No_OBsT33R's LSD Calc UPDATED Version 1.1

[Adrenaline]

I'll go in depth in a sec, but power is always equal between both wheels at all times in open, closed, and LSDifferentials.

You need to explain, how you can say that they both have the same amount of power at the wheels.

 

[No_OBsT33R]

You need to explain, how you can say that they both have the same amount of power at the wheels.

No I don't, because Its not my job to teach you the basics if you don't know already.

Him hint

It's similar to how a lead car in NASCAR has to use more power than the car right behind him but they go the same speed, same principal you figure it out.

If your questions come in interrogation form it's your fault if your told anything or ignored. I don't have to do a good damn thing to prove anything to anybody in virtual Lala land. Especially people who seam to have fun trashing or digging at somebody only to play the victim like little girls when you finally irritate them to the point their frustration gets the better of them. GTFO!!! Grow a set.

 

[Adrenaline]

So you can't prove your theory.
I disproved it with physics.
I can only assume that means we've come to an agreement that your theory is wrong, since you're giving up.
Thank you for the clarification.

 

[No_OBsT33R]

So you can't prove your theory.
I disproved it with physics.
I can only assume that means we've come to an agreement that your theory is wrong, since you're giving up.
Thank you for the clarification.

What a joke!! To clarify you haven't disproved anything, you need help buddy. Show me the physics calculations you used to disprove it. Just saying "Physics" doesn't meant anything other than your a Tool, and clearly show your motivations are not to get to the truth of what's going on, but rather to prove me wrong. "Physics" proves me right, I can show it with the calc, you talk a lot of crap, but haven't shown anything at all, other than your great at running your mouth through the safety net of the Internet. Hmmmmmm.

Get Bent Adrenalin

 

[Adrenaline]

I can't simplify this scenario any further than I have:

A car with an open diff
The same tires
The same brakes
The same asphalt beneath said tires
and
The same power being applied to each wheel (your theory)

Does a burn out and only a single tire spins...

Explain how this is possible.

You can't, because it ISN'T possible.
It is physically, logically and mathematically impossible
for both tires to have the same force acting against them
with the same power being applied to them
and end in 2 different results.

 

[No_OBsT33R]

I can't simplify this scenario any further than I have:

A car with an open diff
The same tires
The same brakes
The same asphalt beneath said tires
and
The same power being applied to each wheel (your theory)

Does a burn out and only a single tire spins...

Explain how this is possible.

You can't, because it ISN'T possible.
It is physically, logically and mathematically impossible
for both tires to have the same force acting against them
with the same power being applied to them
and end in 2 different results.

I've repeated over and over the don't have the same forced acting on them, when the do the car is going straight if you don't understand how in a comer the left and right sides don't have the sane forced acting on them, what's the point of responding to you? All your doing is trying to prove me wring but clearly misunderstand what's being displayed with the calc.

Use your "Physics" to prove the calcs results wrong, or stop being a tool only out to prove me wrong. Bring something tangible to the table other than your stupid scenarios.

It's possible you must not drive a real car. I used do a burn outs with my old Protege all the time. Open diff one wheel smoking away the other doing F-all.

 

[No_OBsT33R]

However yeah, the car with an open diff WILL spin out one tire and not go forward very quickly until the spinning tire slows down or the car inches forward enough to get moving.

You must not drive a real car, if you have ever driven in the mud or on ice. With an open diff you would know one wheel spinning means the other doesn't do much.


Forget it, you know what **lolol** you don't need any help **chuckes** This guy doesnt even drive a real car.

Adrenalin when do you get your license?

 

[Adrenaline]

Okay, let's drop it to kindergarten levels.

I'll go in depth in a sec, but power is always equal between both wheels at all times in open, closed, and LSDifferentials.

"Power is always equal between both wheels at all times in an open diff"

That's what I'm focusing on right now.

We're taking a real world example, that directly disproves your theory.
A 1 wheel burnout disproves your theory that both tires have the same power at all times. That's it.

