is no.
Holding the wheel isn't any different than both wheels rolling under static friction. The differential is still going to divide the torque equally between two wheels.

Wait, what? no.

If the left wheel is completely immobilized, and the right wheel is able to hold 1000lbft of torque then the motor only needs to generate more than 1000lbft of torque to cause the right wheel to spin. the right wheel is not going to hold infinite torque through an open differential simply because the left wheel is immobilized.

I don't think you actually understand how an open differential works and have only ever looked at the math. Please watch this and get back to me. Pay close attention at the 6:00 mark as it is a visual representation of what I am describing.

 

[TIppy]

This is what I am saying, on both counts. It is smart engineering to back off from the absolute maximum capability of any system relative to duty cycle. This whole thread started because the 90kW battery pack was clearly too close to it's maximum capability and some of us like to drive our cars to 11 more then Tesla expected.

Decreased max hp or max torque isn't something fiksegts would detect as an increase in acceleration at higher speeds.

[QUOTE="Also, are you really thinking that the motors are going from 0 to maximum amperage in an instant? Clearly, they wouldn't do this. There has to be a ramp, even if it's a split second.[/QUOTE]

The ramp up is very quick and probably the same on p and non-p cars. This is probably to lower the lash up forces and may actually be a longer ramp on more powerful cars. I doubt the few hundredths of a second ramp is what fiksegts is noting.

 

[TIppy]

Wait, what? no.

If the left wheel is completely immobilized, and the right wheel is able to hold 1000lbft of torque then the motor only needs to generate more than 1000lbft of torque to cause the right wheel to spin. the right wheel is not going to hold infinite torque through an open differential simply because the left wheel is immobilized.

I don't think you actually understand how an open differential works and have only ever looked at the math. Please watch this and get back to me. Pay close attention at the 6:00 mark as it is a visual representation of what I am describing.

Perhaps the fact that the wheels are rotating is complicating this. Let's chain the car to a tree and bring the torque up slowly. If the motor generates 50 ft-lbs of torque, the output of the diff while be 462 ft-lbs and will be evenly split as 231 ft-lbs to each wheel. Remember an open diff divides the torque evenly between the two driven wheels.
motor 125 ft-lbs : diff 1155 ft-lbs: each wheel 578 ft-lbs
motor 250 ft-lbs : diff 2310 ft-lbs: each wheel 1155 ft-lbs

1155 ft-lbs / 1.125 ft is 1027 lbs of driving force even though the wheels aren't turning.

Because we're chained to a tree the wheels aren't rotating since the torque to the individual wheels is less than needed to break the wheel loose. This is no different than holding one wheel from rotating.

 

[JRP3]

Remember an open diff divides the torque evenly between the two driven wheels.

Then what does a Locker differential do?

 

[TIppy]

Then what does a Locker differential do?

It provides a certain amount of torque to the wheels even if one of them has no traction. Up to a certain level of torque difference it behaves like a solid axle. In this particular case, it would likely distribute the torque close to equally to both wheels. But it is possible that one of the wheels could get more than half of the torque.

 

[JRP3]

Though it seems somewhat counter intuitive TIppy is correct

An open (or unlocked) differential always provides the same torque (rotational force) to each of the two wheels, on that axle.

Locking differential - Wikipedia

In open differential the distribution is fixed at 50-50. Both the wheels get the same amount of torque.

https://www.quora.com/Transmission-...oes-this-differ-from-the-locking-differential

 

[jerjozwik]

has anyone logged above 430kW in ludicrous mode only in 17.10.42 or higher?

 

[fiksegts]

Tesla has updated and increased performance on many different models in the past via software updates, but on this model it's not possible? just looking at the 108 MPH trap says a lot as well...

This is what I am saying, on both counts. It is smart engineering to back off from the absolute maximum capability of any system relative to duty cycle. This whole thread started because the 90kW battery pack was clearly too close to it's maximum capability and some of us like to drive our cars to 11 more then Tesla expected.

Also, are you really thinking that the motors are going from 0 to maximum amperage in an instant? Clearly, they wouldn't do this. There has to be a ramp, even if it's a split second.

 

[TIppy]

Tesla has updated and increased performance on many different models in the past via software updates, but on this model it's not possible? just looking at the 108 MPH trap says a lot as well...

There might be more max performance available, but your comment was that the acceleration felt soft at the low end verses what came later. You felt they were holding it back at the low end. But the power curve you provided doesn't support this. It shows that the acceleration( torque) is constant up to max power.

 

[TJtv]

Just because the acceleration is constant does not mean that its achieving the maximum acceleration possible. The graph he showed had a linear of increase of KW as speed increased(up to about 45 mph), and this produces the constant acceleration. If the slope of the KW vs speed line was higher this would also produce constant acceleration - but it would be a bigger acceleration.

 

[fiksegts]

indeed they are holding it back.... any other Tesla that traps 108 runs well into the 12's... the car just slugs off the initial hit.... vbox G shows a consistent pull, no initial G spike and drop off line every other Tesla I've tested...

There might be more max performance available, but your comment was that the acceleration felt soft at the low end verses what came later. You felt they were holding it back at the low end. But the power curve you provided doesn't support this. It shows that the acceleration( torque) is constant up to max power.