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Non-HLSD question

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Old 12-06-2003, 11:07 PM
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Non-HLSD question

I have an '03 6sp without the HLSD option. I know that it is supposed to send more power to the wheel with more grip. I also test drove a 6sp Max with this option when shopping and could definitely feel the difference coming out of curves and starting in a turn. It had definite advantages in those situations. I saw no real advantage to it for accelerating quickly in a straight line, however(at least one I could feel). My question is whether in a car without HLSD, like mine, are both front wheels being turned by the engine or is it just one of the front wheels that gets all the power all of the time? And if both are being powered in a non-HLSD car, is it a 50/50 split of the power to each wheel all the time? Also, what happens when one drive wheel begins to slip in a non-HLSD car?
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Old 12-06-2003, 11:50 PM
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i believe both wheels get power until one starts to slip then the inner wheel gets all the power. im sure someone else has a more technical answer.

will
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Old 12-07-2003, 03:49 AM
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This is kind of wrong, with a non-HLSD tranny its just a standard differential. That means the wheel with the least amount of traction is going to spin the most. So that means if one wheel was on dry pavement and the other is on ice, its going to spin like crazy on the wheel on ice. The car wont move.

There is no advantage there. Not to mention we have unequal drive shafts which usually cause the driver's side wheel to spin more if you in a staight line spin and causes torque steer.

Its only really a 50/50 split when both wheels have traction, after that, it is a 0/100 split. Cause its an open diff.

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Old 12-07-2003, 09:48 AM
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The odd part is when I do a really heavy burnout I smoke both the tires and get sideways and according to my VIN I dont have the HLSD. I have noticed that when the burnout is light or unintentional to begin with it is one wheel. I have also noticed that going around corners its easy to spin one wheel. I know this means I do not have the HLSD, but why do I have a two tire burnout when I really get on it?
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Old 12-07-2003, 01:19 PM
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Originally Posted by bgates1654
The odd part is when I do a really heavy burnout I smoke both the tires and get sideways and according to my VIN I dont have the HLSD. I have noticed that when the burnout is light or unintentional to begin with it is one wheel. I have also noticed that going around corners its easy to spin one wheel. I know this means I do not have the HLSD, but why do I have a two tire burnout when I really get on it?
Using BidDog's explanation, the torque is evenly distributed 50/50 when both tires have equal traction. So when you're do a burnout, both tires are probably losing the same amount of traction at the same time which in turn doesn't distribute the torque to the wheel that is spinning more since both wheels still have the same (lack of) traction.
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Old 12-07-2003, 03:36 PM
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Like dude above said, the reason why it pulls left initially when you burning out, its cause the driver's side shaft is almost 1/2 the length, thats torque steer. Thats why it does that. Then the passenger side starts slipping cause they both get about even traction while slipping. But its still pulling left.

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Old 12-07-2003, 06:45 PM
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Actually... last time I did a burnout it went to the right not the left. It has gone both ways.
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Old 12-07-2003, 08:48 PM
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Driving in the snow and ice the last few days, I've noticed that when the car starts to lose traction, it always pulls to the right.
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Old 12-07-2003, 09:09 PM
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Originally Posted by BigDogJonx
Its only really a 50/50 split when both wheels have traction, after that, it is a 0/100 split. Cause its an open diff.

Dixit
That's not true. It's always exactly a 50/50 split. When one wheel is on ice you can put virtually no torque to that wheel. As a result you can only put the same amount of torque to the other wheel. That tiny put of torque you can put to the wheel that has grip is not enough to turn it and move the car forward.
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Old 12-07-2003, 09:11 PM
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Originally Posted by BigDogJonx
Like dude above said, the reason why it pulls left initially when you burning out, its cause the driver's side shaft is almost 1/2 the length, thats torque steer. Thats why it does that. Then the passenger side starts slipping cause they both get about even traction while slipping. But its still pulling left.

Dixit
The car has equal length half shafts. It shouldn't tend to pull to any particular side on a flat surface if each wheel is equally loaded. However roads aren't flat, nor is the car equally loaded under most circumstances.
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Old 12-07-2003, 09:16 PM
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You right, its not a 0/100 split, its more like 0/50 split. Cause its not transfering power. I meant to write that.

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Old 12-07-2003, 09:27 PM
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Originally Posted by BigDogJonx
You right, its not a 0/100 split, its more like 0/50 split. Cause its not transfering power. I meant to write that.

Dixit
No, really, it's still 50/50, not 50/0. What you're overlooking is that when a wheel is spinning free the engine isn't really putting out any torque regardless of RPM. Think of it as reving the car in neutral, or with the clutch pressed in. The engine doesn't puts out virtually no torque because there's no resistance to it's motion. With a wheel spinning it's the same way. That's why when a wheel starts spinning the RPMs suddenly jump. There's nothing much resisting a change in RPM.
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