 

[No_OBsT33R]

Okay, let's drop it to kindergarten levels.

"Power is always equal between both wheels at all times in an open diff"

That's what I'm focusing on right now.

We're taking a real world example, that directly disproves your theory.
A 1 wheel burnout disproves your theory that both tires have the same power at all times. That's it.

No moron because the spinning wheel is putting ZERO power down moving the car forward, and so the wheel with grip gets ZERO power to move the car....... It's just spinning the tire away the car not getting any power put down.

In your scenario the spinning wheel has ZERO useable power, since an open diff only gives as much as the weakest tire can handle, in this case it's ZERO. As the spinning wheel cannot put any power down (ZERO) because both wheels are going to get equal power the wheel with grip gets ZERO power.


USEABLE Power! If it's spinning, how much power is it using to move the car???? ZERO!

 

[Adrenaline]

The wheel not spinning has 0power, the wheel spinning has the entire power output available. Whether that power is 'useable' or not means nothing.
The point is that the tires don't have equal power all the time like you claim.
So, that disproves your theory; plain and simple.

 

[No_OBsT33R]

The wheel not spinning has 0power, the wheel spinning has the entire power output available. Whether that power is 'useable' or not means nothing.
The point is that the tires don't have equal power all the time like you claim.
So, that disproves your theory; plain and simple.

WRONG!

In an Open Diff The tire spinning has ZERO useable power and the wheel with grip gets the same ZERO. IT MEANS EVERYTHING, the car doesn't get any acceleration force, that means nothing to you?????

It also disproved your stupid conclusion to your scenario. Like I said the tire smokes the car doesn't move, you Dumbass.

If you disagree with this simple FACT you need to STFU and let the adults speak, you DO NOT understand the "Physics", plain and simple.

I'm glad you CLEARLY displayed your lack of understanding. Your an incredible Bull salesman.

 

[Adrenaline]

Whether or not the car can 'use' the power, has nothing to do with if it's there or not.

The tire has 400hp to it, just because that power isn't being used to push the car forward (burn out) doesn't mean that all of the sudden the power isn't still being transferred through the tire.

 

[No_OBsT33R]

Whether or not the car can 'use' the power, has nothing to do with if it's there or not.

The tire has 400hp to it, just because that power isn't being used to push the car forward (burn out) doesn't mean that all of the sudden the power isn't still being transferred through the tire.

Yes it does when the tool states "useable power" it means everything, I'm talking about USEABLE power if your talking about something else STFU, GTFO. Your just starting crap over things you don't understand (or you would of UNDERSTOOD I'm talking about USEABLE power.) your just trying to prove me wrong. Got no life or anything better to do?

But really it sounds like backtracking. Answer me this, it proves your scenario wrong doesn't it.....

 

[articzap]

Yes it does when the tool states "useable power" it means everything, I'm talking about USEABLE power if your talking about something else STFU, GTFO. Your just starting crap over things you don't understand (or you would of UNDERSTOOD I'm talking about USEABLE power.) your just trying to prove me wrong. Got no life or anything better to do?

But really it sounds like backtracking. Answer me this, it proves your scenario wrong doesn't it.....

Define usable power? And with that stated, wouldn't different tire compounds limit the usable power, not just the LSD setting?

 

[Rotary Junkie]

I don't know if it's because I'm presently sick as a dog or what but Noobsteer's counterarguments are presently confusing the hell out of me...

How... No, no, not worth trying to figure out how he's managed it...

Edit: I will say though that his definition of "usable power" seems to change according to what he needs it to mean... At the start of the thread it seemingly had something to do with initial torque and power loss due to LSD slippage (which doesn't happen)... I don't know. I don't want to know.

 

[dr_slump]

Ok, one question:
The tool states as example 200ft-lb. Are these 200ft-lb the power delivered to the wheel which has grip while the other tire is in the air (without grip)?

 

[No_OBsT33R]

Define usable power? And with that stated, wouldn't different tire compounds limit the usable power, not just the LSD setting?

Power used to move car forward, power converted into forward motion.

Yes many factors effect how much power can be used including tire compound.

RJ the useable power comes in from the drive shaft and is effected by the int setting. The more it takes to overpower the int clutch the less is leftover and transferable into useable power to move the car forward.

 

[articzap]

Power used to move car forward, power converted into forward motion.

Yes many factors effect how much power can be used including tire compound.

RJ the useable power comes in from the drive shaft and is effected by the int setting. The more it takes to overpower the int clutch the less is leftover and transferable into useable power to move the car forward.

Who said its a clutch LSD?

 

[No_OBsT33R]

Ok, one question:
The tool states as example 200ft-lb. Are these 200ft-lb the power delivered to the wheel which has grip while the other tire is in the air (without grip)?

If a tire is in the air it has 0 useable power as the wheel is not contacting the ground, in an open diff, the tire with grip would get the same 0 power to move the car.

An LSD will retain some power and the wheel with grip will be able to move the car forward.

Notice that even when you input 0 as the amount of useable power the wheel with grip still gets some.

The calc can't recreate a wheel loosing contact & lifting in the air, it shows only what happens with 2 wheels on the ground and the LSD activated.

If a tire breaks out (spins out, slides on ice whatever) then it's useable power becomes 0. THIS is the tire that has 0 useable power (as you input in the calc) So IMAGINE one of the 2 sides disappearing.

This is an excel calc, that's all, (and early release on top of it) it has some limitations.

Who said its a clutch LSD?

Nobody.

 

[dr_slump]

If a tire is in the air it has 0 useable power as the wheel is not contacting the ground, in an open diff, the tire with grip would get the same 0 power to move the car.

An LSD will retain some power and the wheel grip will be able to move the car forward.

Yeah I know how a LSD works. But a LSD in GT5 is never a complete open one. Your Calc states different numbers of power, but you say it's for both wheels with grip. Where is the lost power gone? Magic?
(That's why I thought it shows the power delivered to the wheel with grip while the other is in the air.)

 

[EivlEvo]

Adrenalin. Why can't you seem to post in any thread without denigrating someone? Everything you have to say is just so "ur wrong, and I'm right. If you don't understand why, you must be stupid". Why? If you have such knowledge of these topics, maybe you should TEACH us about them instead of telling everyone they're stupid for being wrong?

 

[No_OBsT33R]

Yeah I know how a LSD works. But a LSD in GT5 is never a complete open one. Your Calc states different numbers of power, but you say it's for both wheels with grip. Where is the lost power gone? Magic?

This is an excel calc. It ONLY shows results for an activated LSD, it can't simulate a wheel breaking out in the presented results, but it's implied when you state 0 useable power, that a wheel must have 0 useable power.

 

[Rotary Junkie]

Yeah I know how a LSD works. But a LSD in GT5 is never a complete open one. Your Calc states different numbers of power, but you say it's for both wheels with grip. Where is the lost power gone? Magic?

Supposedly it's gone to slippage in the LSD clutches or something...

Which doesn't happen because that's not how LSDs work, but whatever.


I think I may have a new theory as to what initial torque actually is/does... It's not preload (even though that would make sense), I'm thinking it's the percentage of the accel/decel % values that it "kicks in" at...

So I'll use default values here to demonstrate my point...

The diff will stay "open" until a certain amount of torque is applied to it, at which point it will lock to IT x accel/decel value and ramp up from there as more torque is applied. Higher IT means it locks later and harder.

W/ default under accel it'd go from open to a minimum lock % of .1x.4 or 4% lock and ramp up to 40% lock under accel. Initial of 60 would mean an initial lock "hit" of 24% and ramp up from there. 60/60 would be initial lock of 36% and ramp up to 60%.

It'd reasonably explain what I've noticed with it... All I know for sure is Noobsteer has something we don't and he needs to share.

 

[dr_slump]

Supposedly it's gone to slippage in the LSD clutches or something...

Which doesn't happen because that's not how LSDs work, but whatever.

I think I may have a new theory as to what initial torque actually is/does... It's not preload (even though that would make sense), I'm thinking it's the percentage of the accel/decel % values that it "kicks in" at...

In real life it's definitely preload.
In GT5 it feels the same for me. When you drive trough a corner without braking or throttle, you can feel a difference.

 

[No_OBsT33R]

If we have a car with 500ft-lb and one tire goes in the air (or hits ice and slips, hits a patch of leaves, looses traction due to WT, whatever the reason may be.) this is what the GT5 LSD will do.

Keep in mind one wheel spinning out or in the air,

With an open diff the wheel with grip get 0ft-lb.

LSD;

With settings of 10/40 (int/accel)

The wheel with grip gets 90ft-lb

With settings of 15/40

The wheel with grip gets 85ft-lb

With settings of 15/45

The wheel with grip gets 95ft-lb

If there were a Closed locked Diff the wheel with grip gets 250ft-lb





Good luck with figuring out LSD's

 

[dr_slump]

If we have a car with 500ft-lb and one tire goes in the air (or hits ice and slips, hits a patch of leaves, looses traction due to WT, whatever the reason may be.) this is what the GT5 LSD will do.;

Keep in mind one wheel spinning out or in the air, the GT5 LSD will;

With an open diff the wheels get 0ft-lb

LSD;

With settings of 10/40 (int/accel)

The wheel with grip gets 90ft-lb

With settings of 15/40

The wheel with grip gets 85ft-lb

With settings of 15/45

The wheel with grip gets 95ft-lb

With a Closed locked Diff the wheel with grip gets 250ft-lb

That's what I'm talking about the whole time!

 

[Rotary Junkie]

Way to change units there... 500hp at what RPM?

Let's say you meant 500 ft-lbs... Open diff the loaded tire would get whatever drag there is on the unloaded tire. 10/40 by my math would get 100 ft-lbs (or 115 based on another thought).

Speaking of that "other thought"... If initial = preload, it means accel/decel values are then based off the remaining slip. So 10% preload leaves 90% "slip"; 40 accel would be 40% of 90% tacked on for 46% max lock, etc etc... But I'm liking my other theory a little too much.

Edit: He edited, disregard first line.

 

[No_OBsT33R]

That's what I'm talking about the whole time!

So have I those are all results out of the calc

 

[No_OBsT33R]

Way to change units there... 500hp at what RPM?

I take it from the following you got it was a typo

Let's say you meant 500 ft-lbs... Open diff the loaded tire would get whatever drag there is on the unloaded tire. 10/40 by my math would get 100 ft-lbs (or 115 based on another thought).

Your not even sure what thought your on. Lol

With 100 as your result your not factoring in the torque required to overpower the int setting.

Use the same math to 15/40 & 15/45

You will begin to see the error in your math (maybe you won't) you almost have it, but not quite. As you admittedly have multiple variations on your calculations with varying results.

Speaking of that "other thought"... If initial = preload, it means accel/decel values are then based off the remaining slip. So 10% preload leaves 90% "slip"; 40 accel would be 40% of 90% tacked on for 46% max lock, etc etc... But I'm liking my other theory a little too much.

This is on the right track good show, it's clicking!

I almost thought it wasn't going to happen, but this portion shows MUCH improvement.

Hats off to RJ, made my day.

Da_Slump too, it seems like you guys are getting it.

 

[articzap]

The more it takes to overpower the int clutch the less is leftover and transferable into useable power to move the car forward.

Who said its a clutch LSD?

Nobody.

There's to many variables aside from the differential to claim what power is "usable".

I think this calculator would be awesome if you found a way to calculate how much wheel torque difference causes the LSD to do its thing. And how exactly the LSD shifts the power around. 👍